A Transition to Advanced Mathematics

FOURTH

EDITION

A TRANSITION TO

ADVANCED MATHEMATICS Douglas Smith University of North Carolina at Wilmington

Maurice Eggen Trinity University

Richard St. Andre Central Michigan University

Brooks/Cole Publishing Company

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Library of Congress Cataloging-in-Publication Data Smith, Douglas, [date] A transition to advanced mathematics / Douglas Smith, Maurice Eggen, Richard St. Andre. - 4th ed. p. cm. Includes index. ISBN 0-534-34028-8 I. Mathematics. I Eggen, Maurice, [date]. II. St. Andre, Richard, [date]. III. Title. QA37.2.S575 1996 51O-dc21 96-47266 CIP

PREFACE

To the First Edition "I understand mathematics but Ijust can't do proofs." Our experience has led us to believe that the remark above, though contradictory, expresses the frustration many students feel as they pass from beginning calculus to a more rigorous level of mathematics. This book developed from a series of lecture notes for a course at Central Michigan University that was designed to address this lament. The text is intended to bridge the gap between calculus and advanced courses in at least three ways. First, it provides a firm foundation in the major ideas needed for continued work. Second, it guides students to think and to express themselves mathematically-to analyze a situation, extract pertinent facts, and draw appropriate conclusions. Finally, we present introductions to modern algebra and analysis in sufficient depth to capture some of their spirit and characteristics. We begin in Chapter I with a study of the logic required by mathematical arguments, discussing not formal logic but rather the standard methods of mathematical proof and their validity. Methods of proof are examined in detail, and examples of each method are analyzed carefully. Denials are given special attention, particularly those involving quantifiers. Techniques of proof given in this chapter are used and referred to later in the text. Although the chapter was written with the idea that it may be assigned as out-of-c1ass reading, we find that most students benefit from a thorough study of logic. Much of the material in Chapters 2, 3, and 4 on sets, relations, and functions will be familiar to the student. Thus, the emphasis is on enhancing the student's ability to write and understand proofs. The pace is deliberate. The rigorous approach requires the student to deal precisely with these concepts. Chapters 5, 6, and 7 make use of the skills and techniques the student has acquired in Chapters I through 4. These last three chapters are a cut above the earlier chapters in terms of level and rigor. Chapters I through 4 and anyone of Chapters 5,6, or 7 provide sufficient material for a one-semester course. An alternative is to choose among topics by selecting, for example, the first two sections of Chapter 5, the first three sections of Chapter 6, and the first two sections of Chapter 7. iii

iv

Preface

Chapter 5 begins the study of cardinality by examining the properties of finite and infinite sets and establishing countability or uncountability for the familiar number systems. The emphqsis is on a working knowledge of cardinality-particularly countable sets, the ordering of cardinal numbers, and applications of the Cantor-Schraeder-Bernstein Theorem. We include a brief discussion of the Axiom of Choice and relate it to the comparability of cardinals. Chapter 6, which introduces modern algebra, concentrates on the concept of a group and culminates in the Fundamental Theorem of Group Homomorphisms. The idea of an operation preserving map is introduced early and. developed throughout the section. Permutation groups, cyclic groups, and modular arithmetic are among the examples of groups presented. Chapter 7 begins with a description of the real numbers as a complete ordered field. We continue with the Heine-Borel Theorem, the Bolzano-Weierstrass Theorem, and the Bounded Monotone Sequence Theorem (each for the real number system), and then return to the concept of completeness. Exercises marked with a solid star (*) have complete answers at the back of the text. Open stars (n) indicate that a hint or a partial answer is provided. "Proofs to Grade" are a special feature of most of the exercise sets. We present a list of claims with alleged proofs, and the student is asked to assign a letter grade to each "proof' and to justify the grade assigned. Spurious proofs are usually built around a single type of error, which may involve a mistake in logic, a common misunderstanding of the concepts being studied, or an incorrect symbolic argument. Correct proofs may be straightforward, or they may present novel or alternate approaches. We have found these exercises valuable because they reemphasize the theorems and counterexamples in the text and also provide the student with an experience similar to grading papers. Thus, the student becomes aware of the variety of possible errors and develops the ability to read proofs critically. In summary, our main goals in this text are to improve the student's ability to think and write in a mature mathematical fashion and to provide a solid understanding of the material most useful for advanced courses. Student readers, take comfort from the fact that we do not aim to turn you into theorem-proving wizards. Few of you will become research mathematicians. Nevertheless, in almost any mathematically related work you may do, the kind of reasoning you need to be able to do is the same reasoning you use in proving theorems. You must first understand exactly what you want to prove (verify, show, or explain), and you must be familiar with the logical steps that allow you to get from the hypothesis to the conclusion. Moreover, a proof is the ultimate test of your understanding of the subject 1l1atter and of mathematical reasoning. We are grateful to the many students who endured earlier versions of the manuscript and gleefully pointed out misprints. We acknowledge also the helpful comments of Edwin H. Kaufman, Melvin Nyman, Mary R. Wardrop, and especially Douglas W. Nance, who saw the need for a course of this kind at CMU and did a superb job of reviewing the manuscript. We thank our reviewers: William Ballard of the University of Montana, Sherralyn Craven of Central Missouri State University, Robert Dean of Stephen F. Austin State University, Harvey Elder of Murray State University, Hoseph H. Oppenheim of San Francisco State University, Joseph Teeters of the University of

Preface

v

Wisconsin, Dale Schoenefeld of the University of Tulsa, Kenneth Slonnegar of State University of New York at Fredonia, and Douglas Smith of University of the Pacific. And we wish to thank Karen St. Andre for her superb and expeditious typing of the manuscript.

To the Fourth Edition The intent of this edition is to continue the evolution of the text as an introduction to both foundational topics of logic, sets, relations, and functions and to the rigor of mathematical thinking and writing. Each edition has brought improvements through incorporation of many valuable suggestions from reviewers, instructors, and students. We are grateful for the responses we have received. For previous editions we found that many instructors followed the development of the text, covering most of Chapters I through 4, much of the treatment of cardinality in Chapter 5, and, as time permits, topics from Chapter 6 or 7. One common approach to Chapter 5 has been to treat the definitions and results of the first two sections on finite and infinite sets, then Cantor's Theorem and facts about countable sets. This approach has been facilitated by gathering countability results in Section 5.3 and shifting the Axiom of Choice to the end of the chapter. New and revised explanations, examples, and exercises appear throughout this edition of the text. This is especially so for Chapter I, which continues to receive our greatest attention as we strive to provide allowance for variance in instructor preferences. A sixth section has been added to Chapter I, providing additional examples of proofs. There is a new section on orders and partial orderings in Chapter 2. Discussions of real analysis concepts in Chapter 7 are tied more closely to students' experience in calculus. An accompanying instructor's manual by Kurt Friedman and Greg Perkins is available to adopters of the text. We would like to thank our reviewers for the second and third editions: Mangho Ahuja, Southeast Missouri State University; David Barnette, University of California at Davis; Harry Coonce, Mankato State University; Michael 1. Evans, North Carolina State University; Benjamin Freed, Clarion University of Pennsylvania; Robert Gamble, Winthrop College; Dennis Garity, Oregon State University; Robert P. Hunter, Pennsylvania State University; Jack Johnson, Brigham Young University-Hawaii; Daniel Kocan, State University of New York, Potsdam; James McKinney, California State Polytechnic University; Yves Nievergelt, Eastern Washington University; Victor Schneider, University of Southwestern Louisiana; and Lawrence Williams, University of Texas, San Antonio. We also thank our reviewers for this fourth edition: Gerald Beer, California State University-Los Angeles; Ron Dotzel, University of Missouri; Gerald Farrell, Cal Poly-San Luis Obispo; and Andrew Martin, Morehead State University. We also thank Gary Ostedt, Nancy Shammas, Kelly Shoemaker, Lisa Torri, and everyone else involved at Brooks/Cole as well as Barbara Kimmel, our manuscript editor, for their professional work and friendly encouragement.

Richard St. Andre Douglas Smith

To Karen, Karen, and Karen

CONTENTS

CHAPTER

CHAPTER

CHAPTER

1

Logic and Proofs

1.1 1.2 1.3 1.4 1.5 1.6

Propositions and Connectives 1 Conditionals and Biconditionals 9 Quantifiers 18 Mathematical Proofs 26 Proofs Involving Quantifiers 40 Additional Examples of Proofs 49

2

Set Theory

2.1 2.2 2.3 2.4 2.5 2.6

Basic Notions of Set Theory 59 Set Operations 67 Extended Set Operations and Indexed Families of Sets Induction 85 Equivalent Forms of Induction 97 Principles of Counting 104

3

Relations

3.1 3.2 3.3 3.4 3.5

Cartesian Products and Relations Equivalence Relations 128 Partitions 136 Ordering Relations 141 Graphs of Relations 151

1

59

74

115 115

vii

viii

Contents

C HAP T E R 4

Functions

161

4.1 Functions as Relations 161 4.2 Constructions of Functions 171 4.3 Functions That Are Onto; One-to-One Functions 4.4 Induced Set Functions 188

CHAPTER

179

5

Cardinality

5.1 5.2 5.3 5.4 5.5

Equivalent Sets; Finite Sets 195 Infinite Sets 203 Countable Sets 209 The Ordering of Cardinal Numbers 218 Comparability of Cardinal Numbers and the Axiom of Choice

C HAP T E R 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7

C HAP T E R 7 7.1 7.2 7.3 7.4 7.5

195

Concepts of Algebra: Groups

226

233

Algebraic Structures 233 Groups 241 Examples of Groups 246 Subgroups 251 Cosets and Lagrange's Theorem 258 Quotient Groups 262 Isomorphism; The Fundamental Theorem of Group Homomorphisms 266

Concepts of Analysis: Completeness of the Real Numbers Ordered Field Properties of the Real Numbers 273 The Heine-Borel Theorem 280 The Bolzano-Weierstrass Theorem 290 The Bounded Monotone Sequence Theorem 294 Equivalents of Completeness 303

Answers to Selected Exercises Index 337 List of Symbols 343

307

273

CHAPTER

1

Logic and Proofs

Although mathematics is both a science and an art, special characteristics distinguish mathematics from the humanities and from other sciences. Particularly important is the kind of reasoning that typifies mathematics. The natural or social scientist generally makes observations of particular cases or phenomena and seeks a general theory that describes or explains the observations. This approach is called inductive reasoning, and it is tested by making further observatfons. If the results are incompatible with theoretical expectations, the scientist usually must reject or modify the theory. Mathematicians, too, frequently use inductive reasoning as they attempt to describe patterns and relationships among quantities and structures. The characteristic thinking of the mathematician, however, is deductive reasoning, in which one uses logic to draw conclusions based on statements accepted as true. The conclusions of a mathematician are proved to be true, provided that the assumptions are true. If the results of a mathematical theory are deemed incompatible with some portion of reality, the fault lies not in the theory but with the assumptions about reality that make the theory inapplicable to that portion of reality. Indeed, the mathematician is not restricted to the study of observable phenomena, even though one can trace the development of mathematics back to the need to describe spatial relations (geometry) and motion (calculus) or to solve financial problems (algebra). Using logic, the mathematician can draw conclusions about any mathematical structure imaginable. The goal of this chapter is to provide a working knowledge of the basics of logic and the idea of proof, which are fundamental to deductive reasoning. This knowledge is important in many areas other than mathematics. For example, the thought processes used to construct an algorithm for a computer program are much like those used to develop the proof of a theorem.

1.1

Propositions and Connectives Natural languages such as English allow for many types of sentences. Some sentences are interrogatory (Where is my sweater?), others exclamatory (Oh, no!), and others have a definite sense of truth to them (Abe Lincoln was the first U.S. pres1

2

CHAPTER 1 Logic and Proofs

ident). A proposition is a sentence that is either true or false. Thus a proposition has exactly one truth value: true, which we denote by T, or false, which we denote by F. Some examples of propositions are: (a) (b) (c) (d)

{i is irrational. 1 + 1 = 5. The elephant will become extinct on the planet Earth before the rhinoceros will. Julius Caesar had two eggs for breakfast on his tenth birthday.

We are not concerned here with the difficulty of establishing the actual truth value of a proposition. We readily see that proposition (a) has the value T while (b) has the value F.1t may take many years to determine whether proposition (c) is true or false, but its truth value will certainly be established if either animal ever becomes extinct. If both species (and Earth) somehow survive forever, the statement is false. There may be no way ever to determine what value proposition (d) has. Nevertheless, each of the above is either true or false, hence is a proposition. Here are some sentences that are not propositions: (e) (f)

(g)

What did you say? x 2 = 36. This sentence is false.

Sentence (e) is an interrogative statement that has no truth value. Sentence (f) could be true or false depending on what value x is assigned. We shall study sentences of this type in section 1.3. Statement (g) is an example of a sentence that is neither true nor false, and it is referred to as a paradox. If (g) is true, then by its meaning (g) must be false. On the other hand, if (g) is false, then what it purports is false, so (g) must be true. Thus, (g) can have neither T nor F for truth value. The study of paradoxes such as this has played a key role in the development of modern mathematical logic. A famous example of a paradox formulated by the English logician Bertrand Russell is discussed in section 2.1. Propositions (a)-(d) are simple or atomic in the sense that they do not have any other propositions as components. Compound propositions can be formed by using logical connectives with simple propositions. DEFINITIONS

Given propositions P and Q,

The conjunction of P and Q, denoted P A Q, is the proposition "p and Q." P A Q is true exactly when both P and Q are true. The disjunction of P and Q, denoted P V Q, is the proposition "p or Q." P V Q is true exactly when at least one of P or Q is true. The negation of P, denoted ~P, is the proposition "not P." ~P is true exactly when P is false. If Pis "1 oF 3" and Q is "7 is odd," then P A Q is "1 oF 3 and 7 is odd." P V Q is "1 oF 3 or 7 is odd." ~Q

is "It is not the case that 7 is odd."

1.1

Propositions and Connectives

3

Since in this example both P and Q are true, P A Q and P V Q are true, while -Q is false. All of the following are true propositions: "It is not the case that

f16 > 4."

"ji < f3 or chickens have lips."

"Venus is smaller than Earth or 1 + 4 = 5." "6 < 7 and 7 < 8."

All of the following are false: "Mozart was born in Salzburg and TC is rational." "It is not the case that 10 is divisible by 2." "24 = 16 and a quart is larger than a liter." Other connectives commonly used in English are but, while, and although, each of which would normally be translated symbolically with the conjunction connective. A variant of the connective or is discussed in the exercises.

Example. Let M be "Milk contains calcium" and I be "Italy is a continent." Since M has the value T and I has the value F, "Italy is a continent and milk contains calcium," symbolized I AM, is false; "Italy is a continent or milk contains calcium," I V M, is true; "It is not the case that Italy is a continent," -I, is true. An important distinction must be made between a proposition and the form of a proposition. In the previous example, "Italy is a continent and milk contains calcium" is a proposition with a single truth value (F), while the propositional form P A Q, which may be used to represent the sentence symbolically, has no truth value. The form P A Q is an expression that obtains a value T or F after specific propositions are designated for P and Q (when for instance, we let P be "Italy is a continent" and Q be "Milk contains calcium"), or when the symbols P and Q are given truth values. By the form of a compound proposition, we mean how the proposition is put together using logical connectives. For components P and Q, P A Q and P V Q are two different propositional forms. Informally, a propositional form is an expression involving finitely many logical symbols (such as A and -) and letters. Expressions that are single letters or are formed correctly from the definitions of connectives are called well-formed formulas. For example, (P A (Q V is well-formed, whereas (P V Q-), (-P - Q), and vQ are not. A more precise definition and study of well-formed formulas may be found in Elliot Mendelson's An Introduction to Mathematical Logic (New York, Chapman & Hall, 1987). The truth values of a compound propositional form are readily obtained by exhibiting all possible combinations of the truth values for its components in a truth table. Since the connectives A and V involve two components, their truth tables must list the four possible combinations of the truth values of those components. The truth tables for P A Q and P V Q are

-Q»

4

CHAPTER 1 Logic and Proofs P

Q

T F T F

T T F F

P/\Q

P

Q

pVQ

T F F F

T F T F

T T F F

T T T F

Since the value of ~P depends only on the two possible values for P, its truth table is P

'-P

T

F

F

T

Frequently you will encounter compound propositions with more than two simple components. The propositional form (P /\ Q) V ~R has three simple components P, Q, and R; it follows that there are 23 = 8 possible combinations of truth values. The two main components are P /\ Q and ~R. We make truth tables for these and combine them by using the truth table for V. P

Q

R

P/\Q

-R

(P/\Q)v-R

T F T F T F T F

T T F F T T F F

T T T T F F F F

T F F F T F F F

F F F F T T T T

T F F F T T T T

The propositional form (~Q V P) /\ (R V S) has 16 possible combinations of values for P, Q, R, S. Two main components are ~Q V P and R V S. Its truth table is shown here: P

Q

R

S

-Q

-QvP

RVS

T F T F T F T F T F T F T F T F

T T F F T T F F T T F F T T F F

T T T T F F F F T T T T F F F F

T T T T T T T T F F F F F F F F

F F T T F F T T F F T T F F T T

T F T T T F T T T F T T T F T T

T T T T T T T T T T T T F F F F

(-Qv P)/\(Rv S) T F T T T F T T T F T T F F F F

1.1

Propositions and Connectives

5

Two propositions P and Q are equivalent if and only if they have the same truth value. The propositions "I + I = 2" and "6 < 10" are equivalent (even though they have nothing to do with each other) because both are true. The ability to write equivalent statements from a given statement is an important skill in writing proofs. Of course, in a proof we expect some logical connection between such statements. This connection may be based on the form of the propositions.

DEFINITION Two propositional forms are equivalent if and only if they have the same truth tables.

For example, the propositional forms P V (Q A P) and P are equivalent. To show this, we examine their truth tables. P

Q

QAP

T F T F

T T F F

T F F F

PV(QAP)

T F T F

Since the P column and the P V (Q A P) column are identical, the propositional forms are equivalent. This means that, whatever propositions we choose to use for P and for Q, the results will be equi valent. If we let P be "91 is prime" and Q be "1 + 1 = 2," then "91 is prime" is equivalent to the proposition "91 is prime, or 1 + 1 = 2 and 91 is prime." With these propositions for P and Q, Q is true and both P and P V (Q A P) are false. Thus, we have an instance of the second line of the truth table. Notice that "Two propositions are equivalent" has a different meaning from "Two propositional forms are equivalent." We don't look at truth tables to decide the equivalence of propositions, because a proposition has only one truth value. This makes the question of equivalence of propositions rather easy: all true propositions are equivalent to each other and all false propositions are equivalent to each other. On the other hand, propositional forms are neither true nor false; generally they have the value true for some assignments of truth values to their components and the value false for other assignments. Thus to decide equivalence of propositional forms, we must compare truth tables. Another example of equivalent propositional forms is P and ~ ( ~ P). The truth tables for these two proposi tional forms are shown: -(-P)

P

T F

T F

F T

DEFINITION lent to

~S.

A denial of a proposition S is any proposition equiva-

6

CHAPTER 1 Logic and Proofs

By definition, the negation ~P is a denial of the proposition P, but a denial need not be the negation. A proposition has only one negation but many different denials. The ability to rewrite the negation of a proposition into a useful denial will be very important for writing indirect proofs (see section 1.4).

Example. The proposition P: "n is rational" has negation ~P: "It is not the case that n is rational." Some useful denials are "n is irrational." "n is not the quotient of two integers." "The decimal expansion of n is not repeating." Note that since P is false, all denials of P are true.

Example. The proposition "The water is cold and the soap is not here" has two components, C: "The water is cold" and H: "The soap is here." The negation, ~(C /I. ~H), is "It is not the case that the water is cold and the soap is not here." Some other denials are "Either the water is not cold or the soap is here." "It is not the case that the water is cold and the soap is not here and the water is cold."

It may be verified by truth tables that the propositional forms ~[(C /I. ~H) /I. C] are equivalent to ~(C /I. ~H).

(~C) V

Hand

Note that the negation in the last example is ambiguous when written in English. Does the "It is not the case" refer to the entire sentence or just to the component "The water is cold"? Ambiguities such as this are allowable in conversational English but can cause trouble in mathematics. To avoid ambiguities, we introduce delimiters such as parentheses ( ), square brackets [ ], and braces { }. The negation above may be written symbolically as ~(C /I. ~H). To avoid writing large numbers of parentheses, we use the rule that, first, ~ applies to the smallest proposition following it, then /I. connects the smallest propositions surrounding it, and, finally, V connects the smallest propositions surrounding it. Thus, ~P V Q is an abbreviation for (~P) V Q. The negation of the disjunction P V Q must be written with parentheses ~(P V Q). The propositional form P /I. ~Q V R abbreviates [P /I. (~Q)] V R. As further examples, P V Q /I. R abbreviates P V [Q /I. R]. P V ~Q V ~R abbreviates [P V (~Q)] V ~P V ~Q abbreviates (~P) V (~Q).

(~R).

When the same connective is used several times in succession, parentheses may also be omitted. We reinsert parentheses from the left, so that P V Q V R is really (P V Q) V R. For example, R /I. P /I. ~ P /I. Q abbreviates [(R /I. P) /I. (~P)] /I. Q, whereas R V P /I. ~P V Q, which does not use the same connective consecutively, abbreviates (R V [P /I. ( ~ P)]) V Q. Leaving out parentheses is not required; some propositional forms are easier to read with a few well-chosen "unnecessary" parentheses.

1.1

Propositions and Connectives

7

Some compound propositional forms always yield the value true just because of the nature of their form. Tautologies are propositional forms that are true for every assignment of truth values to their components. Thus a tautology will have the value true regardless of what proposition(s) we select for the components. For example, the Law of Excluded Middle, P V -P, is a tautology. Its truth table is P

F T

T F

T T

We know that "the ball is red or the ball is not red" is true because it has the form of the Law of Excluded Middle.

Example. Show that (P V Q) V ( - P A -Q) is a tautology. We see that the truth table for the propositional form is P

Q

pVQ

-P

-Q

-PA-Q

(PvQ)V(-PA-Q)

T

T T

T T T

F T

F F

F F F

F

T

T T T T

F T

F

F F

Thus (P

V

F

T T

T

Q) V (-P A -Q) is a tautology.

A contradiction is the negation of a tautology. Thus -(P V -P) is a contradiction. The negation of a contradiction is, of course, a tautology. Conjunction, disjunction, and negation are very important in mathematics. Two other important connectives, the conditional and biconditional, will be studied in the next section. Other connectives having two components are not as useful in mathematics, but some are extremely important in digital computer circuit design.

Exercises 1.1 1. Which of the following are propositions? (a) Where are my car keys? (b) Christopher Columbus wore red boots at least once. (c) The national debt of Poland in 1938 was $2,473,596.38. (d) x 2 ;::: 20 (e) Between January 1,2205 and January 1,2215, the population ofthe world will double. (f) There are no zeros in the decimal expansion of n. (g) She works in New York City. (h) Keep your elbows off the table!

* *

*

8

CHAPTER 1

Logic and Proofs

(i)

(j)

2.

* * * *

3.

There are more than 3 false statements in this book and this statement is one of them. There are more than 3 false statements in this book and this statement is not one of them.

Make truth tables for each of the following propositional forms. (a) (c) (e) (g) (i) (k)

P/\~P

P/\(QVR) P /\ ~Q (P /\ Q) V ~Q) (P V ~Q) /\R P /\ P

(b) (d) (f) (h) (j) (I)

PV~P

(P/\Q)V(P/\R) P /\(Qv ~Q) ~(P /\ Q) ~P /\ ~Q (P /\ Q) V (R /\ ~S)

Which of the following pairs of propositional forms are equivalent?

* * * * * * *

(a) (c) (e) (g) (i) (k) (m)

4.

If P, Q, and R are true while S and Tare false, which of the following are true?

* * * * * *

(a) (c) (e) (g) (i) (k)

P/\P,P P/\Q,Q/\P (P/\Q)/\R,P/\(Q/\R) ~P /\ ~Q, ~(P /\ ~Q) (P/\Q)VR,P/\(QVR) ~(P/\Q),(~P)V(~Q)

P/\(QVR),(P/\Q)V(P/\R) Q/\(R/\S) (PVQ)/\(RVS) (~P)V(Q/\~Q) ((~Q) V

S) /\ (Q V S) (PVS)/\(PVT) (~P)/\(Qv~Q)

(b) (d) (f) (h) (j) (I) (n) (b) (d) (f) (h) (j) (I)

pVP,P pVQ,Qv~P ~(P/\Q),~P/\~Q

(P V Q) V R, P V (Q V R) ~(pvQ),(~P)/\(~Q)

(P/\Q)VR,PV(Q/\R) (~P)V(~Q),~(pv~Q)

QV(R/\S) ((~P)V(~Q»V((~R)V(~S»

(~P)V(~Q)

(S /\R) V (S /\ T) ((~T)/\P)V(T/\P) (~R)/\(~S)

5.

Give a useful denial of each statement. (a) x is a positive integer. (Assume that x is some fixed integer.) (b) We will win the first game or the second one. (c) 5;::: 3 (d) 641,371 is a composite integer. (e) Roses are red and violets are blue. (f) x < y or m 2 < I (Assume that x, y, and m are fixed real numbers.) (g) T is not green or T is yellow. (h) Sue will choose yogurt but will not choose ice cream. (i) n is even and n is not a multiple of 5. (Assume that n is a fixed integer.)

6.

P, Q, and R are propositional forms, and P is equivalent to Q, and Q is equiv-

* * * *

* 7.

alent to R. Prove that (a) Q is equivalent to P. (c) ~Q is equivalent to ~P. (e) P V Q is equivalent to Q V R.

(b) (d)

P is equivalent to R. P /\ Q is equivalent to Q /\ R.

UseA: "Horses have four legs," B: "17 is prime," and C: "Three quarters equal one dollar" to write the propositional form of each of the following. Decide which are true. (a) Either horses have four legs or 17 is not prime. (b) Neither do three quarters equal a dollar nor do horses have four legs.

1.2

*

9

17 is prime and three quarters do not equal one dollar. Horses have four legs but three quarters do not equal one dollar.

S.

Let P be the sentence "Q is true" and Q be the sentence "P is false." Is P a proposition? Explain.

9.

The word or is used in two different ways in English. We have presented the truth table for v, the inclusive or, whose meaning is "one or the other or both." The exclusive or, meaning "one or the other but not both" and denoted Q), has its uses in English, as in "She will marry Heckle or she will marry leckie." The "inclusive or" is much more useful in mathematics and is the accepted meaning unless there is a statement to the contrary. (a) Make a truth table for the "exclusive or" connective, Q). (b) Show that A Q) B is equivalent to (A V B) A -(A A B).

* 10.

*

1.2

(c) (d)

Conditionals and Biconditionals

Determine whether each of the following is a tautology, a contradiction, or neither. Prove your answers. (a) (b) (c) (d) (e) (f)

(PAQ)V((-P)A(-Q» -(P A -P) (PAQ) V ((-P) V (-Q» (A A B) V (A A -B) V ((-A) A B) V ((-A) A (-B» (Q A -P) A -(P A R) PV((-Q)AP)A(RVQ)'

Conditionals and Biconditionals The most important kind of proposition in mathematics is a sentence of the form "If P, then Q." Examples include "If a natural number is written in two ways as a product of primes, then the two factorizations are identical except for the order in which the prime factors are written"; "If two lines in a plane have the same slope, then they are parallel"; and "If f is differentiable at Xo andf(xo) is a relative minimum for f, thenf'(xo) = 0." DEFINITIONS Given propositions P and Q, the conditional sentence P =} Q (read "P implies Q") is the proposition "If P, then Q." The proposition P is the antecedent and Q is the consequent. The conditional sentence P =} Q is true whenever the antecedent is false or the consequent is true. Thus, P =} Q is defined to be equivalent to (-P) V Q.

The truth table for P =} Q is P (antecedent)

Q (consequent)

P~Q

T F T F

T T F F

T T F T

10

CHAPTER 1

Logic and Proofs

This table gives P =:} Q the value F only when P is true and Q is false, and thus it agrees with the meaning of "if ... , then ... " in promises. For example, the person who promises, "If Lincoln was the second U.S. president, I'll give you a dollar" would not be a liar for failing to give you a dollar. In fact, he could give you a dollar and still not be a liar. In both cases we say the statement is true because the antecedent is false. One curious consequence of the truth table for P =:} Q is that conditional sentences may be true even when there is no connection between the antecedent and the consequent. The reason for this is that the truth value of P =:} Q depends only upon the truth value of components P and Q, not on their interpretation. For this reason all of the following are true:

!

sin 30° = =:} 1 + 1 = 2. Mars has ten moons =:} 1 + 1 = 2. Mars has ten moons =:} Paul Revere made plastic spoons. and both of the following are false: 1 + 2 = 3 =:} 1 < o. Ducks have webbed feet

=:}

Canada lies south of the equator.

Our truth table definition of =:} is not really unfamiliar; it captures the same meaning for "if ... , then ... " that you have always used in mathematics. We all agree that "If x is odd, then x + 1 is even" is a true statement about any integer x. It would be hopeless to protest that in the case where x is 6, then x + 1 is 7, which is not even. After all, the claim is only that if x is odd, then x + 1 is even. We know that "If (1, 3) and (2, 5) are points on a line L, then the line L has slope 2" is true because (5 - 3)/(2 - 1) = 2. The truth values of the antecedent and consequent depend on what line L we are talking about, but in all cases the value of the conditional sentence is true: In the case that the line L is y = 2x + 1, the antecedent and consequent are both true. This matches the first line of the truth table for P =:} Q, where P =:} Q is true. In some cases, such as the line y = 2x + 4, the antecedent is false and the consequent is true. This matches the second line of the truth table, where P =:} Q is also true. In other cases, such as y = 3x + 1, we find instances of the fourth line of the truth table, since both the antecedent and consequent are false. Again, P =:} Q is true. There is no example of the third line of the truth table for P =:} Q for this sentence; that is why "If (1,3) and (2, 5) are points on a line L, then the line L has slope 2" is true. Note that in the truth table of P =:} Q the only line in which both P and P =:} Q are true is the first line, in which case Q is also true. In other words, if we know that

1.2

Conditionals and Biconditionals

11

both P and P ==> Q are true, then we know that Q must be true. This deduction, called

modus ponens, is one of several we will discuss in section 1.4. Two propositions closely related to the conditional sentence P ==> Q are its converse and its contrapositive.

DEFINITIONS For propositions P and Q, the converse of P ==> Q is Q ==> P, and the contrapositive of P ==> Q is (~Q) ==> (~P).

For the conditional sentence "If a function f is differentiable at xo, then f is continuous at xo," its converse is "If f is continuous at xo, then f is differentiable at xo," whereas the contrapositive is "If f is not continuous at xo, then f is not differentiable at xo." Calculus students know that the converse is a false statement. If P is the proposition "It is raining here" and Q is "It is cloudy overhead," then P ==> Q is true. Its contrapositive is "If it is not cloudy overhead, then it is not raining here," which is also true. However, the converse "If it is cloudy overhead, then it is raining here" is not a true proposition. We describe the relationships between a conditional sentence and its contrapositive and converse in the following theorem.

Theorem 1.1

(a)

(b)

The propositional form P ==> Q is equivalent to its contrapositive (~Q) ==> (~P). The propositional form P ==> Q is not equivalent to its converse, Q ==> P.

Proof. A proof requires examining the truth tables: p

Q

P~Q

~Q

~P

(~Q)~(~P)

Q~P

T

T T

T T

F F

F

T T

T

T

F F

F

T T

F

F

T

T

T T

F T

F

T

F

Comparing the third and sixth columns, we conclude that P ==> Q is equivalent to (~Q) ==> (~P). Comparing the third and seventh columns, we see they differ in the second and third lines. Thus, P ==> Q and Q ==> P are not equivalent. The equivalence of a conditional sentence and its contrapositive will be the basis for an important proof technique developed in section 1.4 (proof by contraposition). However, no proof technique will be developed using the converse because the truth of a conditional sentence cannot be inferred from the truth of its converse. The converse cannot be used to prove a conditional sentence. Closely related to the conditional sentence is the biconditional sentence P {=} Q. The double arrow {=} reminds one of both <= and ==> , and this is no accident, for P {=} Q is equivalent to (P ==> Q) /I. (Q ==> P).

12

CHAPTER 1 Logic and Proofs

DEFINITION For propositions P and Q, the biconditional sentence P <=> Q is the proposition "P if and only if Q." The sentence P <=> Q is true exactly when P and Q have the same truth values.

The truth table for P <=> Q is P

Q

P¢:!?Q

T

T T

T

F T

F

F F

F F T

As a form of shorthand, the words "if and only if' are frequently abbreviated to "iff' in mathematics. The statements "A rectangle is a square iff the rectangle's diagonals are perpendicular" and "I

+7=

6 iff

Ii + f3 = )5"

are both true biconditional sentences, while "Lake Erie is in Peru iff n is an irrational number" is a false biconditional sentence. Any properly stated definition is an example of a biconditional sentence. Although a definition might not include the iff wording, biconditionality does provide a good test of whether a statement could serve as a definition or just a description. The sentence "A diameter of a circle is a chord of maximum length" is a correct definition of diameter because "A chord is a diameter iff the chord has maximum length" is a true proposition. However, the senteqfes "A sundial is an instrument for measuring time" and "A square is a quadrilateralwhose interior angles are right angles" can be recognized as descriptions rather than definitions. Because the biconditional sentence P <=> Q has the value T exactly when the values of P and Q are the same, we can use the biconditional connective to restate the meaning of equivalent propositional forms. That is, The propositional forms P and Q are equivalent precisely when P <=> Q is a tautology. One key to success in mathematics is the ability to replace a statement by a more useful or enlightening one. This is precisely what you do to "solve" the equation x 2 - 7x = -12 by the method of factoring: x 2 -7x= -12<=>X2_7x+ 12=0 <=>(x-3)(x-4)=0 <=> x - 3 = 0 or x - 4 <=> x = 3 or x = 4

=0

1.2

Conditionals and Biconditionals

13

Each statement is simply an equivalent of its predecessor but is more illuminating as to the solution. The ability to write equivalents is crucial in writing proofs. The next theorem contains seven important pairs of equivalent propositional forms. They should be memorized.

Theorem 1.2

For propositions P, Q, and R, (a) (b) (c) (d) (e) (f)

(g)

P <=> Q is equivalent to (P =:} Q) /I. (Q =:} P). -(P /I. Q) is equivalent to (-P) V (-Q). -(P V Q) is equivalent to (-P) /I. (-Q). -(P =:} Q) is equivalent to P /I. -Q. -(P /I. Q) is equivalent to P =:} -Q. P /I. (Q V R) is equivalent to (P /I. Q) V (P /I. R). P V (Q /I. R) is equivalent to (P V Q) /I. (P V R).

You will be asked to give a proof of this theorem in exercise 7. Before plunging into the truth table computations, you should think about the meaning behind each equivalence. For example, parts (b) and (c), later referred to as De Morgan's Laws, are quite similar in form. They distribute negation over conjunction (b) and over disjunction (c). In (b), -(P /\ Q) is true precisely when P /I. Q is false. This happens when one of P or Q is false, or, in other words, when one of -P or -Q is true. Thus, -(P /I. Q) is equivalent to (-P) V (-Q). Another way to say this is, "If you do not have both P and Q, then either you do not have P or you do not have Q." Your reasoning for part (d) should be something like this: If -(P =:} Q) is true, then P =:} Q is false, which forces P to be true and Q to be false. But this means that both P and -Q are true, and so P /I. -Q is true. This reasoning can be reversed to show that if P /I. -Q is true, then -(P =:} Q) is true. We conclude that -(P =:} Q) is true precisely when P /I. -Q is true, and thus they are equivalent. For example, given any fixed triangle ABC, the statement "It is not the case that if triangle ABC has a right angle, then it is equilateral" is equivalent to "Triangle ABC has a right angle and is not equilateral." Recognizing the structure of a sentence and translating the sentence into symbolic form using logical connectives is an aid in determining its truth or falsity. The translation of sentences into propositional symbols is sometimes very complicated because English is such a rich and powerful language, with many nuances, and because the ambiguities we tolerate in English would destroy structure and usefulness if we allowed them in mathematics. Connectives in English that may be translated symbolically using the conditional or biconditional logical connectives present special problems. The word unless is variously used to mean a conditional or its converse or a biconditional. For example, consider the sentence "The Dolphins will not make the play-offs unless the Bears win all the rest of their games." In conversation an explanation can clarify the meaning. Lacking that explanation, here are three of the nonequivalent ways people translate the sentence, using the symbols D: "The Dolphins make the playoffs" and B: "The Bears win all the rest of their games." Dictionaries indicate that

14

CHAPTER 1 Logic and Proofs

the conditional meaning of unless is preferred (the first translation), but the speaker may have meant any of the three: (~B)

==> (~D)

(~D)==>(~B) (~B) ¢=} (~D)

Sometimes a sentence in English explicitly uses a conditional connective but the converse is understood, so that the meaning is biconditional. For example, "I will pay you if you fix my car" and "I will pay you only if you fix my car" both mean "I will pay you if, but only if, you fix my car." Contrast this with the situation in mathematics: "If x = 2, then x is a solution to X 2 = 2x" is not to be understood as a biconditional. Here are some phrases in English that are ordinarily translated by using the connectives ==> and ¢=} • Use P ==> Q to translate: If P, then Q. P implies Q. P is sufficient for Q. Ponly ifQ. Q, ifP. Q whenever P. Q is necessary for P. Q, whenP. Use P ¢=} Q to translate: P P P P

if and only if Q. if, but only if, Q. is equivalent to Q. is necessary and sufficient for Q.

In the following examples of sentence translations, it is not essential to know the meaning of all the words because the logical connectives are what concern us.

Examples.

Assume that Sand G have been specified. The sentence "S is compact is sufficient for S to be bounded"

is translated S is compact ==> S is bounded. The sentence "A necessary condition for a group G to be cyclic is that G is abelian" is translated G is cyclic ==> G is abelian.

1.2

Conditionals and Biconditionals

15

The sentence "A set S is infinite if S has an uncountable subset" is translated S has an uncountable subset ==> S is infinite. If we let P denote the proposition "Roses are red" and Q denote the proposition "Violets are blue," we can translate the sentence "It is not the case that roses are red, nor that violets are blue" in at least two ways: ~(P V Q) or (~P) /\ (~Q). Fortunately, these are equivalent by Theorem 1.2(c). Note that the proposition "Violets are purple" requires a new symbol, say R, since it expresses a new idea that cannot be formed from the components P and Q. The sentence "17 and 35 have no common divisors" shows that the meaning, and not just the form of the sentence, must be considered in translating; it cannot be broken up into the two propositions: "17 has no common divisors" and "35 has no common divisors." Compare this with the proposition "17 and 35 have digits totaling 8," which can be written as a conjunction.

Example.

Suppose b is a real number. "If b is an integer, then b is either even or odd" may be translated into P ==> (Q V R), where P is "b is an integer," Q is "b is even," and R is "b is odd."

Example.

Suppose a, b, and p are integers. "If p is a prime number that divides ab, then p divides a or b" becomes (P /\ Q) ==> (R V S), when P is "p is prime," Q is "p divides ab," R is "p divides a," and S is ''p divides b."

The convention governing use of parentheses, adopted at the end of section 1.1, can be extended to the connectives ==> and ¢=}. The connectives ~, /\, V, ==>, and ¢=} are applied in the order listed. That is, ~ applies to the smallest possible proposition, and so forth, and otherwise parentheses are added left to right. For example, P ==> ~Q V R ¢=} S is an abbreviation for (P ==> [( ~Q V R]) ¢=} S, while P V ~Q ¢=} R==>S abbreviates [PV(~Q)]¢=}(R==>S), and P==>Q==>R abbreviates (P==>Q) ==> R.

Exercises 1.2 1.

*

Identify the antecedent and the consequent for each of the following conditional sentences. Assume that a, b, andfrepresent some fixed sequence, integer, or function, respectively. (a) If squares have three sides, then triangles have four sides. (b) If the moon is made of cheese, then 8 is an irrational number. (c) b divides 3 only if b divides 9.

16

CHAPTER 1 Logic and Proofs

* *

"*

2.

(d) (e) (f) (g) (h) (i)

The differentiability of ! is sufficient for! to be continuous. A sequence a is bounded whenever a is convergent. A function! is bounded if ! is integrable. I + 2 = 3 is necessary for I + I = 2. The fish bite only when the moon is full. A grade point average of 3.0 is sufficient to graduate with honors.

Write the converse and contrapositive of each conditional sentence in exercise I.

3. Identify the antecedent and consequent for each conditional sentence in the following statements from this book. Theorem l.3, section l.3 Theorem 2.4, section 2.1 Theorem 2.20, section 2.6 Theorem 4.3, section 4.2

(a) (c) (e) (g)

Exercise 6(c), section 1.4 Theorem 2.12, section 2.5 Theorem 3.8, section 3.4 Lemma 5.3, section 5.1

4.

Which of the following conditional sentences are true? (a) If triangles have three sides, then squares have four sides. (b) If a hexagon has six sides, then the moon is made of cheese. (c) If7 + 6 = 14, then 5 + 5 = 10. (d) If 5 < 2, then 10 < 7. (e) If one interior angle of a right triangle is 92°, then the other interior angle is 88°. (f) If Euclid's birthday was April 2, then rectangles have four sides. (g) 5 is prime if ji is not irrational. (h) 1 + I = 2 is sufficient for 3 > 6. (i) Horses have four legs whenever September 15 falls on a Saturday.

5.

Which of the following are true? (a) Triangles have three sides iff squares have four sides. (b) 7 + 5 = 12 iff 1 + 1 = 2. (c) b is even iff b + 1 is odd. (Assume that b is some fixed integer.) (d) 5 + 6 = 6 + 5 iff7 + I = 10. (e) A parallelogram has three sides ifd7 is prime. (f) The Eiffel Tower is in Paris if and only if the chemical symbol for helium is H. < {IT + iff < {IT - jf6. (g) jf6 +

6.

Make truth tables for these propositional forms.

* * *

* *

J13

* *

"*

(b) (d) (f) (h)

(a) (b) (c) (d) (e) (f)

Jf2 J13 - Jf2

P~(Q/\P).

(( ~P)

~

Q) V (Q ¢=} P). (Q ¢=} P). (P V Q) ~ (P /\ Q). (P /\ Q) V (Q /\R) ~ P V R. (~Q) ~

[(Q~S)/\(Q~R)]~[(pvQ)~(SVR)].

7.

Prove Theorem 1.2 by constructing truth tables for each equivalence.

S.

Rewrite each of the following sentences using logical connectives. Assume that each symbolf, n, x, S, B represents some fixed object.

1.2

*

(a)

If f has a relative minimum at Xo and if !'(xo) =

* *

9.

*

Conditionals and Biconditionals

o.

17

f is differentiable at xo, then

If n is prime, then n = 2 or n is odd. A number x is real and not rational whenever x is irrational. Ifx=lorx=-I,thenlxl=1. f has a critical point at Xo iff !'(xo) = 0 or !'(xo) does not exist. (f) S is compact iff S is closed and bounded. (g) B is invertible is a necessary and sufficient condition for det B *- O. (h) 6::::: n - 3 only if n > 4 or n > 10. (i) x is Cauchy implies x is convergent. (b) (c) (d) (e)

Show that the following pairs of statements are equivalent. (a) (P V Q) =} R and ~R=} (~P /\ ~Q) (b) (P /\ Q) =} Rand (P /\ ~R) =} ~Q (c)

P=}(Q/\R)and(~QV~R)=}~P

(d) (e)

P =} (Q V R) and (P /\ ~R) =} Q (P =} Q) =} Rand (P /\ ~Q) V R P <=> Q and (~P V Q) /\ (~Q V P)

(f)

10.

Give, if possible, an example of a true conditional sentence for which (a) the converse is true. (b) the converse is false. (c) the contrapositive is false. (d) the contrapositive is true.

11.

Give, if possible, an example of a false conditional sentence for which (a) the converse is true. (b) the converse is false. (c) the contrapositive is true. (d) the contrapositive is false.

12.

Give the converse and contrapositive of each sentence of exercise 8(a), (b), (c), and (d). Tell whether each converse and contrapositive is true or false.

13.

The inverse, or opposite, of the conditional sentence P =} Q is ~ P =} ~Q. (a) Show that P =} Q and its inverse are not equivalent forms. (b) For what values of the propositions P and Q are P =} Q and its inverse both true? (c) Which is equivalent to the converse of a conditional sentence, the contrapositive of its inverse, or the inverse of its contrapositive?

14.

Determine whether each of the following is a tautology, a contradiction, or neither. (a) [(P =} Q) =} P] =} P.

* *

* *

(b) (c) (d) (e) (f) (g) (h) (i)

P<=>P/\(pvQ). P=} Q<=> P /\ ~Q. ~ P=}[P=}(P=}Q)]. P /\ (Q V ~Q) <=> P. [Q /\ (P =} Q)] =} P. (P <=> Q) <=> ~(~P V Q) V (~P /\ Q). [P =} (Q V R)] =} [(Q =} R) V (R =} P)].

(j)

P /\ (P <=> Q) /\ ~Q. (P V Q) =} Q =} P.

(k) (I)

[P =} (Q /\ R)] [P =} (Q /\ R)]

=} =}

[R =} (P =} Q)]. R =} (P =} Q).

18

1.3

CHAPTER 1

Logic and Proofs

Quantifiers Unless some particular value of x has been assigned, the sentence "x 2: 3" is not a proposition because it is neither true nor false. When the variable x is replaced by certain values (for example, 7), the resulting proposition is true, while for other values of x (for example, 2) it is false. This is an example of an open sentence or predicate-that is, a sentence containing one or more variables-which becomes a proposition only when the variables are replaced by the names of particular objects. For notation, if an open sentence is called P and the variables are Xl, xz, ... ,xk, we write P(xJ, Xz, ... , xk), and in the case of a single variable x, we write P(x). The sentence "Xl is equal to X2 + X3" is an open sentence with three variables. If we denote this sentence by P(xJ, Xz, x3), then P(7, 3,4) is true since 7 = 3 + 4, while P(l, 2, 3) is false. The collection of objects that may be substituted to make an open sentence a true proposition is called the truth set of the sentence:t Before the truth set can be determined, we must know what objects are available for consideration. That is, we must have specified a universe of discourse. In many cases the universe will be understood from the context. However, there are times when it must be specified. For the sentence "x < 5," with the universe all the natural numbers, the truth set is {I, 2, 3, 4}, while with the universe all integers, the truth set is { ... , -2, -1,0, 1,2,3, 4}. For a sentence like "Some intelligent people admire musicians," the universe is presumably the entire human population of Earth. Let Q(x) be the sentence "x 2 = 4." With the universe specified as all real numbers, the truth set for Q(x) is {2, -2}. With the universe the set of natural numbers, the truth set is {2}. With a particular universe in mind, we say two open sentences P(x) and Q(x) are equivalent iff they have the same truth set. Example. The sentences "3x + 2 = 20" and "x = 6" are equivalent regardless of what number system we use as the universe. On the other hand, "x 2 = 4" and "x = 2" are equivalent when the universe of discours~js the set of all positive integers but are not equivalent for the universe consisting of all integers. An open sentence P(x) is not a proposition, but Pea) is a proposition for any a in the universe. Another way to construct a proposition from P(x) is to modify it with a quantifier. DEFINITIONS For an open sentence P(x) with variable x, the:sentence (Vx)P(x) is read "for all x, P(x)" and is true precisely when the truth set for P~)

is the entire universe. The symbol V is called the universal quantifier. The sentence C3x)P(x) is read "there exists x such that P(x)" and is true precisely when the truth set for P(x) is non empty. The symbol .3 is called the existential quantifier. 'The reader unfamiliar with the basic concepts of set theory is encouraged to read the first few pages of section I of chapter 2 before proceeding.

1.3

Quantifiers

19

Before we can say (Vx)A(x) is true, we must know a great deal about the truth set of A (x)-it must be exactly the universe. On the other hand, we know (3x)A(x) is true as soon as we know that Pea) is true for some object a in the universe. We may say that (3x)A(x) is true even when we do not know a specific object in the truth set of A(x), only that there is one.

Examples.

Let the universe be all real numbers.

(Vx)(x::::: 3) is false, since there are many real numbers x for which x::::: 3 is not

true. ('Ix) (Ix I > 0) is false. The truth set of Ixl

> 0 contains almost all reals but does not contain O. (Vx)(x + 2> x) is true. (Vx)(x < -2 V -2::; x < 5 V x::::: 5) is true. Examples.

Again, let the universe be all real numbers.

(3x)(x::::: 3) is true because the truth set of x::::: 3 contains 3, 7.02, and many

other real numbers. (3x)(lxl = 0) is true. The nonempty truth set of Ixl = 0 contains only the elementO. (3x)(x 2 = -1) is false since x 2 = -1 is never true for real numbers x. We agree that "All apples have spots" is quantified with V, but what form does it have? If we use A(x) to represent "x is an apple" and Sex) to represent "x has spots," and we consider the universe to be the set of all fruits, should we write (Vx)[A(x) /I. Sex)] or (Vx)[A(x) => Sex)]? First, (Vx)[A(x) /I. Sex)] says "For all objects x in the universe, x both is an apple and has spots." Since we don't really intend to say that every fruit is a spotted apple, this is not the meaning we want. Our other choice, ('Ix) [A (x) => Sex)], says "For all objects x, if x is an apple, then x has spots," or "If a fruit is an apple, then it has spots," which is the meaning we seek. In general, a sentence of the form "All P(x) are Q(x)" should be symbolized (Vx)[P(x) => Q(x)]. Now consider "Some apples have spots." Should this be symbolized (3x)[A(x) /\ Sex)] or (3x)[A(x) => Sex)]? The first translates as "There is an object x such that it is an apple and has spots," and thus (3x)[A(x) /I. Sex)] is correct. On the other hand, (3x)[A(x) => Sex)] reads "There is an object x such that, if it is an apple, then it has spots." Thus, (3 x)[A (x) => Sex)] is not a correct symbolic translation, as it does not ensure the existence of apples with spots. In general, a sentence of the form "Some P(x) are Q(x)" should be symbolized (3x)[P(x) /\ Q(x)].

Examples. The sentence "For every odd prime x less than 10, x 2 + 4 is prime" means that if x is prime, and odd, and less than 10, then x 2 + 4 is prime. It is written symbolically as ('Ix) [x is prime /I. x is odd /I. x

< 10 => x 2 + 4 is prime].

The sentence "Some functions defined at 0 are not continuous at 0" can be written symbolically as (3f)[/ is defined at 0/1./ is not continuous at 0].

20

CHAPTER 1 Logic and Proofs

You should be alert for hidden quantities, because sometimes statements have the meaning of a quantified sentence even when the words/or all or there exists are not present.

Example. The sentence "Some real numbers have a multiplicative inverse" could be symbolized (3x)(x is a real number Ax has a real multiplicative inverse).

However, "x has an inverse" means some number is an inverse for x (hidden quantifier), so a more complete symbolic translation is (3x)[x is a real number A (3y) (y is a real number Axy

= 1)].

Example. The sentence "Some people dislike taxes" may be symbolized in different ways depending on how it is interpreted (another case where English allows for imprecision). The "some people" phrase must be symbolized with an existential quantifier. So our first symbolization is (3x)(x is a person Ax dislikes taxes).

A slight ambiguity could arise because of the phrase "dislikes taxes." Since there is no word modifying "taxes," the statement probably refers to a dislike of all taxes. The preferred interpretation i3 (3x)[(x is a person A(Vy)(y is a tax

=:}

x dislikes y)].

However, the statement may refer to a dislike of certain taxes: (3x)[(x is a person A(3y)(y is a tax Ax dislikes y)].

We note that this is a symbolization of the more explicit statement "Some people dislike some taxes." Let us consider the translation of "Some integers are even and some are odd." One correct translation is (3x)(x is even) A (3x)(x is odd), because the first quantifier (3x) extends only as far as the word even. After that, any variable (even x again) may be used to express "some are odd." It is equally correct and sometimes preferable to write (3x)(x is even) A (3y)(y is oddf,lbut it would be wrong to write (3x)(x is even A x is odd). Why? You will be reminded in chapter 2 that the symbolism "x E A" stands for "The object x is an element of the set A." Since this occurs so frequently in combinations with quantifiers, we adopt two abbreviations. The sentence "Every x E A has the property P" could be restated as "If x E A, then x has property P" and symbolized by (Vx) [x E A=:} P(x)]. This is abbreviated to (Vx E A)P(x). Likewise, "Some x E A has the property P" can be restated as "There is an x that is in A and that has property P" and symbolized by (3x)[x E A A P(x)]. This is abbreviated by (3x E A)P(x).

Examples. We take the universe to be all living things in these examples. Let P be the set of all people. The sentence "Some people are tall" may be symbolized

1.3 Quantifiers

21

(3x)(x E P I\x is tall)

or, more compactly, (3x E P)(x is tall).

The sentence "Everyone has some faults" is translated either by (\;fx)(x E P ==> x has some faults)

or (\;fx E P)(x has some faults).

Notice that we do not attempt to introduce a quantifier to symbolize "some faults" because faults are not in the universe, which contains only living things.

Example. Let N denote the set of natural numbers and ~ denote the set of real numbers. "For every natural number there is a real number greater than the natural number" may be symbolized by (\;fn)[n EN==> (3r)(r E ~ and r

>

n)]

or (\;fn E N)(3r E ~)(r

>

n).

DEFINITION Two quantified sentences are equivalent for a particular universe iff they have the same truth value in that universe. Two quantified sentences are equivalent iff they are equivalent in every universe.

Example. (\;fx) (x > 3) is equivalent to (\;fx)(x ::::: 4) in the universe of integers because both are false. In another universe, consisting of all real numbers larger than 3.4, these two sentences are not equivalent because (\;fx) (x > 3) is true but (\;fx) (x::::: 4) is false. Again, it is necessary to make a distinction between a sentence and its logical form. With universe the set of all integers, the sentence "All integers are odd" is an instance of the logical form (\;fx)P(x), where P(x) is "x is odd." The form (\;fx)P(x) is neither true nor false but becomes false in this case when "x is odd" is used for P(x). Two logical forms of quantified sentences are equivalent if the truth of one implies the truth of the other, and conversely, for every possible meaning of the open sentences in every universe. For example, the forms (\;fx)[P(x) 1\ Q(x)] and (\;fx)[Q(x) 1\ P(x)] are equivalent. Another pair of equivalent sentences is (\;fx) [P(x) ==> Q(x)] and (\;fx)[~Q(x) ==> ~P(x)]. There is a natural interplay between negation and the two quantifiers \;f and 3.

Theorem 1.3

If A (x) is an open sentence with variable x, then

(a)

~(\;fx)A(x)

(b)

~(3x)A(x)

is equivalent to Ox) ~ A(x). is equivalent to (\;fx) ~ A(x).

22

CHAPTER 1 Logic and Proofs Proof.

(a)

The sentence

~(Vx)A(x)

is true

iff (Vx)A(x) is false iff the truth set of A(x) is not the universe iff the truth set of ~ A (x) is nonempty iff (.3x)

(b)

~ A(x)

is true.

Thus, ~(Vx)A(x) is true if and only if (.3x) ~ A(x) is true, so the propositions are equivalent. The proof of this part is exercise 3. There is a proof similar to part (a) and another proof that uses part (a). _

Example. Find a denial of "All primes are odd." Use the natural numbers as the universe. The sentence may be symbolized (Vx)(x is prime ==>x is odd).

The negation is ~(Vx)(x

is prime ==> x is odd)

which by Theorem 1.3(a) is equivalent to

Ch)[ ~(x is prime ==> x is odd)]. By Theorem 1.2(d) this is equivalent to (.3x)[x is prime /\ ~(x is odd)].

Thus, a denial is "There exists a number that is prime and not odd" or "Some prime number is even."

Example. Find a denial of "Every positive real number has a multiplicative inverse." Let the universe be the set of all real numbers. The sentence is symbolized by (Vx)[x>O==>(.3y)(xy7' 1)]. <, l

The negation is ~(Vx)[x>O==>(.3y)(xy= 1)]

which may be rewritten successively as (.3x)~[x>O==>(.3y)(xy= 1)]

(.3x)[x> 0 /\ ~(.3y)(xy = 1)] (.3x)[x

>

O/\(Vy)(xy

* I)].

In English the last of these statements would read "There is a positive real number for which there is no multiplicative inverse."

Quantifiers

1.3

23

Example. Let I be a function whose domain is the real numbers. Find a denial of the definition of "I is continuous at a." Use the reals as the universe. A translation of the definition is (Ve) (.0 > 0 ==> (3J)[J > 0/\ (Vx)(lx - al < J ==> I/(x) - l(a)1 < .0)]). After several applications of Theorems 1.2 and 1.3 (you are expected to carry out the steps), we find a denial (3.0)(.0

> 0 /\ (VJ)[J > 0 ==> (3x)(lx

- al

< J /\ I/(x)

-

l(a)1 ::::: .0)]).

In words, this denial is "There is a positive .0 such that, for all positive J, there exists Ix - al < J, but I/(x) - l(a)1 ::::: e."

x such that

Example. Find a useful denial (that is, a simplified form of the negation) of the sentence "For every number z, there are two numbers larger than z such that there is no number that is larger than their sum and less than their product." The sentence can be symbolized by (Vz)(3x)(.3y)[x

> z /\y >

Z /\ ~(.3w)(x

+ y < w < xy)]

Beginning with the negation and successively applying Theorems 1.2 and 1.3 we have:

> z /\y > Z /\ ~(.3w)(x + y < w < xy)] (.3z) ~ (.3x)(.3y)[x > z J\y > Z J\ ~(.3w)(x + y < w < xy)] (.3z)(Vx) ~ (.3y)[x > z J\y > Z J\ ~(.3w)(x + y < w < xy)] (.3z)(Vx)(Vy) ~ [x> z J\y > Z J\ ~(.3w)(x + y < w < xy)] ~(Vz)(.3x)(.3y)[x

(.3 z) (Vx)(Vy)[(x

>

(.3z)(Vx)(Vy)[(x

z J\y

>

z) ==> ~ ~ (.3w)(x

+ y < w < xy)]

> z J\y > z) ==> (.3w)(x + y <

w

<

xy)].

This denial can be read as follows: "There is a real number z with the property that for any two larger numbers there is a number that is larger than their sum and less than their product." We will prove the denial in section 1.5.

Example. Find a denial of "Some intelligent people admire musicians." The sentence is symbolized by (.3x)[x is an intelligent person /\(Vy)(y is a musician ==> x admires y)].

By Theorems 1.2 and 1.3, its negation is equivalent to (Vx) ~[x is an intelligent person /\(Vy)(y is a musician ==> x admires y)]

or (Vx)[x is an intelligent person ==> ~(Vy)(y is a musician ==> x a~mires y)]

or (Vx)[x is an intelligent person ==> (.3y) ~(y is a musician ==> x admires y)]

or

24

CHAPTER 1 Logic and Proofs (\;Ix)[x is an intelligent person

=:}

(3y)(y is a musician Ax does not admire y)]

which may be translated into "Every intelligent person has some musician whom he or she does not admire." We often hear statements like the complaint one fan had after a great ball game. The game was fine, he said, but "Everybody didn't get to play." We easily understand the complaint to be that "Not everyone got to play" or "Some team members didn't get into the game." This meaning is obviously intended because if it were true that everybody didn't get to play, there would have been no game. Such imprecision is always to be avoided in mathematics. For example, in a vector space, three vectors x, y, and z are dependent if there exist scalars a, b, and c, not all zero, such that ax + by + cz = O. It would be incorrect to say that the scalars a, b, and c must be all nonzero, because ~(\;Ix)A(x) is not equivalent to (\;Ix) ~ A(x). The last quantifier we consider is a special type of existence.

DEFINITION For an open sentence P(x), the proposition (3! x)P(x) is read "There exists a unique x such that P(x)." The sentence (3! x)P(x) is true when the truth set for P(x) contains exactly one element from the universe, hence the designation of 3! as the unique existence quantifier.

In the universe of natural numbers, (3! x)(x is even and prime) is true, since the truth set of "x is even and prime" contains only the number 2. The sentence "(3! x)(x 2 = 4)" is true when the universe is the natural numbers and false when the universe is the integers. The quantifier 3! is a special case of the more general existential quantifier (see exercise 5). For (3x)P(x) to be true it is unimportant how many elements are in the truth set of P(x), as long as there is at least one. For (3! x)P(x) to be true, the number of elements in the truth set of P(x) is crucial: there must be exactly one. A useful equivalent form for (3! x)P(x) is (3x)(P(x) A (\;Iy)[P(y) =:} x

= y])

(\;Ix)(~P(x) V

*- y])

(; for if there is exactly one x such that P(x), then P(x) is true; and for every y, if P(y) is true, then y must be identical to x, and conversely. Using this form we see that a denial of (3! x)P(x) is (3y)[P(y) Ax

That is, (3! x)P(x) is false when, for every x, either P(x) is false or P(y) is true for some y different from x.

Exercises 1.3 1.

*

Translate the following English sentences into symbolic sentences with quantifiers. The universe for each is given in parentheses. (a) Not all precious stones are beautiful. (All stones)

1.3

"* * * *

"* "*

Quantifiers

25

All precious stones are not beautiful. (All stones) Some isosceles triangle is a right triangle. (All triangles) No right triangle is isosceles. (All triangles) All people are honest or no one is honest. (All people) Some people are honest and some people are not honest. (All people) (g) There is a smallest positive integer. (Real numbers) (h) Between any real number and any larger real number, there is a rational number. (Real numbers) No one loves everybody. (All people) (i) (j) Everybody loves someone. (All people) (k) For every positive real number x, there is a unique real number y such that 2 Y = x. (Real numbers) (I) For every complex number, there is at least one complex number such that the product of the two complex numbers is a real number. (Real numbers, as the coefficients) (m) For every nonzero complex number, there is a unique complex number such that their product is 7r. (Real numbers, as the coefficients) (b) (c) (d) (e) (f)

2.

For each of the propositions in exercise I, write a useful denial, and give a translation into ordinary English.

3.

Give two proofs of Theorem 1.3(b).

4.

Which of the following are true? The universe for each is given in parentheses. (a) (\;fx)(x + x 2: x) (Real numbers) (b) (\;fx)(x + x 2: x) (Natural numbers) (c) (3x)(2x + 3 = 6x + 7) (Natural numbers) (d) (3x)(3 X = x 2) (Real numbers) (e) (3x)(3 X = x) (Real numbers) (f) (3x)(3(2 - x) = 5 + 8(1 - x)) (Real numbers) (g) (\;fx)(x 2 + 6x + 5 2: 0) (Real numbers) (h) (\;fx)(x 2 + 4x + 5 2: 0) (Real numbers) (i) (3x)(x 2 + x + 41 is prime) (Natural numbers) (j) (\;fx)(x 2 + X + 41 is prime) (Natural numbers) (k) (\;fx)(x 3 + 17x 2 + 6x + 1002: 0) (Real numbers)

* * * 5.

* * 6.

* *

Give an English translation for each. The universe is given in parentheses. (a) (Vx)(x 2: I) (Natural numbers) (b) (3! x) (x 2: 0 Ax::; 0) (Real numbers) (c) (\;fx)(x is prime Ax =1= 2=:}x is odd) (Natural numbers) (d) (3! x) (loge X = 1) (Real numbers) (e) ~(3X)(X2 < 0) (Real numbers) (f) (3! x)(x 2 = 0) (Real numbers) (g) (\;fx)(x is odd =:} x 2 is odd) (Natural numbers) Which of the following are true for the universe of all real numbers? (a) (b) (c)

(\;fx)(3y)(x + y = 0). (.3x)(Vy)(x + y = 0). (.3x)(.3y)(x 2 + y2 = -I).

(d)

(\;fx)[x

> 0 =:} (.3y)(y < 0 J\xy > 0)].

26

CHAPTER 1 Logic and Proofs

* *

(e) (f) (g) (h)

(i) (j) (k)

7.

(a)

(b)

('dy)(.3x)(\iz)(xy = Xz). (.3x) (\iy) (X :5 y). ('dy)(.3X)(X :5 y). (.3!y)(y < 0 Ay + 3> 0). (.3! x) ('dy) (x = y2). ('dy)(.3! x) (x = y2). (.3! x)(.3! y)('dw)(w 2 > x - y).

Give a proof of (.3! x)P(x) =:} (.3x)P(x). Show that the converse of the conditional sentence in (a) is false.

S.

Write a symbolic translation of the Mean Value Theorem from calculus.

9.

Write a symbolic translation of the definition of limf(x)

= L.

x~a

Find a useful denial and give an idiomatic English version. 10.

* 11.

* 1.4

Which of the following are denials of (.3! x)P(x)? (a) (b) (c) (d)

('dx) (P(x) ) V ('dx) ( -P(x». ('dx)(-P(x» V (.3y)(.3z)(y =1= z AP(y) AP(z». ('dx)[P(x) =:} (.3y)(P(y) Ax =1= y)]. -('dx)('dy)[(P(x)AP(y»=:}x=y].

Give a denial of "You can fool some of the people all of the time and all of the people some of the time, but you cannot fool all of the people all of the time."

12. Riddle:

What is the English translation of the symbolic statement 'd .3 .3 'd?

Mathematical Proofs A proofis aJustification of the truth of a statement called a theorem. It generally begins with some hypotheses stated in the theorem ane proceeds by correct reasoning to the claimed statement. Along the way it may draw upon other hypotheses, previously defined concepts, or some basic axioms setting forth properties of the concepts being considered. A proof, then, is a logically valid deduction of a theorem, from axioms or the theorem's premises, and may use previously proved theorems. As you write a proof, be sure it is not just a string of symbols. Every step ofyour proof should express a complete sentence. It is perfectly acceptable and at times advantageous to write the sentence symbolically, but be sure to include important connective words to complete the meaning of the symbols. The truth of any statement in a proof can eventually be traced back to some initial set of concepts and assumptions. We cannot define all terms, or else we would have a circular set of definitions. Neither can we prove all statements from previous ones. There must be an initial set of statements called axioms or postulates that are assumed true, and an initial set of concepts called undefined terms, from which new concepts can be introduced by means of definitions and from which new statements (theorems) can be deduced.

1.4

Mathematical Proofs

27

Recall that a tautology is a propositional form that is true for every assignment of truth values to its components. Tautologies playa key role in the construction of proofs. A few of the basic tautologies we shall refer to are (Excluded Middle) P =:} Q <=>

~Q =:} ~P

(Contrapositive)

P V (Q V R) <=> (P V Q) V R P /I. (Q /I. R) <=> (P /I. Q) /I. R

}

P /I. (Q V R) <=> (P /I. Q) V (P /I. R)

}

P V (Q /I. R) <=> (P V Q) /I. (P V R) (P <=> Q) <=> [(P ~(P =:}

=:}

Q) <=> P /I.

Q) /I. (Q =:} P)]

}

~(pvQ)<=>~P/I.~Q

P<=>(~P=:}Q/I.~Q)

[(P =:} Q) /I. (Q =:} R)] =:}

Q) =:} Q

(Distributivity)

~Q

~(P/l.Q)<=>~pv~Q

P /I. (P

(Associativity)

=:}

(De Morgan's Laws) (Contradiction)

(P =:} R)

(Transitivity) (Modus Ponens)

Each may be verified as a tautology by its truth table, but you should also examine each to see that it expresses a logical relationship that is always true. See the discussion following Theorem 1.2 (p. 13). De Morgan's laws are named in honor of the English logician Augustus De Morgan (1806-1873). In writing proofs, a working knowledge of tautologies is helpful for several reasons. One reason is that a sentence whose symbolic translation is a tautology may be used at any time in a proof. For example, if a proof involves a number x, one can at any time correctly assert "Either x = 0 or x =1= 0," since this is an instance of the tautology P V ~P. Most steps in a proof follow from earlier lines or other known results. We may use the modus ponens tautology to make deductions. That is, A statement Q may be deduced as true in a proof if both statements P and P =:} Q have been deduced earlier in the proof.

We refer to this rule of deduction as the modus ponens rule. As a useful extension of the modus ponens rule, we allow P to be a compound statement whose components are either hypotheses, axioms, earlier statements in the proof, or statements of previously proved theorems. Likewise, the conditional sentence P =:} Q may have been deduced earlier in the proof or may be a previous theorem, an axiom, or an instance of a tautology.

Example. Suppose it has been proved that "If F is a finite integral domain, then F is a field." Suppose that one step in a proof is the statement "F is finite" and another step is the statement "F is an integral domain." Then a later step may be the statement "F is a field." We have applied the modus ponens rule to deduce R from

28

CHAPTER 1 Logic and Proofs

(S /\ D) and [(S /\ D) ==> R], where S is "F is finite," D is "F is an integral domain," and R is "F is a field."

Example. Suppose a proof contains the statements (1) "If the crime did not take place in the billiard room, then Colonel Mustard is guilty"; (2) "The lead pipe was not the weapon"; and (3) "Either Colonel Mustard is not guilty or the weapon used was the lead pipe." Then we may deduce that "The crime took place in the billiard room." In this application of the modus ponens rule to deduce Q from P and P ==> Q, P is the conjunction of the three statements from the proof (which we symbolize below), and Q is the sentence deduced. You should verify that P ==> Q (symbolized below) is a tautology, because that fact is what makes this deduction a legitimate use of the modus ponens rule. It is not necessary that the statements P and Q be included when this proof is written out. What matters is that Q follows from the other statements by the modus ponens rule. Statement I is Statement 2 is Statement 3 is

~B==>M

Pis Q is

(~B==>

~L

~MVL

M) /\

~L /\(~M

V L)

B

P==> Q is

If a sentence R appears in a proof and another sentence S is equivalent to R, then S may be deduced in the proof. This replacement of R by its equivalent S is justified because (R /\ (R ¢::} S» ==> S is a tautology. Thus, by modus ponens, S may be asserted. For instance, a proof containing the line "The product of real numbers a and b is zero" could later have the statement "a is zero or b is zero." In this example the equivalence of the two statements comes from our knowledge of real numbers, not from the form of the statements. As another use of replacement, we may deduce the contrapositive of a conditional sentence in a proof. For instance, sinc~~e know "If f'(xo) = 0, then f has a critical point at xo," we could assert in a pro6f that "If f does not have a critical point at xo, thenf'(xo) 0." Replacement can also be applied to components of statements in a proof. For instance, the statement "If x is an integer greater than 2 and x is prime, then x does not divide 32" may be replaced by "If x is an odd prime, then x does not divide 32."

'*

Example. We may not deduce the statement "n is a prime number" from the statements "If n is a prime number, then n is an integer" and "n is an integer," since this could employ only the propositional form ((P ==> Q) /\ Q) ==> P, which is not a tautology. Most of the tautologies on page 27 have a biconditional as the main connective and hence are statements about equivalences. Replacement justifies using those tautologies to write equivalent statements. For example, if one line in a proof is "It is

1.4 Mathematical Proofs

29

not the case that x is prime and x::::: 30," we may use De Morgan's Laws and replacement to conclude that "x is not prime or x < 30." The strategy to construct a proof of a given theorem depends greatly on the logical form of the theorem's statement and on the particular concepts involved. As a general rule, when you write a step in a proof, ask yourself if deducing that step is valid in the sense that some tautology permits you to deduce it. If the step in the proof is valid, you will usually find that the step follows as the result of a deduction based on a tautology. It is not necessary to cite the tautology in your proof. In fact, with practice you should eventually come to write proofs without purposefully thinking about tautologies. What is necessary is that every step be justifiable. Not all steps in a proof must be based on tautologies. Some steps may be axioms for the basic theory in which the discussion resides or previously proved results. Other steps may be hypotheses (a certain type of assumption) given as components in the statement of the result to be proved. Still other steps may be assumptions you may wish to introduce. In a proo/you may at any time state an assumption, an axiom, or a previously proved result. The statement of an assumption generally takes the form "Assume P," to alert the reader to the introduction of a statement not derived from previous statements in the proof. A great deal of care must be exercised in stating an assumption. The most common, the hypothesis, will be discussed later with proof methods. A common error in learning to write proofs is to assume what is to be proved or something that trivially implies what is to be proved. The statement of an axiom is usually easily identified as such by the reader because it is a statement about a very fundamental fact assumed about the theory. Thus, statements of axioms are frequently omitted from proofs, but there are cases (such as the Axiom of Choice in chapter 5) where it is prudent to mention an axiom's statement in every proof where it is employed. Often proof steps are previously proved results. In calculus, for example, to prove that the derivative of sin x is cos x, one generally proves first that lim sin Ax = 1. !lx-O

Ax

This result is then given as one step in the proof of sin' x = cos x. Great latitude is allowed for differences in taste and style among proof writers. Generally, the further you go in mathematics, the less justification you will find given, because with more advanced topics more is expected of the reader. In this text our proofs will be complete and concise, but we shall on occasion insert parenthetical comments (offset by ( ) and in italics) to explain how and why a proof is proceeding as it is. Such comments should not be taken as part of the proof but are inserted to help clarify the workings of the proof. The first method of proof we will examine is the direct proof of a conditional sentence of the form P =:} Q. This implication is false only when P is true and Q is false, so it suffices to show this situation can't happen. The simplest way to proceed, then, is to assume that P is true, and show (deduce) that Q is also true. A direct proof of P =:} Q will have the following form:

30

CHAPTER 1 Logic and Proofs

DIRECT PROOF OF P => Q AssumeP.

Therefore, Q. Thus P=> Q.

•

Some examples we consider in this section actually involve quantified sentences. Since we will consider proofs with quantifiers in the next section, in this section we will imagine that a variable represents some fixed object. That is, when you read "If x is odd, then x + I is even," you should think of x as being some particular integer. Several of our examples are drawn from number theory, the study of the arithmetic of the positive integers. Some of the statements we will prove are statements of things we already know. They may be as obvious as "If x and yare odd integers, then xy is odd." Examples like this are chosen to illustrate the form of a proof in a familiar setting. To prove such a statement, we base our reasoning on the following definitions of even and odd integers: we say that x is even if there exists an integer k such that x = 2k, and x is odd if there exists an integer j such that x = 2j + I (or x = 2j - I). We will also use the idea of divisibility: we say that x divides y if there exists an integer k such that y = kx. Whenever we reason about the integers, we may use these familiar properties of the number system: the fact that the sum and product of any two positive integers are positive integers (closure); the commutative, associative, distributive, and cancellation laws; the additive identity 0; the multiplicative identity I; the existence of additive inverses (negatives); and the fact that for any two positive integers a and b either a < b, a = b, or a> b (trichotomy). Many of these properties are listed on page 85. Since our goal here is to understand proof techniques rather than to develop the theory of numbers from axioms, we may on occasion use more advanced properties (such as the Fundamental Theorem of Arithmetic) in proof examples. These properties will be clearly labeled. We will be ca~eful never to use a theorem to prove that same theorem, because that would be circt'iiar reasoning.

Example. Suppose a, b, and c are integers. Prove that if a divides band b divides c, then a divides c. Proof. (Fora direct proof, we assume the antecedent, which is "a divides band b divides c. " Our goal is to derive the consequent" a divides c" as our last step.) Suppose a, b, and c are integers, and that a divides band b divides c. (We now rewrite these assumptions by using the definition of "divides, " so that we have some equations to work with.) Then b = ak for some integer k, and c = bj for some integer j. (To show that a divides c, we have to write c as a multiple of a.) Therefore, c = bj = (ak)j = a(kj), so a divides c. Thus if a divides band b divides c, a divides c. •

In this example, we did not worry about what would happen if a did not divide b or if b did not divide c. We assumed that both these statements were true, because we need show only that if these hypotheses hold, then the conclusion must also be

1.4

Mathematical Proofs

31

true. The fact that we assumed that the antecedent is true and proceeded step by step to the conclusion is why this type of proof is called direct. Your strategy for developing a direct proof of a conditional sentence should involve these steps:

1. 2. 3. 4.

Determine precisely the antecedent and the consequent. Replace (if necessary) the antecedent with a more usable equivalent. Replace (if necessary) the consequent by something equivalent and more readily shown. Develop a chain of statements, each deducible from its predecessors or other known results, that leads from the antecedent to the consequent. Our next example is another proof of a fact about divisibility.

Example. Let a, b, and c be integers. Prove that if a divides b and a divides c, then a divides b - c. Proof. Suppose a divides b and a divides c. (Now we use the definition of divides.) Then b = an for some integer nand c = am for some integer m. (Notice that we don't assume m = n.) Thus, b - c = an - am = a(n - m). Since n - m is an integer (we use the fact that the difference of two integers is an integer), a divides b -- c. • Our third example, which comes from an exercise in a precalculus class, involves a point (x, y) in the Cartesian plane (figure 1.1) and uses algebraic properties available to students in such a class. This example also has the form of a direct proof.

Example. Prove that if x < -4 andy> 2, then the distance from (x, y) to (I, -2) is at least 6. Proof. Assume that x < -4 and y > 2. Then x-I < - 5, so (x - 1)2 > 25. Also + 2> 4, so (y + 2)2 > 16. Therefore,

y

~(x - 1)2 + (y + 2)2 > ~25 + l6>.j36,

•

so the distance from (x, y) to (I, -2) is at least 6.

y -

(x, y).

-

y=2

I

I

x=-4

I

I

I

I

I

I

-

--(1,-2)

Figure 1.1

I

x

32

CHAPTER 1 Logic and Proofs

How could we discover such a proof? The antecedent x < -4 andy> 2 doesn't give us much to work with, so we examine the conclusion that the distance from (x, y) to (I, - 2) is at least 6. We rewrite that as

J(x - 1)2 + (y + 2)2 ::::: 6, then square both sides and realize that plugging in x < -4 and y > 2 would make the conclusion true. Only after this preliminary work would we be ready to write the proof. Direct proofs of statements of the form P =:} Q are not always so straightforward when either P or Q is itself a compound proposition. To prove P =:} (Q V R), one often proves either the equivalent (P /I. ~Q) =:} R or the equivalent (P /I. ~R) =:} Q. For instance, to prove "If the polynomial/has degree 4, then / has a real zero or/can be written as the product of two irreducible quadratics," we would prove "If / has degree 4 and no real zeros, then / can be written as the product of two irreducible quadratics." To prove (P V Q) =:} R, one could proceed by cases, first proving P =:} R, then proving Q =:} R. This is valid because of the tautology [(P V Q) =:} R] ¢=} [(P =:} R) /I. (Q =:} R)]. The statement "If a quadrilateral has opposite sides equal or opposite angles equal, then it is a parallelogram" is proved by showing both "A quadrilateral with opposite sides equal is a parallelogram" and "A quadrilateral with opposite angles equal is a parallelogram." Here is an example that has two cases.

Example. Suppose n is an odd integer. Then n = 4j n = 4i - I for some integer i. Proof.

+

I for some integer j, or

Suppose n is odd. Then n = 2m - I for some integer m. (We now show

(P V Q) =:} R, where P is "m is even" and Q is "m is odd.")

Case 1. Case 2.

If m is even, then m = 2j for some integer j, and so n = 2(2j) + I = 4j + 1. If m is odd, then m = 2k Sh~1 for some integer k. In this case, n = 2(2k + I) + I = 4k + 3 = 4(k + I) - 1. Choosing i to be the integer k + I, we have n = 4i - 1. •

A proof of a statement symbolized by P =:} (Q /I. R) would probably also have two parts. We first show P =:} Q and then show P =:} R. We would use this method to prove the statement "If two parallel lines are cut by a transversal, then corresponding angles are equal and alternate interior angles are equal." A proof of a statement symbolized by (P /I. Q) =:} R presents no new difficulties. A direct proof is likely to be a good choice because we have the advantage of assuming both P and Q in the beginning of the proof. Consider the two similar statement forms (P =:} Q) =:} Rand P =:} (Q =:} R). They have remarkably dissimilar direct proof outlines. For (P =:} Q) =:} R, we assume P =:} Q and deduce R. We cannot assume P; we must assume P =:} Q. On the other

1.4

Mathematical Proofs

33

hand, in a direct proof of P =:} (Q =:} R), we do assume P and show Q =:} R. Furthermore, after the assumption of P, a direct proof of Q =:} R begins by assuming Q is true as well. This is not surprising since P =:} (Q =:} R) is equivalent to (P /I. Q) =:} R. A second form of prooffor a conditional sentence is the contrapositive proof or proof by contraposition. The idea here is that since P =:} Q is equivalent to its contrapositive, (~Q) =:} (~P), we first give a direct proof of (~Q) =:} (~P) and then conclude P =:} Q. This method works well when the connection between the denials of P and Q is easier to understand than the connection between P and Q themselves. The format of a proof by contraposition is as follows:

CONTRAPOSITION PROOF OF P => Q Suppose ~Q. Therefore, ~P (via a direct proof). Thus, (~Q) =:} (~P). Therefore, P =:} Q.

Example.

•

Let m be an integer. Prove that if m 2 is odd, then m is odd.

Suppose m is not odd. (Suppose ~Q.) Then m is even. Thus, m = 2k for some integer k. (Equivalent statement.) Then m 2 = (2ki = 4k2 = 2(2k 2). Since m 2 is twice the integer 2k 2, m 2 is even. (Deduce ~P.) Thus, if m is even, then m 2 is even; so, by contraposition, if m 2 is odd, then m is odd. •

Proof.

Let x and y be real numbers such that x < 2y. Prove that if 7xy:::; 3x 2 + 2y2, then 3x:::; y.

Example.

Suppose x < 2y. Suppose also that 3x > y. (The second assumption (~Q) begins our proof by contraposition.) Then 2y - x > 0 and 3x - y > O. Therefore, (2y - x)(3x - y) = txy - 3x 2 - 2y2 > 0, so txy > 3x 2 + 2y2. We have shown that if3x > y, then txy > 3x 2 + 2y2. We conclude that if7xy:::; 3x 2 + 2y2, then 3x:::; y .

Proof.

•

A proof by contradiction makes use of the tautology P <=> [( ~ P) =:} (Q /I. ~Q)]. To prove a proposition P, it is sufficient to prove (~P) =:} (Q /I. ~Q). Two aspects about this form of proof are especially noteworthy. First, this method of proof can be applied to any proposition P, whereas direct proofs and proofs by contraposition can be used only for conditional sentences. Second, the proposition Q does not even appear on the left side of the tautology. The idea of proving (~P) =:} (Q /I. ~Q) then has an advantage and a disadvantage. We don't know what proposition to use for Q, but any proposition that will do the job is a good one. This means a proof by contradiction will requir'e a ,"spark of insight" to determine a useful Q. A proof by contradiction has the following form:

34

CHAPTER 1 Logic and Proofs

PROOF OF P BY CONTRADICTION Suppose ~P. Therefore, Q. Therefore, ~Q. Hence, Q A ~Q, a contradiction. Thus, P.

Ii is an irrational number. Suppose that Ii is a rational number. (Assume ~P.) Then Ii = sit, where

Example. Proof.

-

Prove that

sand t are integers. Thus, 2 = s21t 2, and 2t 2 = s2. Since s2 and t 2 are squares, s2 contains an even number of 2's as factors (this is our Q statement), and t 2 contains an even number of2's. But then 2t 2 contains an odd number of2's as factors. Since s2 = 2t 2, s2 has an odd number of 2's. (This is the statement ~Q.) This is a contradiction, because by the Fundamental Theorem of Arithmetic, every natural number can be uniquely expressed as a product of primes. We conclude that is irrational.

Ii

-

As a second example of a proof by contradiction, we give a proof (attributed to Euclid) that there are infinitely many primes. By this we mean that it is impossible to list all the prime numbers from the first one to the kth or last one, where k is a natural number.

Example. Prove that the set of primes is infinite. Proof. Suppose the set of primes is finitt<, that is, has k elements for some natural number k. (Suppose ~P.) Let the primes be Plo pz, P3," "Pk and considerthe number n = (PI P2" 'Pk) + 1. We now use the fact (to be proved in section 2.5) that every natural number n has a prime divisor q, where q > 1. (The Q statement is q > 1.) Since q is a prime and PI, P2,"" Pk is the list of all primes, q is one ofthe Pi and thus q divides the product PI P2' .. Pk' Since q also divides n, q divides n - PI P2'" Pk' (If a divides b and c, then a divides b - c.) But n - PI P2'" Pk = 1 and so q = 1. (This is ~Q.) From _ this contradiction we conclude that the set of primes is infinite. Proofs of biconditional sentences are often based on the equivalence of P ¢::} Q with (P =:} Q) A (Q =:} P). Many proofs of P ¢=} Q will have the following form:

TWO-PART PROOF OF P ¢=} Q (i) Show P =:} Q by any method. (ii) Show Q =:} P by any method. Therefore, P ¢=} Q.

-

1.4

Mathematical Proofs

35

Of course, the two proofs in (i) and (ii) may use different methods. Frequently the proof of one part is more difficult than the other. This is true, for example, of the proof that "The natural number x is prime iff no positive integer greater than I and less than or equal to {x divides x." It is immediate from the definition that "If x is prime, then no positive integer greater than I and less than or equal to {x divides x," but the converse requires a little reasoning. In some cases it is possible to prove a biconditional sentence P {=} Q that uses the "iff' connective throughout. This amounts to starting with P and then replacing it with a sequence of equivalent statements, the last one being Q.

Example. Consider a triangle with sides of length a, b, c. Use the Law of Cosines to prove that the triangle is a right triangle with hypotenuse c if and only if a 2 + b 2 = c 2 (figure l.2).

b

Figure 1.2 Proof. By the Law of Cosines, a 2

+ b2 = c 2 -

2ab cos

e, where eis the angle be-

tween the sides of length a and b. Thus,

a2

Thus, a 2

+ b2 = c 2

iff iff iff

e= 0

2ab cos cos = 0 = 90°.

e

e

+ b 2 = c 2 iff the triangle is a right triangle with hypotenuse c.

•

One form of proof is known as proof by exhaustion. Such a proof consists of examining every possible case. This was our method in Theorem 1.1, where we examined all four combinations of truth values for two propositions. Naturally, the idea of proof by exhaustion is appealing only when there is a small number of cases, or when large numbers of cases can be systematically handled. Care must be taken to ensure that all possible cases have been considered.

Example. For a real number x, prove that

-Ixl :::; x :::; Ixl.

Proof. (Since the absolute value ofx is defined by cases (Ixl if x < 0) this p roof will proceed by cases.} Case 1.

= x if x 2: 0; Ixl = -x

Let x 2: O. Then Ixl = x. Since x 2: 0, we have -x:::; x. Hence, - x :::; x :::; x, which is -Ixl :::; x :::; Ixl in this case.

36

CHAPTER 1 Logic and Proofs

Case 2.

Let x < O. Then Ixl = -x. Since x < 0, x $ -x. Hence, we have x$x$ -x,or -(-x)$x$ -x, which is -Ixl $x$ Ixl.

Thus, in all cases we have

-Ixl $

x

$

Ixl.

-

There have been instances of truly exhausting proofs involving great numbers of cases. In 1976 Kenneth Appel and Wolfgang Haken announced a proof of the Four-Color Theorem. The original version of their proof of the famous Four-Color Conjecture contains 1,879 cases and took 3! years to develop. t The proof techniques presented in this section and the next will be used throughout this text. Additional examples and helpful hints will be given in section 1.6. To help you distinguish among the various techniques developed in this section, we give an example of a statement proved by three different methods.

Example. For given integers x and y, give a direct proof, a proof by contraposition, and a proof by contradiction of If x and yare odd integers, then xy is odd.

Direct Proof. Assume x is odd and y is odd. Then integers m and n exist so that x = 2m + I and y = 2n + I. Thus, xy = (2m + 1)(2n + I) = 4mn + 2m + 2n + I = 2(2mn + m + n) + I. Thus, xy is an odd integer. Proof by Contraposition. (To prove (x is odd /\ y is odd) ==> xy is odd, we show xy is even ==> (x is even V y is even).) Assume xy is even. Thus, 2 is a factor of xy. But since 2 is a prime number and 2 divides the product xy, then either 2 divides x or 2 divides y. (We use the fact, to be proved later, that if a prime divides a product it must divide one of the factors. See exercise 8:crf section 1.6.) We have shown that if xy is even, then either x or y is even. Thus, if x and yare odd, then xy is odd. Proof by Contradiction. Suppose x and yare odd. To prove by contradiction that xy is odd, assume xy is even. Since x and yare odd, then x = 2m + 1 and y = 2n + I for some integers m and n. Thus, xy = (2m + 1)(2n + 1) = 2(2mn + m + n) + I. Then 2(2mn + m + n) is even (because it is divisible by 2) and the next integer 2(2mn + m + n) + I is even (because xy is even). This is impossible. Because the supposition that xy is even leads to a contradiction, we conclude that xy is odd. By now you may have the impression that, given a set of axioms and definitions of a mathematical system, any correctly stated proposition in that system can ~The Four-Color Theorem involves coloring regions or countries on a map in such a way that no two adjacent countries have the same color. It states that four colors are sufficient, no matter how intertwined the countries may be. The fact that the proof depended so heavily on the computer for checking cases raised questions about the nature of proof Verifying the 1,879 cases required more than 10 billion calculations. Many people wondered whether there might have been at least one error in a process so lengthy that it could not be carried out by one human being in a lifetime. Haken and Appel's proof has since been improved, and the Four-Color Theorem is accepted; but the debate about the role of computers in proof continues.

1.4 Mathematical Proofs

37

be proved true or proved false. This is not the case. There are important examples in mathematics of consistent axiom systems (so that there exist structures satisfying all the axioms) for which there are statements such that neither the statement nor its negation can be proved. It is not a matter that these statements are difficult to prove or that no one has yet been clever enough to devise a proof; it has been proved that there can be no proof of either the statement or its negation within the system. Such statements are called undecidable in the system because their truth is independent of the truth of the axioms. The classic case involves the fifth of five postulates that Euclid (circa 300 B.C.) set forth as his basis for plane geometry: "Given a line and a point not on that line, exactly one line can be drawn through the point parallel to the line." For centuries some thought Euclid's axioms were not independent, believing that the fifth postulate could be proved from the other four. It was not until the 19th century that it became clear that the fifth postulate was undecidable. There are now theories of Euclidean geometry where the fifth postulate is assumed true and nonEuclidean geometries where it is assumed false. Both are perfectly reasonable subjects for mathematical study.

Exercises 1.4 1.

U sing truth tables, verify that each of the basic tautologies of this section is a tautology.

2.

Analyze the logical form of each of the following statements and construct just the outline of a proof by the given method. Since the statements may contain terms with which you are not familiar, you should not (and perhaps could not) provide any details of the proof. (a) Outline a direct proof that if (G, *) is a cyclic group, then (G, *) is abelian. (b) Outline a direct proof that if B is a nonsingular matrix, then the determinant of B is not zero. (c) Outline a proof by contraposition of the statement in (a). (d) Outline a proof by contraposition of the statement in (b). (e) Suppose A, B, and C are sets. Outline a direct proof that if A is a subset of Band B is a subset of C, then A is a subset of C. (f) Outline a direct proof that if the maximum value of the differentiable function/(x) on the closed interval [a, b] occurs at xo, then either Xo = a or Xo = b or !'(xo) = O. (g) Outline a proof by contradiction that the set of natural numbers is not finite.

* *

3.

It is a theorem of linear algebra that if A and B are invertible matrices, then the product AB is invertible. As in exercise 2, outline (a) a direct proof of the theorem. (b) a direct proof of the converse of the theorem. (c) a proof of the theorem by contraposition. (d) a proof of the converse of the theorem by contraposition. (e) a proof of the theorem by contradiction. (f) a proof of the converse of the theorem by contradiction.

38

CHAPTER 1 Logic and Proofs

4.

* 5.

"* 6.

* *

Let x and y be integers. Prove that (a) if x and yare even then x + y is even. (b) if x and yare even then xy is divisible by 4. (c) if x and yare odd then x + y is even. (d) if x is even and y is odd, then x + y is odd. (e) if x is even and y is odd, then xy is even. Let a and b be real numbers. Prove that (a) labl = lallbl. (b) la - bl = Ib - al. (c)

I~I

(d) (e)

la + bl :5 lal + Ihi. lal:5 b iff -b:5 a :5 b.

=

i~i,forb*O.

Suppose a, b, e, and d are positive integers. Prove that (a) 1 divides a and a divides a. (b) if a divides b, then a:5 b. (c) if a divides b, then a divides be. (d) if ab = 1, then a = b = 1. (e) if a divides band b divides a, then a = b. (f) if a divides band e divides d, then ae divides bd. (g) if ab divides e, then a divides e. (h) ae divides be if and only if a divides b. Prove by contradiction that if n is a natural number, ~l > ~2' ~\ n+ n+ 2 Prove by cases that if n is a natural number, n + n + 3 is odd.

7.

(a)

8.

In this exercise the proofs involve verifying an algebraic expression. In such proofs (and many others) it is a good idea to "work backward." That is, to show that an equation is true, decide what other equation it could be proved from, and what that equation could be proved from, and so forth. After doing such preliminary work, remember to write your proof forwards, so that your conclusion is the statement to be proved. (a) Let x and y be positive real numbers. Prove that

(b)

x; y?:.[xY.

(b) 9.

Where in the proof is it essential that x and yare positive? Suppose a right triangle has hypotenuse e and legs a and b. Prove that if the area of the triangle is 2 then the triangle is isosceles.

!e

Recall that except for degenerate cases, the graph of Ax 2 + Bxy

+ Cy2 + Dx + Ey + F =

IS

an ellipse iff B2 - 4AC < 0, a parabola iff B2 - 4AC = 0, a hyperbola iff B2 - 4AC > 0.

°

1.4

*

Mathematical Proofs

39

(a)

Prove that the graph of the equation is an ellipse whenever A > C > B

(b)

Prove that the graph of the equation is a hyperbola if AC < 0 or B< C<4A <0. Prove that if the graph is a parab?la, then BC = 0 or A = B 2/(4C).

>0.

(c) Proofs to Grade

10.

*

Problems with this title throughout this book ask you to analyze an alleged proof of a claim and to give one of three grades. Assign a grade of A (excellent) if the claim and proof are correct, even if the proof is not the simplest or the proof you would have given. Assign an F (failure) if the claim is incorrect, if the main idea of the proof is incorrect, or if most of the statements in it are incorrect. Assign a grade of C (partial credit) for a proof that is largely correct but contains one or two incorrect statements or justifications. Whenever the proof is incorrect, explain your grade. Tell what is incorrect and why. (a) Suppose m is an integer. . Claim. If m 2 is odd, then m is odd. "Proof." Assume m is odd. Then m = 2k + I for some integer k. Therefore, m 2 = (2k + 1)2 = 4k2 + 4k + I = 2(2k 2 + 2k) + I, which is odd. Therefore, if m 2 is odd, then m is odd. _ (b) Suppose m is an integer. Claim. If m 2 is odd, then m is odd. "Proof." Assume that m 2 is not odd. Then m 2 is even and m 2 = 2k for some integer k. Thus 2k is a perfect square; that is, is an integer. If is odd, then = 2n + 1 for some integer n, which means m 2 = 2k = (2n + 1)2 = 4n 2 + 4n + I = 2(2n2 + 2n) + 1. Thus m 2 is odd, contrary = m must be even. Thus if m 2 is not to our assumption. Therefore _ odd, then m is not odd. Hence if m 2 is odd, then m is odd. (c) Suppose t is a real number. Claim. If t is irrational, then 5t is irrational. "Proof." Suppose 5t is rational. Then 5t = pi q where p and q are integers and q =1= O. Therefore, t = pl(5q) where p and 5q are integers and 5q =1= 0, so t is rational. Therefore, if t is irrational, then 5t is irrational. _ (d) Suppose a, b, and c are integers. Claim. If a divides b and a divides c, then a divides b + c. "Proof." Suppose a divides b and a divides c. Then for some integer q, b = aq and c = aq. Then b + c = aq + aq = 2aq = a(2q), so a divides _

12k

fii5

12k

p

*

b+~

*

(e)

Suppose x is a positive real number. Claim. The sum of x and its reciprocal is greater than or equal to 2. That is, I x+-2:2 x

"Proof." x2

-

2x

+

Multiplying by x, we get x 2 12:0

(x - 1)22: O.

+ 1 2: 2x. By algebra,

40

CHAPTER 1 Logic and Proofs

Any real number squared is greater than or equal to zero, so x

*

(f)

(g)

1.5

+ k:::=:: 2 is

true. Suppose x and yare integers. Claim. If x and yare even then x + y is even. "Proof." Suppose x and y are even but x + y is odd. Then, for some integer k, x + Y = 2k + I. Therefore, x + y + (-2)k = I. The left side of the equation is even because it is the sum of even numbers. However, the right side, I, is odd. Since an even cannot equal an odd, we have a contradiction. Therefore, x + y is even. Suppose a, b, and c are integers. Claim. If a divides both band c, then a divides b + c. "Proof." Assume that a does not divide b + c. Then there is no integer k such that ak = b + c. However, a divides b, so am = b for some integer m; and a divides c, so an = c for some integer n. Thus am + an = a(m + n) = b + c. Therefore k = m + n is an integer satisfying ak = b + c. Thus the assumption that a does not divide b + c is false, and a does divide b + c. -

Proofs Involving Quantifiers Most theorems in mathematics are quanti!~ed sentences, even though the quantifier may not actually appear in the statement. For example, proving "If x is an odd integer, then x + I is even," as we did in section 1.4, actually involves a quantifier, since the sentence has the symbolic translation (Vx)(x is odd =:} x + I is even). We suppressed the importance of quantifiers in that section in order to concentrate on basic proof forms. In this section we present methods of proof for theorems of the form (3x)P(x), (Vx)P(x), and (3 !x)P(x). The validity of the proof techniques still depends on employing statements that are always true. There are several ways to prove existence theorems-that is, propositions of the form C3x)P(x). The most direct, called a constructive proof, is to name or describe some object in the universe that actually makes P(x) true. For example, the statement "2 is a prime" is a proof of the theorem "There is an even prime integer." Here is another example.

Example. Prove that there exists a natural number whose fourth power is the sum of 3 fourth powers. Proof:

20615673 4

= 26824404 + 153656394 + 187967604 .

-

The question of whether any nth power is a sum of fewer than n nth powers was raised by Leonhard Euler (1707-1783) and solved in 1968 by L. 1. Lander and Thomas Parkin, who used a computer search to find a fifth power that is the sum of 4 fifth powers. The example given for n = 4 was discovered in 1987 by Noam Elkies. Euler, whose work extended into almost every area of mathematics, was one of the greatest mathematicians of all time.

1.5

Proofs Involving Quantifiers

41

It is also possible to give a proof of (3x)P(x) by contradiction. The proof has the following form:

PROOF OF (3x) P(x) BY CONTRADICTION Suppose ~(.3x)P(x). Then (Vx) ~ P(x).

Therefore, Q /I. ~Q, a contradiction. Hence, ~(.3x)P(x) is false; so (.3x)P(x) is true.

-

The heart of such a proof generally involves making deductions from the universally quantified statement (Vx) ~ P(x). We will consider such reductions later in this section. Other proofs of existence theorems (.3x)P(x) show that there must be some object for which P(x) is true, without ever actually producing a particular object. Both Rolle's Theorem and the Mean Value Theorem from calculus are good examples of this. Here is another.

Example.

Prove that the polynomial rex)

= x 71

-

2x 39

+ 5x -

0.3

has a real zero.

Proof. (What we must show has the form (.3 t) (t is real and ret) = 0).) By the Fundamental Theorem of Algebra, rex) has 71 zeros that are either real or complex. Since the polynomial has real coefficients, its complex zeros come in pairs (by the Complex Root Theorem). Hence, there is an even number of nonreal roots, and that leaves an odd number of real roots. Therefore, rex) has at least one real root.

-

Sometimes a statement to be proved has the form (.3x)P(x) ==} Q. As a first step, we assume (.3x)P(x). However, just that some object x in the universe has the property P(x) does not give us much to work with to construct a proof. A useful next step is to name some particular object that has the property and use the property of the object to derive Q.

Example. The graph of x 2 + y2 = r2 is a circle with center (0,0) and radius r. Prove that if one of the x-intercepts of the circle has rational coordinates, then all four intercepts have rational coordinates. Proof:

Suppose an x-intercept (a, 0) of the circle has rational coordinates. Then r2, so a 2 = r2 and a = ±r. Then the other x-intercept is (-a, 0). To find the y-intercepts, we solve 02 + y2 = r2 and find y = ±r = ±a. Therefore, the four intercepts are (±a, 0) and (0, ± a), all of which have rational coordinates. -

a 2 + 02

=

42

CHAPTER 1

Logic and Proofs

It is often necessary to prove that a statement of the form (Vx)P(x) is false. Since -(Vx)P(x) is equivalent to (3x) - P(x), this amounts to a proof that (3x) - P(x) is true. Any object t in the universe for which -pet) is true is called a counterexample to (Vx)P(x). For example, f(x) = Ixl is a counterexample to "Every function continuous at 0 is differentiable at 0." The number 2 is a counterexample to "All primes are odd." Some counterexamples have eluded mathematicians for centuries before being observed. To prove a proposition of the form (Vx)P(x), we must show that P(x) is true for every object x in the universe. A direct proof of this is done by letting x represent an arbitrary object in the universe, and showing that P(x) is true for that object without using any special properties of the object x. Then, since x is arbitrary, we can conclude that (Vx)P(x) is true. Thus, a direct proof of (Vx)P(x) has the following form:

DIRECT PROOF OF (\Ix) P(x) Let x be an arbitrary object in the universe. (The universe should be named or its objects described.) Hence, P(x) is true. Since x is arbitrary, (Vx)P(x) is true.

-

The open sentence P(x) will often be a combination of other open sentences, and thus a deduction of P(x) will require the selection of an appropriate proof technique.

Example.

If x is an even integer, then x 2 is an even integer.

Proof. (The statement has the form (Vx)P(x), where the universe is the integers and P(x) denotes "lfx is even, then x 2 is even." Thefirst step is to let x represent an arbitrary object in the universe.) Let x be an integer. (To attempt a direct proof of a conditional, we assume the antecedent.) Suppose x is even. (Now we apply the definition of even to x and show that x 2 is even.) Then x = 2k for some integer k. Therefore, x 2 = (2k)2 = 2(2k 2). Since 2k2 is an integer, x 2 is even. _ It is important to remember in a direct proof of a statement of the form (Vx)P(x) that nothing other than "Let x be an arbitrary object in the universe" may be assumed about x. After this statement, the proof proceeds having a fixed x in mind, but all other properties about x must be derived from either x belonging to the universe or from assumptions, axioms, or previous results. A beginner may attempt to prove a universal statement by simply giving an example. The example chosen may have special properties not shared by other objects in the universe. Although an example may be enlightening as to the nature of a correct proof, this approach cannot constitute a valid proof. Many of the statements we considered in section 1.4 should have been treated as universally quantified statements. See, for example, the proof that if x is a real

1.5

Proofs Involving Quantifiers

43

number, then -Ixl :::; x:::; Ixl. In that section we were viewing the symbol x as a fixed but unknown number. This saved us the trouble of worrying about the generalized statement that the property is true for every real number x, so we could concentrate on the basic form of the proof. From now on we will use the style suggested in the preceding box for a direct proof of (Vx)P(x). A similar proof outline can be used for a statement of the form ~(:3x)P(x), which we know is equivalent to (Vx) ~ P(x). For example, although we proved in section 1.4 that every odd integer can be written in the form 4j - 1 or 4k + 1, we show now that no number can be written in both these forms.

Example. There is no odd integer that can be expressed in the form 4j - 1 and in the form 4k

+ 1 for integers j and k.

Proof. Suppose n is an odd integer, and suppose n = 4j - 1 and n = 4k + 1 for integers j and k. Then 4j - 1 = 4k + 1, so 4j - 4k = 2. Therefore, 2j - 2k = 1. The left side of this equation is 2U - k), which is even, but 1 is odd. This is a contradiction. The method of proof by contradiction is often used to prove statements of the form (Vx)P(x). The form of the proof is as follows:

PROOF OF ('fix) P(x) BY CONTRADICTION Suppose ~(Vx)P(x). Then (3x) ~ P(x). Let t be an object such that ~P(t). Therefore, Q /\ ~Q. Thus, (:3x) ~ P(x) is false; so its denial (Vx)P(x) is true.

-

The following example of a proof by contradiction comes from an exercise in a trigonometry class. It uses algebraic and trigonometric properties available to students in the class.

Example. If 0 < x

<~, sinx

+ cosx >

1.

Proof. Suppose there is a real number t, with 0 < x < ~, such that sin t + cos t :::; 1. Both sin x and cos x are positive on the interval (0, ~), so we have 0 < sin t + cos t :::; 1. Thus 0 < (sin t + cos t)2:::; 1, so sin 2 t + 2 sin t cos t + cos 2 t :::; 1. But sin 2 t + cos 2 t = 1 and 2 sin t cos t 2: 0 because sin t and cos t are both positive, so sin 2 t + 2 sin t cos t + cos 2 t > 1. This is a contradiction. The last proof method we give is for the :3 ! quantifier. The standard technique for proving a proposition of the form (3 !x)P(x) is based on its equivalent expression in the form (:3x)P(x) /\ (Vy) (Vz)[(P (y) /\P(z)) ==} y = z).

44

CHAPTER 1 Logic and Proofs

PROOF OF (3! x) P(x} (i) Prove that (3x)P(x) is true by any method. (ii) Assume that t J and t 2 are objects in the universe such that pet J} and P(t2} are true. Therefore, tJ = t 2· We conclude (3! x)P(x).

Example.

•

Prove that the polynomial rex) = x - 3 has a unique zero.

Proof. (i) (ii)

Since r(3} = 3 - 3 = 0, 3 is a zero of rex}. Suppose that t J and t2 are two zeros of rex}. Then r(tJ) = 0 = r(t2}. Therefore, t J - 3 = t2 - 3. Thus t J = t2.

Therefore, rex) = x - 3 has a unique zero.

Example.

•

Every nonzero real number has a unique multiplicative inverse.

Proof. (The theorem is symbolized (Vx)[x =1= 0 =:} (3 !y)(y is real and xy = 1)].) Let x be a real number. (We start this way because the statement to be proved is quantifiedwith V.) Suppose x =1= O. (Assume the antecedent.) We must show thatxy = 1 for exactly one real number y. (We need two steps: one to show that x has an inverse and one to show that x cannot have two different inverses.) (i)

(ii)

Let y = t. (This is a constructive proof that there exists an inverse. We assume for this proof that the quotient of two real numbers is a real number, provided that the divisor is not zero.) Then y is a real number since x =1= 0, and xy = x(t) = § = 1. Thus, x has a multiplicative inverse. Now suppose y and z are two real multiplicative inverses for x. ( This y is not necessarily the y = i in part (i). ) Then xy = 1 and xz = 1. Thus xy = xz, and xy - xz = x(y - z} = O. Since x =1= O,y - z = O. Therefore, y = z.

Thus, every nonzero real number has a unique multiplicative inverse.

•

Many statements have more than one quantifier, so we must deal with each quantifier in succession.

Example. There is a real number with the property that for any two larger numbers there is another real number that is larger than the sum of the two numbers and less than their product. Proof: (A symbolic form of this statement is (3 z)(Vx}(Vy)[(x > z Ay > z) =:} (3 w}(x + Y < w < xy)]. We must choose a z so that the statement will be true for all x andforall y. Notice that x + y is not always smaller thanxy-for example, ifx and yare 1.1 and 1.2. Let's suppose x
1.5

Proofs Involving Quantifiers

45

be less than xy ifwe had x> 2. Similar reasoning would work in the case y :S x, as long as y > 2. We begin the proofby choosing z to be the number 2.) Choose z = 2. Let x and y be any two real numbers such that x > z and y > z. Then x + y :S 2· max {x, y} < xy. If we now choose w to be the average (x + y + xy)/2 of x + y and xy, we have x + y < w < xy. •

Example. For every natural number n, there is a natural number M such that for all natural numbers m > M,

1..<J... m

Proof.

3n'

(With the natural numbers as the universe the statement may be symbolized (Vn)(3M)(Vm)(m > M

==>1.. < J...). m

3n

We consider quantifiers in order from the left.) Let n be a natural number. Choose M to be 3n. Let m be a natural number, and suppose m > M. Then m > 3n, and 3mn > 0, so dividing by 3mn we have k < (The choice of 3n for M is the result of some scratchwork, working backwardfrom the intended conclusion k < -tn.) •

tn.

Great care must be taken in proofs that contain expressions involving more than one quantifier. Here are some manipulations of quantifiers that permit valid deductions.

1. 2. 3. 4. 5. 6.

(Vx)(Vy)P(x,y) ¢=} (Vy)(Vx)P(x,y). (3x)(3y)P(x,y) ¢=} (3y)(3x)P(x,y). [(Vx)P(x) V (Vx)Q(x)] ==> (Vx) [P(x) V Q(x)]. (Vx)[P(x) ==> Q(x)] ==> [(Vx)P(x) ==> (Vx)Q(x)]. (Vx) [P(x) /\ Q(x)] ¢=} [(Vx)P(x) /\ (Vx)Q(x)]. Ox)(Vy)P(x, y) ==> (Vy)(3x)P(x, y).

You should convince yourself that each of these is a logically valid conditional or biconditional. For example, the last on the list is always true because if (3x) (Vy) P(x, y) is true, then there is (at least) one x that makes P(x, y) true no matter what y is. Therefore, for any y, (3x)P(x,y) is true because the one x exists. All six rules may be proved formally. It is important to be aware of the most common incorrect deductions making' use of quantifiers. We list four here and show by example that each is not valid.

1.

2.

Ox)P(x) ==> (Vx)P(x) is not valid. If the universe is all integers and P(x) is the sentence "x is odd," then peS) is true and P(8) is false. Thus, (3x)P(x) is true and (Vx)P(x) is false, so the im-

plication fails. (Vx)[P(x) V Q(x)] ==> [(Vx)P(x) V (Vx)Q(x)] is not valid. We use the example in I and also let Q(x) be "x is even." Then it is true that "All integers are either odd or even" but false that "Either all integers are odd or all integers are even."

46

CHAPTER 1 Logic and Proofs 3.

4.

[('dx)P(x) => ('dx)Q(x)] => ('dx) [P(x) => Q(x)] is not valid. The example in 2 can be used here. Because ('dx)P(x) is false, ('dx)P(x) => ('dx)Q(x) is true. However, ('dx) [P(x) => Q(x)] is false. ('dy)(3x)P(x, y) => C3x)(Vy)P(x, y) is not valid. This is probably the most troublesome of all the possibilities for dealing with quantifiers. Let the universe be the set of all married people and P(x, y) be the sentence "x is married to y." Then ('dy)(3x)P(x,y) is true, since everyone is married to someone. But (3x)(Vy)P(x, y) would be translated as "There is some married person who is married to every married person," which is clearly false.

Exercises 1.5 1.

* *

*'

2.

*

Prove that (a) there exist integers m and n such that 2m + 7n = 1. (b) there exist integers m and n such that 15m + 12n = 3. (c) there do not exist integers m and n such that 2m + 4n = 7. (d) there do not exist integers m and n such that 12m + 15n = 1. (e) if there exist integers m and n such that 15m + 16n = I, then there exist integers rand s such that 3r + 8s = 1. (f) if there exist integers m and n such that 12m + 15n = 1, then m and n are both positive. (g) if the odd natural number m has the form 4k + 1 for some integer k, then m + 2 has the form 4j - I for some integer j. (b) if m is an odd integer, then m2 = 8k + 1 for some integer k. (Hint: use the fact that k(k + 1) is even for any integer k.) (i) if m and n are integers and mn = 4k - 1 for some integer k, then m or n is of the form 4j - 1 for some integer j. Prove that, for all integers a, b, e, and d, (a) if a divides b and a divides e, then for all integers x and y, a divides bx + ey. (b) if a divides b - 1 and a divides e - 1, then a divides be - 1. (c) if a divides b, then for all natural numbers n, an divides bn. (d) if there exist integers m and n such that am + bn = 1 and d > 1, then d does not divide a or d does not divide b. (Hint: use part (a). Also compare this statement to exercises 1 (c) and 1 (d).)

3.

Prove that if every even natural number greater than 2 is the sum of two primes, then every odd natural number greater than 5 is the sum of three primes.

4.

Prove that if p is a prime number and p =I- 3, then 3 divides p2 + 2. (Hint: When p is divided by 3, the remainder is either 0, 1, or 2. That is, for some integer k, p = 3k or p = 3k + lor p = 3k + 2.)

5.

Provide either a proof or a counterexample for each of these statements. (a) For all positive integers x, x 2 + X + 41 is a prime. (b) ('dx)(3y)(x + y = 0). (Universe of all reals) (c) ('dx) ('dy) (x > 1 Ay > 0 => yX > x). (Universe of all reals) (d) For integers a, b, e, if a divides be, then either a divides b or a divides e.

1.5

Proofs Involving Quantifiers

47

(e)

"* "*

For integers a, b, c, and d, if a divides b - c and a divides c - d, then a divides b - d. (f) For all positive real numbers x, x 2 - X > o. (g) For all positive real numbers x, 100x 2 > O. (h) For every positive real number x, there is a positive real number y less than x with the property that for all positive real numbers z, yz ::::: z. (i) For every positive real number x, there is a positive real number y with the property that if y < x, then for all positive real numbers z, yz ::::: z.

6.

The conjugate of a complex number a + bi is a + bi = a - bi. (a) Prove that the conjugate of the sum of any two complex numbers is the sum of their conjugates. (b) Prove that the product of any complex number and its conjugate is a real number. (c) Prove that the conjugate of the conjugate of a + bi is a + bi.

7.

Let a, b, and d be positive integers. Suppose that (i) d divides a and d divides b, and also (ii) for every integer c, if c divides a and c divides b, then c divides d. Then we say that d is the greatest common divisor of a and b, and write d = OCD (a, b). (For example, OCD (12, 15) = 3.) Let a, b, and c be positive integers and d = OCD (a, b). Prove that (a) if c divides a and c divides b, then c :::; d. (b) d = a if and only if a divides b.

*

(c)

if c divides a and c divides b, then OCD

(d) OCD

"*

(e) (f)

"*

(g)

8.

* *

(9.,c !!.) =~.c c

(~, ~) = 1.

OCD (ac, bc) = c . OCD(a, b). It can be shown that OCD (a, b) can be written in the form ax + by for some integers x and y. Use this fact to prove Euclid's Lemma: If a divides bc and OCD (a, b) = 1, then a divides c. Use Euclid's Lemma to prove that if OCD (a, b) = 1, then ab divides c if and only if a divides c and b divides c.

Prove that (a) for every natural number n, k:::; 1. (Hint: Use the fact that n::::: 1 and divide by the positive number n.) (b) there is a natural number M such that for all natural numbers n > M, k< 0.13. (c) for every natural number n, there is a natural number M such that 2n < M. (d) there is a natural number M such that for every natural number n, k< M. (e) there is no largest natural number. (f) there is no smallest positive real number. (g) for every real number e > 0, there is a natural number M such that for all natural numbers n > M, k< [;. (h) for every real number e > 0, there is a natural number M such that if m > n > M, then k- J, < e.

48

CHAPTER 1

Proofs to Grade

Logic and Proofs

9.

*

*

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Every polynomial of degree 3 with real coefficients has a real zero. "Proof." We note that the polynomial p(x) = x 3 - 8 has degree 3, real coefficients, and a real zero (x = 2). Thus the statement "Every polynomial of degree 3 with real coefficients does not have a real zero" is false, and hence its denial, "Every polynomial of degree 3 with real coefficients has a real zero," is true. (b) Claim. There is a unique polynomial whose first derivative is 2x + 3 and which has a zero at x = I. "Proof." The antiderivative of 2x + 3 is x 2 + 3x + C. If we let p(x) = x 2 + 3x - 4, then p'(x) = 2x + 3 and p(l) = O. So p(x) is the desired polynomial. (c) Claim. For all integers a and b, if a divides b, then for all natural numbers n, an divides b n. "Proof." Let a and b be integers, and assume a divides b. Then a = kb for some integer k. Then an = (kb)n = knb n, so an divides b n. _ (d) Claim. There exists an irrational number r such that rft is rational. "Proof." If is rational, then r = is the desired example. Otherwise, is irrational and (j3ft)ft = (13)2 = 3, which is rational. or is an irrational number r such that rft is Therefore either rational. (e) Claim. Every real function is continuous at x = O. "Proof." Since either a sentence or its negation is true, we know that every real function either is continuous at x = 0 or is not continuous at x = O. Thus every real function is continuous at x = 0 or every real function is not continuous at x = O. The latter half of this sentence is false, sincef (x) = x 2 is a real function that is continuous at x = O. Since the sentence is a disjunction, the first half is true. Thus every real function is continuous at x = O. (f) Claim. If x is prime, then x + 7 is composite. "Proof." Let x be a prime number. If x = 2, then x + 7 = 9, which is composite. If x 0/= 2, then x is odd, so x + 7 is even and greater than 2. In this case, too, x + 7 is composite. Therefore, if x is prime, then x + 7 is composite. (g) Claim. For all irrational numbers t, t - 8 is irrational. "Proof." Suppose there exists an irrational number t such that t - 8 is rational. Then t - 8 =~, where p and q are integers and q 0/= O. Then t = ~ + 8 = p~8q, with p + 8q and q integers and q 0/= O. This is a contradiction because t is irrational. Therefore, for all irrational numbers t, t - 8 is irrational. (h) Claim. For real numbers x and y, if xy = 0 then x = 0 or y = O. "Proof."

13ft

Case 1. Case 2.

13ft 13 13ft

If x = 0, then xy = Oy = O. Ify = 0, then xy = xO = O.

13

1.6

(i)

In either case, xy = O. Claim. For real numbers a and b,

Additional Examples of Proofs

49

•

labl = lal Ibl.

"Proof." Case 1. Case 2.

Suppose a and b ab 2: O. Thus labl Suppose a and b ab 2: O. Thus labl

Therefore, in all cases

1.6

are nonnegative. Then

lal

=

a,

Ibl =

b, and

= ab = lal Ibi. are negative. Then lal = -a, = ab = (-a)(-b) = lallbl.

labl = lal Ibi.

Ibl

=

-b, and

•

Additional Examples of Proofs Sections 1.4 and 1.5 showed the basic forms of proof and provided examples. In this section we offer additional examples that may help your understanding of the process of constructing proofs. For most people the hardest part of writing a proof is knowing where or how to start, so we'll give special attention to the organization of proofs and the first lines of our examples. The most important thing is to make a start-almost any start. Once you've begun you may get stuck and need to begin again with a different approach, but often the first attempt will give you some ideas that can be useful in a new approach. Writing a proof is not done by staring at a statement to be proved until a full-blown proof pops into your head. It's done step by step. Here is a summary of approaches you should try when you begin a proof.

1. Make sure you understand what the assumptions are and what conclusion is to be proved. Most theorems have the form of an implication. The antecedent gives you the hypotheses to work with; your goal is to reach the consequent. To understand the antecedent and the consequent, you must also know the definitions of any technical terms that appear in the statement. It often happens that rewriting the defined terms in the form of an equation or formula or some other expression gives you the idea for how the proof will work or the material to work with. 2. To prove a statement that can be written in the form "If P, then Q," first consider a direct proof. Usually there will be one or more universal quantifiers, which may be implicit. Begin with the first step, "Assume P." P may be a conjunction of several statements, so we assume all of these statements are true. After writing your assumptions and possibly rewriting the assumptions in a more useful form, try working backward from your intended conclusion. Leave some space for the middle of your proof, and write the conclusion as your last line. Then think about what would make the conclusion true. You might rewrite the conclusion or find a suitable statement from which your conclusion follows. The idea is to try to reason forward from your assumption and also backward from your conclusion until you join them. At the middle of your proof, you will have steps that follow from the antecedent of the statement and from which the consequent follows. This makes a complete proof. 3. Don't be overly concerned with naming different types of proofs and devising "formulas" for writing proofs of each type. Here are the basic proof types:

50

CHAPTER 1 Logic and Proofs

To prove a statement of the form P ==> Q try a direct proof first. When a direct proof fails, consider a proof by contrapositive, especially when Q has the form of a negation. That is, try a direct proof of (~Q) ==> (~P). If both the direct and contrapositive methods fail, try the method of contradiction. That is, let your first step be to assume P and ~Q, then aim for some contradiction. There are two other observations about proving statements of the form P ==> Q that can be helpful. If P has the form PI V P2 , then to prove (~ V P2 ) ==> Q it is usually best to give a proof by cases. That is, try to show that (i) PI ==> Q and (ii) P2 ==> Q. When Q has the form QI V Qb a good first step is to assume P and ~QI (or assume P and ~Q2 if that is more convenient). This method is valid because (P /\ ~QI) ==> Q2 is equivalent to P ==> (QI V Q2)' A few other types of proofs are standard. To prove a biconditional statement of the form P <=> Q, you should consider first an "iff' proof in which you construct a list of equivalent statements linking P and Q. But usually, and especially when P and Q are complicated, you will need to prove both P ==> Q and Q ==> P. To prove an existence theorem, you may be able to construct or guess an object that has the desired property. If not, you may still be able to prove existence without producing an actual object, perhaps by contradiction. To prove a uniqueness theorem, you may choose one of two approaches. You may either assume that there are two objects with the property and prove that they are equal, or assume that there are two different objects with the property and arrive at a contradiction. You must also show that some object has the property. One more major type of proof, the inductive proof, will be considered in Chapter 2. There are two kinds of first steps for proofs that must be avoided. The first is to assume the result you wish to prove or any statement equivalent to that result. Regardless of what follows in your proof, this technique is invalid. Here is an example of an attempt to prove a false statement by this incorrect method.

"Claim."

-2 = 2.

Assume that -2 = 2. Then squaring both sides yields (-2)2 = 2 2, or 4 = 4. Since we have deduced a true statement (4 = 4), our assumption must be • true. Therefore, -2 = 2.

"Proof."

A second common error, whether in the first step or later in a proof, is to choose a particular value from the universe as an example to prove a universally quantified statement. One cannot prove, for example, that "If x is prime, then x + 8 is prime" by showing that the sentence is true when x = 3. Computation or analysis of the statement with particular values may give you insight about how a proof may proceed, but showing that a statement is true for one or many examples does not constitute a proof unless you explain why your reasoning in the examples must work for every object in the universe. We tum now to examples of the standard types of proofs. Our first examples are direct proofs.

1.6

Example.

For every odd integer n, 6n 2 + 5n

Additional Examples of Proofs

51

+ 4 is odd.

Proof. (We must show ('dn)(n is odd =:} 6n 2 + 5n + 4 is odd).) Let n be an odd integer. (It might help to write n in a different form, using the definition of odd.) Then n = 2k + I for some integer k. (Looking ahead to the conclusion, we see that we must write out 6n 2 + 5n + 4.) Thus 6n 2 + 5n + 4 = 6(2k + 1)2 + 5(2k + I) + 4 = 24k2 + 24k + 6 + 10k + 5 + 4 = 24k2 + 34k + 15 = 2(l2k 2 + 17k + 7) + l. Therefore, 6n 2 + 5n + 4 is odd. • Example. Prove that if two non vertical lines are perpendicular, then the product of their slopes is -I. Proof.

(A symbolic translation of this statement is ('d LI )('d L 2) [(LI is a line /\ L2 is a line /\ LI and L2 are not vertical /\ LI and L2 are perpendicular) =:} (slope of L I) times (slope of L 2) is -I]. This form suggests the first two sentences of the proof) Let LI and L2 be lines. Suppose LI is perpendicular to L2 and neither is vertical. Let al and a2 be the angles of inclination of Ll and L20 respectively. (See figure 1.3.) Since neither line is vertical, the slope of LI is tan (al) and the slope of L2 is tan (a2)' (This is the definition of slope for a nonverticalline.) We may suppose al is the larger angle (if not, simply relabel the lines). Since LI is perpendicular to L2 we have al = a2 + 90°. Thus

tan (al) = tan (a2 + 90) = -cot (a2) -I (Trigonometric identities) tan (a2)' Therefore, tan (al) tan (a2) = -l. That is, the product of the slopes of the lines • is -l. Our next example is a proof of a statement of the form P =:} (Q V R). Thus it might be proved by assuming P and ~Q, and showing R (that is, a direct proof that if x + y is irrational and x is rational, then y is irrational.) However, since the contrapositive is so easy to state by changing not-irrational to rational, a proof by contrapositive works well.

y

---r----~--~------~~----~x

Figure 1.3

52

CHAPTER 1 Logic and Proofs

Example. For real numbers x and y, if x + y is irrational, then either x or y is irrational. Proof. Assume that neither x nor y is irrational. Then both x and yare rational, so they can be written in the form x = ~ and y = f, where p, q, r, and s are integers. Therefore, x + y = ~ + f = ps:srq . Since ps + rq and qs are integers, x + y is a rational number. We have shown that if x and yare rational, then x + y is rational. We conclude that if x + y is irrational, then either x or y is irrational. Example. Let u and v be real numbers. If the point (u, v) is on the graph of the functionf(x) = x 2 - 4x + 6, then (u, v) is not on the graph of g(x) = 2x - 4. Proof.

(We might try a direct p roof by assuming that (u, v) is on the graph off, but it would be difficult to show that (u, v) is not on the graph of g. The contrapositive of the statement has the same difficulty. A proof by contradiction seems best.) (A denial of the conditional P =:} Q can be written as P A ~Q. This form tells us what assumption to make as the first step of the proof) Assume that (u, v) is on the graph of f and also on the graph of g. Then v = u 2 - 4u + 6 and v = 2u - 4. Then u 2 - 4u + 6 = 2u - 4, so u 2 - 6u + 10 = O. But the equation u 2 - 6u + 10 = 0 has no real solutions, because the discriminant 36 - 4(1)(10) = -4 < O. This is a contradiction. We conclude that if (u, v) is on the graph of the function f, then it is not on the graph of g.

The method of proof by cases is very common in proofs about absolute values, since the definition of absolute value is based on the fact that for a real number x, either x 2: 0 or x < O.

Example. Proof. Case 1.

Prove that for all real numbers x 2: I,

31x - 21 :::; 4. x

Let x be a real number such that x 2: I. We consider two cases. Suppose x - 22: O. Then Ix - 21 = x - 2. (The proofthatfollows was developed by working backwardfrom the desired conclusion that 31x;21 :::; 4. The scratch work that was used to discover this proof is shown after the proof) Since x 2: 2, -6 :::; x. Therefore, 3x - 6:::; 4x. We may divide both sides of the last inequality by x without changing the sense of the inequal-

Case 2.

ity because x is positive, and thus we have 3 (xx- 2 ) :::; 4. Suppose x - 2 < O. Then I :::; x (by hypothesis) and x < 2. Thus Ix - 21 = 2 - x. (Again, we present a proof that was worked out in advance, backward. The scratchwork used to find this part of the proof is also shown after the proof) Since x 2: 1, x 2:~. Thus 6 - 3x:::; 4x, and, using . .. 3(2-x) 4 the f act that x IS positive, -x-:::; . -

Here is the scratch work that was used to discover the proofs.

1.6

Case 1.

Case 2.

Additional Examples of Proofs

53

We want 3(\-2):::; 4. This could come from 3(x - 2):::; 4x (if x > 0) or 3x - 6 :::; 4x or -6:::; x. We can get -6:::; x from x> 0, and x > 0 follows from x - 2 2: 0, the hypothesis of case 1. We want 3 (2x-x) :::; 4. This could come from 3(2 - x):::; 4x, if x> O. To get this, we need 6 - 3x:::; 4x or 6 :::; 7x or ¥:::; x. We can get both x 2: ¥ and x > 0 from the hypothesis x 2: 1.

Our next example deals with the least common multiple of positive integers a and b. If a divides nand b divides n, we say that n is a common multiple of a and b. If (i) m is a common multiple of a and band (ii) for all n, if n is common multiple of a and b, then m divides n, then we say that m is a (the?) least common multiple of a and b, and write m = LCM (a, b). To verify that we can call m the LCM, we prove that every pair of positive integers has only one LCM.

Example.

For all positive integers a and b, LCM(a, b) is unique.

Proof. Let a and b be positive integers. Suppose mj and m2 are two LCMs for a and b. Since mj is an LCM and m2 is a common multiple of a and b, mj divides m2' Thus mj ::; m2' Similarly, m2 divides mj, so m2::; mj' Therefore, mj = m2' • A proof by contradiction begins by assuming the negation of the claim, followed by a restatement as a more useful denial.

Example.

The polynomialf(x)

= x 3 + 5x 2 + 2x

+

1 has no positive real zeros.

Proof. Assume it is not the case thatf(x) has no positive real zeros. That is, assume there exists a real number r such that r > 0 andf(r) = O. Since r > 0, each of the terms r 3, 5r2, 2r, and 1 is positive. Thus fer) = r3 + 5r2 + 2r + I > O. This is a contradiction to fer) = O. Therefore, f(x) has no positive real zeros. • Our final example in this section is a proof of Rolle's Theorem from calculus. This theorem, which has three main hypotheses about a function!, is an existence theorem because its conclusion is the existence of a number c in the interval (a, b) that makes the derivative zero. The geometric meaning of the theorem is that the graph of any function meeting all the conditions has at least one horizontal tangent between a and b. The proof constructs a value for c in one case, but not in the other two cases. In calculus texts this theorem follows the Extreme Value Theorem and the Local Maximum Theorem, so we may use those results in our proof.

Example (Rolle's Theorem). Suppose f(x) is a function defined on the closed interval [a, b Jt such that (i) fis continuous on [a, b); (ii) f is differentiable on (a, b) i; and (iii) f(a) = feb) = O. Then there exists at least one number c between a and b such that the derivative !'(c) = O. '[a, bland (a, b) denote closed and open intervals of real numbers. See section 2.1.

54

CHAPTER 1 Logic and Proofs

a

b

c

(b) Two c's with horizontal tangent lines.

(a) One c with horizontal tangent line.

Figure 1.4

Proof. (We start by stating the several hypotheses.) Let f be a function defined on [a, b]. Suppose f is continuous on [a, b], differentiable on (a, b), and f(a) = feb) = o. We consider three cases: Case 1. Suppose f(x) > 0 for some x in [a, b]. We apply the Extreme Value Theorem, which says that if f is continuous on [a, b], then f attains its absolute maximum value on [a, b] at some point c in the interval. Since the maximum value of f is positive and f(a) = feb) = 0, c cannot equal a or b. Thus c is in (a, b). By the Local Maximum Theorem (Iff (c) is the maximum value of a continuous functionf andf'(c) exists, thenf'(c) = 0.), either f'(c) = 0 or f'(c) does not exist. Since f is differentiable on (a, b), f'(c) does exist. Thus f'(c) = O. Case 2. Supposef(x) < 0 for some x in [a, b]. Then by the Extreme Value Theorem f attains its absolute minimum value on [a, b] at some point c of the interval. Since the minimum value of f is negative and f(a) = feb) = 0, c does not equal a or b. Thus c is in (a, b). By the Local Minimum Theorem either f'(c) = 0 or f'(c) does not exist. Since f is differentiable on (a, b), f'(c) does exist. Thus f'(c) = O. (Note thatfor afunction like the one shown infigure I.4(b), both cases I and 2 apply. Usually a proof by cases is constructed so that the cases are mutually exclusive, but this is not required. What is essential is that a proof account for all possibilities. Our third case below covers the only remaining possibility.) Case 3. Suppose f(x) = 0 for all x in [a, b]. Then f'(x) = 0 for all x in (a, b). (The derivative of a constant function is zero.) In this case we may choose any c in (a, b)-say, c = b , the average of a and b. Then c is in (a, b) and f'(c) = O. •

ai

Exercises 1.6 1.

Prove that (a) for all integers n, 5n 2 + 3n + 1 is odd. (b) for all odd integers n, 2n2 + 3n + 4 is odd.

1.6

(e) (d) *: (e)

2.

55

the sum of 5 consecutive integers is always divisible by 5. if two nonverticallines have slopes whose product is - I, then the lines are perpendicular. if x and yare rational numbers with x < y, there exists a rational number between x and y.

Let I be the line 2x + ky = 3k. Prove that (a) if k =I- -6, then I does not have slope (b) for every real number k, I is not parallel to the x-axis. (e) there is a unique real number k such that I passes through (1,4).

l

3. (a) (b)

(e) (d)

4.

Additional Examples of Proofs

(a)

(b)

Prove that if x is rational and y is irrational, then x + y is irrational. Prove that there exist irrational numbers x and y such that x + y is rational. Prove that for every rational number z, there exist irrational numbers x and y such that x + y = z. Prove that for every rational number z and every irrational number x, there exists a unique irrational number y such that x + y = z. Let (x, y) be a point inside the circle with center at the origin and radius r. Prove that the line passing through (x, y) and (r, 0) is not perpendicular to the line passing through (x, y) and ( - r, 0). Prove that except for two points on the circle, if (x, y) is on the circle with center at the origin and radius r, then the line passing through (x, y) and (r, 0) is perpendicular to the line passing through (x, y) and (- r, 0). Which two points are the exceptions?

5.

Prove that (a) every point on the line y = 6 - x is outside the circle with radius 4 and center (-3, 1). (b) Prove that there exists a three-digit natural number less than 400 with distinct digits, such that the sum of the digits is 17 and the product of the digits is 108. (e) Use the Extreme Value Theorem to prove that if/does not have a maximum value on the interval [5, 7], then/is not differentiable on [5,7]. (d) Use Rolle's Theorem to show that x 3 + 6x - 1 = 0 does not have more than one real solution.

6.

Prove that

7.

*

(a)

for all nonnegative real numbers x,

(b)

If -2 < x < 1 or x> 3, then

.

12x-11 1::5 2.

x+

(x - l)(x (x - 3)(x

+ 2) > O. + 4)

Prove or disprove: (a) Every point inside the circle (x - 3)2 + (y - 2)2 = 4 is inside the circle x 2 + y2 = 41. (b) If (x,y) is inside the circle (x - 3)2 + (y - 2)2 = 4, then x - 6 < 3y. (e) Every point inside the circle (x - 3)2 + (y - 2)2 = 4 is inside the circle (x - 5)2 + (y + 1)2 = 25.

56

CHAPTER 1 Logic and Proofs 8.

*

"*

Prove that for every prime p and for all natural numbers a and b, (a) GCD(p, a) = p iff p divides a. (b) GCD(p, a) = 1 iff P does not divide a. (c) if p divides ab, then p divides a or p divides b.

9. Prove that if the natural number q has the property that q divides a or q divides b whenever q divides ab, then q is prime.

10. Let a, b, c, and n be natural numbers and LCM(a, b)

"* "* "* Proofs to Grade

11.

*

(a) (b) (c) (d) (e) (f) (g)

= m. Prove that if a divides nand b divides n, then m :s n. LCM(a, b) = b iff a divides b. LCM(a, b) :s abo ( b) if c divides a and c divides b, then LCM f:!., - =~. c c c for all natural numbers n, LCM(an, bn) = n' LCM(a, b). if GCD(a, b) = 1, then LCM(a, b) = abo LCM(a, b) . GCD(a, b) = abo

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. For all natural numbers n, GCD(n, n + 1) = 1. "Proof." (i) 1 divides nand 1 divides n + 1. (ii) Suppose c divides n and c divides n ~ 1. Then 1 divides C. Therefore, GCD(n, n + I) = I. (b) Claim. For all natural numbers n, GCD(2n - 1, 2n + 1) = I. "Proof." Obviously I divides both 2n - 1 and 2n + 1. Suppose c divides 2n - 1 and 2n + 1. Then c divides their sum, 4n, so c also divides 4n 2 . Furthermore, c divides their product, 4n 2 - 1. Since c divides 4n 2 and 4n 2 - 1, c divides 4n 2 - (4n 2 - 1) = 1. Therefore, I is the greatest common divisor. (c) Claim. For all real numbers x, Ix - 21 < x 2 - 2. "Proof." Consider x = 5. 15 - 21 = 3 < 23 = 52 - 2. (d) Claim. If x is any real number, then either n - x is irrational or n + x is irrational. "Proof." It is known that n is an irrational number; that is, n cannot be written in the form ~ for integers a and b. Consider x = n. Then n - x = 0, which is rational, but n + x = 2n. If 2n were rational, then 2n = ~ for some integers a and b. Then n = Cfb)' so n is rational. This is impossible, so 2n is irrational. Therefore either n - x or n + x is irrational. (e) Claim. If x is any real number, then either n - x is irrational or n + x is irrational. "Proof." It is known that n is an irrational number; that is, n cannot be written in the form ~ for integers a and b. Let x be any real number. Suppose both n - x and n + x are rational. Then since the sum of two rational numbers is always rational, (n - x) + (n + x) = 2n is rational. Then 2n = ~ for some integers a and b. Then n = C;b)' so n is rational. This is impossible. Therefore, at least one of n - x or n + x is irrational.

1.6

(f)

*

(g)

Additional Examples of Proofs

57

Claim. There is a unique 3 digit number whose digits have sum 8 and product 10. "Proof." Let x, y, and z be the digits. Then x + y + z = 8 and xyz = 10. The only factors of 10 are 1, 2, 5, and 10, but since lOis not a digit, the digits must be 1,2, and 5. The sum of these digits is 8. Therefore, 125 is the only 3 digit number whose digits have sum 8 and product 10. • Claim. There is a unique set of three consecuti ve odd numbers that are all prime. "Proof." The consecutive odd numbers 3, 5, and 7 are all prime. Suppose x, y, and z are consecutive odd numbers, all prime, and x =1= 3. Then y = x + 2 and z = x + 4. Since x is prime, when x is divided by 3, the remainder is 1 or 2. In case the remainder is 1, then x = 3k + 1 for some integer k 2: 1. But then y = x + 2 = 3k + 3 = 3(k + 1), so y is not prime. In case the remainder is 2, then x = 3k + 2 for some integer k 2: 1. But then z = x + 4 = 3k + 2 + 4 = 3(k + 2), so z is not prime. In either case we reach the contradiction that y or z is not prime. Thus x = 3 and y = 5, z = 7. Therefore, the only three consecutive odd primes are 3, 5, and 7. •

CHAPTER

2

Set Theory

Much of mathematics is written in terms of sets. We assume that you have had some experience with the basic notions of sets, unions, and intersections. Sections I and 2 consist of a brief review of sets and operations and present the notation we shall use. Section 2 provides the opportunity for you to prove some set-theoretical results. In section 3 we extend the set operations of union and intersection and encounter indexed collections of sets. Sections 4 and 5 deal with inductive sets and proof by induction. Basic techniques of counting and some applications of these techniques appear in the optional section 6.

2.1

Basic Notions of Set Theory We shall understand a set to be a specified collection of objects. The objects in a given set are called the elements (or members) of the set. By saying that a set is a specified collection we mean that, for any object, there must be a definite yes or no answer to the question whether the object is a member of the set. In general, capital letters are used to denote sets, and lowercase letters to denote objects. If the object x is an element of set A, we write x EA; if not-that is, if ~(x E A)-we write x t!. A. For example, if B is the set of all signers of the Declaration of Independence, we write John Hancock E B and Joe Slobotnik t!. B. Sets can be described in words, such as "the set of odd integers between 0 and 12," or the elements may be listed, as in {I, 3, 5, 7, 9, II}, or even partially listed, as in {I, 3, 5, ... , II}. Listing the elements of sets is often impractical and sometimes impossible. To designate most sets, we will use the following notation: {x: P(x)},

where P(x) is a one-variable open sentence description of the property that defines the set. For example, if P(x) is

"x is an odd integer between 0 and 12," then the set {x: P(x)} is {I, 3, 5,00', II}. 59

60

CHAPTER 2 Set Theory

The variable x in {x: P~)} is a dummy variable in the sense that any letter or symbol serves equally well. If P(x) is the sentence "x is a real number and x 2 - 6x = 0," then {x: P(x)}, {y: P( y)}, {a: P(a)} all designate the set {O, 6}. A word of caution: Trouble may arise unless some care is taken with the open sentence P(x) in {x: P(x)}. The best example of this situation is the famous Russell paradox (see exercise 18), which first appeared in a 1902 letter written by the British mathematician and philosopher Bertrand Russell (1872-1970). Russell, pointing out a flaw in the powerful new theory of infinite sets (see chapter 5) developed by Georg Cantor (1845-1918), showed that it is not true that for every open sentence P(x), there corresponds a set {x: P(x)}. The resolution of Russell's and other paradoxes involved making a distinction between sets and arbitrary collections of objects. Sets may be defined within the framework of the Zermelo-Fraenkel system of axiomatic set theory,1" which evolved from the work of Ernst Zermelo (1871-1953) and Abraham Fraenkel (1891-1965). Their axioms assert, for example, that a collection of two objects forms a set (Axiom of Pairing) and that the collection of all subsets of a set is a set (Axiom of Powers). Under their system, known paradoxes such as Russell's are avoided. It is not our purpose here to carry out a formal study of axiomatic set theory, but to introduce the most basic and useful concepts of sets. All our discussions of sets are consistent with the Zermelo-Fraenkel system of axiomatic set theory. Special notation will be used for the following sets of numbers: N = {I, 2, 3,00 .}, the natural numbers 7L = {oo., -3, -2, -1,0, 1,2, 3,00.}, the integers Q = the set of rational numbers ~ = the set of real numbers In addition, we will use the conventional interval notation for certain collections of the real numbers. For a, b E ~ with a < b, we will use these notations: (a, b) = {x: x E ~ and a < x < b} is the open interval from a to h. [a, b] = {x: x E ~ and a ::5 x ::5 b} is the closed interval from a to h. (a, 00) = {x: x E ~ and x> a} and (-00, a) = {x: x E ~ and x < a} are open

rays. [a, 00) = {x: x E ~ and x rays.

2:

a} and (-00, a] = {x: x E ~ and x

::5

a} are closed

The half-open (or half-closed) intervals [a, b) and (a, b] are defined similarly. Note that we are not defining a number "00"; we are simply using the infinity symbol as a convenience in ray designation. One should be careful not to confuse (1,6) with the set {2, 3, 4, 5}, since (1,6) is defined as the set of all real numbers between 1 and 6 and contains, for example, loglO (15), and A second word of caution: Recall that the universe of discourse is a collection of objects understood from the context or specified at the outset of a discussion and that all objects under consideration must belong to the universe. Some ambiguity

2,13,

¥.

t A complete study of the foundations of set theory from the Zermelo-Fraenkel axioms may be found in Notes on Set Theory by Y. N. Moschovakis (Springer-Verlag, 1994). The subject of set theory is still active today, with many unsolved and interesting problems.

2.1

Basic Notions of Set Theory

61

may arise unless the universe is known. For example, membership in the set A = {x: X 2 - 6x = o} depends on an agreed-upon universe. For the universe of real numbers, A is {O, 6}, but A is {6} for the universe of natural numbers.

DEFINITION Let 0 = {x: x oF x}. Then 0 is a set with no elements and is called an empty set.

Be certain that you understand that we have defined a set. The open sentence x oF x is false for any object x; that is, x E 0 is false for every object x. We can define other empty sets, such as {x: x E IR and x = x + I} and x: x E N and x < OJ.

However, we shall see later in this section that these are all the same, so there is in fact only one empty set.

DEFINITION Let A and B be sets. We say that A is a subset of B iff every element of A is also an element of B. In symbols, A ~ B <=> (\;Ix)(x E A ==> x E B).

Venn diagrams may be used to display some simple relationships between sets. For example, sets A and B such that A ~ B is depicted in figure 2.1.

c0=V

Universe

Figure 2.1 Since the statement A ~ B is defined with the universal quantifier (\;Ix), a direct proof of A ~ B has the following form:

DIRECT PROOF OF A k B Let x be any object. (Show that x E A ==> x E B.) Suppose x EA. Thus x EB. Therefore A

~

B.

•

In practice, the first sentences of this outline are often combined into one: "Suppose x is any object in A," or "Suppose x EA."

62

CHAPTER 2 Set Theory

Theorem 2.1

For any set A, 0~A.

(a) (b)

A

~A.

Proof. (a) LetA be any set. (We now give a direct proofthat 0 ~A.) Letxbe any object. Because its antecedent is false, the conditional sentence (x E 0 => x E A) is true. Therefore, 0 ~ A. (b) Let B be any set. (To prove B ~ B, we must show that, for all objects x, if x E B, then x E B.) Let x oe any object. Then x E B => x E B is true. (Here we use the tautology P => P.) Therefore, (Vx)(x E B => x E B), and so B ~ B. • For a given set, the subsets 0 and A are called improper subsets of A, whereas any subset of A other than 0 and A is called a proper subset of A. To denote that B is a proper subset of A, some authors write Be A. Other authors write B~A, a combination meaning that B ~ A and B A.

*"

Theorem 2.2

LetA, B, C be sets. If A Proof.

~

Band B ~ C, then A ~ C.

•

Exercise 12.

We have seen that a set may ~e described in different ways. Often it becomes necessary to know whether two descriptions of sets do in fact yield the same set. Intuitively, two sets A and B are "equal" ifthey contain exactly the same elements. We might say, then, that A

= B iff

(Vx)(x EA <=>x EB).

However, the latter is equivalent to (Vx)(x EA=>x EB) and (Vx)(x EB=>x EA),

which is longer but more natural to use in proving that A

DEFINITION

Let A and B be sets. Then A

= B.

= B iff A ~ Band B ~ A.

According to this definition, the task of proving an equality between two sets A and B is accomplished by showing that each set is a subset of the other, that is, by showing (i) (ii)

A B

~B, ~A,

and then concluding A = B. The proofs of (i) and (ii) need not be by the same methods. Occasionally, A = B can be proved by a chain of equivalent statements showing that x E A iff x E B.

2.1

Basic Notions of Set Theory

Example. Prove that X = Y where X = {x: x is a solution to x 2 Y = {-I, I}.

Proof. We must show (i) Y

-

1=

63

o} and

~ X and (ii) X ~ Y.

We show Y ~ X by individually checking each element of Y. By substitution, we see that both I and -I are solutions to x 2 - I = O. Thus Y ~ X. Next, we must show X ~ Y. Let 1 EX. Then, by definition of X, 1 is a solution to x 2 - I = O. Thus 12 - I = O. Factoring, we have (t - 1)(1 + 1) = O. This product is 0 exactly when 1 - I = 0 or 1 + 1 = O. Therefore, 1 = 1 or 1 = -I. Thus if 1 is a solution, then t = 1 or 1 = -1; so 1 E Y. This proves X ~ Y.

(i)

(ii)

By (i) and (ii), X

~

Y and Y

~X;

•

soX = Y.

We are now in a position to prove that there is only one empty set, in the sense that any two empty sets are equal.

Theorem 2.3

If A and B are sets with no elements, then A = B. Since A has no elements, the sentence (\fx)(x E A ~ x E B) is true. Therefore, A ~ B. Similarly, (\fx)(x E B ~ x E A) is true, so B ~ A. Therefore, by definition of set equality, A = B. •

Proof.

It should now be clear that if we want to prove A ~ B, the simplest case would be that A is empty. This is because if A is empty, then A ~ B by Theorem 2.1. In the chapters that follow are some more interesting cases where special care must be taken to be sure that a statement about a set is true, even when the set happens to be empty. Next we define a set whose elements are themselves sets. One of the axioms of set theory asserts that for every set A, the collection of all subsets of A is also a set.

DEFINITION Let A be a set. The power set of A is the set whose elements are the subsets of A and is denoted g'>(A). Thus g'>(A)

Example.

Let A =

{a,

b,

= {B: B ~A}.

c}. Then

g'>(A) = {0,

{a},

{b},

{c}, {a,

b},

{a, c},

{b,

c}, A}.

You should always remember that the elements of the set g'>(A) are themselves sets, specifically the subsets of A. Also, in working with sets whose elements are sets, it is important to recognize the distinction between "is an element of' and "is a subset of." To use A E B correctly, we must consider whether the object A (which happens to be a set) is an element of the set B, whereas A ~ B requires determining whether all objects in the set A are also in B.

64

CHAPTER 2 Set Theory

Example. Let X = HI, 2, 3}, {4, 5}, 6}. Then X is a set with three elements, namely, the set {I, 2, 3}, the set {4, 5}, and the number 6. The set H4, 5}} has one element; it is {4, 5}. All of the following are true: 6EX

and

{4, 5} E X,

but

4 ff. X.

{4} ff. {4, 5} but {4} C {4, 5}. {4, 5} i. X because 5 ff. X. {6} C X but {6} ff. X. {6} E !Jl(X)

but

{6}

i. !Jl(X).

H4, 5}} C X because {4, 5} EX. {4, 5} ff. !Jl(X) but H4, 5}} E !Jl(X).

o C X,

so

0 E !Jl(X),

and

{0} C !Jl(X).

HI, 2, 3}}, H4, 5}}, {6}, HI, 2, 3}, {4, 5}}, HI, 2, 3}, 6}, H4, 5}, 6}, X}.

!Jl(X) = {0,

Notice that for the set A = {a, b, c} in the example above, A has three elements and !Jl(A) has 23 = 8 elements. As we see in the next theorem, this observation can be generalized to any set of n elements. It is for this reason that 2A is sometimes used to denote the power set of A.

Theorem 2.4

If A is a set with n elements, then !Jl(A) is a set with 2n elements.

Proof. (The number of elements in !Jl(A) is the number of subsets of A. Thus to prove this result, we must count all the subsets of A.) If n = 0, that is, if A is the empty set, then !Jl(0) = {0}, which is a set with 2° = I elements. Thus the theorem is true for n = O. Suppose A has n elements, for n:::=:: 1. We may write A as A = {Xl, X2,"" x n }. To describe a subset B of A, we need to know for each Xi E A whether the element is in B. For each Xi' there are two possibilities (Xi E B or Xi ff. B), so there are 2 . 2 . 2 ..... 2 (n factors) different ways of making a subset of A. Therefore!Jl (A) has 2n elements. (The counting rule used here is called the Product Rule. See Theorem 2.21 and the second example following that theoremfor further explanation.) •

Theorem 2.5

Let A and B be sets. Then A C B iff !Jl(A) C !Jl(B).

Proof. (This is a good example of a biconditional theorem for which a two-part proofis easier than an ifjproof) (i)

We must show that A C B implies !Jl(A) C !Jl(B). Assume that A C B, and suppose X E !Jl(A). We must show that X E !Jl(B). But X E !Jl(A) implies X CA. Since X C A and A C B, then X C B by Theorem 2.2. But X C B implies X E !Jl(B). Therefore, X E !Jl(A) implies X E !Jl(B). Thus !Jl(A) C !Jl(B).

2.1 (ii)

Basic Notions of Set Theory

65

We must show that QJ>(A) ~ QJ>(B) implies A ~ B. Assume that QJ>(A) ~ QJ>(B). By Theorem 2.1, A ~ A; soA E QJ>(A). Since QJ>(A) ~ QJ>(B) , A E QJ>(B). There• fore A ~ B.

The second half of the proof above could have been done by showing directly that if x E A, then x E B. Such proofs that concentrate on the whereabouts of individual members of sets are often called "element-chasing" proofs. The given proof is preferable because it makes use of a theorem we already know. Both proofs are correct. When you write proofs, you may choose one method of proof over another because it is shorter, or easier to understand, or for any other reason. In proving that A ~ B, we do not assume that A has elements. We begin by supposing that x E A and use that supposition to show that x E B. The phrase "Let x EA" is usually reserved for a situation in which it is known that A has at least one element. For example, in the proof that QJ>(A) ~ QJ>(B) in part (i) of Theorem 2.5, we can be sure that QJ>(A) is not empty because 0 E QJ>(A) for every set A.

Exercises 2.1 1. Write the following sets by using the set notation {x: P(x)}. (a) The set of natural numbers strictly less than 6 (b) The set of integers whose square is less than 17

* *

"*

(c) (d) (e) (f)

[2, 6] (-1,9] [-5, -I)

The set of rational numbers less than -I

2. Write each of the sets in exercise I by listing (if possible) all its elements. 3. True or false?

* * * * *

(a) (c) (e)

NCO. N ~!R. [!, ~] ~ (!, ~).

(b) (d) (f)

(g) (i)

[7, 10] ~!R. [7, 10) ~ {7, 8, 9, 1O}.

(h) (j)

4. True or false? 0E {0, {0}}. * (a) (c) {0} E {0, {0}}. * (e) H0}} E {0, {0}}. * (g) For every set A, 0 EA. * (i) {0, {0}} ~ H0, {0}}}. * (k) {I, 2, 3} ~ {I, 2, 3, {4}}.

*

(b) (d) (f) (h)

(j) (I)

O~E.

[!,~] ~ O. !R C O. [2,5] = {2, 3, 4, 5}. (6, 9] ~ [6, 10).

o ~{0, {0}}. {0} ~ {0, {0}}. H0}} ~ {0, {0}}. For every set A, {0} ~ A. {I, 2} E HI, 2, 3}, {I, 3}, I, 2}. H4}} ~ {I, 2, 3, {4}}.

66

CHAPTER 2 Set Theory

5.

* *

Give an example, if there is one, of sets A, B, and C such that the following are true. If there is no example, write "Not possible." (a) A C B, B ~ C, and A C C. (b) A C B, B C C, and C CA. (c) A iB, B ~ C, and A C C. (d)

ACB,BiC,andA~C.

6. Write the power set, r;IP(X), for each of the following sets. (a) X = {o, £:', D} (b) X = is, {sH (c) X = {0, {a}, {b}, {a, bH (d) X = {I, {2, {3}}} (e) X={1,2,3,4} 7. List all of the proper subsets for each of the following sets.

* *

* *

{{0H

{I}

{0, {0H

(f)

{o,

£:',

D}

9.

True or false? (a) (c) (e) (g) (i)

(b) (d) (f)

{0} E r;IP({0, {0}}}. 0 C r;IP({0, {0}}}. {{0H C r;IP({0, {0}}}.

3 E Q.

(b)

{3} E r;IP(Q).

(j)

{3} C r;IP(Q). {{3H C r;IP(Q).

0 E r;IP( {0, {0H). {{0H E r;IP({0, {0}}}. {0} C r;IP({0, {0}}}.

10.

(I) {{3H E r;IP(Q). (k) {3} C Q. LetA = {x: P(x)} and B = {x: Q(x)}. (a) Prove that if ('tfx) (P(x) ==> Q(x», then A C B. (b) Prove that if ('tfx)(P(x) <=> Q(x», then A = B.

11.

Prove that if x

t!. B and A C B, then x t!. A.

12. Prove Theorem 2.2. 13. Let X = {x: P(x)}. Are the following statements true or false? (a)

*

(b) (d)

Give an example, if there is one, of each of the following. If there is no example, write "Not possible." (a) A set A such that r;IP(A) has 64 elements. (b) Sets A and B such that A C B, and r;IP(B) C r;IP(A). (c) A set A such that r;IP(A) = 0. (d) Sets A, B, and C such that A C B, B C C, and r;IP(A) C r;IP(C).

*

"*

0 {I,2}

8.

* * * * * *

"*

(a) (c) (e)

(b) (c)

IfaEX,thenP(a). If pea), then a EX. If ~P(a), then a t!.X.

14. Prove that if A C B, B C C, and C C A, then A = Band B = C. 15. Prove that X = Y, where X = {x: x E Ilhndxis a solution tox 2 - 7x + 12 = o} and Y = {3, 4}. 16. Prove that X = Y, where X = {x E Z: Ixl::5 3}and Y = {-3, -2, -I, 0, 1,2, 3}. 17. Prove that X = Y, where X = {x E N: x 2 < 30} and Y = {I, 2, 3,4, s}. 18. (Russell paradox) A logical difficulty arises from the idea, which at first appears natural, of calling any collection of objects a set. A set B is ordinary if

2.2

Set Operations

67

B tt. B. For example, if B is the set of all chairs, then B tt. B, for B is not a chair. It is only in the case of very unusual collections that we are tempted to say that a set is a member of itself. (The collection of all abstract ideas certainly is an abstract idea.) Let X = {x: x is an ordinary set}. Is X E X? Is X tt. X? What should we say about the collection of all ordinary sets? Proofs to Grade

19.

*

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If X = {x E N: x 2 < I4} and Y = {I, 2, 3}, then X = Y. "Proof." Since 12 = 1 < 14,22 = 4 < 14, and 32 = 9 < 14, X = Y. • (b) Claim. If A, B, and C are sets, and A ~ Band B ~ C, then A ~ C. "Proof." Let A = {I, 5, 8}, B = {I, 4,5,8, IO}, and C = {I, 2,4,5, 6,8, IO}. Then A ~ B, and B ~ C, and A ~ C. • (c) Claim. If A, B, and C are sets, and A ~ Band B ~ C, then A ~ C. "Proof." Suppose x is any object. If x E A, then x E B, since A ~ B. If x E B, then x E C, since B ~ C. Thus, x E C. Therefore, A ~ C. • (d) Claim. If A, B, and C are sets, and A ~ Band B ~ C, then A ~ C. "Proof." If x E C, then, since B ~ C, x E B. Since A ~ B and x E B, it follows that x EA. Thus x E C implies x EA. Therefore, A ~ C. • (e) Claim. If A is a set, A ~ peA). "Proof." Assume A is a set. Suppose x E A. Then x ~ A. Thusx E 0">(A). Therefore, A ~ 0">(A). • (f) Claim. If A is a set, A ~ 0">(A). "Proof." Assume A is a set. Suppose x EA. Then {x} ~ A. Thus {x} E 0">(A). Therefore, A ~ 0">(A). • (g) Claim. If A and B are sets and 0">(A) ~ 0"> (B) , then A ~ B. "Proof."

XEA=:}lxl~A x E 0">(A) =:} =:}

x E 0">(B)

=:}

x

~B

=:}xEB.

*

2.2

Therefore, x E A =:} x E B. Thus A ~ B. • (h). Claim. If A ~ Band B C, then A C. "Proof." Suppose A ~ Band B C. Then there exists x E B such that x tt. c. Since x E B, x E A by definition of subset. Thus x E A and x tt. c. Therefore, Art c. •

rt

rt

rt

Set Operations In this section we present the most common ways to combine two sets to produce a third. These binary operations on sets (that is, operations on exactly two sets) will be generalized in the next section.

68

CHAPTER 2 Set Theory

DEFINITIONS

Let A and B be sets. The union of A and B is defined by AU B

= {x: x E A or x E B}.

The intersection of A and B is defined by A

nB=

{x: x E A and x E B}.

The difference of A and B is defined by A - B = {x: x E A and x

t!. B}.

The set A U B can be thought of as a new set formed from A and B by choosing as elements the objects that appear in at least one of A or B; A n B consists of objects that appear in both; A - B contains those elements of A that are not in B (see the Venn diagrams in figures 2.2, 2.3, and 2.4). In case A n B = 0, we say A and B are disjoint.

Example.

For A = {I, 2, 4, 5, 7} and B = {I, 3, 5, 9}, AU B :: 11,2,3,4,5, 7, 9}, AnB- 1,5}, A - B = 2, 4, 7}, B - A = 3,9}.

Example.

For C = [I, 4) and D = (2, 6], CUD = [1, 6], enD = (2, 4), and C - D = [I, 2]. The sets [I, 2] and (2,4) are disjoint. Notice that it follows from the definitions of the set operations that xt!.AUB

means

xt!.Aandxt!.B.

xt!.AnB

means

x

xt!.A-B

means

x

t!. A or x t!. B. t!. A or x E B.

The set operations obey certain rules that at times allow us to simplify our work, replace an expression by an equivalent one, or make conclusions. Several such laws have been gathered together in the next theorem. Some of these will seem very apparent, especially if you visualize them with a Venn diagram. Although such a diagram can lead you toward a proof, the diagram itself does not constitute a proof. Those parts of Theorem 2.6 not proved are left as exercise 8.

AUB

AnB

A-B

Figure 2.2

Figure 2.3

Figure 2.4

2.2

Theorem 2.6

Set Operations

69

Let A, B, and C be sets. Then (a) (b) (c) (d) (e) (f) (g) (h) (i)

(j) (k) (I) (m) (n) (0) (p) (q) (r)

A ~A UB. A nB~A. An0=0. AU0=A. AnA =A. A UA =A. A UB=BUA. A nB=BnA. A-0=A. 0-A=0. A U (B U C) = (A U B) U C. A n (B n C) = (A n B) n c. An (B U C) = (A B) U (A C). AU (B n C) = (A U B) n (A U C). A ~ B iff A U B = B.

n

A ~ B iff A n B = A. If A ~ B, then A U C If A ~ B, then A n C

n

~ ~

B U C. B n c.

Proof. (b) (f)

(h)

We must show that, if x E A n B, then x EA. Suppose x E A n B. Then x E A and x E B. Therefore, (because P A Q =:} p) x EA. We must show that x E A U A iff x EA. By the definition of union, x E A U A iff x E A or x EA. This is equivalent to x EA. Therefore, A U A = A. (This biconditional proof uses the definition of intersection and the equivalence ofP A Q and Q AP.)

xEAnB iff x E A and x E B iff x E B and x E A iff x EB nA. (m)

(As you read this proof, watchfor the steps in which the definitions of union and intersection are used (twofor each). Watch also for the use of the equivalencefrom Theorem (l.2(f).) x EA n (B U C) iff x E A and x E B U C iff x E A and (x E B or x E C) iff (x E A and x E B) or (x E A and x E C) iff x E A n B or x E A n C iff x E (A n B) U (A n C).

(p)

Therefore A n (B U C) = (A n B) U (A n C). (The statement A ~ B if.! A n B = A requires separate proofs for each implication. We make use of earlier parts ofthis theorem.) First, assume that A ~ B. We must show that A n B = A. If x E A, then from the hypothesis A ~ B, we have x E B. Therefore x E A implies x E A and x E B, so x E A n B. This

70

CHAPTER 2 Set Theory

shows that A ~A n B, which, combined with A n B ~A from part (b) of this theorem, gives A n B = A. Second, assume that A n B = A. We must show that A ~ B. By parts (b) and (h) of this theorem, we have B n A ~ Band B n A = A n B. Therefore, An B ~ B. By hypothesis, An B = A, soA ~ B. • Recall that the universe of discourse is a collection of objects understood from the context or specified at the outset of a discussion and that all objects under consideration must belong to the universe.

DEFINITION If U is the universe and B ~ U, then we define the complement of B to be the set iJ = U - B.

Thus, iJ is to be interpreted as all those elements of the universe that are not in B (figure 2.5).

Figure 2.5 For the set B = {2, 4, 6, 8}, iJ = {1O, 12, 14, 16, ... } if the universe is all even natural numbers, while iJ = {l, 3, 5, 7, 9, 10, II, 12, 13, ... } if the universe is all natural numbers. For the universe ~, if B = (0, 00), then iJ = (-00, 0]. If A = {x: x is rational}, then A = {x: x is irrational}. The complement is nothing more than a short way to write set difference in the case where the first set is the universe. Complementation is a unary operation on sets because it applies to a single set. The complement operation obeys several rules.

Theorem 2.7

Let U be the universe, and let A and B be subsets of V. Then (a) (b) (c) (d) (e)

A=A. A UA = u. AnA=0. A - B =A niJ. A C B iff iJ C A.

(f) (g) (h)

A UB=A nB.} De Morgan's Laws AnB=A UB. A n B = 0 iff A ~ iJ.

~

r--./

--_

_

2.2 Set Operations

Proof. (a) (e)

A

71

A

By definition of the complement x E iff x fI. A iff x E A. Therefore = A. (For this part ofthe theorem, we give two separate proofs. The first proof has two parts, and its first part is used to prove its second part.) First proof of (e). We prove that if A ~ B then iJ ~A. Assume that A ~ B. Suppose x E iJ. Then x fI. B. Since A ~ B and x fI. B, we have x fI. A. (This is the contrapositive of A~B.) Thus xEA. Therefore, iJ~A. For the second part of this proof, we show that iJ ~ A implies A ~ B. Assume that iJ ~ A. Then by the first part of this proof, A ~ iJ. Therefore, using part (a), A ~ B. Combining the two parts of this proof, A ~ B iff iJ ~ A. Second proof of (e). (This proof makes llse of the fact that a conditional sentence is equivalent to its contrapositive.) A~B

iff (Vx)(x EA =>x E B) iff (Vx)(x fI. B => x fI. A) iff (Vx)(x E iJ => x E A) iff iJ ~A. '---'

(f)

The object x is a member of A U B iff x is not a member of A U B iff it is not the case that x E A or x E B iff x fI. A and x fI. B (see Theorem 1.2(c») iff x E A and x E iJ iffx EA n iJ.

•

The proofs of the remaining parts are left as exercise 9.

Exercises 2.2 1. Let A = {I, 3, 5, 7, 9}, B = {a, 2, 4, 6, 8}, C {I, 2, 3, 5, 6, 7, 8, 9, 1O}. Find

* * * * *

2.

* *

= {I, 2, 4,5, 7, 8},

(a) (c) (e)

AU B A- B (A - B) - C

(f)

A

00

~nC)nD

00

An~uC)

(i)

(AnB)U(AnC)

(j)

(AUB)-(CnD)

and D =

An B

(b)

(d) A - (B - C)

n (C n D)

Let U be the set of all integers. Let E, D, P, and N be the sets of all even, odd, positive, and negative integers, respectively. Find (c) D-E (a) E - P (b) P - E (f) N (d) P - N (e) P (g)

E

* *

(h) (k)

* *

b U

(j) U - P 3. Let the universe be all real numbers. Let A and D = (5, 00). Find (a) (c) (e)

AU B BUC An B

* *

(b) (d) (f)

(i) (I)

E-N

o

= [3, 8), B = [2, 6],

AU C BUD BnC

C

= (1, 4),

72

CHAPTER 2 Set Theory

* 4.

*

*

(g) (i) (k) (m)

(j)

A-B D-A

(I) (n)

(A U C) - (B

(h)

AnD B-D A B - (A U C)

jj

n D)

Let V = {I, 2, 3} be the universe for the sets A = {I, 2} and B = {2, 3}. Find (a) (c) (e)

g>(A) g>(A) n g>(B) g>(A) - g>(B)

* *

(b) (d) (f)

g>(A) g>(A) U g>(B) g>(A) - g>(B)

5.

Let A, B, C, and D be as in exercise 1. Which pairs of sets are disjoint?

6.

Let V, E, D, P, and N be as in exercise 2. Which pairs of sets are disjoint?

7.

Let A, B, C, and D be as in exercise 3. Which pairs of sets are disjoint?

8.

Prove the remaining parts of Theorem 2.6.

9.

Prove the remaining parts of Theorem 2.7.

10.

Let A, B, and C be sets. Prove that A C B iff A - B = 0. (b) Prove that if A C B U C and A n B = 0, then A C C. (c) Prove that C CA n B iff C CA and C C B. (d) Prove that if A C B, then A - C C B - C. (e) Prove that (A - B) - C = (A - C) - (B - C). (f) Prove that if A C C and B C C, then A U B C C.

*: (a)

* 11.

*

Let A, B, C, and D be sets with C C A and DeB. (a) Prove that C n DCA n B. (b) Prove that CUD CA U B. (c) Prove that if A and B are disjoint, then C and D are disjoint. (d) Prove that D - A C B - C.

12. Let A, B, C, and D be sets. Prove that if A U B C CUD, A n B = 0, and C CA, then BCD. 13. Let A and B be sets. (a) Prove that g>(A n B) = g>(A) n g>(B). You may use exercise lO(c). (b) Prove that g>(A) U g>(B) C g>(A U B). (c) Show by example that set equality need not be the case in part (b). Under what conditions on A and B is g>(A U B) = g>(A) U g>(B)? (d) Show that there are no sets A and B such that g>(A - B) = g>(A) - g>(B). 14. Provide counterexamples for each of the following. (a) If A U C C B U C, then A C B. (b) If An C C B n C, then A C B. (c) If (A - B) n (A - C) = 0, then B n C = 0.

*

*

* * *

(d) (e) (f)

g>(A) - g>(B) C g>(A - B). A - (B - C) = (A - B) - (A - C).

A - (B - C)

= (A - B) - C.

15. Define the symmetric difference operation Ll on sets by A Ll B (A - B) U (B - A). Prove that (a)

ALlB=BLlA.

(b)

ALlB=(AUB)-(AnB).

=

2.2 (c)

16.

17.

*

'* *

*

* *

= 0.

(d)

73

Ail0=A.

Let the universe be all real numbers. For B!: JR, define B* = B U {O}. Prove that (a) B !: B*. (b) For all B!: JR, B* oF 0. (c) (e) (g)

Proofs to Grade

A ilA

Set Operations

B=B*iffOEB. (AUB)*=A*UB*

B* = B - {o}

(d) (f)

(B*)* = B* (A n B) * = A * n B*

Assign a grade of A (correct), C (partially correct) or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If A !: B, then A - C !: B - C. "Proof." Assume A !: B. Suppose x EA. Then x E B, since A!: B. Let C be any set. Then x E A and x tt. c. Then x E B and x tt. c. Thus x E A - C and x E B - C. Therefore, A - C !: B - C. • (b) Claim. If A!: B, then A - C !: B - C. "Proof." Assume A !: B. Then x E A and x E B. Suppose A - C. Then x E A and x tt. c. Since x E B and x tt. C, B - C. Therefore, A - C !:

B-C.

•

Claim. A!: B<=>A n B = A. "Proof." Assume that A !: B. Suppose x E A n B. Then x E A and x E B, sox EA. This shows that A n B = A. Now assume that A n B = A. Suppose x EA. Then x E A n B, since A = A n B; and, therefore, x E B. This shows that x E A implies x E B, and so A !: B. • (d) Claim. A n 0 = A. "Proof." We know that x E A n 0 iff x E A and x E 0. Since x E 0 is false, x E A and x E 0 iff x EA. Therefore, x E A n 0 iff x E A; that is, A n0=A. • (e) Claim. If An B oF 0 and B n C oF 0, then A n C oF 0. "Proof." Assume A n B oF 0 and B n C oF 0. Since A n B oF 0, there exists x such that x E A n B; thus x EA. Since B n C oF 0, there exists x E B n C; thus x E C. Hence x EA and x E C. Therefore, x EA n C, which shows A n C oF 0. • (f) Claim. AnA = 0. "Proof." (We show each side is a subset of the other.) By Theorem 2.1 (a), 0 is a subset of every set. Thus 0 !: A n A. Now let x E A n A. Then x E A and x E A. Thus x E A and ~(x E A). Therefore, x oF x. Hence, by the definition of 0, x E 0. Therefore, A n A !: 0. • (g) Claim. rJ>(A - B) - {0} !: rJ>(A) - rJ>(B). "Proof." Suppose x E rJ>(A - B) - {0}. ThenxE rJ>(A - B) andx oF 0. Since 0E rJ>(A) and 0 E rJ>(B), 0 tt. rJ>(A) - rJ>(B). Since x E rJ>(A - B), x E rJ>(A) - rJ>(B). Therefore, rJ>(A - B) - {0} !: rJ>(A) - rJ>(B). • (h) Claim. A UA = U. "Proof." For every object x in the universe U, (x E A) V (x tt. A) is true. • (i) Claim. A ilA = 0 [exercise 15(c)). "Proof." By part (b) of exercise 15, AilA = (A UA) - (A nA). By Theorem 2.6, parts (e) and (f), (A U A) - (A Ii A) = A-A. Butx E A - A (c)

74

CHAPTER 2 Set Theory

iff both x E A and x ft. A iff x E 0. (Each is false.) Therefore, A ~ A = 0 .

*

(j)

Claim. If A ~ B, then A U B = B. "Proof." Let A ~ B. Then A and B are related as in this figure.

•

Since A U B is the set of elements in either of the sets A or B, A U B is the shaded area in this figure.

•

Since this is B, A U B = B.

2.3

Extended Set Operations and Indexed Families of Sets A set of sets is often called a family of sets. We will use script letters to denote fam2, 3}, {3, 4, 5}, {5, 6}, {6, 7, 9, is a family ilies of sets. For example, .il = of four sets. The collection .il = Ha, 00): a E ~} is a family of closed rays. In this section we extend the definitions of union and intersection to families of sets, so that the union and intersection of two sets will be a special case. If .il is a family of sets, the intersection over .il will be the set of all elements common to all sets in .il, while the union over .il will consist of those objects appearing in at least one of the sets in .il (figure 2.6). Note the relationships between U and :3 and between n and 'if in the next definition. This occurs because an element of the union is a member of at least one set in the family, whereas an element of the intersection is in every set in the family.

HI,

IOH

2.3

Extended Set Operations and Indexed Families of Sets

u A

n A

s1= {R,S, T, U}

AE.1I1

75

AEsd

Figure 2.6

DEFINITIONS Let .ii be a family of sets. The union over .ii is defined by

UA =

{x: (3A E .s4)(x E A)} = {x: (3A)(A E .s4 Ax E A)}

AEdi

and the intersection over .ii is

n A={x: (VA E.ii)(xEA)} = {x: (VA)(AE.ii=:}xEA)}.

AEdi

Example. Let.ii = {{a, b, c}, {b, c, d}, {c, d, e,f}} be a family of three sets. Then U A = {a, b, c, d, e,f}, since each of the elements a, b, c, d, e, f, appears in at

n

AEdi

least one of the three sets in .ii. Also A = {c}, since c is the only object appearing in every set in .ii. AEdi

Example. Let.ii be the set of all subsets of the real line of the form (-a, a) for E.ii. Then some positive a E IR. For example, (-I, I) E.ii and A = IR and A = {o} (see figure 2.7).

(-Ii, Ii)

n

U

AEdi

AEdi

c

~ (-1T,

0(-2,2)

c ~(-I,

0

I)

I

I

I

I

I

I

-3

-2

-I

0

2

3

Figure 2.7

1T)

76

CHAPTER 2 Set Theory

Theorem 2.8

For every set B in a family.
n A ~B. AE.sil

(b)

B~

U A. AE.sil

Proof. (a) Let .
n

A, then x E A for every

AE.sil

(Notice that the set A in the last sentence is a dummy symbol. It stands for any set in the family. The set B is in the family.) In particular, x E B. Therefore A ~ B.

n

AE.sil

(b)

The proof that B is a subset of the union over .
•

You might be tempted to conclude from Theorem 2.8 that

(see exercise 4). However, the next example shows this is not always the case.

Example. Let.
n

AE.sil

Also there is no real number that is in some member of .
U A = 0. Here AE.sil

we have an example where

n A ~ U A is false. This points out the need for cauAE.sil

AE.sil

tion when dealing with a family of sets when the family may be empty.

Example. Let CJ3 be the family of sets of the form (- 5, n], where n is a positive even integer. Some members of the family CJ3 are A2 = (- 5, 2], A4 = (- 5, 4], and A28 = (-5, 28]. Then A = (-5,00) and A = (-5, 2].

U

AE~

n

AE~

It is often possible, as in the previous example, to associate a'kind of identifying tag, or index, with each set in a family. For the family CJ3 and set (- 5, 2], the index used above is 2, and 28 is the index for the set (- 5, 28]. With this method of indexing, the set of indices for CJ3 is the set of all positive even integers.

2.3 Extended Set Operations and Indexed Families of Sets

77

DEFINITIONS Let Ll be a nonempty set. Suppose for each a E Ll, there is a corresponding set Au.' Then the family of sets .il = {Au: a Ell} is an indexed family of sets. Each a Ellis called an index, and Ll is called an indexing set.

Indexing is a familiar concept in everyday life. For each house on the 800 block of Elm Street in Pleasant Valley, the collection of residents at that address at a given time is a set. As an index for the set of these sets, we may use {80l, 802, 803, ... , 820}, the set of house numbers in the 800 block. That is, 804 is the index for the set AS04 of residents of 804 Elm Street.

Example. Let Ll= N. For all nE N, letA n = {n, n + 1, 2n}. ThenA, = {I, 2},A 2 = {2, 3, 4}, A3 = {3, 4, 6}, and so forth. The set with index 10 is AIO = {I 0, II, 20}.

There is no real difference between a family of sets and an indexed family. Every family of sets could be indexed by finding a large enough set of indices to label each set in the family. An indexing set may be finite or infinite, as well as ordered (like the integers) or unordered (like the complex numbers).

Example. For the sets Al = {I, 2, 4, 5}, A2 = {2, 3, 5, 6}, and A3 = {3, 4, 5, 6}, the index set has been chosen to be Ll = {I, 2, 3}. The family .il indexed by Ll is .il = {AI' Az, A 3} = {Ai: i Ell}. The family.il could be indexed by another set. For instance, if r = {IO, 21, n}, and AIO = {I, 2, 4, 5}, A2J = {2, 3, 5, 6}, and An = {3, 4, 5, 6}, then {Ai: i E Ll} = {Ai: i E q. Example. Let Ll = {O, 1,2,3, 4}. For each x E Ll, let Ax = {2x + 4,8, 12 - 2x}. Then Ao = {4, 8, 12}, Al = {6, 8, 1O}, A2 = {8}, A3 = {6, 8, 1O}, and A4 = {4, 8, 12}. The indexing set has five elements but the indexed family.il = {Ax: x E Ll} has only three members, since A, = A 3 and Ao = A 4.

As this last example demonstrates, it is not necessary that different indices correspond to different sets in an indexed family, nor do the number of elements in the member sets have to be the same. The notation for unions and intersections over indexed families of sets is given in the next definition. Again, note the close relationships between U and :3 and between n and V.

78

CHAPTER 2 Set Theory

DEFINITIONS

If.ii

= {Au: a E L\} is an indexed family of sets, then

U Au = U A = {x: (.3a E uEL1.

L\)(x EAu)}

AEsd

and

n

nA

Au =

uEL1.

= {x: (Va E L\)(x EAu)}.

AEsd

= {n, n + I, 2n}, we have

In the example above, with index set N and each An

U An =

N.

nEN

To convince someone that 27 E

U An' we need only point out some index n nEN

(26 or 27 will do) such that 27 E All" Since there does not exist x such that for all

n

n E N, x E Am we conclude that

An = 0.

nEN

Example. LetL\ = {I, 2, 3},A 1 = {I, 2, 4, 5},A 2 = {2, 3, 5, 6},A 3 = {3, 4,5, 6}, and.ii

= {Ai: i E L\}. Then

n

Ai

= {5} and

U Ai = {I, 2, 3,4,5, 6}. iEL1.

iEL1.

Example. Let L\ be the set of positive reals. For each a E L\, let Ha = (-a, a). Let .ii = {Ha: a E L\}. Then, as we have seen before, Ha = {O} and Ha = IR. aEL1. aEL1.

n

U

i]'

Example. For each real number x, define Bx = [x 2 , x 2 + I]. Then B- 1/ 2 = [t, Bo = [0, I], and BIO = [100, 101]. This is another example in which we have different indices representing the same set. For example, B-2 = B2 = [4, 5]. Here the index set is IR, Bx = 0, and Bx = [0,00).

n

U

xEIR

xEIR

Example. Let A be any nonempty set. For each a EA, let Xa = {a}. Then

U Xa = A. If A has more than one element, then

n

Xa

= 0.

aEA

aEA

With index set N, the union over the family .ii = {Ai: i E N} is sometimes

U Ai rather than U Ai, and the intersection over .ii is written n Ai' This 00

written

i=1

00

iEN

notation is useful if we want to work with such sets as A3

n A4 n As, and so forth.

2

i=1

U Ai = Al U A 2 , i=1

n S

Ai

i=3

=

2.3

Example.

Extended Set Operations and Indexed Families of Sets

For each n E N;Iet An

=

{n, n

79

+ 1, n2 }. For the family .sa = {An: n E N}

UA;=N ;=1 3

UA; = {I, 2, 3, 4, 9} ;=1 6

UA; = {4, 5, 6, 7, 16,25, 36}

;=4 4

nA; = {4} ;=2

Families of sets, whether indexed or not, obey rules similar to those stated for two sets in Theorem 2.7. The next theorem gives a statement of De Morgan's Laws for indexed families and also restates Theorem 2.8 for indexed families.

Theorem 2.9

Let .sa (a)

=

{Au: a E L\} be an indexed collection of sets. Then

n

Au ~ AfJ for each fJ E L\.

uE~

(b)

AfJ ~

U Au for each fJ E L\. uEA

(c) (d)

(lA:'= uEA UAU'}

uEA

,..-." _ Au -

U

uEA

n

_

(De Morgan's Laws)

Au·

uEA

Proof. The proofs of parts (a) and (b) are similar to those for Theorem 2.8 and are left for exercise 5(a). (c)

xE

[fA:' UEA

iffx 1$

n

Au

uEA

iff it is not the case that for every a E L\, x E Au iff for some fJ E L\, x 1$ AfJ iff for some fJ E L\, x E iffx E

U Au' uEA

AfJ

80

CHAPTER 2 Set Theory

Therefore,

W. uEL1. UAu· =

uEL1. (d)

(One proof ofpart (d) is very similar to that given for part (c) and is left as exercise 5(b). However, since part (c) has been proved, it is permissible to use it, as follows.) ~

~

UAu= UAu uEL1. uEL1.

(because A =

A)

~

= nAu uEL1. = nAu uEL1.

(by part (c)) (because

A= A)

DEFINITION The family s'l = {Au: a E ~} of sets is pairwise disjoint iff for all a and Pin ~, if Au =1= A p, then Au nAp = 0.

Two questions are commonly asked about this definition. The first question is why we bother with such a definition when we could more easily talk about a disjoint family satisfying

The answer is that the two ways of talking about disjointness of families of sets are not the same, and the situation covered by the definition given above is by far the more interesting. A pairwise disjoint family containing at least two different sets always satisfies the condition that the intersection over the whole family is empty (exercise 14). One reason the pairwise disjoint idea is important will appear in the study of partitions (chapter 3, section 3). The fact that the two ideas of disjointness are not the same can be seen from the following examples. Let C€ = {AI, A 2, A 3}, where Al = {I, 2}, A2 = {2, 3}, and A3 = {4, 5}. Then A = 0, but the family is not pairwise disjoint because A I n A2 =1= 0.

n

A='€

For the family s'l = {An: n E N} where, for each n, AI! = (n, (0), the family is An = 0. In this case not only are there some pairs of not pairwise disjoint, but

n

nEN

sets that are not disjoint, but for every pair AI!> Am' with, say n < m, the intersection An n Am = (m, (0) = Am. The second common question about the definition of pairwise disjoint families has to do with the need for including the phrase" ... whenever Au =1= AP" in the definition. This is necessary because we do not require that two sets with different indices be different. The phrase "whenever a =1= P" is also insufficient for the same reason.

2.3

Extended Set Operations and Indexed Families of Sets

81

Exercises 2.3 1.

* * * * *

"*

2. 3.

Find the union and intersection of each of the following families or indexed collections. (a) Let .91 = HI, 2, 3, 4, 5}, {2, 3, 4, 5, 6}, {3, 4, 5, 6, 7}, {4, 5, 6, 7, S}}. (b) Let .91 = HI, 3, 5}, {2, 4, 6}, {7, 9, II, 13}, {S, 10, 12}}. (c) For each natural number n, let An = {I, 2, 3, ... , n}, and let .91 = {An: n EN}. (d) For each natural number n, let Bn = N - {I, 2, 3, ... , n}, and let CJA = {Bn: n EN}. (e) Let .91 be the set of all sets of integers that contain 10. (f) Let AI = {I}, A2 = {2, 3}, A3 = {3, 4, 5}, ... , AIO = {1O, II, ... , 19}, and let .91 = {An: n E {I, 2, 3, ... , 10}}. (g) For each natural number, letA n = (0, k), and let .91 = {An: n EN}. (h) Let IR+ = (0, 00). For r E IR+, let Ar = [-n, r), and let .91 = {A r: r E IR+}. (i) For each real number r, let Ar = [Irl, 21rl + I], and let .91 = {A r: r E IR}. (j) For each n E N, let Mn = {... , -3n, -2n, -n, 0, n, 2n, 3n, ... }, and let .!lit = {Mn: n EN}. (k) For each natural number n 2: 3, let An = [k. 2 + k) and .91 = {An: n 2: 3}. (I) For each n E 71., let Cn = [n, n + I) and let C(6 = {Cn: n E 71.}. (m) For each n E 71., let An = (n, n + I) and .91 = {An: n E 71.}. (n) For each n E N, let Dn = (-n, k) and CZlJ = {Dn: n EN}. Which families in exercise I are pairwise disjoint? Prove the remaining part of Theorem 2.S. That is, prove that if .91 is a family of sets and B E .91, then B ~ A.

U

AEsd

4.

"*

Prove that for every non empty family .91 of sets,

n

A~

AEsd

5.

(a) (b)

6.

Let .91 = {Au: a E

*

U A. AEsd

Prove parts (a) and (b) of Theorem 2.9. Give a direct proof of part (d) of Theorem 2.9 that does not use part (c).

L\} be a family of sets and let B be a set. Prove that (a) B n U Au = U (B n Au)·

(b)

uEL1.

uEL1.

uEL1.

uEL1.

BUn Au = n UAu)· (B

7.

Let .91 = {Au: a

*

(a)

(

U Au) n ( U Bp) as a union of intersections. uEL1.

(b)

E L\} and let CJA = {Bp: PEl}. Use exercise 6 to write pEr

(n Au) U ( n uEL1.

pEr

Bp) as an intersection of unions.

82

CHAPTER 2 Set Theory

8. Let A = {An: a E il} be a family of sets, and let B be a set. Is each of the following true or false? (a)

(n

B-

An)

=

nEA

(b)

*

(e)

B- (

AJ.

nEA

An) - B

=

nEA

(d)

(B-A n)·

U An) = U (B ilEA

(n

n

nEA

n

(An - B).

nEA

(U An) -

B

=

nEA

U (An -

B).

nEA

9. Let d be a family of sets, and suppose 0 E d. Prove that

n

A = 0.

AE.s4

10. If d

*

(a)

= {An:

a E il} is a family of sets and if r ~ il, prove that

UAll ~ UAn·

nEr

nEA

nEA

nEr

11. Let d be a family of sets (a) Suppose B ~ A for every A E d. Prove that B ~

* *

n

A.

AE.s4

(b)

(e)

What is the largest set X such that X ~ A for all A E d? That is, find the set X such that (i) X ~ A for all A E d; and (ii) if V ~ A for all A E d, then vex. Suppos~A ~ D for every A E d. Prove that A ~ D.

U

AE.s4

What is the smallest set Y such that A ~ Y for all A E d? That is, find the set Y such that (i) A ~ Y for all A E d; and (ii) if A ~ W for all A E d, then Y ~ W. 12. Let X = {I, 2, 3, 4, ... , 20}. Give an example of each of the following: (a) a family d of subsets of X such that A = {I} and A = X. (d)

n

U

AE.s4

(b)

AE.s4

a family 00 of pairwise disjoint subsets of X such that 00 has 4 elements and B=X.

U

BE(Jd

(e)

a family «6 of pairwise disjoint subsets of X such that «6 has 20 elements and C =x.

U

CE~

13. Give an example of an indexed collection of sets {An: a E il} such that each An ~ (0, I), and for all aandpE il, An nAp *" but nAn = 0.

o

nEA

14. Prove that, if d = {An: a E il} is pairwise disjoint and d contains at least two An = 0. sets that are not equal, then

n

nEA

Extended Set Operations and Indexed Families of Sets

2.3

15. 16.

Let.ii be a family of pairwise disjoint sets. Prove that if 'ZA family of pairwise disjoint sets.

~

83

.ii, then 'ZA is a

Let.ii and 'ZA be two pairwise disjoint families of sets. Let C(6 = .ii n 'ZA and .ii U 'ZA. (a) Prove that C(6 is a family of pairwise disjoint sets. (b) Give an example to show that Cj) need not be pairwise disjoint.

Cj) =

(c)

U A and U B are disjoint, then Cj) is pairwise disjoint.

Prove that if

AE.sil

17.

BE@,

Let.ii = {Ai: i E N} be a family of sets and k, m be natural numbers with k ::; m. Prove that k+1 k (a) Ai = UAi UAk+Jo

U

i=1

i=1

k+1 (b)

n

Ai

i=1

k

= nAi nAk+I' i=1

m

*

(c)

UAi~UAi'

i=k

i=1 m

(d)

(e)

(f)

nAi~nAi' i=1

i=k

k

m

i=1

i=1

m

k

UAi~UAi' nAi~nA;. ;=1

18.

;=1

Suppose.ii = {Ai: i E N} is a nested family of sets. That is, suppose that for all i, j E N, if i ::; j, then Aj ~ Ai'

n k

(a) (b)

Prove that, for every kEN, Prove that

U Ai = AI'

Ai

=

A k·

i=1

;=1

19.

*

Give an example of a nested family {Ai: i E N} (see problem 18) for each condition. (a)

n

Ai = [0, I].

i=1

(b)

nAi=(O,I). i=1

(c)

nAi = {o, I}. i=1

84

CHAPTER 2 Set Theory

Proofs to Grade

20.

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a)

n

For any indexed family {Au: a E ~},

Claim.

Au

~

uEA

Choose any ApE {Au: a E ~}. Then

"Proof,"

n

Au

~Ap

Ap~

and

uEA

UAu. uEA

Therefore, by transitivity of set inclusion,

n

Au

~

uEA

*

(b)

Claim.

U Au· uEA

If Au ~ B for all a E~, then

U Au·

•

uEA

U Au ~ B. uEA

UAu. Then, since

"Proof," Suppose x E Therefore,

U A ~ B.

Au

uEA

•

uEA

(c)

Claim.

~ B for all a E ~, x E B.

For any indexed family {Au: a E ~},

U

Au

~

uEA

n

Au.

uEA

U Au. Then x E Au for every a E ~, which implies x E Au for at least one a E ~ since ~ *" 0. Therefore, x E Au. "Proof," Suppose x E

n

uEA

uEA

(d)

Claim.

For any indexed family {Au: a E ~}, Let ~

n

Au

~

ue-A

•

U Au· uEA

= {r, s, t}, Ar = {a, b, c, d}, As = {b, c, d, e}, At = {c, d, e,f}. Then Au = {c, d} ~ {a, b, c, d, e,f} = U Au· • "Proof,"

n

uEA

(e)

Claim.

"Proof,"

For any indexed family {Au: a E ~}, Assume

n

fI.

uEA Au

~

uEA

Au

uEA X

n

rt. U Au·

U Au. uEA

Then for some x E

uEA

uEA

a E A Therefore, x

n n

fI. Au for every a E ~. But since x E

Au, x E

uEA

for every a E A This is a contradiction, so we conclude

Au

uEA

(f)

Au,

U Au· Since x fI. U Au, it is not the case that x E Au for some uEA

*

n

uEA

Claim.

U [n, n + 1) =

R

~

Au

U Au. uEA

•

n=1

"Proof," Let x E R Choose a natural number y such that y :5 x < Y + 1. Thus x E [y, y + 1). Therefore, x is an element of

U [n, n + 1). Since [n, n + 1) ~ ~ for all n E 1\1, U [n, n + 1) ~ R n=1 Therefore, U [n, n + 1) = R • n=1

n=1

2.4

2.4

Induction

85

Induction The most familiar number system is the system of natural numbers, N = {I, 2, 3, 4, ... }. In 1899 Giuseppe Peano (1858-1932) set forth five axioms, based on the notion of successors, that serve as a complete description of the natural number system. The first four axioms assert that (i) I is a natural number, (ii) every natural number has a unique successor, which is also a natural number, (iii) no two natural numbers have the same successor, and (iv) I is not a successor for any natural number. When combined with the fifth axiom (which leads to the induction property discussed later), these axioms are sufficient to derive all the familiar properties of the natural numbers. Since the development of all these properties takes more time than we can devote to the subject, we will state them without proof.

1.

Successor properties

I is a natural number. Every natural number x has a unique successor x I is not the successor of any natural number.

2.

+ I.

Closure properties

For each pait of natural numbers, the sum x + y is a natural number. For each pair of natural numbers, the product xy is a natural number.

3.

Associativity properties

For all x, y, and z in N, x + (y + z) = (x For all x, y, and z in N, x(yz) = (xy)z.

4.

+ y) + z.

Commutativity properties

For all x and y in N, x + Y = Y + x. For all x and y in N, xy = yx.

5.

Distributivity properties

For all x, y, and z in N, x( y + z) = xy For all x, y, and z in N, (y + z)x = yx

6.

+ xz. + zx.

Cancellation properties

For all x, y, and z in N, if x + z = Y + z, then x = y. For all x, y, and z in N, if xz = yz, then x = y.

7.

Order properties

For all x, y, zEN, x < y iff there is wEN such that x x ::; y iff x < y or x = y.

+w

= y.

x < y and y < z implies x < z. x ::; y and y ::; x implies x = y. if x < y, then x + z < Y + z and xz

< yz.

Peano's fifth axiom leads to another characteristic property of the natural number system, the induction property.

86

CHAPTER 2

Set Theory

Principle of Mathematical Induction (PMI) If S is a subset of N with these two properties: (i) (ii)

I E S, for all n E N, if n E S, then n

+ I E S,

then SEN. A set of natural numbers S with the property that whenever n E S, then n + I E S is called an inductive set. The set {5, 6, 7, 8, ... } is inductive, as is the set {IOO, 101, 102, 103, ... }. We leave it as an exercise to show that Nand 0 are inductive sets. The set {I, 3, 5, 7, 9, ... } is not inductive because, for example, 7 is a member but 8 is not. Many sets of natural numbers have the inductive property, but only one set is inductive and contains I. By the Principle of Mathematical Induction, that set is N. In this and the next section we shall see that there are three forms of induction, each useful in certain situations. The PMI in the form given is basic to our understanding of the natural numbers: the natural numbers consist of I, then I + I = 2, then 2 + 1 = 3, ... , and so on. The PMI allows us to do two important things: first, to make inductive definitions; second, to prove that some properties are shared by all natural numbers. An inductive definition is a means to define an infinite set of objects that can be indexed by the natural numbers: that is, there is a first object, a second object, a third, and so on. Inductive definitions follow the form and spirit of the PMI. The first object is defined. Then the (n + I )st object is defined in terms of the nth object. The PMI ensures that the set of all n for which the corresponding object is defined is N.

Example. The factorial of a natural number may be defined inductively. You have probably seen the noninductive definition n!=n·(n-I)· ····3·2·1

The inductive definition of n! proceeds as follows: (i) (ii)

Define I! = I. For n E N, define (n

+ I)! =

(n

+ l)n!.

Let S be the set of n for which n! is defined. First, I E S because of part (i). Second, S is inductive because if n E S, then n! is defined and hence, by part (ii), (n + I)! is also defined. Thus n + I E S. By the PMI, S = N. In other words, the set of numbers for which the factorial is defined is N, so n! has been defined for all natural numbers. The inductive definition make clear the relationship between the factorial of a number and the factorial of the next number; if you happen to know that II! = 39,916,800, then you compute 12! = 12· II! = 479,001,600. In some settings, an inductive definition is called a recursive definition.

2.4

Induction

87

Example. Sets may also be defined inductively. Suppose we let 0 be the collection of natural numbers defined by (i) (ii)

lEO. If n E 0, then n + 2 E O.

The set 0 is, of course, the set of odd natural numbers, which has the familiar noninductive definition 0 = in: n = 2k - 1 for some kEN}. Mathematical induction is a natural technique for proving statements like the one in the following example.

Example. We will prove that for every natural number n, I + 3 + 5 + ... + (2n - I) = n2 . The correctness of this statement can be verified for, say, n = 6, by adding I + 3 + 5 + 7 + 9 + II = 36. A proof by induction that the property holds for all n proceeds as follows: Proof. Let S = {n E N: I + 3 + 5 + ... + (2n - 1) = n2 }. (We have defined S to be the set of natural numbersfor which the statement is true. Our aim is to show that the statement is true for every natural number by showing that S = N.) (i) (ii)

I = 12 , so I E S. Let n be a natural number such that n E S. Then for this n, I + 3 + 5 + ... + (2n - I) = n2 . (We have not assumed what is to be proved. We assume only that some n is in S to show that n + 1 E S follows.) Therefore, by adding 2(n + 1) - I to both sides, we have I

(iii)

+ 3 + 5 + ... + (2n -

I)

n2 + 2(n + I) - I (Compare the statementfor n andfor n = n2 + 2n + I =(n+ 1)2.

+ [2(n + I) - I]

=

This shows that n + I E S. By the PMI, S = N. That is, for every natural number n, I +(2n - 1) = n2 .

+ 1.)

+ 3 + 5 + ... •

Step (ii) of the preceding proof is nothing more than a direct proof of the statement symbolized by (Vn)(n E S =:} n + I E S). Compare this with (Vn)(n E S), the statement we are trying to prove. To make a direct proof, we assume n E S (for some particular n); this is not the same as assuming (Vn)(n E S). Every proof using the PMI can be written in the following form:

PROOF USING THE PRINCIPLE OF MATHEMATICAL INDUCTION Let S = {n E N: the statement is true for n}. (i) (ii) (iii)

Show that I E S. Show that S is inductive (that is, for all n, if n E S then n By the PMI, S = N.

+

I E S).

88

CHAPTER 2 Set Theory

Recall that the distributive law says that for any numbers m, XI> and X2, + x2) = mx] + mx2' It is also true that no matter how many numbers are involved,thenumbersm(xl +x2+ Oo. + xn) andmx] +mx2+ Oo. +mxnareequal.We would normally accept this statement as being obviously true, as we did in the last example. However, as another illustration of how the PMI is used, we will give a proof ofthis generalization ofthe distributive law. Since + is defined only for two numbers, we must be certain to make a sum Xl + X2 + ... + Xn of n numbers meaningful by thinking of it as a sum of two numbers. We use the following inductive definition: m(xl

For n ::::: 3, Xl

+ X2 + ... + Xn =

(Xl

+ X2 + ... + Xn-l) + Xn·

This definition assures us that a sum of n numbers has been unambiguously defined for every natural number n. The sum of one number is understood to be the number; we know all about the sum of two numbers; Xl + X2 + x3 is the sum of Xl + X2 and x3; Xl + x2 + x3 + X4 is the sum of Xl + X2 + X3 and X4; and so on. We are now prepared to prove the generalized distributive law.

Example. m(xl

For every list

Proof.

X2,"" Xn

of n numbers, and every number m,

Let S be the set of natural numbers with the desired property. That is, S=

(i) (ii)

XI>

+ X2 + ... + xn) = mXl + mx2 + '" + mxw {n E N: for every list XI> Xz,Oo', Xn of n numbers, m(xl + X2 + ... + xn) = mxl + mx2 + Oo. + mx n}.

= mxl for every list of 1 number, so 1 E S. Suppose n E S, and consider the situation for a list XI> X2,00., numbers. Then

m(xl)

X n +]

of n

m(x] + X2 + ... + x n+]) = m[(xl + X2 + ... + xn) + Xn+l] (by definition) = m(x] + X2 + ... + xn) + mxn+l (by the distributive law) = (mxl + mx2 + ... + mx n) + mxn+l (by the assumption that n = mxl + mX2 + ... + mxn + mXn+I' (by definition)

(iii)

+1

E S)

Therefore, n + 1 E S. By the Principle of Mathematical Induction, S = N. That is, the generalized distributive law holds for the sum of n numbers, for every n E N. •

In actual practice most people, when they write induction proofs, do not define the set S of all numbers that satisfy the property in question. If the property is expressed with the open sentence Pen), the induction proof takes the following form: PROOF OF ('fix EN) Pen) BY INDUCTION (i) (Basic step) Show that 1 E S. (ii) (Inductive step) Show that for all natural numbers n, if Pen) is true, then P{n + l) is true. (iii) (Conclusion) By steps (i) and (ii) and the PMI, Pen) is true for all nEN.

2.4

Induction

89

The assumption "if pen) is true" in part (ii) is called the hypothesis of induction. Every good proof by induction will use the hypothesis of induction to show that the property P is true for n + 1.

Example.

For all n E N, n

+ 3 < Sn2.

Proof. (i) (ii)

1 + 3 < S . 12, so the statement is true for 1. Assume that for some n, n + 3 < Sn 2. Then (n

+ 1) + 3 = n + 3 + 1 < Sn2 + 1

(by the hypothesis of induction)

and

so (n

(iii)

+ 1) + 3 < S{n + 1f

Thus the property holds for n + 1. By the Principle of Mathematical Induction, n

+ 3 < Sn2 for every n E

N.

-

Example. The polynomial x - y divides the polynomials x 2 - y2 and x 3 - y3 because x 2 - y2 = (x - y)(x + y) and x 3 - y3 = (x - y)(x 2 + xy + y2). This suggests the possibility that for every natural number n, x - y divides xl! - yl!. We prove this by induction.

Proof. (i)

x - y divides xl - Y 1 = X - Y because x - y = 1(x - y), so the statement

(ii)

holds for n = 1. Assume that x - y divides xl! - yl! for some n (this is the hypothesis of induction). We must show that x - y divides xl!+ 1 - yl!+l. We write Xl!+l - yl!+l

= xxI! - yyl! = xxI! = (x - y)xl! + y{xl! -

yxl! + yxl! - yyl! yl!).

Now x - y divides the first term because that term contains the factor (x - y). Also x - y divides the second term because it divides xl! - yl! (by the hypothesis of induction). Therefore, x - y divides the sum; so x - y divides XI!+l _ yl!+l.

(iii)

By the Principle of Mathematical Induction, x - y divides xl! - yl! for every natural number n. -

From our examples you should not jump to the conclusion that all inductive steps in a proof by induction are proved by cleverly adding some amount to one or both sides of an expression in the hypothesis of induction. To help convince you this is so, we present another example of a proof using the PMI.

90

CHAPTER 2 Set Theory

Example. Consider any "map" formed by drawing straight lines in a plane to represent boundaries. Figure 2.8 shows ten countries labeled A through J, formed by drawing four lines in the plane. The problem is to color the countries so that adjoining countries (those with a line segment as a common border) have different colors. This has been done in figure 2.9 using only the two colors white and burgundy. In fact, every map formed by drawing n straight lines can be colored using exactly two colors, and this can be proved by induction on the number of lines. Proof. (i)

(ii)

If a map is made by drawing one straight line, then there are only two countries. Thus every map formed with one line can be colored with two colors. Assume that for some n, every map formed by drawing n lines can be colored with exactly two colors. Now consider a map with n + I lines. Before coloring this map, choose anyone of the lines and label it L. Now color the map as though the line L were not there, using exactly two colors. This can be done, initially, by the hypothesis of induction. (Such a coloring is shown in figure 2.10, with the line L shown as a dashed line. Of course, only part of the plane can be shown.) To color the map with line L, proceed as follows. Call one half-plane determined by L side I, and the other half-plane side 2. Leave

Figure 2.8

Figure 2.9

2.4

Induction

91

.............. ..................... ..............

..............

.............. L

Figure 2.10

..............

..............

..............

.............. ....... .......

..............

.............. .....................

.............. L Figure 2.11 all colors on side 1 exactly as they were but change every color on side 2 to the other color. This gives a coloring to every country in the map with line L. (See figure 2.11.) It remains to verify that adjacent countries in this map with n + 1 lines have different colors. Suppose we have two adjacent countries. There are two cases to consider: (a)

(b)

First, suppose L is the border between the two countries, which means that one country is on side 1 and the other on side 2. Initially, the two countries had the same color because they were parts of the same country in the map with n lines. When L was added to the map, the color of the country on side 2 switched to a different color from the country on side 1. Second, suppose L is not the border between the two countries. Then both countries are either on side 1 or side 2. If both countries are on side 1, they were initially colored differently and remain so when L is added. If both countries are on side 2, their colors were initially different, but are now switched, and still different.

In both cases, the two adjoining countries have different colors.

a

The next example involves computations using trigonometry and complex numbers.

92

CHAPTER 2 Set Theory

Theorem 2.10

(De Moivre's Theorem) Let 0 be any real number. For all n E N, (cos 0 + i sin o)n = cos nO + i sin n().

Proof.

(We will need to recall addition and multiplication of complex numbers and use the following "sum of angles" fo rmulas from trigonometry:

cos (a + fJ) = (cos a)(cos fJ) - (sin a)(sin fJ) sin (a + fJ) = (sin a)(cos fJ) + (cos a)(sin fJ).) (i) (ii)

For n = 1, the equation is (cos 0 + i sin 0) 1 = cos 0 + i sin 0, which is certainly true. Assume that (cos 0 + i sin O)k = cos kO + i sin kO, for some natural number k. Then, using the sum of angles formulas above, (cos 0 + i sin O)k+ 1 = (cos 0 + i sin O)k(cos 0 + i sin 0)

= (cos kO + i sin kO)(cos 0 + i sin 0)

(by the hypothesis of induction)

cos kO cos 0 + i sin kO cos 0 + i cos kO sin 0 + i 2 sin kO sin 0 = (cos kOcos 0 - sin kO sin 0) + i(sin kOcos 0 + cos kO sin 0) (remember that i 2 = -1.)

=

= cos (kO + 0) + i sin (kO + 0) = cos(k + I)O+isin(k+ 1)0. (iii)

By steps (i) and (ii) and the PMI, the (cos 0 + i sin o)n = cos nO + i sin nO is true for all n E N. •

As one more example of the use of the PMI, we will prove a small but useful result, needed in a later theorem, about the comparative sizes of natural numbers. Suppose a and b are any two natural numbers. Even if b is smaller than a, the result says that a can eventually be surpassed by taking natural number multiples of b. One reason for considering this theorem is that it happens to be one of those facts about natural numbers that can be proved without the PMI. We give two proofs, neither of which is "more correct" than the other. Notice once again that the key to an induction proof is in the inductive step where we use an assumption about a natural number n to draw a conclusion about n + 1.

Theorem 2.11

For all natural numbers a and b, there exists a natural number s such that a < sb.

Proof 1. Let a and b be two natural numbers. Then a < a + 1 and since 1 ::5 b, a + 1 ::5 (a + I )b. Therefore, by choosing s to be the natural number a + I, we have that a < sb. • Proof 2. (i) (ii)

(iii)

By induction on b.

If b = I, choose s to be a + 1. Then a < a + I = sb. Suppose the statement is true when b = n, for some natural number n. Then there is some natural number s such that a < sn. Choosing the same s, we have a < sn < s(n + 1), so the statement is true when b = n + 1. By parts (i) and (ii) and the PMI, the statement is true for all natural numbers a~d~ •

2.4

Induction

93

There are statements that are not true for all natural numbers but are true for the numbers in some inductive subset of N. To prove such statements, we need a slightly generalized form of the PMI:

Generalized Principle of Mathematicallndudion Let x be a natural number. If S is a subset of N with these two properties: (i) (ii)

xES, for all n E N with n 2: x, if n E S, then n

+ I E S,

then S contains all natural numbers greater than or equal to x.

Example.

Prove by induction that n 2

-

n - 20> 0 for n > 5.

Proof. (We will use the generalized version o/the PM! starting at x (i) (ii)

=

6.)

For n = 6, 62 - 6 - 20 = 10, which is greater than zero. Assume for some natural number k > 5 that k 2 - k - 20 > O. Then

(k + 1)2 - (k + I) - 20 =

k2

= k2

(iii)

+ 2k + -

I - k - I - 20 k - 20 + 2k

Since k 2 - k - 20> 0 (by the induction hypothesis) and 2k > 0 (since k is a natural number), k 2 - k - 20 + 2k > O. Thus (k + I? - (k + I) - 20> O. By the (generalized) PMI, n 2 - n - 20> 0 is true for all n > 5. •

Exercises 2.4 1. Which of these sets have the inductive property? (a) {20, 21, 22, 23, ... } (b) {2, 4,6,8, 1O, ... } (c) {I, 2, 4, 5, 6, 7, ... } (d) {17} (e) {x E N: x 2 :::; WOO} (f) {l, 2, 3,4,5,6,7, 8} 2. Suppose S is inductive. Which of the following must be true? (a) If n + I E S, then n E S. (b) If n E S, then n + 2 E S. (c) Ifn+I~S,thenn~S. (d) If6ES,thenIIES. (e) 6 E S and II E S. 3. (a) Prove that N is inductive. (b) Prove that 0 is inductive. 4. Prove or find a counterexample for each: (a) If A and B are inductive, then A U B is inductive. (b) If A and B are inductive, then An B is inductive. 5. Use the inductive definition of n! · l'f (n + z)! · d 4' 7' 97! 8! (b) S Imp () Iy a FIn ., ., 96!' 3! 5 r n." (c) From 1O! = 3,628,800 find II! and 9!.

* * *

*

*

94

CHAPTER 2 Set Theory

6.

*

Give an inductive definition for each: in: n = Sk for some kEN}. (b) in: n E Nand n > IO}. (c) in: n = 2k for some kEN}. (d) A set formed as an arithmetic progression {a, a + d, a + 2d, a + 3d, ... }. (e) A set formed as a geometric progression {a, ar, ar2, ar 3 , ••• }.

7.

Give an inductive definition for each:

(a)

n

(a)

UAi, for some indexed family {Ai: i EN}. i=1

(b)

8.

The product

nXi = XI • X2 • X3 . . . . . i=1

xn of n real numbers.

Use the PMI to prove the following for all natural numbers n. (a) (b) (c) (d)

*

n

1+4 + 7 + ... + (3n - 2) = ~n(3n - I). 3+II+19+"·+(8n-S)=4n2 -n. 21 + 22 + 2 3 + ... + 2 n = 2 n + 1 - 2. I·I!+2·2!+3·3!+"·+n·n!=(n+I)!-1. n3 n5 7n. .

(e)

'3 + 5 + 15 IS an mteger.

(f) (g)

(j)

2n;=:: I + n. n3 + Sn + 6 is divisible by 3. 4n - I is divisible by 3. 3n ;=:: I + 2n. 4n + 4 > (n + 4t

(k) (I)

If a set A has n elements, then QJ>(A) has 2n elements. IOn + 3 . 4 n+ 2 + S is divisible by 9.

(h) (i)

nn .

(m) i=/21 (n)

_(2n)!

I) - n! 2n'

(Sum of a finite geometric series)

n-I a(rn - I) Iar i = forr=l=I,n;=::l. i=O r- I (0)

9.

=

I and

.!!:...(Jg) = f dd g + gddf , prove that for all n E N, dd (xn) = nx n- I. dx X x x Use the (generalized) PMI to prove the following: (a) 2 n > n 2 for all n > 4. (b) (n+ I)!>2 n+ 3 forn;=::S. (c) For all n > 2, the sum of the angle measures of the interior angles of a (d)

10.

d (From calculus). Using the differentiation formulas d)x)

convex polygon of n sides is (n - 2) . 180°. r::. I I I I 'In < {l + + + ... + {n for n;=:: 2.

Ii f>

Suppose that a statement P(n) satisfies: (i) P (I) is true.

2.4

Induction

95

+ 2} is true. Is P{n} true for all n E N? Explain.

(ii) if P{n} is true, then P{n

11.

Let PI>

Pz, ... , Pn be n points in a plane with no three points collinear.

Figure 2.12 2

Show that the number of line segments joining all pairs of points is n ~ n. 12.

A decreasing sequence of n natural numbers is a collection a" a2,"" an of natural numbers such that al > a2 > a3 > ... > an' For example, 45, 19, 15,9 is a decreasing sequence of four natural numbers. (a) Use the PMI to show that for every natural number n there is a decreasing sequence of n natural numbers. (b) Give a constructive proof that for every natural number n there is a decreasing sequence of n natural numbers. That is, tell exactly how to make such a sequence for each n.

13.

A puzzle called the Towers of Hanoi consists of a board with 3 pegs and several disks of differing diameters that fit over the pegs. In the starting position all the disks are place on one peg, with the largest at the bottom, and the others with smaller and smaller diameters up to the top disk (figure 2.13). A move is made by lifting the top disk off a peg and placing it on another peg so that there is no smaller disk beneath it. The object of the puzzle is to transfer all the disks from one peg to another.

Figure 2.13 With a little practice, perhaps using coins of various sizes, you should convince yourself that if there are 3 disks, the puzzle can be solved in 7 moves. With 4 disks, 15 moves are required. Use the PMI to prove that with n disks, the puzzle can be solved in 2n - 1 moves.

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CHAPTER 2 Set Theory

Proofs to Grade

14.

In a certain kind of tournament, every player plays every other player exactly once and either wins or loses. There are no ties. Define a top player to be a player who, for every other player x, either beats x or beats a player y who beats x. (a) Show that there can be more than one top player. (b) Use the PMI to show that every n-player tournament has a top player.

15.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. For all n E N, in every set of n horses, all horses have the same color. "Proof." Clearly in every set containing exactly I horse, all horses have the same color. Now suppose all horses in every set of n horses have the same color. Consider a set of n + I horses. If we remove one horse, the horses in the remaining set of n horses all have the same color. Now consider a set of n horses obtained by removing some other horse. All horses in this set have the same color. Therefore all horses in the set of n + I horses have the same color. By the PMI, the statement is true for every n E N. • (b) Claim. For every natural number n, n2 + n is odd. "Proof." The number n = 1 is odd. Suppose n E Nand n2 + n is odd. Then

*

(n

(c)

+ 1)2 + (n + 1) = n2 + 2n + 1+ n + 1 = (n 2 + n) + (2n + 2)

is the sum of an odd and an even number. Therefore, {n + 1f + (n + 1) is odd. By the PMI, the property that n2 + n is odd is tru~ for all natural numbers n. • Claim. For every natural number n, the matrix

[o11l]n = [10 n]1 . "Proof."

Let

Clearly,

[o11]'=[11] 1 0 1 ' so 1 E S. Assume that

[o1 11]n+' =[10 n +1 1].

*

(d)

Then n + 1 E S, so n E S implies n + 1 E S. By the PMI, S = N. Claim. Every natural number greater than 1 has a prime factor.

"Proof." (i) Let n = 2. Then n is prime.

•

2.5

Equivalent Forms of Induction

97

Suppose k has a prime factor x. Then k = xy for some y. Thus + 1 = xy + 1 = (x + 1) (y + 1), which is a prime factorization. (iii) By the PMI, the theorem is proved. '. Claim. For all natural number n;::: 4, 2n < n! "Prooj." 24 = 16 and 4! = 24, so the statement is true for n = 4. Assume that 2n < n! for some n E~. Then 2n + 1 = 2(2n) < 2(n!) :5 (n + 1)(n!) = (n + I)!, so 2n + I < (n + I)!, By the PMI, the statement is true for all n ;::: 4. • (ii)

k

(e)

2.5

Equivalent Forms of Induction To use the PMI successfully, one must be able to show that if n has a certain property, then n + 1 has the property. In some cases there is no apparent connection between the property for n and for n + I. In these cases, there may be a connection between the property for n + 1 and some value less than n. This section will discuss two other forms of induction, which turn out to be equivalent to the PMI. The inductive step of a proof using the PMI assumes only that a property holds for one number n. We can make a much stronger assumption in an alternate form of induction, called complete (or strong) induction. This form begins with the assumption that a property is true for all natural numbers less than n-that is, for every k in {I, 2, 3, ... , n - 1}-and shows that the property holds for n. Care must be taken in the case of n = I, where we take the set {I, 2, 3, ... , n - I} to be the empty set. Because this form employs such a strong assumption, the Principle of Mathematical Induction in section 2.4 is sometimes referred to as "weak induction."

Theorem 2.12

(The Principle of Complete Induction (PCI)) If S is a subset of ~ with the property that for all natural numbers m, if {I, 2, 3, ... , m - I} ~ S, then m E S, then S =~.

Proof. (To show that the PCI holds, we assume its hypothesis and show that (based on the PM/) the conclusion must also hold.) Suppose S is a set such that for all natural numbers m, if {I, 2, 3, ... , m - I} ~ S, then mE S. We use the PMI to prove the statement For every natural number n, {I, 2, 3, ... , n} ~ S. (i)

(ii)

(iii)

1 E S because every natural number less than 1 is in S. (If m = I, the set {I, 2, ... , m - I} is the empty set. Since 0 ~ S is true, the property ensures that 1 E S.) Thus {I} ~ S. Assume that the statement is true for some n EN. That is, assume that {I, 2, ... , n} ~ S. Then by the property of S, we have that n + 1 E S. Therefore {I, 2, ... , n + I} ~ S, so the statement is true for n + 1. By (i), (ii), and the PMI, {I, 2, 3, ... , n} ~ S for every natural number. It follows from (iii) that ~ ~ S, and since S ~ ~, we conclude S = N. •

The Principle of Complete Induction could have been accepted rather than the PMI as the fundamental induction property of the natural numbers because the two

98

CHAPTER 2 Set Theory

are equivalent. We showed in Theorem 2.l2 that the PCI follows from the PMI. We now show the converse.

Theorem 2.13

Suppose that the PCI holds for N. If S is a subset of N with these two properties: (i) (ii)

1 E S, For all n E N, if n E S, then n + 1 E S.

ThenS = N.

Proof. (To show that the PMl holds, we assume the hypotheses of the PMl and show that (based on the PCl) the conclusion must also hold.) Suppose S is a set for which (i) I E Sand (ii) if n E S, then n + I E S. We must show that S = N. (To prove S = N, wefirst show that S satisfies the hypothesis of the PC!; that is we show that for every natural number m {I, 2, 3, ... , m - I} r;;;;, S =} m E S.) Let m be a natural number such that {I, 2, 3, ... , m - I} r;;;;, S. We consider two cases: (a) (b)

Ifm = I, then {I, 2, 3, ... ,m - I} = 0, but by (i), I E S. Ifm> I then from {I, 2, 3, ... , m - I} r;;;;, S,m - 1 E S. But then by (ii) mE S. In either case, {I, 2, 3, ... , m - l} r;;;;, S =} mE S.

Thus by the Principle of Complete Induction, S = N.

•

Notice that the Principle of Complete Induction does not require the separate step of showing that I E S, because the PCI property reduces to "0 r;;;;, S implies I E S" in the case when m = I. In practice, many PCI proofs do include a separate proof of 1 E S. In fact, to prove that for all m, {I, 2, ... , m - I} r;;;;, S implies m E S, special considerations may be necessary when m = 1 or m = 2. This caution may apply to the PMI, too, when m = 2 [see exercise I5(a) of section 2.4]. Let's reconsider the first PMI example from section 2.4, this time using a proof based on the PCI. We note that this proof is not as "natural" as the one using the PMI.

Example.

Prove for all natural numbers that 1 + 3 + 5 + ... + (2n - 1) = n 2.

Let S = {n EN: 1 + 3 + 5 + ... + (2n - I) = n2}. (We must show that {I, 2, 3, ... , m - I} r;;;;, S =} mE S and then conclude S = N by the PCl.) Suppose m is a natural number and {I, 2, 3, ... , m - I} r;;;;, S. In the special case that m = I, we note that I = 12 and so mE S. Otherwise, m - 1 E S (since m-I E{I,2,3, ... ,m-I}r;;;;,S) and therefore 1 +3+5+'" +(2{m-I)-I) = {m - 1)2. Adding 2m - I to both sides yields

Proof.

I + 3 + 5 + ... + (2{m - I) - 1) + (2m - I) =(m-1)2+{2m-I) = m 2 - 2m + 1 + 2m - I =m2 . Thus m E S and we conclude S

=

N by the PCI.

•

Like the PMI, the PCI has a more general form that can be used to prove that some property holds for all natural numbers greater than, say 4. Such a PCI proof would use the set {5, 6, 7, ... , m - I}. If m is 9, this s~ is {5, 6, 7, 8}; if m is 5, this set is 0. The PCI proof would verify that. "

2.5

Equivalent Forms of Induction

99

For all natural numbers m > 4, if {5, 6, ... , m - I} ~ S, then mE S. The next example uses this form of PCI for n > 1. The statement we prove, that every natural number greater than 1 has a prime factor, was used without proof in chapter 1 because we did not yet have the method of induction needed to prove the statement. This is a good example of a statement that is proved easily with the PCI, but very difficult to prove without the PCI.

Example.

Every natural number n > 1 has a prime factor.

(Recall that a number p is prime iff p > 1 and its only lactors in N are 1 and p.) Let S be {n E N: n> 1 and n has a prime factor}. Notice that 1 is not in S, but 2 is in S. Let m be a natural number greater than 1. Assume that for all k E {2, ... , m - I}, k E S. We must show that mE S. If m has no factors other than 1 and m, then m is prime, and so m has a prime factor-itself. If m has a factor x other than 1 and m, then 1 < x < m so xES. Therefore x has a prime factor, which must also be a prime factor of m. In either case, mE S. Therefore, S = {n E N: n> I}, and every natural number greater than 1 has a prime factor. a

Proof.

The Principle of Complete Induction leads to richer inductive definitions, one of which is the definition of the sequence 1, 1,2,3,5,8, 13, ... examined by Leonardo Fibonacci in the 13th century. (See exercise 4.) These numbers have played important roles in applications as diverse as population growth, flower petal patterns, and highly efficient file sorting algorithms in computer science. For each natural number n, the nth Fibonacci number In is defined inductively by

II =

1'/2 = 1, andln + 2 = In+1

+ In

for n:::=:: 1.

We see that!3 = 2'/4 = 3'/5 = 5'/6 = 8, and so on. Inductive proofs of properties of the Fibonacci numbers usually involve the PCI because we need to "reach back" to both In-I and In-2 to prove the property for In- Here is a typical example.

Example. Let a be the positive solution to the equation x 2 = x + 1. (The value of a is (1 + .j5)/2, approximately 1.618.) Prove that In ::=; a n - I for all n:::=:: 1. Proof.

Let m be a natural number and assume for all k E {I, 2, 3, ... , m - I} that

Ik::=; ak-I. In the special cases of m = 1 and m = 2, the inequality 1m ::=; a m - I is true since II = 1 ::=; aD = 1 and 12 = 1 ::=; a 1 = (1 + .j5)/2. For m :::=:: 3, we have 1m = 1m-I + Im-2 ::=; am - 2 + am - 3 (induction hypothesis lor m - 1 and m - 2 E {I, 2, 3, ... , m - I})

= am - 3 (a + 1) = am -\a2 ) =

(since a is a solution to x 2 =

X

+

1, a

+ 1 = a 2)

am-l.

Therefore,fm ::=; am-I. By the PCI, we conclude that In ::=; a n - l for all n

:::=::

1.

a

100

CHAPTER 2 Set Theory

Another property characteristic of N is the Well-Ordering Principle (WOP). Although quite simple to state, the WOP turns out to be a powerful tool for constructing proofs about N.

Theorem 2.14

(The Well-Ordering Principle (WOP)) Every non empty subset of N has a smallest element.

Proof. Let T be a subset of N. We must show that if Tis nonempty, then T has a smallest element. (We will prove the contrapositive.) Suppose T has no smallest element. Let S = N - T. (We will use the PM! to show that S = N and thus

T=0.) (i) (ii)

(iii)

Since I is the smallest element of Nand T has no smallest element, I $. T. Therefore, I must be in S. Suppose n E S. No number less than n belongs to T, for if any numbers less than n were in T, then one of those numbers would be the smallest element of T. Also, n is not in T because n E S. Then n + I cannot be in T or else it would be the smallest element of T. Thus n + I E S. By the PMI, S = N.

Thus if T has no smallest element, T must be the empty subset of N.

-

Theorems 2.12 and 2.13 show that the PMI and PCI are equivalent. Theorem 2.14 shows that the PMI implies the WOP. In exercise 6 you are asked to complete the arguments-that the PCI implies the WOP and the WOP implies both the PMI and PCI. This establishes that all three properties are equivalent. Anyone of the PMI, PCI, or WOP may be used as a basic characteristic of N. They are the first proof methods to turn to whenever it must be shown that a property holds for all natural numbers. In some cases one of the methods is more appropriate than others, but sometimes more than one are suitable. The fact that every natural number greater than 1 has a prime factor, proved above using the PCI, can be proved much more elegantly by using the WOP.

Example.

Every natural number n > I has a prime factor.

Proof. If n is prime, then n is a prime factor of n. If n is composite, then n has factors other than I and n. Let p be the smallest of these factors. We must show that p is prime. Suppose p is composite. Then p has a divisor d, with I < d < p. Since d divides p and p divides n, d divides n. But d < n, and this contradicts the definition of p. Therefore, p is prime. Proofs using the WOP frequently take the form of assuming that some desired property does not hold for all natural numbers. This produces a nonempty set of natural numbers that do not have the property. By working~with the smallest such number, one can often find a contradiction. As another example of this method, we will prove that for every n E N, n + I = I + n. The purpose of this example is not

2.5

Equivalent Forms of Induction

101

to prove that n + I = I + n, but to see how a proof using the WOP is done. So imagine for a moment that we did not know that addition is commutative, and we will show how the statement can be proved (from the associative property) by using the WOP. Example.

For every natural number n, n

+I

=

I

+ n.

"*

Proof. Suppose there exist natural numbers n such that n + I 1 + n. Let b be the smallest such number. Obviously, I + I = I + 1, so b I. Thus b must be of the form b = c + I for some c E N. (See the successor properties for N.) Then (c + I) + I I + (c + I). By the associative property, (c + I) + I (I + c) + I. Subtracting I (from the right side) from each expression, we have c + I I + c. But this is a contradiction because c < band b is the smallest natural number with the property. We conclude that n + 1 = I + n for all natural numbers n. •

"*

"*

"*

"*

As a more interesting application of the WOP, we will prove the division algorithm for natural numbers. An extension of this algorithm to the integers is presented as exercise II.

Theorem 2.15

(The Division Algorithm for N) Let a E Nand bEN, with b :::; a. Then there exist q E Nand r E N U {o} such that a = qb + r, where 0 :::; r < b. The numbers q and r are the quotient and remainder, respectively. Let T = {s E N: a < sb}. By Theorem 2.11, this set is nonempty. Therefore, by the WOP, T contains a smallest element w. Let q = w - I and r = a - qb. Since a 2: b, w> I and q E N. Since q < w, a 2: qb and r 2: O. By definition of r, a = qb + r. Now suppose r 2: b. Then a - (w - I)b 2: b, so a 2: wb. This contradicts the fact that a < wb. Therefore, r < b. •

Proof.

We can combine the Division Algorithm and the WOP to produce one of the most useful theorems in number theory.

Theorem 2.16

For natural numbers a and b, the GCD of a and b may be written as a linear combination of a and b. That is, for d = GCD(a, b), there exist integers x and y such that ax

Proof.

+ by = d.

(We will show that the smallest positive linear combination ofa and b is the GCD of a and b.) Let C = {z: z > 0 and z = ax + by for some integers x and y). Cis a set of natural numbers and C 0 because, for example, a + b E C. Therefore, by the Well-Ordering Principle, the set C has a smallest element m = axo + byo for some integers Xo and yo. We note that m :::; a and m :::; b because the combinations a( I) + b(O) and a(O) + b(l) are also in C. We will show m = d. Since d divides a and d divides b, d divides axo + byo. Thus d divides m. Therefore, d:::; m.

"*

102

CHAPTER 2 Set Theory

Suppose m does not divide a. Then by the Division Algorithm for N, there exist natural numbers q and r such that a = mq + r, where 0 < r < m. (Note that r > 0 since r = 0 would mean that m does divide a.) Then from m= mq = a - r= r= r =

axo + byo, aqxo + bqyo (note that q > 0) aqxo + bqyo (since a = mq + r) a - aqxo - bqyo a(1 - qxo) + b( -qyo).

Therefore, r is a linear combination of a and b, so r E C. But r < m, which is a contradiction to m being the least element of C. Therefore m divides a. Similarly, m divides b. But then m divides d (since d is the GCD). Hence, m ::; d. Therefore, m = d and since mE C, the OCD is a linear combination of a andb. -

Exercises 2.5

"*

1.

Prove that every natural number greater than 3 may be written as a linear combination of the numbers 2 and 5; that is, for all n 2:: 4, there exist integers x and y such that n = 2x + 5y. 2. Let al = 2, a2 = 4, and a n+2 = 5a n+l - 6an for all n 2:: 3. Prove that an = 2 n for all natural numbers n. 3. In this exercise, as in some examples, you are to prove some well-known facts about numbers as a way of demonstrating use of the WOP. Use the WOP to prove the following: (a) If a > 0, then for every natural number n, an > o. (b) For all positive integers a and b, b a + b. (Hint: Suppose for some a there is b such that b = a + b. By the WOP, there is a smallest ao such that, for some b, b = ao + b. Apply the WOP again.)

"*

4.

5.

"*

In 1202, Leonardo Fibonacci (1l70-1250) posed the following famous problem: Suppose a particular breed of rabbit breeds one new pair of rabbits each month, except that a I-month old pair is too young to breed. Suppose further that no rabbit breeds with any other except its paired mate and that rabbits live forever. At 1 month we have our original pair of rabbits. At 2 months we still have the single pair. At 3 months, we have two pairs (the original and their one pair of offspring). At 4 months we have three pairs (the original pair, one older pair of offspring, and one new pair of offspring). (a) Show that at n months, there are fn pairs of rabbits. (b) Calculate the first ten Fibonacci numbers'l,,f2,i3, ... ,fIO. (c) Find a formula for fn+3 - fn+'. Use the PMI to show that each of the following statements about Fibonacci numbers is true: (a) i3n is even and both i3n+ I and i3n+2 are odd for all natural numbers n. (b) OCD(fm fn+ I) = 1 for all natural numbers n.

2.5

*

Equivalent Forms of Induction

103

(c) GCD{/',,fn+2) = 1 for all natural numbers n. (d) II + 12 + h + ... + In = /,,+2 - 1 for all natural numbers n.

6. Use the PCI to prove the following properties of Fibonacci numbers: (a) In is a natural number for all natural numbers n. (b) /',+6 = 41n+3 + In for all natural numbers n. (c) For any natural number a,fal" + la+dn+1 = la+,,+ I for all natural numbers n. (d) (Binet's formula). Let a be the positive solution and fJ the negative solu1-v'S) . to t h e equatIOn . x 2 = x +. I (Th eva Iues are a = -1+v'S tlOn 2 - an d a = -2-' Show for all natural numbers n that

"*

_ an - fJn In - a- fJ .

"*

7.

Using the WOP, prove that {i is irrational.

8.

The Division Algorithm lor Integers: If a and b are integers with b *- 0, then there exist integers q and r such that a = bq + r, where 0 :::; r < Ihi. (a) Find q and r when a = 7, b = 2; a = -6, b = -2; a = -5, b = 3; a = -5, b = -3; a = 1, b = -4; a = 0, b = 3. (b) Use the WOP to prove the Division Algorithm for Integers.

9.

Every positive rational number has an infinite number of expressions of the form E. where a and b are integers and b *- O. For example, 2.6 = =~6, Prove that for every rational number r there is a smallest positive integer n such that r = W

rg, ¥,

*,. .

10. Prove Theorem 2.11 by induction on the natural number a. 11.

In the tournament described in problem 14 of section 2.4, a top player is defined to be one who either beats every other player or beats someone who beats the other player. Use the WOP to show that in every such tournament with n players (n EN), there is at least one top player.

12. Let the "Foobar-nacci" numbers gn be defined as follows: gl = 2, g2 = 2, and g,,+2 = gn+lgn (a) (b) 13.

for all n::=::: 1.

Calculate the first five "Foobar-nacci" numbers. Show that gIl = 2ft!.

Define "third-order" Fibonacci numbers as follows: fl = 1, f2 = 1, f3 = l,andfn+3 = fn+2 (a) (b) (c)

+ fn+1 + fn

forn::=::: 1.

Calculate the first 10 third-order Fibonacci numbers fl' f2' f 3,···, flO' Find a formula for fn+4 - fn+ I' Write a definition for "fourth-order" Fibonacci numbers.

14. Complete the proof of the equivalences of the PMI, PCI, and WOP by

*

,(a)

(b) (c)

using the PCI to prove the WOP. using the WOP to prove the PMI. using the WOP to prove the PCI.

104

CHAPTER 2

Proofs to Grade

Set Theory

15.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. is irrational. "Proof." LetA = {x E N: x 2 = 2}. By the WOP,A has a least element, Xo. Since Xo E N, xo;::: I. Clearly, Xo I. Thus xo;::: 2, hence Xo ;::: 4. • Thus Xo E A is false. Hence A must be empty. (b) Claim. The WOP implies the PCI. "Proof." By exercise 6(a), the PCI implies the PMI. By exercise 14(b), the WOP implies the PMI. Therefore, the WOP implies the PCI. • (c) Claim. For all natural numbers n, 5 divides gn - 3n • "Proof." Suppose there is n E N such that 5 does not divide gn - 3n • Then by the WOP there is a smallest such natural number t. Now t I since 5 does divide gl - 3 1. Therefore t - I is a natural number smaller than t, so 5 divides gl-I - 31 - 1. But then 5 divides g(gl-I - 31 - 1) and 5 divides 3(gl-1 - 31 - 1), so 5 divides their sum which is gl - 31 • This is a • contradiction. Therefore, 5 divides gn - 3 n for all n E N.

Ii

'*

'*

2.6

Principles of (ounting This section informally introduces some basic techniques for counting the elements of finite sets and examines several applications. Omitting this section at this time will not significantly affect your transition to advanced mathematics, since many of the results will be developed carefully in chapter 5. We say that the setA is finite iff it has n elements for some n E N. The number of elements in A is denoted either #(A) or, more briefly, #A. For example, #{a, b} = 2. #{5, 7, 2, 7} = 3.

#(0)

=

o.

The first counting principle we accept for now as being obvious.

Theorem 2.17

(The Sum Rule) If A and B are disjoint finite sets with #A

= m and #B = n, then #(A U B) = m + n.

The Sum Rule is more powerful than it looks. It is proved rigorously (Theorem 5.6) as part of the careful study in chapter 5.H-can also be extended to the union of finitely many finite sets.

Theorem 2.18

If .sil = {A I, A 2, . .. , A k} is a family of distinct pairwise disjoint sets and #Ai = ai for I :::; i :::; k, then k

k

i=1

i=1

#U Ai = L. ai = al + a2 + ... + ak· Proof. then

The proof is by induction on the number k of sets in the family .sil. If k = I,

U.sil = AI> so #(U.sil) = al·

2.6

Principles of Counting

105

Suppose now that the statement is true for some natural number k. Let .ii be a family {AI, A 2 , .•. ,Ak> Ak+d of k + I sets.

#(U.ii) = #(~Ai UAk+I) (by exercise 17 O!S:Clion 2.3). But i~Ai are disjoint, so this number, by the Sum Rule, is #(~ Ai) + #Ak+

Then and Ak+ I

I'

The family {Ai: i = 1, ... , k} has k sets, so by the hypothesis of induction, this number is k

k

L ai + #Ak+ = L ai + ak+ I

i=1

k+ I I =

i=1

L ai'

i=1

By the PMI, the statement is true for every family of k sets, for all kEN.

•

Example. Suppose you are shopping for a used car and have three cars to choose from at Dealer I, 6 cars from Dealer 2, and 5 cars from Dealer 3. You will then have 3 + 6 + 5 = 14 possible choices. (We note that the cars are all different, so the three sets of choices are distinct and pairwise disjoint.) If A and B are not disjoint (see figure 2.14), then #A + #B overcounts #(A U B) by counting twice each element of A n B. Theorem 2.19 states that this overcounting can be corrected by subtracting #(A n B). This theorem will also be proved from a more formal point of view in chapter 5.

Theorem 2.19

For finite sets A and B, #(A U B)

=

#A

+ #B -

#(A

n B).

Example. During one week a total of 46 patients were treated by Dr. Medical for either a broken leg or a sore throat. Of these, 32 had a broken leg and 20 had a sore throat. How many did she treat for both ailments? Letting B be the set of patients with broken legs and S the set of patients with sore throats, the solution is #(B

n S) = #B + #S -

#(B U S)

= 32 + 20 - 46 = 6.

As is easily seen from the Venn diagram in figure 2.15, we could as well solve for the number of patients with a broken leg but no sore throat, or with a sore throat but no broken leg.

B~S

~ Figure 2.14

Figure 2.15

106

CHAPTER 2 SetTheory

Example. It is standard practice in manufacturing quality control to inspect a random sample of items produced to estimate the percentage of defective products. A manufacturer of plastic cups tested 100 items for two types of defects. It was found that 8 had defect A, 10 had defect B, and 3 had both defects. How many inspected cups had neither defect? To solve this problem, we find the number of cups with at least one defect and subtract this number from the total number of cups inspected. Thus the number of cups without defects is 100 - (8

+ 10 - 3) = 100 - 15 = 85.

Theorem 2.19 can be extended to three or more sets by the principle of inclusion and exclusion. The idea is that, in counting the number of elements in the union of several sets by counting the number of elements in each set, we have included too many elements more than once; so some need to be excluded, or subtracted, from the total. When more than two sets are involved, this first attempt at exclusion will subtract too many elements, so that some need to be added back or included again, and so forth. For three sets A, B, C, the principle of inclusion and exclusion states that #(A UBU C) = [#A + #B + #C] - [#(A

n B) + #(A n C) + #(B n C)] + #(A n B n C).

For four sets A, B, C, D, we have #(A U B U CUD) = [#A + #B + #C + #D] - [#(A n B) + #(A n C) + #(A n D) + #(B n C) + #(B n D) + #(C n D)] + [#(A n B n C) + #(A n B n D) + #(A n C n D) + #(B n C n D)] - #(A n B n C n D).

The principle of inclusion and exclusion is often applied to determine the number of elements not in any of several sets (see exercise 5). The next principle is as simple to state as the Sum Rule yet has a vast range of applications.

Theorem 2.20

(The Product Rule) If two independent tasks Tj and T2 are to be performed, and Tj can be performed in m ways and T2 in n ways, then the two tasks can be perfgrmed in mn ways.

Example.

Suppose there are 5 winners in a promotional contest sponsored by a bicycle shop, and each winner is allowed to select one bike from a set of 30 bikes. Then there are 5 . 30 = 150 ways a registration paper might be filled out listing an owner's name (task I) and a bicycle serial number (task 2). The Product Rule, like the Sum Rule, can be extended by induction. (See the next theorem.) The proof is exercise 7.

Theorem 2.21

If k independent tasks T]> T2, ... , Tk are to be performed and the number of ways task ~ can be performed is n;, then the number of ways the tasks can be performed in sequence is k

nn; = njn2···nk·

;=1

2.6

Principles of Counting

107

Example. Suppose a standard license plate has three letters followed by three digits and that the letters I, 0, and Q are not used on standard plates. The number of possible standard plates is 12,167,000 = 23 . 23 . 23 . 10· 10· 10. Example. If the set A has n elements, then forming a subset of A amounts to carrying out n independent tasks, where each task is to decide whether to place the element in the subset. Since each task has two outcomes, there are 2n ways this process can be carried out, so peA) has 2" elements. This argument is simply a restatement of the proof of Theorem 2.4. Example. A two-symbol variable name in the original version of the programming language BASIC must be one letter of the alphabet followed by one decimal digit. Thus the number of two-symbol variable names in BASIC is 260, because there are 26 ways to choose a letter (task I) and 10 ways to choose a decimal digit (task 2). If single-letter variable names are also allowed in BASIC, then both the Product Rule and Sum Rule must be used. Such variable names may be in the set of one letter names (with 26 elements) or letter/digit names (with 260 elements). Since these sets are disjoint, there are 26 + 260 = 286 one-letter or letter/digit variable names. Example. Personalized plates (as compared to standardized ones in a previous example) may have from one to seven letters or digits, and there may be blank spaces between symbols. To determine the number of possible personalized plates (we include in our count the prohibited plates, such as POLICE or those containing obscenities), we must determine the number of plates with each number of symbols. By the Product Rule, the number of plates with one symbol is then 36, the number of plates with two symbols is 36 2 , the number of plates with 3 symbols is (36)(37)(36) = 362 . 37 (because the middle symbol could be a blank space), and so forth. By the Sum Rule, the total number of possible personalized plates is 36

+ 36 2 + 36 2 (37) + ... + 36 2 (37 5 ) = 92,366,150,744.

DEFINITION A permutation of a set with n elements is an arrangement of the elements of the set in some order.

Recall that n! (n-factorial) is defined by O! = 1 n!=n{n-I)!

forn>O

or more informally as n! = n{n - I)(n - 2) ..... 3 ·2· 1.

Theorem 2.22

The number of permutations of a set of n elements is n!.

Proof. See exercise 8.

•

108

CHAPTER 2 Set Theory

Example. Four rodeo contestants entered a bull-riding contest in which the first rider to last 30 seconds would get the $1,000 prize. List all possible orders in which the riders A, B, C, and D might be arranged. ABCD ABDC ACBD ACDB ADBC ADCB

BACD BADC BCAD BCDA BDAC BDCA

CABD CADB CBAD CBDA CDAB CDBA

DABC DACB DBAC DBCA DCAB DCBA

We see that there are 4! = 24 such permutations.

Example. The number of different possible orders in which a radio disk jockey can play the recordings of ten golden oldies is 10! = 3,628,800. If in the bike shop contest example, each winner, in the order determined by the postmarks of the winning entries, is asked to name his or her preference among the bikes, the number of different ways this could be accomplished is 305 (because winners may have the same preferences). However, when prizes are awarded (again in the order of earliest postmarks), no one may select a bike already taken. This is a permutation problem of selecting 5 bikes from 30 and arranging them in some order. This can be done only 30 . 29 . 28 . 27 . 26 ways, as the next theorem shows.

Theorem 2.23

(The Permutation Rule) The number of permutations of any r distinct objects taken from a set of n objects is

n!

{n - r}! Proof.

See exercise 9.

In the preceding example, 30' 30 . 29 . 28 . 27 . 26 = _. 25!

DEFINITION An r-element subset of objects selected from an n-element set is called a combination of n elements taken r at a time. The number of ways such a selection can be made is the number of subsets having r elements that are included in a set of size n. This number is called a binomial coefficient and denoted by (~). The symbol (~) is read "n choose rOO or "n binomial r."

2.6

Principles of Counting

109

Example. The four subsets of {a, b, c, d} having three elements are {a, b, c}, c, d}, and {b, c, d}, so (j) = 4. The set {a, b, c, d} has 6 subsets with two elements, so (i) = 6. The 6 subsets are {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}.

{a, b, d}, {a,

There is only one O-element subset of 0, so we take (8) to be I. In what follows we develop a simple calculation for (~).

Theorem 2.24

(The Combination Rule) Let n, r E 7l.. and 0

::5

r ::5 n. The number of r-element subsets of an n-element set is ( n) r

=

n! r! {n - r}!"

Proof. The elements of an n-element set can be arranged in n! ways. The n elements may also be arranged as follows. We select r objects, which can be done in (~) ways; then order these selected elements, which can be done in r! ways; and finally, order the remaining n - r elements, which can be done in (n - r)! ways. Thus the number of ways to order n objects is n!

= (;) .

r! . (n - r)!,

so

n! ( n) r = r! (n - r)!

•

For example, in a company with 15 employees, the number of ways 5 employees could be selected for bonus pay is

_ 15! _ 15 . 14 . 13 . 12 . II _ ( 15) 5 -5!10!5.4.3.2.1 -3,003. We assume all 5 employees are to be given the same bonus, so there is no ordering in this situation; think of the selection of the 5 employees as being made simultaneously. If on the other hand, the company was distributing 5 different bonus amounts, we would use the Permutation Rule to conclude that 5 elements chosen from a set of 15 elements can be ordered in

15 . 14 . 13 . 12 . II

=

360,360

ways. We select the first employee for the largest bonus (15 ways), a different employee for the next largest bonus (14 ways), and so forth. We might also think of giving the different bonus amounts by arranging 5 se.lected employees in order. The number of ways to give five different bonuses is then the number of combinations times the number of permutations within each combination, or 3,003' 5!, which is again 360,360.

110

CHAPTER 2

Set Theory

Example. In a group of 20 automobiles, 7 have safety defects. How many ways can 5 cars be selected from the 20 so that (a) all fail the safety inspection? (b) two fail the inspection? (a) (b)

If all 5 fail the inspection, then they must all be selected from the 7 unsafe cars. This can be done in G) = 21 ways. The two that fail the inspection can be selected from the 7 unsafe vehicles in G) = 21 ways. The remaining 3 of the set of 5 must be selected from the 13 safe cars. This can be done in ('f) = 286 ways. Thus the number of ways to select 5 cars among which 2 fail the inspection is 21 . 286 = 6,006.

Several relationships exist among binomial coefficients (see exercise 17). Some of these relationships can be proved either algebraically or combinatorially. An algebraic proof will rely on formulas such as those in Theorem 2.24. A combinatorial proof is one that is based on the meaning of the binomial coefficients. Part (a) of the next theorem explains why (~) is called a binomial coefficient.

Theorem 2.25

For n, r E 7l with 0

::5

r

::5

n,

~, (a + bt = %0 (; )arbn- r.

(a)

For a, b E

(b)

(~) + (7) + (~) + ... + (~) = 2n.

(c)

(;)=C~r).

(d)

(;)=(n~I)+(;=Dforr~l.

(Binomial Expansion)

Proof. (a)

(We give a combinatorial prooffor this part.) Since (a+ b) n = (a

+ b)(a + b) .. ·(a + b) .

n times

,

(b)

each term of the product contains one term from each of the n factors (a + b). Since each has a total of n a's and b's, the terms of the product are of the form arb"- r. The coefficient of arb,,-r is the number of times this term is obtained in multiplying out. Since anb n - r is obtained exactly by choosing the term a from r of the factors (a + b), the coefficient is (~). Since 2/l = (I + It, we have by the Binomial Expansion that

(c)

(We give an algebraic prooffor this part.)

(;) = r! (d)

See exercise 16.

(nn~ r)! = (n -

r)!

(nn~(n -

r))! =

C~

r).

•

2.6

Principles of (ounting

111

n=O n

=

I

n=2

2

n=3

3

n=4

4

n=5

5

n=6

6

6

20

15

5

10

10

15

6

Figure 2.16 You are probably familiar with Pascal's triangle, named for the great 17thcentury mathematician and scientist Blaise Pascal (1623-1662). The triangle, shown in figure 2.16, is a simple means for computing the binomial coefficients and was actually used hundreds of years before Pascal studied it. For example, we read from the row for n = 4 that (a

+ b)4 =

la 4 + 1a 3 b

+ §.a 2b 2 + 1ab3 + Ib 4.

The triangle is constructed by beginning with the configuration

and constructing a new row by putting l' s on the far left and right. All other numbers in each row are found by adding the two numbers directly to the left and right in the preceding row. Thus the first lOin the row for n = 5 is the sum of 4 and 6 from the row for n = 4. The triangle has the property that the entry in the nth row and rth "column" has the value (~). The triangle is symmetric about a vertical line down its middle.

Exercises 2.6 1. Find (a)

#{nE;:l:n2 <41}

(b)

#{2, 6, 2, 6, 2} #{x E~: x 2 = -I} #{n E N: n + 1 = 4n - 1O}

(c) (d)

2.

"*

Suppose #A = 24, #B = 21, #(A U B) and #(C - B) = 12. Find (a) (c) (e)

3.

#(A n B) #(B - A)

#C

*

= 37, #(A n C) = 11, #(B - C) = 10, (b) (d) (f)

#(A - B) #(B U C) #(A U C)

How many natural numbers less than 1 million are either squares or cubes of natural numbers?

112

CHAPTER 2

Set Theory

4.

Of the four teams in a softball league, one team has four pitchers and the other teams have three each. Give the counting rules that apply to determine each of the following. (a) The number of possible selections of pitchers for an all-star team, if exactly four pitchers are to be chosen. (b) The number of possible selections if one pitcher is to be chosen from each team. (c) The number of possible selections of four pitchers, if exactly two of the five left-handed pitchers in the league must be selected. (d) The number of possible orders in which the four pitchers, once they are selected, can appear (one at a time) in the all-star game.

S.

Among the 40 first-time campers at Camp Forlorn one week, 14 fell into the lake during the week, 13 suffered from poison ivy, and 16 got lost trying to find the dining hall. Three of these campers had poison ivy rash and fell into the lake, 5 fell into the lake and got lost, 8 had poison ivy and got lost, and 2 experienced all three misfortunes. How may first-time campers got through the week without any of these mishaps?

6.

(a) (b)

(c)

If you have 10 left shoes and 9 right shoes and do not care whether they match, how many "pairs" of shoes can you select? A cafeteria has 3 meat selections, 2 vegetable selections, and 4 dessert selections for a given meal. If a meal consists of one meat, one vegetable, and one dessert, how many different meals could be constructed? There are 3 roads from Abbottville to Bakerstown, 4 roads from Bakerstown to Cadez, and 5 roads from Cadez to Detour Village. How many different routes are there frornAbbottville through Bakerstown and then Cadez to Detour Village?

7.

Prove Theorem 2.21 by induction on the number of tasks.

8.

Prove Theorem 2.22 by induction on the number of elements in the set.

9.

Prove Theorem 2.23 using a combinatorial argument.

10.

Find the number of ways seven school children can line up to board a school bus.

11.

Suppose the seven children of exercise 10 are three girls and four boys. Find the number of ways they could line up subject to these conditions. (a) The three girls are first in line. (b) The three girls are together in line. (c) The four boys are together in line. (d) No two boys are together.

*

*'

12. The number of four-digit numbers that can be formed using exactly the digits 1,3,3,7 is less than 4!, because the two 3's are indistinguishable. Prove that the number of permutations of n objects, m of which are alike, is n!/m!. Generalize to the case when ml are alike, m2 others are alike, and so forth. 13. Among ten lottery finalists, four will be selected to win individual amounts of $1,000, $2,000, $5,000, and $10,000. In how many ways may the money be distributed?

2.6

Principles of Counting

113

14.

From a second-grade class of II boys and 8 girls, 3 are selected for flag duty. (a) How many selections are possible? (b) How many of these selections have exactly 2 boys? (c) Exactly I boy?

15.

Among 14 astronauts training for a Mars landing, 5 have advanced training in exobiology. If 4 astronauts are to be selected for a mission, how many selections can be made in which 2 astronauts have expertise in exobiology?

16. Prove Theorem 2.25 as follows: *: (a) Prove part (a) by induction on n. (b) Prove part (b) using a combinatorial argument. (c) Prove part (c) using a combinatorial argument. (d) Prove part (d) using a combinatorial argument.

*

17.

18.

Find (a) (b)

(a + b)6 (a + 2b)4

(c) (d)

the coefficient of a3b lO in the expansion of (a the coefficient of a2b lO in the expansion of (a

(a)

Prove combinatorially that if n is odd, then the number of ways to select an even number of objects from n is equal to the number of ways to select an odd number of objects. Give a combinatorial proof of Vandermonde's identity: for n, m, r E 71 with 0 ::5 r ::5 n + m,

*: (b)

(c) Proofs to Grade

19.

Prove that

e:)

+

(n ~ I)

=

+ b)13 + 2b)12

ke:: n·

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. n2 + n n2 - n (a) Claim. For all n 2: 0 , -2- = n + -2-

"Proof." Consider a set of n + I elements, and let one of these elements n22 + -n ways to choose n - I of these elebe x. There are ( nn + _ I = -

I)

ments. Of these, there are (n

~

I) = n ways to the n - Ielements with-

n ) n2-n out choosing x, and ( n _ 2 = - 2 - ways to choose n - I elements n2 + n including x. Therefore, - 2 (b)

Claim.

For n 2: 0,

=

n

2

n - n +2-'

•

114

CHAPTER 2 SetTheory

"Proof."

0= (-1 + It =

!

k=O

(c)

(n)(-l)k(lt- k k

-

Claim. For n ::::: 0, the number of ways to select an even number of objects from n is equal to the number of ways to select an odd number. "Proof." From part (b) of this exercise (the claim made there is correct), we have that

(~) + (~) + ... = (7) + (~) + ... The left side of this equality gives the number of ways to select an even number of objects from n and the right side is the number of ways to select an odd number. -

CHAPTER

3

Relations

Given a set of objects, we may want to say that certain pairs of objects are related in some way. For example, we may say that two people are related if they have the same citizenship or the same blood type, or if they like the same kinds of food. If a and b are integers, we might say that a is related to b when a divides b. In this chapter we will study the idea of "is related to" by making precise the notion of a relation and then concentrating on certain relations called equivalence relations. We finish the chapter with another type of relation, orderings.

3.1

Cartesian Products and Relations The study of relations begins with the concept of an ordered pair, symbolized as (a, b), which has the property that if either of the coordinates a or b is changed, the ordered pair changes. Thus two ordered pairs (a, b) and (x, y) are equal iff a = x and b = y. A more rigorous definition of an ordered pair as a set is given as exercise 12. The adjective ordered is used to describe (a, b) because we wish to distinguish (a, b) from (b, a). For example, although the set {3, 7} = {7, 3}, the ordered pair (3, 7) =1= (7, 3). Even though the same notation (3, 7) is used to represent the ordered pair and the open interval consisting of all real numbers between 3 and 7, there is almost never any confusion between the two meanings. The context should always make clear whether the interval or the ordered pair is meant. Generalizing, we can say that the ordered n-tuples (ab ab ... , an) and (Xb Xb ... ,xn) are equal iff aj = x;for i = 1,2, ... , n. Thus the 5-tuples (3, 7, 1,3,6), (3,7,1,3,8), and (7,3,1,3,6) are all different. An ordered 2-tuple is just an ordered pair; an ordered 3-tuple is usually called an ordered triple. Most of our work in this chapter·is concerned with ordered pairs.

115

116

CHAPTER 3 Relations

DEFINITION Let A and B be sets. The set of all ordered pairs having first coordinate in A and second coordinate in B is called the Cartesian product of A and B and is written A X B. Thus A X B

=

{(a, b): a EA and bE B}.

Some authors refer to A X B as the cross product of A and B. The word Cartesian honors the French philosopher and mathematician Rene Descartes (1596-1650), who is credited with integrating the study of algebra and geometry to create the study of analytic geometry. If (a, b) E A X B, then both a E A and b E B must be true. Thus whenever (a, b) t!. A X B, then either a t!. A or b t!. B.

Example. If A

=

{I, 2} and B = {2, 3, 4}, then

A X B

= {(t, 2), (1,3), (1,4), (2,2), (2,3), (2, 4)}

B XA

= {(2, I), (2,2), (3, I), (3,2), (4,1), (4, 2)}.

whereas

In this example we have (1,2)EAXB

(2, I) t!.A X B {(3, I), (2,2)}!:BXA AXB=I=BXA

Example. In this example we must pay close attention to the punctuation and parentheses. Let A = {a, {x}, 2} and B = {{a}, I}. Then

A X B = {(a, {a}), (a, I), ({x}, {a}), ({x}, I), (2, {a}), (2, I)} and

A X A = {(a, a), (a, {x}), (a, 2), ({x}, a), ({x}, {x}), ({x}, 2), (2, a), (2, {x}), (2, 2)}. By the Product Rule (Theorem 2.20), if A has m elements and B has n elements, then A X B has mn elements (see exercise 17). The Cartesian product of three or more sets is defined similarly. For example, if A, B, and C are sets, then A X B XC = {(a, b, c): a EA and b EB and c E C}. The sets A X B X C, (A X B) X C, and A X (B X C) are different. The first is a set of ordered triples but the second and third are different sets of ordered pairs. In practice this distinction is often glossed over. Some useful relationships between the Cartesian product of sets and the other set operations are gathered in the next theorem.

Theorem 3.1

If A, B, C, and D are sets, then

(a) (b)

A X (B U C) A X (B n C)

= (A X B) U (A X C). = (A X B) n (A X C).

3.1

Cartesian Products and Relations

117

A X 0 = 0.

(c) (d)

(e)

(A X B) n (C X D) (A X B) U (C X D)

(f)

(A X B)

n (B X

= (A n C) <:;::;

X (B n D). (A U C) X (B U D).

A) = (A

n B) X

(A

n B).

Proof. (a)

(Since both A X (B U C) and (A X B) U (A X C) are sets of ordered pairs, their elements have the form (x, y). To show that each set is a subset of the other we use an "if.! argument. ,,)

The ordered pair (x, y) E A X (B U C) iff x E A and y E B U C iff x E A and (y E B or Y E C) iff (x E A and y E B) or (x E A and y E C) iff (x, y) EA X B or (x, y) EA X C iff (x, y) E (A X B) U (A X C).

(e)

Therefore, A X (B U C) = (A X B) U (A X C). If (x, y) E (A X B) U (C X D), then (x, y) EA X B or (x, y) E C X D. If (x,y)EAXB, then xEA and yEB. Thus xEAUC and yEBUD. (Because A <:;::;A U C andB <:;::; B U D.) Therefore, (x, y) E (A U C) X (B U D). If (x, y) E C X D, a similar argument shows (x, y) E (A U C) X (B U D). This shows that (A X B) U (C X D) <:;::; (A U C) X (B U D).

Parts (b), (c), (d), and (f) are exercise 3.

•

Part (e) of Theorem 3.1 cannot be sharpened to equality. That is, there are sets A, B, C, and D such that (A X B) U (C X D) = (A U C) X (B U D) is false (see exer-

cise 4(a).) When we speak of a relation on a set, we identify the notion of "a is related to boo with the pair (a, b). For the set of all people, if Phoebe and Monica were born on the same day of the year, then the pair (Phoebe, Monica) is in the relation "has the same birthday as." Thus a relation may be defined simply as a set of ordered pairs.

DEFINITIONS Let A and B be sets. A relation from A to B is a subset of A X B. If R is a relation from A to B and (a, b) E R, we write aRb, and read this as "a is R-related (or simply related) to b." a 1- b means that (a, b) $. R. If sets A and B remain unchanged throughout a discussion, we simply say R is a relation. Subsets of A X A are called relations onA. If A = {I, 2, 3, 4} and B = {-I, 1,2,4, 5}, then R = {(l, 4), (2,5), (2, -1), (4, I)} is a relation from A to B. We have 1 R 4, 2 R 5, 2 R -1, and 4 R 1. Also, 1 P 5 and 3 p 2. The sets {(5, 2), (5, 3), (4, 2), (-1, 3)}, and {(5, 1)} are relations fromB to A. If A = {I, 2, 3} and B = {a, 5, b, c}, then S = {(l, a), (2, c),(2, 5)} is a relation from A to B. We have 1 Sa, 2 S c, and 2 S 5. Also, l ' c and 3' a. The sets {(a, 2), (5, 3), (b, 2), (c, 3)} and {(c, 1)}arerelationsfromBtoA. Let LTE = {(x, y) E ~ X ~: x:::; y}. Then (2, 5) E LTE, so we write 2 LTE 5. Indeed (x, y) E LTE iff x:::; y, so LTE is the "less than or equal to" relation on ~. The notation 2 LTE 5 is consistent with the notation 2:::; 5.

118

CHAPTER 3 Relations

D~F

I

IM~O

I

I

IW~Y I

[!]GJ[!] 555-1234

I

I

The telephone pictured here defines a relation from the set of digits Ll = I, 2, ... , 9} to the set of 26 letters r = {A, B, C, ... }. The relation R defined by "appear on the same phone button" is a subset of Ll X r containing 24 pairs. The pair (4, G) E R since 4 and G appear on the same button. Likewise, 9 R Y and 6 R Mare true. (3, T) $. R since 3 and T do not appear together. Also I p.. E and 4 p.. P are true. For A = {o, 1,2,3,4,5,6, 7, 8, 9}, let R be the set of pairs of digits which sum to 10. R is a relation on A which may be written as R = {(x, y) E A X A: x + y = 1O.} Consider the relation S on the set N X N given by (m, n) S (k, j) iff m + n = k + j. Then (3, 17) S (12, 8), but (5, 4) is not S-related to (6, IS). Notice that S is a relation from N X N to N X N and consists of ordered pairs whose entries are themselves ordered pairs. For this reason, the description above is somewhat simpler than defining S by writing

{o,

S = {(em, n), (k,j)) E (N X N) X (N X N): m + n = k

+ j}.

Since every subset of A X B is a relation from A to Band 0 ~ A X B for any sets 0 is a relation from A to B. (Some care must be taken in mathematical arguments not to overlook this relation.) Likewise, A X B itself is a relation from A to B. If A has m elements and B has n elements, then A X B has mn elements (see exercise 17); so there are 2m " relations from A to B.

A and B,

DEFINITIONS

The domain of the relation R from A to B is the set

Dom (R) =

{x E A: there exists y E B such that x R y}.

The range of the relation R is the set Rng (R)

= {y E B: there exists x E A such that x R y}.

Thus the domain of R is the set of all first coordinates of ordered pairs in R, and the range of R is the set of all second coordinates. By definition, Dom (R) ~ A and Rng(R) ~B.

3.1

Cartesian Products and Relations

119

For A={1,2,3,4}, B={-I,I,2,4,5}, and R={(i,4), (2,5),(2,-1), (4, I)}, the domain is {I, 2, 4} and the range is {-I, 1,4, 5}. For LTE above, both the domain and range are JR. For the relation "appear on the same phone button," the domain is {2, 3, 4, 5, 6, 7, 8, 9} (the numbers 0 and I have no letters on their buttons), and the range is all letters except Q and Z. Every set of ordered pairs is a relation. If M is any set of ordered pairs, then M is a relation from A to B, where A and B are any sets for which Dom (M) ~ A and Rng(M) ~B.

There are several ways to describe a relation. Again, using A = {I, 2, 3, 4}, B = {-I, I, 2, 4, 5}, and R as above, we first explicitly listed pairs in a set R = { (I, 4), (2, 5), (2, -I), (4, I)}. We could define the ordered pairs in R in a table fonnat:

W 2

5

2

-14

4

1

Another method is to describe R in set notation: we could write R as R = {(x, y) EA X B:

Ix - yl

=

3}.

We could also use a Cartesian (or rectangular) coordinate system for a graph of R as a pictorial or geometric representation of the ordered pairs in R. See figure 3.1.

{s, p, q, {r}} and B = {a, b, c, d}, let R be the relation {(p, a), b)}. Then Dom (R) = {p, q, {r}} and Rng (R) = {a, b, c}. The graph of the relation R = {(p, a), (q, b), (q, c), ({r}, b)} is shown in figure 3.2. Example.

(q, b), (q,

For A =

c), ({r},

B

5 4

•

•

3

Rog(R) = (-1, I,4,5)

2

• -----+~--~~~-+A

234

•

-1

Dom(R) = (l, 2, 4)

Figure 3.1 B d

• • •

c

b

•

a s

p

q

Figure 3.2

(r)

A

120

CHAPTER 3 Relations y 8

---+-----i------t---x -18

18

lng(s) =

~

[-8,8]

-8 I- Dom(S) = [-18,18]

)1

Figure 3.3

Example.

Let S =

{(X, y) E IR X IR:

figure 3.3.

;2: + ~: :::; I}

The graph of S is given in

Relations among more than two sets are subsets of the Cartesian product of those sets. For example, if A is the set of all first names, B is all last names, C is {I, 2, 3, ... , 150}, D is the set of two-letter abbreviations for states, and E is dollar amounts, a relation among A, B, C, D, E might be "the names, ages, and states ofr~si­ dence and prize amounts for all winners in a magazine contest." This would be some subset of A X B X ex D X E, which would best be described in a tabular fashion:

First

Last

Age

State

Prize

Krista Adam Kim Jorge

Maire Powell Anen Martinez

21 54 20 45

MI NY FL KY

$1,000,000 $25,000 $1,000,000 $10,000,000

Management of relations such as these is an important part of computer science known as "relational databases." Most large data collections are viewed as relational databases now. A good example is the data relating to student financial aid in a college administrative office. Such data might be arranged in several tables (relations): a table for student directory information (name, SSN, address, major, etc.), a table for aid eligibility (personal and family financial data, maximum eligibility, etc.) and an accounts table (tuition, fees, payments, aid disbursements).

Example. LetA be any set. The set fA = {(x, x): x E A} is called the identity relation onA. For A = {I, 2},lA = {(1, 1), (2, 2)}. Obviously, for any A, Dom (fA) = A and Rng (fA) = A. The graph of fA is the "main diagonal" of A X A. For the identity relation onA = [-2,00), we can picture only a portion of fA in figure 3.4. Let A = [2, 4] and B = (1, 3) U {4}. Let S be the relation on the reals given by x S y iff x EA. Let Tbe the relation on the reals given by x Ty iff y E B. The graphs

of S and Tare shown in figures 3.5(a) and (b). Note that S = A X IR and T = IR X B. Figure 3.5(c) shows the graph of S n T = (A X IR) n (IR X B) = (A n IR) X (B n IR)

3.1

Cartesian Products and Relations

121

y

Figure 3.4

= A x B. (Note the use of Theorem 3.1 (d) in this computation.) The graph of S U T is given in figure 3.5(d). Another important kind of graph may be used to represent a relation on a setA. To form the directed graph or digraph representing the relation R on A, we think of the objects in A as vertices and think of R as telling us which vertices are connected by edges. The edges are directed from one vertex to another, like arrows. There is an edge from x to y exactly when (x, y) E R. Example. The digraph corresponding to the relation S = {(6, 12), (2,6), (2, 12), (6,6), (12, 2)} on the set V = {2, 5, 6, I2} is shown in figure 3.6. Figure 3.7 shows the digraph of the relation "divides" on the set {3, 6, 9, 12}. y

y

4 3 2

y

,

4 3 2

-------•

,

I ' •

2

"

-~~~-,~.:...\'~

-1

3

-2

5

X

-1

-1

r~--[

1 2 3 4 5

-2

-2 (a)

-

2

~-..;.-

X

1 2 3 4 5

y

x -1

-2 (c)

(b)

(d)

Figure 3.5

)---~12

Figure 3.6

3

Figure 3.7

5

122

CHAPTER 3 Relations

~T;V -~-~ MATH 260

MATH 380

Figure 3.8 It is easy to construct the digraph corresponding to a relation, and vice versa. Consider the digraph in figure 3.8 showing required courses for a mathematics major at Someplace College. Here the arrowheads are omitted, but an edge drawn down from one vertex (course) to another means that the course closer to the top of the graph is a prerequisite for the other. As a further simplification, we have not drawn an edge when there is a path, or a sequence of edges, between the courses. For example, the edge from MATH 121 to MATH 350 is not shown, although the graph shows that MATH 121 is a prerequisite for MATH 350. The relation P corresponding to this digraph contains, for example, (MATH 222, MATH 350) and (MATH 122, MATH 380), but not (STAT 240, MATH 380). The remainder of this section is devoted to methods of constructing new relations from given relations. These ideas are important in the study of relations, but they are particularly important to us in the study of functions, in the next chapter. A function is a relation that satisfies a special property. The two most fundamental methods of constructing new relations from given ones are inversion and composition. Inversion is a matter of switching the order of each pair in a relation.

DEFINITION If R is a relation from A to B, then the inverse of R is R- 1 = {(y, x): (x, y) E R}.

The inverse of the relation (O, b), (1, c), (2, c)} is {(b, I), (c, I), (c, 2)}. For any set A, 1;;1 = I A. The inverse of LTE on IR is the relation "greater than or equal to" since x LTE- 1 y iffy LTEx iffy:s x iff x ~ y. The digraph of the inverse of a relation on a set differs from the digraph of the relation only in that the directions of the arrows are reversed. Figure 3.9 shows the digraphs of Rand R- 1, where R is the relation ~ on the set {0, {I}, {3}, {I, 2}}.

Example. Let EXP2 be the relation on IR given by x EXP2 y iff y = 2x. There are too many points in IR to attempt to draw a digraph, but we can graph the relation in

3.1

Cartesian Products and Relations

123

{l,2}

(3)

(3)

~-l

Figure 3.9 y

y

-2

-1

2

3

EXP2

LOG2 = EXP2- 1

(a)

(b)

Figure 3.10 the Cartesian plane as in figure 3.1O(a). The inverse of EXP2 is familiar from calculus: (x, y) E EXpr 1 iff (y, x) E EXP2 iff x = 2Y iff y = log2 (x). The graph of LOG2, which is the same relation as EXprl, is given in figure 3.1O(b).

Theorem 3.2

Let R be a relation from A to B. (a) (b) (c)

R -I is a relation from B to A. Dom (R- I ) = Rng(R). Rng (R- I ) = Dom(R).

Proof. (a) Suppose (x, y) E R- 1• (We show (x, y) E B X A.) Then (y, x) E R. Since R is a relation from A to B, R ~ A X B. Thus yEA and x E B. Therefore, (x, y) E B X A. This proves R- I ~ B X A. (b) x E Dom(R- I ) iff there exists yEA such that (x, y) E R- I iff there exists yEA such that (y, x) E R iff x E Rng(R). (c) This is similar to part (b). • Given a relation from A to B and another from B to C, composition is a method of constructing a relation from A to C.

124

CHAPTER 3

Relations

DEFINITION Let R be a relation from A to B, and let S be a relation from B to C. The composite of Rand S is

So R

= {(a, c):

there exists b E B such that (a, b) E Rand (b, c) E S}.

Since S 0 R ~ A X C, S 0 R is a relation from A to C. It is always true that Dom(S 0 R) ~ Dom(R), but it is not always true that Dom(S 0 R) = Dom(R). We have adopted the right-to-Ieft notation for S 0 R used in analysis. To determine S 0 R, remember that the relation R is from the first set to the second, and S is from the second set to the third. Let A = {I, 2, 3, 4}, B = {p, q, r, s}, and C = {x, y, z}. Let R = {(l, p), (I, q), (2, q), (3, r), (4, s)} be a relation from A to B, and let S = {(p, x), (q, x), (q, y), (s, z)} be a relation from B to C. These relations are illustrated in figure 3.11. An element a from A is related to an element c of C under S 0 R if there is at least one "intermediate" element b of B-intermediate in the sense that (a, b) E Rand (b, c) E S. For example, since (I, p) E Rand (p, x) E S, then (I, x) E So R. By determining intermediates, which is the same idea as following arrows from A through B to C in figure 3.11, we have So R = {(l, x), (I, y), (2, x), (2, y), (4, z)}. If R is a relation from A to B, and S is a relation from B to A, then R 0 Sand S 0 R are both defined but, owing to the asymmetry in the definition of composition, you should not expect that R 0 S = S 0 R. Even when Rand S are relations on the same set, it may happen that R 0 S =1= S 0 R.

Example. LetR = {(x, y) E ~ Then R

0

X ~: y = x+

S = {(x, y): (x, z) E Sand (z, y) E R for some z E~} = {(x, y): z = x 2 and y = z + I for some z E ~} ={(x,y):y=x 2

S

0

R

0

R

=1=

R

+ I}.

(x, z) E Rand (z, y) E S for some z E ~} + I and y = Z2 for some z E~} = {(x, y): y = (x + 1)2}.

= {(x, y): =

Clearly, S

I} and let S = {(x, y) E ~ X ~: y = x 2 }.

0

{(x, y): z = x

S, since x 2

+

1 is seldom equal to (x

+ 1)2.

A B C

Q::?€JO Figure 3.11

Cartesian Products and Relations

3.1

125

The last theorem of this section collects several results about inversion, composition, and the identity relation. We prove only part (b) and the first part of (c), leaving the rest as exercise 14.

Theorem 3.3

Suppose A, B, C, and D are sets. Let R be a relation from A to B, S be a relation from B to C, and T be a relation from C to D. (a)

(R-1)-1

(b) (c) (d)

T

IB (S

0 0 0

(S

0

=

R

R)

= (T 0 S)

R; that is, composition is associative.

0

R = Rand R 0 IA = R R)-l = R- 1 0 S-l.

Proof. (b) The pair (x, w) E To (S

0

R) for some x E A and wED

iff (3z E C)[(x, z) E So Rand (z, w) E T] iff (3z E C)[(3y E B)((x, y) E Rand (y, z) E S) and (z, w) E T] iff (3z E C)(3y E B)[(x, y) E Rand (y, z) E Sand (z, w) E T] iff (3y E B)(3z E C)[(x, y) E Rand (y, z) E Sand (z, w) E T] iff (3y E B) [(x, y) E Rand (3z E C)((y, z) E Sand (z, w) E T)] iff Oy E B)[(x, y) E Rand (y, w) E To S] iff (x, w) E (T (c)

0

S)

0

R.

Therefore, T 0 (S 0 R) = (T 0 S) 0 R. (We first show that IB 0 R ~ R.) Suppose (x, y) E IB 0 R. Then there exists z E B such that (x, z) E Rand (z, y) E lB' Since (z, y) E I B, Z = y. Thus (x, y) E R (since (x, y) = (x, z) E R). Conversely, suppose (p, q) E R. Then (q, q) E IB and thus (p, q) E IB 0 R. Thus IB 0 R = R. •

Exercises 3.1 1. List the ordered pairs in A X Band B X A in each case: (a) A = {I, 3, 5}, B = {a, e, k, n, r}. (b) A = {I, 2, {I, 2}}, B = {q, {t}, n}.

*

2. 3.

*

{0, {0}, {0, {0}}}, B = {(0, {0}), {0}, ({0}, 0)}. = {(4, I), (2, 3)}.

(c)

A =

(d)

A = {(2, 4), (3, I)}, B

Let A and B be nonempty sets. Prove that A X B = B X A iff A = B. Is the statement true if one of A or B is empty? Complete the proof of Theorem 3.1 by proving (b) (c) (d) (f)

A X (B

n C) = (A X B) n (A X

A X 0 = 0. (A X B) n (C X D) = (A (A X B)

n (B X

A)

C).

n C) X

= (A n B)

n D). n B).

(B

X (A

126

CHAPTER 3 Relations 4.

Give an example of sets A, B, and C such that (a) (b) (c)

5.

* *

(A X B) U (C X D) (A U C) X (B U D). (C X C) - (A X B) (C - A) X (C - B). A X (B X C) (A X B) X C.

*

Let Tbe the relation {(3, I), (2, 3), (3, 5), (2, 2), (1, 6), (2, 6), (1, 2)}. Find (a) Dom (T). (b) Rng (T). (c) T- 1• (d) (T-1)-I.

6. Find the domain and range for the relation Won IR gi ven by x W y iff

* * *

"*

(a) (c)

y

= 2x + I.

y

= x 2 + 3.

(b) (d) (f)

y = l/x 2 .

(e)

y = ~. y < x 2.

(g)

Ixl < 2 or y = 3.

(h)

y

Ixl <

*x.

2 and y = 3.

7.

Sketch a graph of each relation in exercise 6.

8.

The inverse of R = {(x, y) E IR X iR: y = 2x

+

form R- 1 = {(X, y) E IR X IR: y

the set of all pairs (x, y) subject

* * * * *

* * *

l

(X~2)}

(h)

Rs={(X,Y)EIRX lR:y=

(i) (j)

R9 = {(x, y) E P X P: y is the father of x} RIO = {(x, y) E P X P: y is a sibling of x} Rll ={(x,y)EPXP:ylovesx}

= (Cl, 5), (2,2), (3,4), (5, 2)}, S = {(2, 4), {(I, 4), (3,5), (4, I)}. Find

Let R T = (a) (c) (e) (g)

10.

(x ; I)

to some condition. Use this form to give the inverses of the following relations. In (i), (j), and (k), P is the set of all people. (a) R1={(x,y)EIRXIR:y=x} (b) R 2 ={(x,y)EIRX lR:y= -5x+2} (c) R3 = {Cx, y) E IR X IR: y = 7x - 1O} (d) R4 = {(x, y) E IR X IR: y = x 2 + 2} (e) Rs = {(x, y) E IR X IR: y = -4x 2 + 5} (f) R6 ={(x,y)EIRX iR:y<x+ I} (g) R 7 ={(x,y)ElRxlR:y>3x-4}

(k)

9.

=

I} may be expressed in the

R 0 S To S So R R 0 (S

* *

0

T)

(b) (d) (f) (h)

(3,4), (3, I), (5, 5)}, and

RoT R 0 R ToT (R 0 S)

0

T

Find these composites for the relations defined in exercise 8. (a) Rl 0 Rl (b) Rl 0 R2 (c) R2 0 R2 (d) R2 0 R3

*

W

~o~

(f)

~o~

00

~o~

00

~o~

(i) (k)

R6 Rs

(j) (I)

R6 Rs

~

~o~

0 0

R4 Rs

~)~o~

*

0 0

R6 Rs

3.1

"*

(0) (q)

Cartesian Products and Relations (p)

R9 0 R9 Rll 0 R9

RIO

11. Give the digraphs for these relations on the (a) (b) (c) ::5 (d) (e) =1= (f)

0

127

R9

set {I, 2, 3}. S={(i,3),(2,1)} S-I,whereS={(i,3),(2,1)} So S, whereS = {(i, 3), (2, I)}

12. Let A = {a, b, c, d}. Give an example of relations Rand S on A such that (a)

13.

(SOR)-l =l=S-1 oR-I.

(b)

RoS=I=SoR

Let Sand R be relations on a set A. (a) Which of these two statements must be true: Rng (S) (b) (c)

~

Rng (S

0

R)

or

Rng (S

0

R) ~ Rng (S)?

Give an example to show that the other statement may be false. Show by example that one of Dom (R) ~ Dom (S 0 R) or Dom (S 0 R) ~ Dom (R) may be false. Give an example of non empty relations Rand S on the setA = {a, b, c, d} such that R 0 Sand S 0 R are empty.

14. Complete the proof of Theorem 3.3.

15. One way to define an ordered pair in terms of sets is to say (a, b) ={{a}, {a, b}}. Using this definition, prove that (a, b) = (x, y) iff a = x and b = y. 16. Prove that (A X B) X C = A X (B X C) is false unless one of the sets is empty. 17. Prove that if A has m elements and B has n elements, then A X B has mn elements. 18. (a) Let R be a relation from A to B. For a E A, define the vertical section of R at a to be Ra = {y E B: (a, y) E R}. Prove that Ra = Rng (R).

U

aEA

(b)

Let R be a relation from A to B. For a E A, define the horizontal section of R at a to be Ifi = {x E A: (x, b) E R}. Prove that Ifi = Dom (R).

U

bEE

Proofs to Grade

19. We may define ordered triples in terms of ordered pairs by saying that (a, b, c) = ((a, b), c). Use this definition to prove that (a, b, c) = (x, y, z) iff a = x and b = y and c = z. 20. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

* W

a~~

0xmUC=0xqu~xq

"Proof." x E (A X B) U C iff x E A X B or x E C iff x E A and x E B or x E C iff x E A X C or x E B X C

*

illxE0xqu~xq.

(b)

• Claim. IfA~BandC~D,thenAXC~BXD. "Proof." Suppose A xci. B X D. Then there exists a pair (a, c) E A X C with (a, c) t!. B X D. But (a, c) E A X C implies that a E A and c E C, whereas (a, c) t!. B X D implies that a t!. Band c t!. D.

128

CHAPTER 3 Relations

(c)

However, A ~ Band C ~ D, so a E Band c ED. This is a contradiction. Therefore, A X C ~ B X D. • Claim. If A X B = A X C and A =1= 0, then B = e. "Proof." Suppose A X B = A X e. Then A X B=A X C

A

*

(d)

(e)

(f)

(g)

3.2

A

so

B=

e.

•

Claim. If AX B = A X C and A =1= 0, then B = e. "Proof." To show B = C, suppose bE B. Choose any a EA. Then (a, b) EA X B. But since A X B=A X C, (a, b) EA X e. Thus bEe. This proves B ~ e. A proof of C ~ B is similar. Therefore, B = e. • Claim. Let Rand S be relations from A to B and from B to C, respectively. Then So R = (R 0 S)-I. "Proof." The ordered pair (x, y) E So Riff (y, x) E R 0 S iff (x, y) E (R 0 S)-I. Therefore, S 0 R = (R 0 S)-I. • Claim. Let R be a relation from A to B. Then IA ~ R -loR. "Proof." Suppose (x, x) E IA- Choose any y E B such that (x, y) E R. Then, (y, x) E R- 1• Thus (x, x) E R- 1 0 R. Therefore, IA ~ R- 1 0 R . • Claim. Suppose R is a relation from A to B. Then R -loR ~ IA' "Proof." Let (x, y) E R- 1 0 R. Then for some z E B, (x, z) E Rand (z, y) E R- 1• Thus (y, z) E R. Since (x, z) E Rand (y, z) E R, x = y. • Thus (x, y) = (x, x) and x E A, so (x, y) E IA .

Equivalence Relations We shall see in these next three sections that there are many properties that make relations interesting. Each of the three properties set forth in the next definition is important in its own right. Relations that possess all three properties are particularly valuable.

DEFINITIONS

Let A be a set and R be a relation on A.

R is reflexive on A iff for all x E A, x R x. R is symmetric iff for all x and yEA, if x R y, then y R x. R is transitive iff for all x, y, and z E A, if x R y and y R z, then x R z.

For a relation R on a nonempty set A, only the reflexive property actually asserts that some ordered pairs belong to R. A proof that R is reflexive must then show x R x for all x EA. Since the identity relation onA is the setlA = {(x, x); x EA}, R is reflexive on A iff IA ~ R. Of course, if A is empty, then R is trivially reflexive on A because x R x for all x E A. To show R is not reflexive on A requires proving there is some x E A such that x, x. Symmetry and transitivity are defined by conditional sentences, so most proofs involving these properties are direct proofs. Neither property requires that R contain

3.2

Equivalence Relations

129

any ordered pairs. In fact, the empty relation 0 is a symmetric and transitive relation on any set A. Since the denial of "If x R y, then y R x" is "x R y and not y R x," a relation R is not symmetric if there are elements x, yEA such that x R y and y p. x. Likewise, R is not transitive if x R y, y R z, and x p. Z hold for some x, y, z E R.

Example. For the set B = {2, 5, 6, 7}, let SI = {(2, 5), (5,6), (2, 6)}, S2 = {(2, 5), (2, 2)}, and S3 = {(5, 6)}. All of SI, S2, and S3 are transitive relations on B. They are not reflexive on B and not symmetric. Notice especially that there do not exist x, y, z E B such that x S3 y and y S3 z; therefore, the conditional sentence if x S3 y and y S3 z, then x S3 z is true. Example. Let R be the relation "is a subset of' on !JP(E), the power set of 7l... R is reflexive on !JP(7l..) since every set is a subset of itself. R is transitive by Theorem 2.3. Notice that {I, 2} <:;::; {I, 2, 3} but {I, 2, 3} r;t {I, 2}. Therefore, R is not symmetric. Example. Let STNR designate the relation {(x, y) E 7l.. X 7l..: xy > O} on 7l... In this example, x STNR x for all x in 7l.. except the integer 0; hence the relation STNR is not reflexive on 7l... STNR is symmetric since, if x and yare integers and xy > 0, then yx > O. STNR is also transitive. To verify this, we assume that x STNR y and y STNR z. Then xy > 0 and yz > O. If y is positive, then both x and z are positive; so xz > O. If y is negative, then both x and z are negative; so xz > O. Thus in either case, x STNR z. Therefore, STNR is symmetric, transitive, and not reflexive on 7l.., whence the acronym that names the relation. The three properties of reflexivity, symmetry, and transitivity can be nicely characterized by properties of the digraph: A relation is reflexive iff every vertex of the digraph has a loop (an edge from the vertex to itself). A relation is symmetric iff between any two vertices of the digraph there are either no edges or both edges. A relation is transitive iff whenever there is an edge from x to y and an edge from y to z, there must be an edge from x to z.

Example. Figure 3.12 shows the digraphs of three relations onA = {2, 3, 6}. Figure 3.12(a) is the digraph of the relation "divides," which is reflexive on A because

2

(a) divides

(b)

~

Figure 3.12

3

~ (c) S

130

CHAPTER 3 Relations

every integer divides itself. Figure 3.12(b) is the digraph of "::=:::," which is also reflexive onA. Figure 3.12(c) is the digraph of the relation S, where x S y iff x + y > 7. Since 2 + 2 = 4, (2, 2) tt. s, S is not reflexive, and there is no loop at 2. In the digraph of a symmetric relation, if there is an edge from vertex x to y, then there must be an edge from y to x. In figure 3.12 only the digraph of S satisfies this condition; S is a symmetric relation. The graphs in figure 3.12(a) and (b) show that both "divides" and "::=:::" are transitive relations. The relation S is not transitive because 2 S 6 and 6 S 3, but 2 is not S-related to 3. The identity relation fA on any setA has all three properties. It is, in fact, the relation "equals," because x fA y iff x equals y. Equality is a way of comparing objects according to whether they are the same. Equivalence relations, defined next, are a method of grouping objects according to whether they share a common trait; they are "equal" in the sense that they share the trait. For example, if Tis the set of all triangles, we might say two triangles are "the same" (equivalent) if they are congruent. This generates the relation R = {(x, y) E TXT: x is congruent to y} on T, which is reflexive on T, symmetric, and transitive. The notion of equivalence, then, is embodied in these three properties.

DEFINITION A relation R on a set A is an equivalence relation on A iff R is reflexive on A, symmetric, and transitive.

For the set P of all people, let L be the relation on P given by x L y iff x and y have the same last name. Then L is an equivalence relation on P (if we make the assumption that everyone has a last name). We have Lucy Brown L Charlie Brown, James Madison L Dolly Madison, and so on. The subset of P consisting of all people who are L-related to Charlie Brown is the set of all people whose last name is Brown. This set contains Charlie by reflexivity. It also contains Sally Brown, James Brown, Buster Brown, Leroy Brown, and all other people who are like Charlie Brown in the sense that they have Brown as a last name. The same is true for the Madisons-all people L-related to Dolly Madison form the set of all people with last name Madison.

DEFINITION Let R be an equivalence relation on A. For x E A, the equivalence class of x determined by R is the set x/R

= {y E A: x R y}.

This is read "the class of x modulo R" or "x mod R." The set of all equivalence classes is called A modulo R and is denoted AIR = {xIR: x EA}.

3.2 Equivalence Relations

131

The equivalence class of Charlie Brown modulo L is the set of all people whose last name is Brown. Furthermore, Buster BrownlL is the same as Charlie Brown/L.

Example. The relation H = {(l, I), (2, 2), (3, 3), (l, 2), (2, I)} is an equivalence relation on the set A = {I, 2, 3}. Here I/H = 2/H = {I, 2} and 3/H = {3}. Thus A/H

= HI, 2}, {3}}.

Example. The relation 0 on IR given by x 0 y iff x 2 = y2 is an equivalence relation on IR. In this example 2/0 = {2, -2}. Notice that -n/O = {-n, n} and 0/0 = {O}. For any x E IR, x/0 = {x, -x}. Example. Two integers have the same parity iff they are either both even or both odd. Let R = {(x, y) E 7l.. X 7l..: x and y have the same parity}. R is an equi valence relation on 7l.. with two equivalence classes, the even integers E and the odd integers D. If x is odd, x/R = D, while if x is even, x/R = E. Thus 7l../R = {E, D}. The digraph of an equivalence relation has a striking property. Consider the relation S on the set A = {2I, 64, 82, 113,247, 1042} given by x S y iff x and y have the same number of digits. The digraph of this equivalence relation (figure 3.13) looks like three separate digraphs. A subset of the vertex set of a graph together with all the edges connecting vertices in the subset, is called a subdigraph. In this case the three subdigraphs on the vertex sets {21, 64, 82}, {113, 247}, and {1042} can be described as follows. Each subdigraph has all possible edges connecting its vertices. (A digraph with all possible edges is called complete.) Furthermore, no edge connects a vertex in one of the three subsets with a vertex in another subset. The digraph in figure 3.13 thus is a union of three components, each of which is a complete subdigraph. The three components are the three equivalence classes, {21, 64, 82}, {II3, 247}, and {1042}. What we have seen is that the equivalence relation S gives us a way of walling off, or partitioning, the set A into disjoint subsets. We will return to this idea in the next section. The last example we give of an equivalence relation has many applications, such as methods for encoding information. Recall that an integer m divides an integer n iff there exists an integer k such that n = km. For a fixed integer m *- 0, let be the relation on 7l.. given by

=m

x

=m Y iff m divides (x -

y).

1042

a

~igure

3.13

132

CHAPTER 3 Relations

We shall see in Theorem 3.4 that =m is an equivalence relation on 7l... Another notation for x =m Y is x = y (mod m), which is read "x is congruent to y modulo m." We read x/=m as "the equivalence class of x modulo m" or simply "the class of x mod m." The set of equivalence classes for =m is denoted 7l..m, called "7l.. mod m." As an example, let's concentrate on congruence modulo 3. We see that 4 = 1 (mod 3) because 3 divides 4 - I. Also, 10= 16 (mod 3), since 3 divides 10 - 16; that is, 10- 16 = 3k for the integer k = -2. However, 5 ¢ -6 (mod 3) since 3 fails to divide II. Next let us calculate all the equivalence classes of 7l.. 3 . For x E 7l.., the equivalence class of x modulo 3 is {y E 7l..: x = y (mod 3)}. For 0, we see 0 = 0 (mod 3), 0 = 3 (mod 3), and 0 = -6 (mod 3). It can be shown that 0/=3 = {... , -12, -9, -6, -3, 0,3,6,9, 12, ... }, the set of all multiples of3. Similarly 1/=3 = {... , -8, -5, -2, I, 4, 7, 1O, ... } and 2/=3 = {... , -10, -7, -4, -1,2,5,8, II, 14, ... }. There are no other equivalence classes. Thus 7l..3 = {0/=3' 1/=3' 2/=3}' The notation x/=m we have used is rather cumbersome. Usually the equivalence class of x modulo m is written x or [x]. This notation works well when the modulus m remains unchanged throughout a discussion. The disadvantage is that if we are dealing with, say, classes modulo 6 as well as with classes modulo 3, the symbol T could have two different meanings. As an equivalence class modulo 6,

T = {... , -5, 1,7, 13, ... } but as an equivalence class modulo 3,

T = {... , -5, -2, 1,4,7, 10, 13, ... }. Theorem 3.4

The relation =m is an equivalence relation on the integers. Furthermore, 7l.. m has m distinct elements, 0,1, 2, ... , and (m - I).

Proof. It is clear that =m is a set of ordered pairs of integers and, hence, is a relation on 7l... We shall first show that =m is an equivalence relation. (i) To show reflexivity on 7l.., let x be an integer. We show that x = x (mod m). Since m . 0 = 0 = x - x, m divides x-x. Thus =m is reflexive on 7l... (ii) For symmetry, suppose x = y (mod m). Then m divides x - y. Thus there is an integer k so that x - y = km. But this means that -(x - y) = -(km), or that y - x = (-k)m. Therefore, m divides (y - x), so that y = x (mod m). (iii) Supposex=y (modm) andy =Z (mod m). Thus m divides both x - y and y - z. Therefore, there exist integers hand k such that x - y = hm and y - z = km. But then h + k is an integer;andx - z = (x - y) + (y - z) = hm + km = (h + k)m. Thus m divides x - z, so x = z (mod m). Therefore, =m is transitive. Now that = m is known to be an equivalence relation, it follows that x =m y iffi = Y (exercise 7). The remainder of the proof shows there are exactly m distinct equivalence classes by showing that

7l..m = {O, 1, 2, ... , (m

- I)}.

First, each k for k = 0, 1,2, ... , (m - 1) is an equivalence class and hence in 7l.. m. Now suppose i E 7l..nl' By the Division Algorithm, there exist integers q and r such that x = mq + r with 0:::::; r:::::; m - I. Thus mq = x - r. Therefore, x = r (mod m)

3.2

Equivalence Relations

133

and, by exercise 7, it follows that x = r. Thus x E {O, T, 2, ... , (m - I)}, and the claim is verified. We will know that 7l.m has exactly m elements when we show that 0, T, 2, ... , (m - I) are all distinct. Suppose k = r where 0:::; r:::; k:::; (m - I). Then k == r (mod m), and thus m divides k - r. But 0:::; k - r:::; m - I, so k - r = O. Therefore, k = r, and the m equivalence classes are distinct. -

Exercises 3.2 1. Indicate which of the following relations on the given sets are reflexive, which are symmetric, and which are transitive. (a) {(l,2)}onthesetA={I,2} (b) :::;onN W =ooN ~
* *

*

* *

*:

134

CHAPTER 3 Relations

(f)

(From calculus) Let R be the equivalence relation on the set of all differentiable functions defined by f R g ifff and g have the same first deri vative, that is,!' = g'. Name three elements in each of these classes: x 2 /R, (4x 3

(g)

5.

+

lOx)/R. Describe x 3 /R and 7/R.

For the set X = {m, n, p, q, r, s}, let R be the relation on 'Jl(X) given by ARB iff A and B have the same number of elements. List all the elements in {m}/R; in {m, n,p, q, r}/R. How many elements are in X/R? How many elements are in 'Jl(X)/R?

Which of these digraphs represent relations that are (i) reflexive, (ii) symmetric, (iii) transitive?

*

(a)

*

(c)

(b)

(d)

3

4

6.

Calculate the equivalence classes for the relation of (a) congruence modulo 5. (b) congruence modulo 8. (c) congruence modulo 1. (d) congruence modulo 7.

7.

For the equivalence relation ==m' prove that (a) if x ==m y, then x = y. (b) if x = y, then x ==m y. (c) ifX ny oF 0, then x = y.

8.

Consider the relations Rand Son N defined by x R y iff 2 divides x x S y iff 3 divides x + y. (a) Prove that R is an equivalence relation. (b) Prove that S is not an equivalence relation.

*

+ y and

3.2 Equivalence Relations 9.

Suppose that Rand S are equivalence relations on a set A. Prove that R an equivalence on A.

135

n S is

10.

Let R be a relation on the set A. (a) Prove that R U R- 1 is symmetric. (R U R- 1 is the symmetric closure of R.) (b) Prove that if S is a symmetric relation on A and R k S, then R - 1 k S.

11.

Let R be a relation on the set A. Define TR = {(x, y) E A X A: for some n E N there exists ao = x, alo a2,"" an = yEA such that (ao, al)' (alo a2), (az, a3)"'" (a n-Io an) E R}. (a) Prove that TR is transitive. (TR is the transitive closure of R.) (b) Prove that if S is a transitive relation on A and R k S, then TR k S.

*

12. The properties of reflexivity, symmetry, and transitivity are related to the identity relation and the operations of inversion and composition. Prove that (a) R is reflexive iff fA k R. (b) R is symmetric iff R = R- 1. (c) R is transitive iff R 0 R k R.

*

13.

Prove that if R is a symmetric, transitive relation on A and the domain of R is A, then R is reflexive on A.

14. The complement of a digraph has the same vertex set as the original digraph, and an edge from x to y exactly when the original digraph does not have an edge from x to y. The two digraphs D and jj shown below are complementary. Call a digraph symmetric (transitive) iff its relation is symmetric (transitive).

--D

r~ (a) (b)

Proofs to Grade

Show that the complement jj of a symmetric digraph D is symmetric. Show by example that the complement of a transitive digraph need not be transitive.

*: 15.

Let L be a relation on a set A that is reflexive and transitive (but not necessarily symmetric). Let R be the relation defined on A by x R y iff xL y and y Lx. Prove that R is an equivalence relation.

16.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If the relation R is symmetric and transitive, it is also reflexive. "Proof." Since R is symmetric, if (x, y) E R, than (y, x) E R. Thus (x, y) E Rand (y, x) E R, and since R is transitive, (x, x) E R. Therefore, R is reflexive. • (b) Claim. If the relations Rand S are symmetric, then R n S is symmetric. "Proof." Let R be the relation of congruence modulo 10 and S the relation of congruence modulo 6 on the integers. Both Rand S are symmetric. If (x, y) ERn S, then 6 and 10 divide x - y. Therefore, 2, 3, and 5 all divide x - y, so 30 divides x - y. Also if 30 divides x - y, then 6

136

CHAPTER 3 Relations

(c)

*

3.3

(d)

and 10 divide x - y, so R n S is the relation of congruence modulo 30. Therefore, R n S is symmetric. Claim. Ifthe relations Rand S are symmetric, then R n S is symmetric. "Proof." Suppose (x, y) ERn S. Then (x, y) E R and (x, y) E S. Since Rand S are symmetric, (y, x) E Rand (y, x) E S. Therefore, (y, x) E RnS. Claim. If the relations Rand S are transitive, then R n S is transitive. "Proof." Suppose (x, y) ERn Sand (y, z) ERn S. Then (x, y) E R and (y, z) E S. Therefore, (x, z) ERn S. -

Partitions If R is an equivalence relation on a nonempty set A and x E A, then the set of elements in A that are R-related to x is the equivalence class of x. Looking back at examples from section 3.2, you may see that every equivalence class is nonempty, that every equivalence class is a subset of A, that the union of the equivalence classes is A, and that the classes of two elements x and yare either equal (when x R y) or have no elements in common (when x P. y). These observations are summarized in the next theorem.

Theorem 3.5

Let R be an equivalence relation on a nonempty set A. Then (a)

For all x E A, x E x/R. (Thus x/R

(b)

x/R~AforallxEA.

(c)

A=

=1=

0.)

U x/R. xEA

(d)

x Ry iff x/R = y/R.

(e)

x P. y iff x/R n y/R

= 0.

Proof. (a)

(b) (c)

Since R is reflexive on A, (x, x) E R. Thus x E x/R. This is from the definition of x/R. First x/R ~ A becauseeachx/R ~A. To prove A ~

U

xEA

By part (a), t E t/R ~

U x/R. Thus A = U x/R. xEA

(d)

(i)

(ii)

U x/R, supposet EA. xEA

xEA

Suppose x R y. To show x/R = y/R, we first show x/R ~y/R. If z E x/R, then x R z. From x R y, by symmetry, y R x. Then, by transitivity, y R z. Thus z Ey/R. The proof that y/R ~x/R is similar. Suppose x/R = y/R. Since y E y/R, y E x/R. Thus x R y.

3.3 (e)

Partitions

137

If x/R n y/R = 0, then (since y E y/R) y tt. x/R. Thus x P y. Finally, we show x p y impliesx/R n y/R = 0. (We prove the contrapositive.) Suppose x/R n y/R"* 0. Let k E x/R n y/R. Then x R k and y R k. Therefore, x R k and k R y. Thus x R y. •

(i) (ii)

If R is an equivalence relation on a setA, Theorem 3.5 tells us that the setA may be thought of as a union of a collection of non empty subsets that are pairwise disjoint. In other words, imposing an equi valence relation on A produces a set of equi valence classes, so that every element of A is in exactly one class and any two classes are either equal or disjoint. This leads to the concept of a partition of a set.

DEFINITION Let A be a set and .
"*

If X E .
U

n Y = 0.

XEdl

Partitions are a very common way that we organize the world around us. The United States is partitioned in several ways: by postal zip codes, by state boundaries, by time zones, etc. The set of magnetizable spots on the surface of a computer disk are partitioned into a collection of concentric circles called tracks. A chessboard is partitioned into two sets-the squares of one color, and the squares of the other color.

Examples. The set {{O}, {I, -I}, {2, -2}, ... } is a partition of 7l. (see figure 3.l4). The two element set of sets {E, D}, where E and D are the sets of even and odd integers, respectively, is a different partition of 7l. (figure 3.l5).

~

2 -2

4 -4

6 -6

8 -8

-1

3 -3

5 -5

7 -7

I

Figure 3.14

E"

2

A partition of 71.

-2

4

-4

6

-6

8

-8

-2

3

-4

5

-6

7

-8

Figure 3.15

)

A partition of 71.

j

138

CHAPTER 3 Relations

The collection HI}, {2}, {3}, ... } is a partition of N. In fact, Hx}: x EA} is always a partition of a nonempty set A. At the other extreme, {A} is a not-so-interesting partition of a set A. For each nEE, let Gn = [n, n + 1). Then {Gn : nEE} is a partition of ~. Theorem 3.5 may be restated to say that every equivalence relation on a set determines a partition of that set. The relation T = {(5, 7), (7, 6), (6, 6), (5, 5), (7, 7), (7,5), (6, 5), (5, 6), (6,7), (4,4)} is an equivalence relation on the setA = {4, 5, 6. 7} with equivalence classes {4} and {5, 6, 7}, so the corresponding partition is A/T =

H4}, {5, 6, 7}}. We shall soon see that every partition in turn determines an equivalence relation. Thus each concept may be used to describe the other. This is to our advantage, for we may use partitions and equivalence relations interchangeably, choosing the one that lends itself more readily to the situation at hand. The method of producing an equivalence relation from a partition is based on the idea that two objects will be said to be equivalent iff they belong to the same member of the partition. For example, letA be {l, 2, 3,4,5} and ~ = HI, 2}, {3}, {4, 5}} be a partition of A. To make an equivalence relation Q on A from B, we note that since {I, 2} E~, 1 is related to 1 and 2, 2 is related to 1 and 2. Also 3 is related to 3, and so forth. Therefore, Q = {(l, 1), (1, 2), (2, 1), (2,2), (3, 3), (4,4), (4, 5), (5, 4), (5, 5)} is the equivalence relation associated with ~. The next theorem asserts that this method of using a partition to define a relation always produces an equivalence relation and, furthermore, that the set of equivalence classes of the relation is the same as the original partition. Before taking up the theorem, one more example may be instructive. The state of Missouri is divided (partitioned) into three telephone area codes (figure 3.16). That is, the set A of all telephones in Missouri is partitioned into three subsets, A 314 , A 417 , and A S16 ' How can we use this fact to define an equivalence relation on the set A? Obviously, we should say that two telephones in A are related iff they are in the same partition element, that is, iff they have the same area code.

Theorem 3.6

Let ~ be a partition of the set A. For x and yEA, define x Q y iff there exists C E such that x E C and y E C. Then (a)

Q is an equivalence relation on A.

(b)

A/Q =~.

Figure 3.16

~

3.3

Partitions

139

Proof. As a special case, observe that if A is empty, then the only partition of A is the empty family CIA. In this case Q will be the empty relation on A, and A/Q = = CIA, as promised. Assume now that A is nonempty.

o

(a)

(b)

We prove Q is transitive and leave the proofs of symmetry and reflexivity on A for exercise 9. Let x, y, z EA. Assume x Q y and y Q z. Then there are sets C and D in CIA such that x, y E C and y, zED. Since CIA is a partition of A, the sets C and D are either identical or disjoint; but since y is an element of both sets, they cannot be disjoint. Hence, there is a set C ( = D) that contains both x and z, so that x Q z. Therefore, Q is transitive. We first show A/Q ~ CIA. Let x/Q E A/Q. Since CIA is a partition of A, choose BE CIA such that x E B. We claim x/Q = B. If y E x/Q, then x Q y; there is some C E CIA such that x E C and y E C. Since either C = B or C n B = 0, and x E C n B, y E B. On the other hand, if y E B, then x Q y, and so y E x/Q. Therefore, x/Q = B. To show CIA ~A/Q, let BE CIA. As an element of a partition, B"* 0. Choose any t E B; then we claim B = t/Q. If s E B, then t Q s, sos E t/Q. On the other hand, if sEt / Q, then t Q s; so sand t are elements of the same member of CIA, which must be B. •

Example. LetA = {I, 2, 3, 4}andCIA = HI}, {2, 3}, {4}}. TheequivalencerelationQ associated with the partition CIA is {(I, I), (2, 2), (2, 3), (3, 2), (3, 3), (4, 4)}. The equivalence classes of this relation are I/Q = {I}, 2/Q = {2, 3} = 3/Q, and'4/Q = {4}, so the set of equivalence classes is precisely CIA. Example. For 7l., let s'l be the partition {Ao, A I, A 2 , A 3} where Ao = { ... , - 8, -4, 0,4,8, ... }, Al = {... , -7, -3, 1,5, 9, ... }, A2 = {... , -6, -2,2,6, 1O, ... }, and A3 ={ ... , -5, -1,3,7, 11, ... }. For integers x and y, we see x and yare in the same set Ai iff x = 4k I + i and y = 4k2 + i, for some integers k I and k2 or, in other words, iff x - y is a multiple of 4. Thus the equivalence relation association with the partition s'l is our old friend, congruence modulo 4.

Exercises 3.3 1.

Describe some partitions of the set of students at a university.

2.

Describe the partition for each of the following equivalence relations. (a) For x, y E IR, x R y iff x - yEll.. (b) For n, mE 7l., n R m iff n + m is even. (c) For x, y E lR,x R y iffsinx = siny.

* 3.

'*

4.

Let C = {i, -I, - i, I}, where i is chosen so that i 2 = -I. The relation R on C given by x R y iff xy = :±: I is an equivalence relation on C. Give the partition of C associated with R. Let C be as in exercise 3. The relation Son

ex

C given by (x, y) S (u, v) iff

xy = uv is an equivalence relation. Give the partition of ex C associated with S.

140

CHAPTER 3 Relations 5.

* * 6.

For each a E lR, let Aa = {(x, y) E lR X lR: y = a - x 2 }. (a) Sketch a graph of the set Aa for a = - 2, -I, 0, I, and 2. (b) Prove that {Aa: a E lR} is a partition of lR X lR. (c) Describe the equivalence relation associated with this partition.

7.

List the ordered pairs in the equivalence relation onA = {I, 2, 3, 4, 5} associated with these partitions: (a) {{I,2},{3,4,5}} (b) {{I},{2}, {3,4}, {5}} (c) {{2,3,4,5},{I}}

*

*

Describe the equivalence relation on each of the following sets with the given partition. (a) N, HI}, {2, 3}, {4, 5, 6, 7}, {8, 9, 10, II, 12, 13, 14, 15}, ... }. (b) 71, {... ,{-2}, {-I}, {O}, {I}, {2}, {3, 4, 5, ... }}. (c) lR, {(-(X), 0), {O}, (0, oo)}. (d) lR, {... , (-3, -2), {-2}, (-2, -I), {-I}, (-1,0), {O}, (0, I), {I}, (1,2), {2}, (2, 3), ... }. (e) 71, {A, B}, where A = {x E 71: x < 3} and B = 71- A.

8.

Partition the set D = {I, 2, 3, 4, 5, 6, 7} into two subsets: those symbols made from straight line segments only (like 4), and those that are drawn with at least one curved segment (like 2). Describe or draw the digraph of the corresponding equivalence relation on D.

9.

Complete the proof of Theorem 3.6 by proving that if 'ZJ3 is a partition of A, and x Q y iff there exists C E 'ZJ3 such that x E C and y E C, then (a) Q is symmetric. (b) Q is reflexive on A.

10.

Let R be a relation on a setA that is refle~ve and symmetric but not transitive. Let R(x) = {y: x R y}. [Note that R(x) is the same as x/R except that R is not an equivalence relation in this exercise.] Does the set.il = {R(x): x EA} always form a partition of A? Prove that your answer is correct.

11.

Repeat exercise 10, assuming R is reflexive and transitive but not symmetric.

12.

Repeat exercise 10, assuming R is symmetric and transitive but not reflexive.

13.

Let A be a set with at least three elements. (a) Is there a partition of A with exactly one element? (b) If 'ZJ3 = {B], B2} is a partition of A with B] Bz, is {S]. S2} a partition of A? Explain. (Here Si = A - Bi.) What if B] = B2? (c) If'ZJ3 = {B]. B2, B3} is a partition of A, is {S]. Sz, S3} a partition of A? Explain. Consider the possibility that two or more of the elements of'ZJ3 may be equal. (d) If'ZJ3 = {B]. B2} is a partition of A, C€] is a partition of Bb and C€2 is a partition of B2, and B] B2, prove that C€] u C€2 is a partition of A.

*

'*

'*

Proofs to Grade

14.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If.il is a partition of a set A and 'ZJ3 is a partition of a set B, then .il U 'ZJ3 is a partition of A U B.

"Prooj." (i)

If X E .il U 'ZJ3, then X E .il or X E 'ZJ3. In either case X

'* 0.

3.4 (ii)

(iii)

*

3.4

(b)

Ordering Relations

141

If XEd U CJ3 and Y E d U CJ3, then XEd and Y E d, or XEd and Y E CJ3, or X E CJ3 and Y E d, or X E CJ3 and Y E CJ3. Since both d and CJ3 are partitions, in each case either X = Yor XnY=0. Since X = A and X = B, X = A U B. •

U

U

U

XEd!

XEi'll

XEd!Ui'll

Claim. If CJ3 is a parti tion of A, and if x Q y iff there exists C E CJ3 such that x E C and y E C, then the relation Q is symmetric. "Proof." First, x Q y iff there exists C E CJ3 such that x E C and y E C. Also, y Q x iff there exists C E CJ3 such that y E C and x E C. Therefore, x Q y iffy Qx. •

Ordering Relations Equivalence relations are an important type of relation but are by no means the only type of relation worthy of study. Other types of relations allow us to impose an order on sets so that we may speak of some elements as being "less than or equal to" other elements. In this section we discuss some of the properties and definitions of orderings. The number systems N, Z, and IR have a familiar order relation:::; . We know many facts about this relation, such as "3 :::; 3" and "from x :::; 5 and 5 :::; y, we can conclude x:::; y." The set inclusion relation, <;::; , on the power set QJ>(A) for a given set A, provides a different ordering that relates some members of QJ>(A). For example, if A = {I, 2, 3, 4}, then we agree that {2, 3} is "less than or equal to" {2, 3, 4} because {2, 3} <;::; {2, 3, 4}. In section 2.1 we proved that the relation <;::; is reflexive and is transitive on QJ>(A). Notice that not all subsets of A are comparable. For example, we cannot compare {2, 3, 4} and { 1, 2, 3} because neither is a subset of the other. For orderings, we have in mind relations like:::; on N, Z, and IR that are reflexive, rather than relations like < that are not reflexive. Transitivity is another essential feature of orderings. As with the inclusion relation, we permit the situation where two objects in the set are not comparable. However, we wish to avoid the situation where there are two different objects each "less than or equal to" the other. For example, the relation Y "is the same age (in years) or younger" on a set Wof people is certainly reflexive and transitive, and it serves to order W if no two people in W have the same age. But if two people in Ware 19 years of age, then each of them is Y-related to the other, so the relation Y does not sort the people of W into order. To put this another way, it is acceptable in an ordering for two objects to be unrelated, but if two objects are related, they can't both be less than the other. The property that we require of an ordering is called anti symmetry .

DEFINITION. A relation R on a set A is antisymmetric if, for all x, yEA, if x R y and y R x, then x = y.

142

CHAPTER 3 Relations

Example. The natural ordering:::; on the set numbers, and a :::; band b :::; a, then a = b.

~

is anti symmetric: if a and b are real

Example. For any set A, the set inclusion relation for QJ>(A), ~ , is anti symmetric. This follows directly from the definition of set equality, since X ~ Y and Y ~ X means that X = Y. Example. For the set of natural numbers N, the relation D, where a D b iff a divides b, is anti symmetric. If a and b are natural numbers and a D b, then a :::; b. Likewise b D a implies b :::; a. From these we know that a = b. The relation "divides" on the set of all nonzero integers is not anti symmetric. In this case 1 divides - 1 and - 1 divides 1, but 1 *- - 1. We leave as an exercise to show that if R is antisymmetric, then x R y and x *- y imply x. That is, the only possible symmetry in an anti symmetric relation is that an object may be related to itself. An anti symmetric relation R is not necessarily reflexive. There may be no x such that x R x, or x R x may be true for some x and false for others. For example, in a military organization with the relation "can give orders to," no soldier can give orders to him or herself. It is important for a chain of command that this relation be anti symmetric, so that no two people can give orders to each other. We now define a partial ordering on a set A.

y,

DEFINITION A relation R on a setA is a partial order (or partial ordering) for A if R is reflexive, anti symmetric, and transitive. The set A is called a partially ordered set, or poset.

The following relations are partial orders: :::; on N, :::; on E, and:::; on on QJ>(A), for any set A, "divides" on N.

~,

~

On the set H of all people (living or deceased), the relation A given by x A y iff x = y or x is an ancestor ofy, is a partial order. Under this relation, two people may not be A -related to each other, but if x A y and y A x, then x

= y.

Example. The relation Won N given by x Wy iff x:::; y and x + y is even, is a partial order. For example, 2 W 4,4 W 6,6 W 8, ... , and 1 W 3,3 W 5,5 W 7, ... , but we never have m W n where one of the numbers m, n is odd and the other is even. We verify that W is a partial order as follows: Reflexivity: Let x E N. Then x + x = 2x is even and x :::; x, so x W x. Antisymmetry: Suppose x Wyand y W x. Then x + y is even, x :::; y, and y :::; x. By anti symmetry of :::; on N, x = y.

3.4

Ordering Relations

143

Figure 3.17

+ y is even, y :::; z and + z is even. By transitivity of:::; on N, X:::; z. Also, x + z is even because x + z = (x + y) + (y + z) + (-2y) is the sum of three even numbers. Therefore, x W z.

Transitivity: Suppose x Wyand y W z. Then x :::; y, x y

Example. Suppose A is a set and R is a reflexive, transitive relation. Further suppose that aRb, b R c, and eRa. A portion of the digraph of R is given in figure 3.17. The chain of relationships aRb, b R c, eRa is called a closed path (of length 3) in the digraph (see the next section for more about paths in graphs). It is closed because as we move from node to node along the path of relationships, we end up back at a. From aRb and b R c, we have aRc by transitivity. But eRa is also true, so we have both aRc and eRa, which means that R is not antisymmetric. Partial orders can never contain closed paths like the one shown in figure 3.17. More generally, the digraph of a partial order contains no closed path of length 2 or more.

Theorem 3.7

If R is a partial order for a set A and x R Xl, Xl

Xl

R X2, X2 R X3, ... , Xn R x, then x =

= X2 = X3 = ... = Xn-

Proof. (We prove this by induction on n.) For n = I, we have X R Xl and Xl R x. By anti symmetry, we conclude that X = Xl. Now suppose that whenever X R XI,Xl R XZ,X2 R x3, ... ,xk R x, then X = Xl = X2 = X3 = ... = Xk for some natural number k, and suppose that X R Xl, Xl R Xz, X2 R x3,· .. , xk R xk+ l> xk+l R x. By transitivity (applied toxk R Xk+ I andXk+1 R x) we have Xk R x. From X R xl> Xl R Xz, ... , xk R X and the hypothesis of induction, we have X = Xl = X2 = ... = Xk. Since Xk = x we have X R xk+ I and xk+ I R x, so X = xk+l. Therefore, X = Xl = X2 = ... = Xk+l· •

DEFINITION Let R be a partial ordering on a set A and let a, b E A with a 1= b. Then a is an immediate predecessor of b if aRb and there does not exist c E A such that a 1= c, b 1= c, aRc and c R b.

In other words, a is an immediate predecessor of b if aRb and no other element lies "between" a and b.

144

CHAPTER 3 Relations

Example. For A = {I, 2, 3,4, 5}, partially order 0'>(A) by ~, set inclusion. For a = {2, 3, 5}, there are three immediate predecessors: {2, 3}, {2, 5}, and {3, 5}. The empty set has no immediate predecessor. Also, 0 is the only immediate predecessor for {3 }. We have {4} ~ {2, 4, 5}, but {4 } is not an immediate predecessor of {2, 4, 5} because {4} ~ {4, 5} and {4, 5} ~ {2, 4, 5}. Let M = {I, 2, 3, 5, 10, 15, 30} be the divisors of 30, and let D be the relation "divides." D is a partial order for M whose digraph is given in figure 3.18(a). We can simplify the digraph significantly. First, since every vertex must have a loop, we need not include them in the digraph. Also, since there are no closed paths, we can orient the digraph so that all edges point upward; thus we may eliminate the arrowheads, assuming that each edge has the arrowhead on the upper end. We can also remove edges that can be recovered by transitivity. For example, since there is an edge from 2 to 10 and another from 10 to 30, we do not need to include the edge from 2 to 30. In other words, we need only include those edges that relate immediate predecessors. The resulting simplified digraph, figure 3.18(b), is called a Hasse diagram of the partial order D.

Example. Let us return to 0'>(A) ordered by ~, where A = {I, 2, 3}. A Hasse diagram for 0'>(A) is given in figure 3.19. It bears a striking resemblance to figure 3.18(b) for good reason. Except for the naming of the elements in the sets, the orderings are the same. In fact, it can be shown that every partial order is "the same" as the set inclusion relation on subsets of some set. Although we need the concepts of chapter 4 to make precise what we mean by "same," exercise 24 outlines how one might start to show this.

(b)

(a)

Figure 3.18

3.4

Ordering Relations

145

Figure 3.19

DEFINITIONS Let R be a partial order for A and let B be any subset of A. Then a E A is an upper bound for B if for every b in B, bRa. Also, a is called a least upper bound (or supremum) for B if (1) a is an upper bound for B, and (2) a R x for every upper bound x for B.

Similarly, a E A is a lower bound for B if for every b in B, aRb. Finally, a is called a greatest lower bound for B if (1) a is a lower bound for B, and

(2) for every lower bound x for B, x R a. We write sup (B) to denote the supremum of Band inf (B) to denote the infimum of B.

Example. Consider QP(A) with set inclusion as our partial ordering, where we let A = {I, 2, 3, 4, 5, 6, 7,8,9, In}. Let B = HI, 4, 5, 7}, {I, 4, 7, 8}, {2, 3, 4, 7}}. An upper bound for B is {I, 2, 3, 4, 5, 6, 7, 8}; another one is {I, 2, 3,4,5,8,9, In}. The least upper bound for B is sup (B) = {I, 2, 3,4,5,7, 8}. Both 0 and {4} are lower bounds for B. The greatest lower bound for B is inf (B) = {4, 7}. You should notice that sup(B) is the union of the sets in Band inf(B) is the intersection of the sets in B in the above example. This is true in general: for any set A with a collection of subsets ordered by set inclusion, and B ~ QP(A), sup (B) = X and inf (B) = X. See exercise 16. XE03

n

U

XE03

Example. Here are least upper bounds and greatest lower bounds for some subsets of IR with the usual ordering:::; : for A = [0,4), sup (A) = 4 and inf (A) = 0 for B = {I, 6, 3, 9, 12, -4, 1O}, sup (B) = 12 and inf(B) = -4

146

CHAPTER 3 Relations for e = {2k: kEN}, sup(e) does not exist and inf (e) = 2 for D = {r k : kEN}, sup (D) = and inf(D) does not exist

!

Example. Let A = {I, 2, 4, 5, 10,20, ... , 1000} be the set of all divisors of 1,000 and D be the relation "divides" on A. Let B = {la, 20, 25, lOa}. Both 500 and 1,000 are upper bounds for B; the least upper bound is 100. The greatest lower bound for B is 5. Note that for "divides," the least upper bound is the LCM (least common multiple) and the greatest lower bound is the GCD (greatest common divisor). Theorem 3.8

If R is a partial order for a set A, and B !: A, then if the least upper bound (or greatest lower bound) for B exists, then it is unique.

Proof. We prove this for sup(B) and leave the proof of the statement for inf(B) to the exercises. Suppose that x and yare both least upper bounds for B. (We prove that x = y.) Since x and yare least upper bounds, then x and yare upper bounds. Since x is an upper bound and y is a least upper bound, we conclude y R x. Likewise, since y is an upper bound and x is a least upper bound, we conclude x R y. From x R y and y R x, we conclude x = y by anti symmetry. We have seen examples of sets B where, when they exist, the least upper and greatest lower bounds for B may be in B or may not be in B.

DEFINITION Let R be a partial order for a set A. Let B !: A. If the greatest lower bound for B exists and is an element of B, it is called the smallest (or least) element of B. If the least upper bound for B is in B, it is called the largest (or greatest) element of B.

The usual ordering of the number systems has the property that, for any x and y, either x:::; y or y :::; x. A partial ordering with this property is called linear.

DEFINITION A partial ordering R on A is called a linear order (or total order) on A if for any two elements x and y of A, either x R y or y R x.

Examples. QP(A) with set inclusion, where A = {I, 2, 3} is not a linear order because the two elements {I, 2} and {I, 3} cannot be compared. Likewise, the relation "divides" is not a linear order for N because 3 and 5 are not related (neither divides the other). If R is a linear order on A, then by anti symmetry , if x and y are distinct elements of A, x R y or y R x (but not both). The Hasse diagram for a linear ordered set is a set of points on a vertical line.

3.4

Ordering Relations

147

Not every subset of a linearly ordered set has a smallest or largest element. The integers with::; is linearly ordered but B = {I, 3, 5, 7, ... } has neither upper bounds nor a least upper bound. Likewise, {-2, -4, -8, -16, -32, ... } has no greatest lower bound (and hence no smallest element.) The set of all integer multiples of 3 has no upper bounds and no lower bounds.

DEFINITION Let L be a linear ordering on a set A. L is a well ordering on A if every nonempty subset B of A contains a smallest element.

In chapter 2 we proved the Well-Ordering Principle from the Principle of Mathematical Induction. In our current terms, the WOP says that the natural numbers are well ordered by ::; . The integers, 7L, on the other hand, are not well ordered since, as we have seen, {- 2, -4, - 8, -16, - 32, ... } is a nonempty subset that has no smallest element.

Exercises 3.4 1.

* * *

Which of these relations on the given set are antisymmetric? (a) A = {I, 2, 3, 4, 5}, R = (cl, 3), (1, I), (2,4), (3,2), (5,4), (4, 2)}. (b) A = {I, 2,3,4, 5}, R = (cl, 4), (1,2), (2,3), (3,4), (5,2), (4,2), (I,3)}. (c) 7L, x R y iff x 2 = y2. (d) JR, x R y iff x::; 2Y • (e) A = {I, 2, 3, 4}, R is given in the digraph:

(f)

A = {I, 2, 3, 4}, R is given in the digraph:

IX] 4 ,

3

y imply y p x.

2.

Show that if R is antisymmetric, then x R y and x

3.

Give an example of a relation R on a set A which is anti symmetric and such that x R x for some, but not all, x in A.

=1=

148

CHAPTER 3 Relations

4.

Give an example of a relation S on a set A = {a, b, c, d} such that S is transitive, antisymmetric, and irreflexive (that is, x R x is false for all x in A).

5.

Show that the relation R on N given by aRb iff b = 2k a for some integer k ~ 0 is a partial ordering.

6.

Define the relation on JR X JR by (a, b) R (x, y) iff a:5 x and b:5 y. Prove that R is a partial ordering for JR X JR.

7.

Let C be the complex numbers. Define (a + bi) R (c + di) iff a 2 c 2 + d 2 . Is R a partial order for C? Justify your answer.

S.

Let A be a partially ordered set, called "the alphabet." Let W be the set of all "words" of length two-that is, all combinations of two letters of the alphabet. Define the relation :5 on Was follows: for XIX2 E Wand YIY2 E W, XIX2 :5 Y lYz iff (i) Xl :5 x2 or (ii) Xl = X2 and Yl :5 Y2. Prove that :5 is a partial ordering for W (called the lexicographic ordering, as in a dictionary).

9. For any poset A with a relation R, define the relation x Show that r is a transitive relation on the set A.

r Y

+ b 2 :5

iff x R Y and x

=1=

y.

10.

Draw the Hasse diagram for the poset QP(A) with the relation set inclusion, where A = {a, b, c, d}.

11.

Use your own judgment about which tasks should precede others to draw a Hasse diagram for the partial order among the tasks for each of the following complex jobs. (a) To make his special stew, Fubini must perform 9 tasks:

*

tl: t2: t3: t4: t5: t6: ti t8: t9: (b)

wash the vegetables cut up the vegetables put vegetables in cooking pot cut up the meat brown the meat in a skillet add seasoning to the skillet add flour to the skillet put the skillet ingredients in the pot cook the stew for 30 minutes

To back a car out of the garage, Kim must perform 11 tasks: t 1: put the key in the ignition t2: step on the gas t3: check to see if the driveway is clear t4: start the car t5: adjust the mirror t6: open the garage door t7: fasten the seat belt t8: adjust the position of the driver's seat t9: get in the car t 10: put the car in reverse gear tll: step on the brake

12.

For each Hasse diagram, list all pairs of elements in the relation on the indicated set.

3.4

*

(a)

A

= {a, b, c}

a~/

(b)

b

c

13.

* 14.

*

"*

15.

16.

A

b

= {a, b, c, d} ~d

~/ a

Ordering Relations (c)

A

149

= {a, b, c, d}

b~d,\

c

a

c

LetA be a nonempty set and let 0'>(A) be partially ordered by set inclusion. Show that (a) If BE 0'>(A) and x E B, then B - {x} is an immediate predecessor of B. (b) If BE 0'>(A) and x$. B, then B is an immediate predecessor of B U {x}.

3) 1

1(5,3)

Let H be the set of all squares having sides of (0, positive length that are within the rectangle (5,0) whose vertices are (0, 0), (5, 0), (0, 3), and (5, 3). (0,0) H is partially ordered by set inclusion, k . (a) Does every subset of H have an upper bound? a least upper bound? (b) Does every subset of H have a largest element? (c) Does every subset of H have a lower bound? a greatest lower bound? (d) Does every subset of H have a smallest element? Let K be the set of all squares that are within the L-shaped region at the right. K is partially ordered by set inclusion, k. (a) Does every subset of K have an upper bound? a least upper bound? (b) Does every subset of K have (c) Does every subset of K have (d) Does every subset of K have

(0,6)

I

1(4,6)

. L - - - - - - - . (10,3)

(4,3) (0, 0)

1

(10, 0)

a largest element? a lower bound? a greatest lower bound? a smallest element?

For A, a collection of subsets ordered by set inclusion, and B k A, prove that the least upper bound of B is B and the greatest ~lower bound for B is B.

U

BE~

17.

For a poset A with ordering Rand B k A, give a useful denial of (a) "a is the least upper bound of B." (b) "a is the greatest lower bound of B."

18.

For what sets A is 0'>(A) with set inclusion a linear ordering?

19.

Which are linear orders on N? Prove your answers. (a) T, where m Tn iff m < 2n (b) V, where m Vn iff m is odd and n is even, or m and n are even and m < n, or m and n are odd and m < n. (c) S = {em, n): m, n E N, m:::; nand m =1= 5} U {em, 5): mEN} (d) T = {em, n): m, n E N, m:::; nand n =1= 5} U {em, 5): mEN}

20.

Prove that each of these relations is a well ordering: (a) the relation Vin exercise 19(b). (b) the relation S in exercise 19(c).

n

BE~

150

CHAPTER 3 Relations

21.

In determining whether a given relation is a well ordering, it is not necessary to verify all the conditions for a linear order as well as the additional condition for a well ordering: (a) Prove that a partial order R on a set A is a well ordering iff every nonempty subset of A has a first element. (b) Prove that a relation R on a set A is a well ordering iff every nonempty subset B of A contains a unique element that is R-related to every element of B.

22.

Prove that every subset of a well-ordered set is well ordered.

23.

P is fine (a) (b) (c)

"* 24.

a preorder for a set A if P is a reflexive and transitive relation of A. Dea relation E on A by x E y iff x P y and y P x. Show that E is an equivalence relation on A. Now define a relation p on A/E by x P Yiff x P y. Show that p is a relation on A/E. (That is, prove that the definition of p does not depend on the representative x from x: if x E z and yEw, then x p y iffz P w.) (d) Show that p is a partial order.

(This exercise will set in place the pieces necessary to prove that every partial ordering is in a sense the same as the set inclusion relation on a collection of subsets of a set.) Let A be a set with a partial order R. For each a E A, let Sa = {x E A: X R a}, the section ofa. Let ~ = {Sa: a E A}. Then ~ is a subset ofQJ>(A) and thus may be partially ordered by C. (a) Show that if aRb, then Sa C Sb' (b) Show that if Sa C Sb, then aRb. (c) From (a) and (b), try to convince yourself that the Hasse diagram for A is

(d) Proofs to Grade

25.

*

the same as that for ~. Show that if B C A and x is the least upper bound for B, then Sx is the least upper bound for {Sb: bE B}, which is a subset of~.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Let A be a set with a partial order R. If C C Be A and sup( C) and sup (B) exist, then sup (C) ::5 sup (B). "Proof." sup (B) is an upper bound for B. Therefore, sup(B) is an upper bound for C. Thus sup (C) ::5 sup (B). • (b) Claim. Let A be a set with a partial order R. If B C A, u is an upper bound for B, and u E B, then sup(B) exists and u = sup (B). "Proof." Since u E B, u::5 sup (B). Since u is an upper bound, sup (B) ::5 u. Thus u = sup (B). • (c) Claim. For A, B C ~ with the usual ::5 ordering, sup (A U B) = sup (A) + sup (B). "Proof." If x E A U B, then x E A or x E B. Thus x::5 sup (A) or x ::5 sup (B). In either case, x::5 sup (A) + sup (B). Therefore, sup (A U B)::5 sup (A) + sup (B). On the other hand, both A C A U Band B C A U B, so by part (a) of this exercise, sup (A) ::5 sup (A U B) and sup (B) ::5 sup (A U B), which is enough to conclude that sup (A) + sup (B) ::5 sup (A U B). •

3.5

3.5

Graphs of Relations

151

Graphs of Relations We have seen several ways to represent relations on a set-as graphs of points corresponding to ordered pairs in a Cartesian coordinate system, in tables listing tuples, and as digraphs indicating ordered pairs by arrows from one point to another. In this optional section we present another way a relation may be represented, with something called simply a graph. Informally, a graph is a collection of points, called vertices, and lines, called edges, connecting some of the vertices. These edges are not directed as in a digraph, but we still may determine whether two vertices are related. Thus a graph is similar to, but not the same as, a digraph. The study of graphs has applications in computer science, statistics, operations research, linguistics, chemistry, genetics, electrical networks, and other areas. Graph theory is also a significant branch of mathematics in its own right. We begin with an example representing conversations at a party. The five people at the party are Doc, Grumpy, Sneezy, Dopey, and Happy. Rather than listing the ordered pairs in the relation S on this set of people defined by x S y iff x had a conversation with y, we describe the relation with a graph (figure 3.20). This graph has an edge connecting two vertices x and y exactly when x is S-related to y. It can be seen from the graph that Doc spoke with each of the others except Dopey and that Grumpy had a conversation only with Doc and Happy. The graph does not show anything about where the party-goers stood, how long they talked, or whether they had more than one conversation. Exactly the same information about whether two people did or did not have a conversation is conveyed by either of the graphs in figure 3.21. We say the three graphs in figures 3.20 and 3.21 are isomorphic. The word isomorphic is derived from Latin with iso meaning "the same" and morph meaning "form or structure."

Sneezy Grumpy ~--r--+--~ Doc

Dopey "'-_ _ _ _......;:>M'

Figure 3.20

Sneezy

Doc Happy

- = : : ; ' - - - - - - -.......

Grumpy

Figure 3.21

Happy ~===-----......::::::» Doc

152

CHAPTER 3

Relations Speaking more precisely, we say two graphs (V, E) and (V', E') are isomorphic iff there is a one-to-one correspondence between the vertices in Vand V' so that there is an edge in E joining two vertices of V exactly when there is an edge in E' joining the corresponding vertices of V'. Because the concept of a one-to-one correspondence is not discussed until the next chapter, we will not pursue the question of whether two given graphs are isomorphic. It is enough to know that a graph remains essentially the same graph when it is drawn with its vertices rearranged or even if the vertices are renamed. The purpose of this section is to discuss some properties of graphs and relate them to equivalence relations and partitions. Possibly because there are so many applications for graph theory, there is an even greater variation in the terminology for graphs than in most branches of mathematics. There is also variation in the types of graphs studied. Some types of graphs contain loops, i.e., edges from vertices to themselves. Multigraphs are graphs that allow more than one edge between two vertices. Such graphs are useful to represent, for example, alternate routes (edges) between cities (vertices). We will deal only with simple graphs, which do not allow multiple edges or loops. A much more complete treatment of graphs may be found in G. Chartrand, Introductory Graph Theory (1985, Dover Publications, Inc., New York).

DEFINITIONS A graph G is a pair (V, E) where V is a nonempty set and E is a set of unordered pairs of distinct elements of V. An element of V is called a vertex and an element of E is an edge.

An edge between vertices u and v is usually written as uv rather than as the set {u, v}. There is no direction implied in the notation uv; the expression vu represents the same edge. Vertices u and v are adjacent iff uv is in E. The edge uv is said to be incident with u and with v. The degree of a vertex u is the number of edges incident with u. Finally, the size of graph G(V, E) is the number of edges (#(E)) and the order of the graph is the number of vertices (#(V)). The order of our conversation graph is 5 and the size is 7. Doc has degree 3, meaning he held conversations with three other people. Dopey and Sneezy are adjacent, whereas Dopey and Doc are not. The definition of a graph G(V, E) allows for E to be empty. Such a graph is a null graph. Figure 3.22 is a null graph of order 5. Here, every vertex is isolatedhas degree zero. A graph in which every pair of distinct vertices are adjacent is called a complete graph. If G(V, E) has order n and is complete, then every vertex has degree n - 1. The complete graph of order n is denoted Kn- Figure 3.23 shows the complete graphs of order 5 and less. If you look again at figure 3.20, you will see that the degrees of the vertices are 3, 3, 4, 2, and 2. The sum of the degrees is 14, which is even. The explanation for the fact that the sum of the degrees of the vertices of a graph is even is the same as

3.5

Graphs of Relations

153

2- 3

1-

-4

5 -

Figure 3.22

-

2

2

l~

4~ Figure 3.23 the explanation for the fact that if a group of people shake hands, the total number of hands shaken must be even.

Theorem 3.9

(a) (b)

(The Handshaking Lemma). For each graph G, the sum of the degrees of the vertices of G is even. For every graph G, the number of vertices of G having odd degree is even.

Proof. (a)

(b)

Each edge is incident with two vertices. Thus the sum of the degrees of the vertices is exactly twice the number of edges. Therefore, the sum is even. Obviously, the sum of the degrees of the vertices that have even degree is an even number. If there were an odd number of vertices with odd degree, then the sum of all the degrees would be odd. •

For Km each vertex has degree n - I. Thus the sum of the degrees of the vertices is n(n - I). By Theorem 3.9(a), this number is twice the number of edges, so the number of edges in Kn is ~n(n - I).

154

D

CHAPTER 3 Relations

Dopey

Happy

ST

Sneezy

Doc

Happy

Dopey ~ Sneezy

HappY~DOC

Figure 3.24 A subgraph (VI, E I ) of the graph (Vz, E 2 ) is a graph such that VI ~ V2 and EI ~ E 2 · Since (VI' E I ) is itself a graph, the edges in EI must join vertices of VI' (An edge cannot terminate at a point that is not a vertex). Figure 3.24 shows three subgraphs of our graph of conversations. A walk in a graph G is a finite alternating sequence of vertices and incident edges, beginning and ending with a vertex. (By a finite sequence we mean that the objects listed could be labeled 1,2, 3, ... ,n, for some natural number n.) What we call a walk is called by some authors a path, an edge-sequence, a route, a trail, or a chain. A walk can be described by giving the sequence of adjacent vertices. A walk va, v], Vz, . .. , Vm is said to traverse the vertices in the sequence. We call Vo the initial vertex of the walk and v m the terminal vertex. The length of a walk is the number of edges in the walk. Listed below are some examples of walks in the graph shown in figure 3.25. WI:

6,3,5

w2:

3, 5

w3:

2,5,3,4,5

W4:

1,6,3,6,5,2, 1

ws:

1,2,3,4,5,6,1

The walk W3 originates at vertex 2, has length 4, and terminal vertex 5. A walk in which all vertices traversed are distinct is called a path. Both WI and W2 are paths. A walk that originates and terminates at the same vertex is said to be closed. Although all the vertices in a path must be distinct, we make an exception when the only repeated vertices in a closed walk are the first and last; we call this a closed path. A

2

6 5

Figure 3.25

3.5 Graphs of Relations

155

closed walk with distinct edges is called a cycle. The walk W4: 1,6,3,6,5,2, I is closed but is not a cycle. The walk Ws: I, 2, 3, 4, 5, 6, 1 is a cycle and also a closed path. The cycle W6: 5, 3, 6, 1, 2, 3, 4, 5 is not a closed path. If we think of a graph as representing a map of cities (vertices) and connecting roads (edges), then it would be important to a road inspector to find cycles in the graph, because the inspector would prefer to drive along roads only once. A salesman might look for closed paths, preferring to visit cities exactly once. It seems clear that if there is a way to drive from one city to another, then there ought to be a way that avoids going through the same city twice. The next theorem states this fact and tells us that the length of a path or closed path is limited.

Theorem 3.10

(a) (b)

If there is a walk originating at v and terminating at u in a graph G, then there is a path from v to u. If G has order n, then the length of a path in G is at most n - 1. The length of a closed path is at most n.

Proof. (a)

Suppose v, VI> Vz, ... , u is a walk from v to u, with v i= u. If the walk is not a path, then some vertex appears twice in the sequence. Let x be the first such vertex. Then the walk contains at least one closed walk of the form x, Vj' Vj+I>"" V m , x.

(b)

Delete the vertices Vj' Vj + 1> .•• , Vm' X from the sequence v, v I, vz, . .. , u. If the result is a path, we are done. Otherwise another such repeated vertex can be found and the deletion process repeated. Since we delete at least one vertex each time and there was a finite number of vertices in v, VI, Vz, ... , u, eventually no more vertices can be deleted, so this process must result in a path from v to u. In the case v = u, the same process is applied to delete all repetitions of vertices except the initial and terminal vertex. The result is a closed path. Consider a path in the graph G, where G has n vertices. If the path has length t, then there are t + I vertices traversed by the path. In a path that is not closed all vertices are distinct, so there are at most n vertices traversed. Thus t + 1 ::5 n and the path has length at most n - I. In a closed path the initial and terminal vertices are the same, and there is no other repeated vertex. Thus if the closed path has length t, there are t distinct vertices. Therefore, if G has n vertices, the length of a closed path is at most n. -

Let u be a vertex in graph G. The vertex v is said to be reachable (or accessible) from u iff there is a path from u to v. If v is reachable from u, then the number of edges in a path of minimum length from u to v is called the distance from u to v and denoted by d(u, v). For any vertex u, we say u is reachable from u and d(u, u) = O. A graph G is said to be connected if every vertex is reachable from every other vertex. Otherwise G is disconnected.

156

CHAPTER 3

Relations c ......- -.... q

h

'~ I

j

Figure 3.26 In the graph with vertex set V = {a, b, c, q, e,f, g, h, i, i, k} shown in figure 3.26, the vertex c is reachable only from the vertices c, q, g, h, i, and k. Thus the graph in figure 3.26 is disconnected. We have d(q,i) = 3, d(c, k) = 2, d(k, q) = 2, and d(c, q) = 1. If we think of a graph as representing cities and roads, then the graph of figure 3.26 might represent roads and cities on three islands. By our definition, every vertex of a graph is reachable from itself. Also, if v is reachable from u, then by reversing the order of edges in a path from v to u, we see that u is reachable from v. Finally, if v is reachable from u and w is reachable from v, then by following the path from u to v and then the path from v to w, we see that there is a walk and therefore a path from u to w. Thus reachability is an equivalence relation on a graph. Each equivalence class in the partition determined by the reachability relation is called a component of the graph. For a vertex v in a graph G(V, E), the component containing v is C(v) =

{u E vlu is reachable from v}.

Two vertices are in the same component iff one is reachable from the other. When a graph is pictured as in figure 3.26, determining components is a simple matter; they are connected subgraphs that stand out from the other components (the three "islands" as we referred to them): C(a) C(c) C(e)

= C(b) = Clf) = {a, b,f}, = C(g) = C(h) = C(j) = C(k) = C(q) = = C(i) = {e, i}.

{c, g, h,i, k, q},

The properties of components are collected in the next theorem, which tells us that the components of a graph are a collection of maximally connected subgraphs that form a partition of the graph.

Theorem 3.11

For each vertex v in a graph G, let C (v) be the component of v in G. That is, C (v) is a subgraph of G whose vertex set is all vertices reachable from v and whose edges are all edges of G joining these vertices. Then (a) (b) (c) (d)

C(v) = C(w) iffw is reachable from v. If C(v) oF C(w), then no vertex is in both C(v) and C(w). For each v, C(v) is connected. For each component C(v), if D(V', £1) is a subgraph of G and C(v) is a proper subset of V', then D is disconnected.

3.5

Graphs of Relations

157

Proof. (a), (b)

(c)

(d)

These parts follow from the fact that the vertex set of each component C(v) is the equivalence class of v under the reachability relation. See exercise 15. Let x and y be any two vertices in C(v). Then both x and yare reachable in G from v, so y is reachable from v, and v is reachable from x. Since all the edges needed to reach x and y from v are also in C(v), y is reachable from x in C(v). Therefore, C(v) is connected. We must show that if G' = (V', E') is a subgraph of G, and V' properly contains the vertices of any component C(v), then G' is disconnected. Suppose u is a vertex in V' that is not in C(v). Then u is not reachable from v in G, so there can be no path from v to u in G'. Thus G' is disconnected. -

Theorem 3.11 tells us that every vertex belongs to exactly one component, and that the collection of components is pairwise disjoint. Further, components are maximally connected subgraphs. It follows that every isolated point forms a component, and that a graph is connected iff it has exactly one component.

Exercises 3.5 1.

List the degrees of the vertices of each of these graphs. Verify Theorem 3.9 in each case. a

c

a

a

c

e

d

b

-----<E~--- d

e (b)

(a)

2.

If possible, give an example of a graph with order 6 such that

(a) (b) (c) (d) (e) (f)

3.

the vertices have degrees 1, 1, 1, 1, 1, 5 (a star graph). . the vertices have degrees 1, 1, 1, 1, 1, 1. the vertices have degrees 2,2,2,2,2,2. the vertices have degrees 1,2,2,2,3,3. exactly two vertices have even degree. exactly two vertices have odd degree.

If possible, give an example of a graph

(a) (b) (c) (d) 4.

d .....----~e (c)

with order 6 and with order 4 and with order 3 and with order 6 and

size 6. size 6. size 6. size 3.

Find all subgraphs of the accompanying graph G (a) with two vertices. (b) with three vertices.

158

CHAPTER 3 Relations 5.

The complement (; of a graph G(V, E) is the graph with vertex set V in which two vertices are adjacent iff they are not adjacent in G. Give the complements of these graphs.

(a)

z

•

•

(b)

(c)

(d)

6. For the graph shown, give (a) all paths of length 4 and initial vertex g. (There are eight.) all cycles of length 6 and initial vertex c. (There are eight.) a path oflength 7. a walk of length 4 that is not a path.

a

7. Give an example of a graph with 6 vertices having de-

e

(b) (c)

(d)

grees 1, 1,2,2,2,2 that is (a) connected.

*

c

(b)

g

disconnected.

8. Give an example of a graph with 6 vertices having (b) two components. (a) one component. (c) three components. (d) six components. 9. Prove that in every simple graph of order n 2:: 2 there are two vertices with the same degree. 10. Prove that every cycle has length 2:: 3. 11. Give an example of a graph with order 6 such that (a) two vertices u and v have distance 5. (b) for any two vertices u and v, d(u, v) :5 2. 12. Verify these properties for the distance between vertices in a graph: (a) d(u, v) 2:: 0 (b) d(u, v) = 0 iff u = v (c)

13.

d(u, w) :5 d(u, v)

+ d(v, w)

Let u and r be vertices in a graph such that d(u, r) a vertex w such that d(u, w) + (d(w, r) = d(u, r).

2::

2. Show that there exists

14. An edge e of a connected graph is called a bridge iff, when e is removed from the edge set, the resulting subgraph is disconnected. For example, the bridges in the graph below are {I, 2}, {2, 3}, and {5, 6}. Give an example of a connected graph of order 7 (a) with no bridges. (b) with one bridge. (c) with 6 bridges.

3.5 Graphs of Relations

Proofs to Grade

159

15.

Prove parts (a) and (b) of Theorem 3.11. Keep in mind that C(v) and C(w) are graphs; to be the same they must have the same vertices and the same edges.

16.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Every closed path is a cycle. "Prooj." Let x, Yl, Y20oo.,Yn' X be a closed path in a graph. Since no vertex other than the initial and terminal vertex is repeated, all edges are distinct. Therefore, X,Yl, Y20 ... ,Yn, X is a closed walk with distinct edges. That is, the closed path is a cycle. (b) Claim. If v and ware vertices in a graph such that d(u, v) = 3 and d(u, w) = 4, then d(v, w) ::5 7. "Prooj." Suppose d(u, v) = 3 and d(u, w) = 4. Then there is a path u, Xlo X20 v from u to v with length 3 and a path u, Ylo Y20 Y3, w from u to w with length 4. Then v, X2, x 10 U, Yl, Y2, Y3, W is a path from v to w with length 7. The distance from v to w is the length of the shortest path from v to wand there is a path of length 7, so d(v, w) ::5 7. -

"*

CHAPTER

4

Functions

The general notion of a function should be familiar to you through previous courses in algebra, trigonometry, and calculus. Intuitively, a function is thought of as a rule of correspondence between two sets that assigns to each object in the first set exactly one object from the second set. The statement "Distance is a function of time" means that there is a rule according to which the distance an object has traveled is associated with the time elapsed. The idea of a function as a rule is useful but not precise enough for a careful study. Unfortunately, the idea of a function as a rule gives the impression that a function must be given by a formula, which is not always the case. The word function was first used by G. W. Leibniz in 1694.1. Bernoulli defined a function as "any expression involving variables and constants" in 1698. The familiar notation f(x) was first used by Euler in 1734. It is only relatively recently that it has become standard practice to treat a function as a relation with special properties. This is possible because the rule that makes an object in one set correspond to an object from a second set may be thought of as producing ordered pairs. The basic properties of functions, some operations on functions, and induced set functions will be presented in this chapter and used throughout the remainder of the book.

4.1

Functions as Relations DEFINITION satisfies (i) (ii)

A function

f from A to B is a relation from A to B that

Dom (f) = A. If (x, y) Ef and (x, z) Ef, then y = z.

In the case where A = B, we say

f is a function on A. 161

162

CHAPTER 4

Functions

A function/from A to B may also be called a mapping of A to B. We write /: A ~ B, and this is read 'jmaps A to B" or 'jis a function from A to B." As required by the definition, the domain of/is A. The set B is called the codomain off. As with any relation, the range of/is Rng (f) = {v: (.3u)(u, v) E/}. No restriction is placed on the sets A and B. They may be sets of numbers, ordered pairs, functions, or even sets of sets of functions. LetA = {I, 2, 3}, B = {2, 5, 6}. The sets

= {(1, 2), (2, 5), (3, 6), (2, 6)} r2 = {(1, 2), (2,6), (3, 5)} r3 = {(l, 5), (2,5), (3, 2)}

rl

are all relations from A to B. Since (2, 5) and (2, 6) are ordered pairs in rl with the same first coordinate, and 5 =1= 6, rl is not a function from A to B. Both r2 and r3 satisfy the conditions (i) and (ii), so they are functions from A to B. Now consider the relation r4 = {(I, 2), (3, 6)}. Is r4 a function? Is it a function from A to B? The answer to the second question must be no; since Dom(r4) ={1, 3} =1= A, r4 is not a function from A to B. However, r4 does satisfy condition (ii) for functions, so the answer to the first question is yes; r4 is a function from { 1, 3} to B. Every relation that satisfies condition (ii) is a function from its domain to its range. Let G = {(x, y) E N X N: y = x + 2}. Then some of the ordered pairs in G are (5, 7), (32, 34), and (517, 519). There are no ordered pairs of the form (x, y) and (x, z) where y =1= z, because y = z = x + 2 is the only possible second coordinate in a pair with first coordinate x. Thus G is a function from N to N. Indeed, since Rng (G) = {3, 4, 5, 6,00 .}, G is a function from N to any set that includes {3, 4, 5, 6,00.}. Let H = {(x, y) E 7L X 7L: x 2 + y2 = 2}. Then H is not a function, since (1, 1) E Hand (1, -1) E H, in violation of condition (ii). To verify that a given relation/from A to B is a function from A to B, it must be shown that every element of A appears as a first coordinate of exactly one ordered pair in f The fact that each a E A is used at least once as a first coordinate makes Dom (f) = A; the fact that a is used only once fulfills condition (ii). It is condition (ii) of the definition that makes / into a rule of correspondence. The rule associated with/is that when (x, y) E/, then y is the unique object that corresponds to x. Having (1, 6) and (1, 6) in a function is allowed (since writing an object twice in a set adds nothing to the set), but having (1, 6) and (1, 5) in / is not allowed, for this would give us two "answers" to the question "What corresponds to I?" Consider the rule that assigns to the nth letter of the alphabet the number n. This rule determines a function containing, for example, the ordered pairs (B, 2), (H, 8), and (T, 20). The rule that assigns to each of the digits 1 through 9 the letter(s) of the alphabet that appear on the corresponding button on a standard telephone is not a function from {I, 2,00', 9} to the alphabet. The digit 1 is not in the domain because there are no letters on the 1 button, and the rule violates condition (ii) because J, K, and L are all assigned to 5. It is often useful to think of a function both as a rule (so long as the rule assigns to each element of the domain exactly one element of the

4.1

Functions as Relations

163

codomain) and as a set of ordered pairs (where the second coordinate of a pair is the one object associated with the first coordinate). Some authors use other words for the sets we have called the range and codomain. In particular, you may find texts that call B the range of a function from A to B, even when B is not the set of second coordinates of the function. In some settings' one says that f is a function from A to another set even when Dom (I) is a proper subset of A. It is worth noting that the definition of a function says a great deal about first coordinates but almost nothing about the second coordinates of ordered pairs. It may happen that some elements of the codomain are not used as second coordinates, or that some elements of the codomain are used as second coordinates more than once. The relation r3 above, for example, has Rng (r3) = {2, 5} oF B, and both (I, 5) and (2, 5) are pairs in r3' Functions that are "one-to-one" or map "onto" their codomains satisfy certain conditions on their second coordinates. They will be discussed in section 4.3.

DEFINITIONS Letf: A-+B. If (x, y) Ef, then we writef(x) = y, and say that y is the value of f at x. Also, y is the image of x under f, and x is a pre-image of y under f. Elements of A are sometimes called arguments of the function f.

Each argument of a function has exactly one image, but elements of the codomain may have several pre-images or none at all. A functionf: A -+ B may be expressed as f

= {(x,f(x»:

x E

A}.

Many functions may be described conveniently by specifying the domain and a rule. The function with domain N given by f(x) = ~~f, for example, contains the and (5, because f(l) = and the image of 5 is f(5) = ordered pairs (I, It is important to distinguish between the symbols f andf(x). Technically, it is incorrect to speak of "the function f(x)." The symbol f denotes a set of ordered pairs, whereas f(x) is simply an element of the range of f. Specifically,f(x) is the element of Rng (f) that corresponds to the element x of Dom (f). Thus for the expression f(x) to be meaningful, f must be the name of a function, x must represent an element of Dom (f), and f(x) must be thought of as an element of Rng (f).

-t)

t)

-t

t-

Example. Let F = {(x, y) E 71. X 71.: y = x 2}. Then F: 71. -+ 71.. The domain of F is 71., the codomain of F is 71., and Rng (F) = {a, 1,4,9, 16, 25, ... }. The image of 4 is 16, the value of F at - 5 is 25, F (12) = 144, and both 3 and - 3 are pre-images of 9. An element of the codomain that has no pre-image is 6. The function F is given by the rule F(x) = x 2 .

Example. The empty set 0 is a function on 0. Moreover, if f: A -+ B and anyone of f, A, or Rng (f) is empty, then all three are empty. (See exercise 12).

164

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Functions

We next present several examples of functions that commonly occur in mathematics. You should remember them by name.

Example. Let K: IR-+ IR be defined by K(x) = 3 for every x E IR. Then K = {(x, 3): x E IR}. This function has range {3}. The only element of the codomain IR that has a pre-image is 3, and every element of IR is a pre-image of 3. K is an example of a constant function-that is, it is a function whose range consists of a single element. Example.

Assume that a universe U has been specified, and that A I} by

~

U. Define

XA: U -+ {O,

if x EA if x E U -A. Then XA(X) is called the characteristic function of A. For example, if A = [1, 4), with the universe being the real numbers, then XA (x) = 1 iff 1 ::5 x < 4. Figure 4.1 is a graph of X[I, 4)'

Example. One generalization of the characteristic function is the step function. Suppose C(6 = {Co: t5 E ~} is a partition of a set A, each Co is an interval, and for each J E ~, bois in the set B. Define f: A -+ B by for x E Co, letf(x) = boo As an example, let A = [1, 5] with C I = [1, 2], C2 = (2, 4), and C3 = [4, 5]. Choose real numbers b l = 3, b2 = 4, b3 = 2. The graph of the corresponding step function is given in figure 4.2.

Example. The best-known step function is called the greatest integer function. (It may also be called the floor function.) It has domain IR and codomain E. For x E IR, G/(x)

=

[x]

= the largest integer n so that n ::5 X.

Specifically, [5.4] = 5, [fi] = 1, [10] = 10, [-t] = -1, [-n] = -4, and [-0.14] -1. The graph of this function looks like an infinite staircase. y 4

y

2

• -3 -2 -1

3

X[I,4)

-1

3

4

-

2

:)

2

C>---"""

5

-

x

2

Figure 4.1

3

Figure 4.2

4

5

=

4.1

Functions as Relations

165

Example. A function with domain N may be called an infinite sequence, or simRather ply a sequence. The function a given by a(n) = khas range {1,!,~,!, than writing a(n) = k it is more convenient to call a(n) the nth term of the sequence a, and to denote the nth term by an' Thus the 51 st term of the sequence a is a51 = 5\' This sequence of reciprocals may also be written as a = (1, ~, !, ... ). The sequence c = (-1, 1, -1, I, -I, 1, ... ) has nth term Cn = ( - I t A sequence may also be a constant function, as for example the sequence b, where bn = 7 for every n E N.

!, ... }. !,

Example. If R is an equivalence relation on the set X, then the function from X to X/R that sends each a E X to aiR is called the canonical map. As an example, let R be the relation of congruence modulo 5 on 7L, and let I be the canonical map. The images of 9 and - 3 are 1(9) = 9/R = { ... , -6, -1,4,9, 14, ... },

1(-3) = -3/R = {... , -8, -3,2,7, 12, ... }. The equivalence classes under the last-name relation L on the set P of all people with last names (see section 3.1) are sets of people all having the same last name. Under the canonical map I from P to P /L, every person corresponds to his or her equivalence class. Thus I (Charlie Brown) is the set of all people with last name Brown and/(Charlie Brown) = I(Buster Brown). The canonical map is a natural function to consider, and it plays an essential role in the development of many mathematical structures. Rules of correspondence between congruence classes have interesting features. Consider for example the equivalence classes {0/==4' 1/==4,2/==4' 3/==4} of 7L4 . Consider the rule I given by l(x/==4) = 2X/==IO' Under this rule,

1(0/==4) = 0/==10 1(1/==4) = 2/==10 1(2/==4)

=

4/==10

1(3/==4) = 6/==10' However, 0 and 4 are in the same equivalence class modulo 4 and, according to the rule, 1(4/==4) = 8/==10' That is, the rule actually assigns both 0/==10 and 8/==10 to the class 0/==4' Therefore, lis not a function. In cases where an object has more than one representative (as the set 0/==4 can be represented by 0/==4,4/==4' -4/==4' 8/==4"") and a supposed "function" assigns different corresponding values depending on the representative, we say that "the function is not well defined," meaning that it is not really a function. In fact, in this example, all of 0/==10' 8/==10' 2/==10,6/==10' and 4/==10' are the "images" of 0/==4'

166

CHAPTER 4

Functions

Examples. Let A be any set and, for each x E A, let fA (x) = x. Then fA is a function on A with range A, called the identity function on A. In chapter 3 the relation fA was called the identity relation on A. If A ~ B, then the function i: A --+ B given by i(x) = x for every x E A is the inclusion map from A to B. It is clear that i = {(x, x): x E A} = fA' but i is thought of as a function from A to B whereas fA is thought of as a function from A to A. By a real-valued function we mean a function whose codomain is a subset of In most applications we assume that the domain of a real-valued function is also a subset of ~. For functions whose domain and range are subsets of ~, the domain is often left unspecified. It is assumed then that the domain is the largest possible subset of the reals. For example, since is a real number iff x 2: -I, the rule G(x) = defines a function with unspecified domain [-1,00). Likewise, the domain of the function H given by H(x) = si~x is ~ - {k7r: k E Z}. For a functionfwhose domain is a collection of ordered pairs, we shall write f(x, y) instead of the correct but cumbersome f((x, y». For the multiplication function f: Z X Z --+ Z that associates the product xy with the pair of integers (x, y), we writef(x, y) = xy. Thusf(3, 2) = 6,1(-10, 4) = -40, etc. ~.

F+T

F+T

Example. For a relation S from A to B we may consider two projection functions, 7r I and 7rz, where

We define 7rl (a, b) = a for all (a, b) E S. Given an ordered pair, the action of 7rl is to pick the first coordinate of the pair. In a similar fashion, 7r2(a, b) = b. Figure 4.3 shows the images of (6,4) under 7rl and 7r2 on the relation ~ X ~. The projection function 7rl on a vertical line in ~ X ~ will send every point on the line to the x-intercept. All other lines are projected by 7rl to the horizontal axis (i.e., have the horizontal axis as range).

y 7T2(6,4) = 4

4

(6,4)

--------------, I I I I

3

2

I I I 7T,(6,4) = 6 --~--L--L--~--L--L--~~---L--~x

2

3

4

5

Figure 4.3

678

4.1

Functions as Relations

167

In determining whether two functions are equal, some authors consider the codomain. With this approach the inclusion map from { I } to { I, 2} is not the same function as the inclusion map from { 1} to { 1, 3 }. In some studies there are good reasons to distinguish between the function F: IR --+ IR given by F(x) = x 2 and the function G: IR--+ [0, 00) given by G(x) = x 2 , but we shall not adopt this approach. Since functions are sets, we already have a definition of function equality: f = g means f C g and g Cf. Since the functions F and G above are equal as sets, F = G. Consider two functions f and g where f(x) = and g(x) = x-I. These functions are not equal, because (-1, -2)Eg and (-1, -2)$.f. The functions f: {-2, 3} --+ {4, 9} given by f(x) = x 2 and g: {-2, 3} --+ N given by g(x) = x + 6 are equal because both are identical to the set {( - 2, 4), (3, 9)}. A very natural and useful way to express the idea that two functions are equal is to assert that they should have the same domain (so that they act on the same objects) and that for each object in the common domain the functions should agree.

:r;;/

Theorem 4.1

Two functions (i) (ii)

f and g are equal iff

Dom (f) = Dom (g), and for aU x E Dom (f),f(x) = g(x).

Proof. (We prove that conditions (i) and (ii) hold whenf = g. The converse is left as exercise 13.) Assume f = g. (i)

(ii)

Suppose x E Dom (f). Then (x, y) Ef for some y and, sincef = g, we have (x, y) E g. Therefore x E Dom (g). This shows Dom (f) C Dom (g). Similarly, Dom (g) C Dom (f); so Dom (f) = Dom (g). Suppose x E Dom (f). Then for some y, (x, y) Ef. Since f = g, (x, y) E g. Therefore,f(x) = y = g(x). •

Exercises 4.1 1.

*

*

Which of the foUowing relations are functions? For those relations that are functions, indicate the domain and a possible codomain. (a) R[ = {CO, 6), (6, D), (0, n), (n, U), (U, O)} (b) R2 = {(1, 2), (1, 3), (1,4), (1,5), (1, 6)} (c) R3 = (O, 2), (2, I)} (d) R4 = {(x, y) E IR X IR: x = siny} (e) Rs = {(x, y) E N X N: x :::=;y} (f) R 6 ={(x,y)EZXZ:y2=x} (g) R7 ={(x,y)EIRXIR:y=x 2 +2x+l}

168

CHAPTER 4

Functions (h) (i)

(j)

2.

* *

R 8 ={(x,y)ENXN:y2=x} R9 = {(l, 1), (2,2), (3,3), (4,3), (5, 3)} RIO = {(0, {0}), ({0}, 0), (0,0), ({0}, {0})}

Identify the domain, range, and a possible codomain for each of the following functions.

! l)}

(a)

{(X, y) E IR X IR: y =

(b)

{(x, y) E IR X

(c)

{(x,y)ENXN:y=x+5}

(d)

{(x,

(e)

{(X, y) EIR XIR: y = _I_} cosx

(f)

{(x, y) E IR X IR: y = xt\lx)}

(g)

{(x,y)ENXN:y=13}

(h)

{(X, y) E IR X IR: y =

(i)

{(X, y) E IR X IR: y = (~: -=- ~)}

(j)

{ (x, y) E E X E: y =

IR:

(x

y = x2

y) E IR X IR: y

+ 5}

= tan x}

(e'

+/-X)}

(~: -=- 2~)}

3.

Give a relation r from A = {5, 6, 7} to the set B = {3, 4, 5} such that (a) r is not a function. (b) r is a function, but not a function from A to B. (c) r is a function from A to B, with Rng (r) = B. (d) r is a function from A to B, with Rng (r) oF B.

4.

For the real functionfgiven by f(x) = x 2 - 1, (a) what is the image of 5 under f? (b) what is a pre-image of 15? (c) find all pre-images of 24. (d) what argument of f is associated with the value 20? (e) what is the value of f at -I? (f) what is a pre-image of -1O?

* 5.

*

Assuming that the domain of each of the following functions is the largest possible subset of IR, find the domain and range of (a)

f(x) = x 2 - 7x + 12 x-3

(b)

f(x) = 2x

+5

(c)

f(x) = _1-

(d)

f(x) =

~5

- x

(e)

f(x)

(f)

f(x)

~x+ n

= ~5 - x + ~x - 3

= ~x + 2 + ~-2 - x

4.1 6.

7.

*

"* 8.

*

9. 10.

*

169

Show that the following relations are not functions on IR. (a) {(x, y) E IR X IR: x 2 = y2} (b)

{(x,y)EIRXIR:x 2 +y2=1}

(c)

{(x, y) E IR X IR: x = cosy}

(d)

{(x, y) E IR X IR: y

= [x}

Give an example of a sequence x such that (a) all terms of x are negative. (b) foralln,O<xn<xn+I' (c) XI = 3, and for all n, Xn < Xn+1 < 4. (d) the range of x has exactly 3 elements. Which of the following are functions from the indicated domain to the indicated codomain? For those relations that are not functions, show that the rule is not well-defined by naming an equivalence class that is assigned two different answers.

= X/==6' = (x + 1)/==6' (c) I: Z.3 --+ Z.6, given by 1(x/==3) = (2X)/==6' (a) I: Z.3 --+ Z.6, given by I(X/==3) (b) I: Z.6 --+ Z.6' given by 1(x/==6)

(d) I: Z.4 --+ Z.6' given by I(X/==4) = (2x

*

Functions as Relations

+ 1)/==6'

I: Z.3 --+ Z.4' given by1(x/==3) = x/==4' (f) I: Z.4 --+ Z.z, given by I(X/==4) = (3X)/==2' (e)

Let the universe be IR and A (a) XA (b) Xt

= [1, 3). Sketch the graph of (c)

X{I/2}

(d)

XN

Let V be the universe and A ~ V with A oF 0, A oF U. Let XA be the characteristic function of A. (a) What is {x E V: XA(X) = I}? (b) What is {x E V: XA(X) = o}? (c) What is {x E V: XA(x) = 2}?

=

?

~ ~ x3

Explain why the functions I(x)

12.

(a) (b)

13.

Complete the proof of Theorem 4.1. That is, prove that if (i) Dom (f) = Dom (g) and (ii) for all x E Dom (f), I(x) = g(x), then 1= g.

14.

For the canonical map I: Z. --+ Z.6' find (a) 1(3) (b) (c) a pre-image of 3/==6 (d)

*

15.

*

and g(x)

= 3 - x are not equal.

11.

Prove that 0: 0 --+ 0. Prove that if I: A --+ B and anyone of f, A, or Rng (f) is empty, then all three are empty.

the image of 6 all pre-images of 1/==6

Let S be a relation from A to B. Let nl and n2 be the projection functions. In terms of S, find (a) Rng (nl) (b) Rng (n2)

170

CHAPTER 4 Functions 16.

A metric on a set X is a function d: X (i) (ii) (iii) (iv)

*

17.

* Proofs to Grade

18.

d(x, y) ;:::::

X X --+ IR

o.

such that for all x, Y, z EX,

d(x, y) = 0 iff x = y. d(x, y) = d(y, x). d(x, y) + d(y, z) ;::::: d(x, z).

Prove that each of the following is a metric for the indicated set. (a) X = N, d(x, y) = Ix - yl

=

!fl)

=

{o

X

(c)

X = IR X IR, d«x, y), (z, w»

(d)

X = IR X IR, d«x, y), (z, w»

if'I>,

d(x, y)

1

if x = y if X'F Y

(b)

= ~"(x-_-Z-)"2+-(y-_-w""""")2 = Ix - zl + Iy - wi

Suppose that #A = m and #B = n. We have seen that #(A X B) = mn and that there are 2 mn relations from A to B. Find the number of relations from A to B that are (a) functions from A to B. (b) functions with one element in the domain. (c) functions with two elements in the domain. (d) functions whose domain is a subset of A. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a)

The functions I(x) = 1 + 1. and g(x) = x + 1 are equal. x x "Prooj." The domain of each function is assumed to be the largest possible subset of IR. Thus Dom (f) = Dom (g) = IR - {a}. For every x E IR - {a} we have Claim.

1

I(x) = 1 + x

(b)

(c)

x +1 = x- + -1 = - = g(x). x

x

x

Therefore, by Theorem 4.1,/ = g. • Claim. If h: A --+ Band g: C --+ D, then hUg: A U C --+ BUD. "Prooj." Suppose (x, y) E hUg and (x, z) E hUg. Then (x, y) E h or (x, y) E g, and (x, z) E h or (x, z) E g. If (x, y) E h and (x, z) E h, then y = z. Otherwise, (x, y) E g and (x, z) E g; so again y = z. Therefore, hUg is a function. Also, Dom (h U g) = Dom (h) U Dom (g) = A U C, so hUg: A U C--+ BUD. • Claim. The rule that assigns to each equivalence class X/=4 in Z4 the class (x + 1)/=2 in Z2 is a function. "Prooj." Suppose (4=4' y/=2) and (4=4' Z/=2) are two ordered pairs in the relation determined by the nile. We must show that y/=2 = Z/=2· According to the rule, y/=2 = (x, + 1)/==2 and Z/=2 = (X2 + 1)/==2 for some x" X2 in the class X/=4· Since Xl and X2 are in the same equivalence class (mod 4), x, - X2 = 4k for some integer k. Therefore (Xl + 1) - (x2 + 1) =4k = 2(2k), sox, + 1 andx2 + 1 • are in the same equivalence class (mod 2). ThenY/=2 = Z/=2.

4.2

4.2

Constructions of Functions

171

Constructions of Functions In this section we consider several methods for constructing new functions from given ones. The operations of composition and inversion of relations were discussed in chapter 3. Since every function is a relation, the operations of composition and inversion are performed in the same way as they were for relations. Thus if F: A --+ B, then the inverse of the function F is the relation

F- 1 = {(x, y): (y, x) E F}. It is necessary to say the relation F- 1 because the inverse of a function is a relation, but might not be a function. Conditions under which F- 1 is a function will be given in the next section.

Example. For the function F = {(x, y): y = 2x to be real numbers), the inverse of F is

+ I} (where x and yare understood

F- 1 = {(x, y): (y, x) E F} ={(x,y):x=2y+

I}

={(X,y):y=(X;

I)}

which is also a function. However, for the function G = {(x, y): y = x 2 }, we have

G- 1 = {(x, y): (y, x) E G}

= {(x, y): = {(x, y):

x = y2} y = ±{x} which is not a function.

As we saw in chapter 3, if F: A --+ Band G: B --+ C, the composite of F and G is the relation Go F = {(x, z) E A X C: for some y E B, (x, y) E F and (y, z) E G}.

Example.

A composite of the two functions F and G, above, is the relation Go F = {(x, z): for some y E IR, (x, y) E F and (y, z) E G} = {(x,y): (.3yEIR)(y=2x+ I andz=y2)} = {(x, z): z = (2x + 1)2}.

We can take advantage of the fact that each element of the domain of a function has a unique image. This will greatly simplify the notation for composition of functions. For any functions Hand K, (x, z) E K 0 H iff for some y, (x, y) E Hand (y, z) E K. This can be restated as H(x) = y and K(y) = z. This means that (x, z) E K 0 H iff z = K(H(x)). The simplification, then, is that (K 0 H)(x) = K(H(x)). For example, the composite Go F for the functions given above could be computed as follows: (G

0

F)(x) = (G(F(x)) = G(2x

+ I) =

(2x

+ 1)2

172

CHAPTER 4

Functions

Notice that the first function applied in composition is the function on the right, which is closer to the argument variable.

Example.

+ 6x, then H(x 2 + 6x) =

If H(x) = sinx and K(x) = x 2 (H

K)(x) = H(K(x)) =

0

sin (x 2

+ 6x)

+6

sinx

and (K

0

H)(x) = K(H(x)) = K(sinx) = sin 2 x

The previous example shows that H 0 K and K composition of functions is not commutative.

0

H need not be equal. Thus

Example. In this example we consider functions on the sets &:.6 and &:'3' Denote by 0, 1, 2, 3, 4, Sthe six elements of &:.6 and by [0], [1], [2] the three elements of &:'3. Let H be the function from &:.6 to &:.3 given by H(O) = H(3) = [0], H(T) = H(4) = [1], and H(2) = H(S) = [2], Let K be the function (([0], [0]), ([1], [2]), ([2], [1])) from &:.3 to &:'3' Then K 0 H(4) = K(H(4)) = K([l]) = [2] and K 0 H(O) = K([O]) = [0]. Incidentally, notice that K- 1 is a function. In fact, K- 1 = K. The composite K 0 K is (([0], [0]), ([1], [1]), ([2], [2])), which is the identity function on &:'3'

Theorem 4.2

If F: A -+ Band G: B -+ C, then G 0 F: A -+ C. Thus the composite of functions is a function ~hose domain is the domain of the first function applied.

(d

0 F is a relation from A to C, and we know that Dom (G 0 F) !: A and Rng (G 0 F) !: c. To show G 0 F is a function from A to C, we must show (i) A!: Dom (G 0 F); and (ii) if(x, y) EGo F and (x, z) EGo F, then y = z.)

Proof.

(i)

(ii)

Suppose x EA. Since A = Dom (F), there is bE B such that F(x) = b. But B = Dom (G), so there is c E C such that G(b) = c. Then c = G(b) = G(F(x)) = (G 0 F)(x), so x E Dom (G 0 F). Therefore, A!: Dom (G 0 F). Assume (x, y) EGo F and (x, z) EGo F. Then there is u E B such that (x, u) E F and (u, y) E G; and there is v E B such that (x, v) E F and (v, z) E G. Since F is a function, (x, u) E F and (x, v) E F imply u = v. Then because G is a function and (u, y) E G and (v, z) = (u, z) E G, y = z. This shows G 0 F is a function.

It has already been proved in chapter 3 that composition of relations is associative, and this result applies to functions as well. Similarly, forming the composite of a function with the appropriate identity function yields the function again. These results are restated here especially for functions in order to emphasize their importance and demonstrate the use of functional notation in their proofs.

Theorem 4.3

The composition of functions is associative. That is, if f: A -+ B, g: B -+ C, and h: C -+ D, then (h 0 g) 0 f = h 0 (g 0 f).

Proof. (The idea behind this proof is the characterization of equal functions in Theorem 4.1. The proof also uses the result from Theorem 4.2 that the domain of a

4.2

Constructions of Functions

173

composite is the domain of the first function applied.) The domain of each function isA, by Theorem 4.2. Ifx EA,then((h 0 g) of)(x) = (h 0 g)(f(x» = h(g(f(x))) = h((g 0 f)(x» = (h 0 (g 0 f)(x). •

Theorem 4.4

Letf: A -+ B. Thenf

0

IA

=f

and IB

0

f

= f.

Proof. Dom (f 0 IA) = Dom (fA) = A = Dom (f). If x E A, then (f 0 IA)(x) = f (IA (x»

Theorem 4.5

= f (x). Therefore f

Let f: A-+ B with Rng (f) fof-I = Ie·

0

I A = f. The proof that!B 0 f

= f is left as exercise 4.

•

= c. If f- I is a function, then f- I 0 f = IA and

Proof. Suppose f: A -+ Band f- I is a function. Then Dom (f-I 0 f) = Dom (f) (by Theorem 4.2). Thus Dom (f) = A = Dom (fA)' Suppose x E A. From the fact that (x,f(x» Ef, we have (f(x), x) Ef-I. Therefore, (f-l 0 f)(x) = f-I(f(x» = x = I A (x). This proves that f- I 0 f = I A' The proof of the second part of the theorem is left as exercise 5. • It is sometimes desirable to create a new function from a given one by removing some of the original ordered pairs. You may recall having seen this done for the sine function with domain ~. When this function is restricted to the domain [T' ~], the result is usually referred to as the Sine function (with a capital S), abbreviated = sin = ~, but Sin, and is the principal branch of the sine function. Then Sin Sin (23lt ) is undefined.

m m

DEFINITION Letf: A -+ B, and let D !:;A. The restriction ofjto D, denoted f D, is

I

{(x, y): (x, y) Ef and xED}.

We note that the restriction of any function to D will be a function with domain D. Iff and g are mappings and g is a restriction off then we say f is an extension of g. In this notation the function Sin is sin I[-It/2,lt/2]' The graphs of sine and Sine are shown in figure 4.4. y

y

--T---~--~--~---T-+X

Figure 4.4

174

CHAPTER 4 Functions y 5 4 3

y

/

4 3

FI[I,2)

•

FI I -2,-1 0, 12)

2

2

-3 -2 -1

•

5

x

-1

-3 -2 -1 ·-1

-2

-2

2

3

4

•

-3

2

3

4

x

-3

Figure 4.5 Let A = {I, 2, 3, 4}, B = {a, b, c, d}, and g = (cl, a), (2, a), (3, d), (4, c)}. Then g I{2} = {(2, a)}, g IA = g, and g I{l, 4} = (cl, a), (4, c)}.

Example.

Let F: ~ -+ ~ be given by F(x) = 2x + 1. Figure 4.5 shows the graphs of F I[I, 2] and F I {-2, -1,0,1, 2}' Let hand g be functions. Is h n g always a function? Is hUg always a function? Consider, for example, h = (cl, 2), (5,7), (3, -9)} and g = (cl, 8), (5,7), (4, 8)}. Then h n g = {(5, 7)}, which is indeed a function. In general, if x is in the domain of both functions, and g(x) = hex) = y, then (x, y) E h n g. An object that is not in both domains or for which g(x) =1= hex) will not be in Dom (h n g). It turns out that h n g is a function (see exercise 8), but this function can just as easily be expressed by restricting the domain of either g or h. The situation regarding hUg is much more interesting and useful. First, for the functions given above, hUg is not a function because (l, 2) and (l, 8) are both in hUg. If we are careful to be sure that two functions hand g have disjoint domains, however, we can make a new function that is an extension of both hand g by putting them together "piecewise." The proof of Theorem 4.6, which states this result, is left as exercise 9. See exercise II for a generalization stating that hUg is a function when hand g agree on the intersection of their domains.

Theorem 4.6

Let hand g be functions such that Dom (h) hUg is a function with domain AU B.

= A, Dom (g) = B, and A n B = 0. Then

For example, if g: {I, 2, 3} -+ {a, b, c} is the function {(l, b), (2, a), (3, c)} and h={(4,d)}, then hUg: {I,2, 3,4}-+{a, b, c,d} is the function {(l, b), (2, a), (3, c), (4, d)}. Let hand g be given by hex) = x 2 and g(x) = 6 _. x. Then hi (-00, 2] and g (2, x)

I

have disjoint domains, so their union/is a function that is an extension of each. It is not an extension of h or g. See figure 4.6.

4.2

Constructions of Functions

175

y

Figure 4.6 The function

I can also be described as follows: if x::=; 2 if x> 2

I(x) = {x2 6-x

Functions can also be constructed piecewise from three or more functions. For example, if

K(x)

=

l

x+I sinnx

if x <-1 if -1 ::=; x::=; 0

+3 x-3

if 0 <x < 3

4

if x

x

2::

3,

then K is a function with domain IR (figure 4.7). Since each of the four rules used to define K are functions on their corresponding domains, to check that the relation given is a function, it is necessary to check only that the conditions given on the

y

4

•

3 2

Figure 4.7

K

176

CHAPTER 4

Functions

right are mutually exclusive. That is, once we verify (by a quick observation) that the four parts of the domain are pairwise disjoint, we can be sure that K is a function. The vertical line test, which says that a graph represents a function from the reals to the reals as long as no vertical touches the graph more than once, is useful as long as all the graph can be seen. It is not a rigorous proof that a given relation is a function. We have considered inversion, composition, restriction, extension, and union as means of constructing new functions from old ones. Other methods, especially for functions from the reals to the reals, may be found in exercises 13 and 17-19.

Exercises 4.2 1.

* * *

*

"*

Find fog and g ° f for each pair of functions f and g. Use the understood domains for f and g. (a) (b)

(c) (d) (e) (f)

f(x) = 2x + 5, g(x) = 6 - 7x f(x) = x 2 + 2x, g(x) = 2x + 1 f(x) = sinx, g(x) = 2X2 + 1 f(x) = tan x, g(x) = sinx f(x) = {(t, r), (s, r), (k, l)}, g(x) = {(k, s), (t, s), (s, k)} f(x) = {(l, 3), (2,6), (3,5), (4,2), (5, 2)}, g(x) = {(l, 5), (2,3), (3,7), (4,3), (5, 4)}

= ~: ~,

(g)

f(x)

(h)

f(x) = 3x

(i)

f(x) = {x2X+ 1

(J')

f(x)

+ 2,

g(x) = x 2 + 1 g(x) =

if x if x

= {x2X2 + 3

::5

Ixl 0

> 0'

if x < 3 if x:::::: 3'

g(x) = {2X - x

()

g x

if x if x

::5

{7 - 2x x + 1

=

-1

>- 1 if x ::5 2 if x> 2

2.

Find the domain and range of each composite in exercise 1.

3.

For which of the following functions f is the relation f- I a function? When f- I is a function, write an explicit expression for f- I (x). Use the understood domain for each function. (a) f(x) = 5x + 2 (b) f(x) = 2x 2 + 1 (c) f(x) = x + 1

*

*

(d) f(x) = sinx

1

f(x) = 1 -x

(g)

4.

*

e:+ 3

(e)

f(x) =

(h)

f(x) = -x

+3

x+2 I-x

(f)

f(x) = -

(i)

f() -x x =3x-4

-x

Prove the remaining part of Theorem 4.4. That is, prove that if f: A -+ B, then 18

0

f=f·

S.

Prove the remaining part of Theorem 4.5. That is, prove that if f: A -+ B with Rng (f) = C, and if f- I is a function, thenf ° f- I = 1c.

6.

Let f(x) = 4 - 3x with domain ~ and A = the functions f A, f [-I, 3], f (2, 4], and f

I I

I

{I, 2, 3, 4}. Sketch the graphs of

I{6}' What is the range of fiN?

4.2 Constructions of Functions 7.

*

8.

177

Describe two extensions oflwith domain IR for the function 1={(x,y)ENXN:y=x 2} (b) 1={(x,y)ENXN:y=3} 1={(x,y)E[-I, I]X[-I, I]:y=-x}

(a) (c)

Prove that, if I and g are functions, then I

n g is a function by showing that

In g = glA where A = {x: g(x) = I(x)}. 9.

'*

Prove Theorem 4.6.

10.

Letlbe a function with domain D, and let g be an extension of I with domain A. Then by definition, I = g I D and DCA. Let i be the inclusion mapping from D to A given by i(x) = x for all xED. Prove that I = g 0 i.

11.

Let h: A -+ B, g: C -+ D and suppose E = A n c. Prove hUg is a function from A U C to BUD if and only if hiE = g!e-

12.

For each pair of functions hand g, determine whether hUg is a function. In each case sketch a graph of hUg. (a) h: (-00, 0] -+ IR, hex) = 3x + 4

*

g: (0, 00) -+ IR, (b) (c)

(d) (e)

13.

h: g: h: g: h: g: h: g:

14.

* *

= .!.

x [-I,oo)-+IR, h(x)=x 2 + 1 (-00, -I]-+IR, g(x)=x+3 (-00, I] -+ IR, hex) = Ixl [0, 00) -+ IR, g(x) = 3 - Ix - 31 (-00,2]-+ IR, hex) = cosx [2, 00) -+ IR, g(x) = x 2 (-00, 3) -+ IR, hex) = 3 - x (0, 00) -+ IR, g(x) = x + 1

Let 11: IR ---+ IR and Iz: IR ---+ IR. Define the pointwise sum II +12 and pointwise product 11 . Iz as follows: 11 + 12 11 '/2

'*

g(x)

= {(a, c + d): (a, c) Ell and (a, d) EIz} = {(a, cd): (a, c) Ell and (a, d) E/2}'

(a) (b)

Prove II + 12 and II .Iz are functions with domain IR. Show that (fl + Iz)(x) = II (x) + 12 (x) and that

(c)

11 (x) . 12(x). Let I(x) = 2x + 5, g(x) = 6 - 7x, and hex) = 3x 2 - 7x (f + g)(x), (f . g)(x), (f + h)(x), and (g . h)(x).

(fl' Iz)(x) =

+ 2. Compute

Let / be an interval of the real line, and let I be a real-valued function with / C Dom (f). We say that I is increasing on / iff for all x, y E /, if x < y, then I(x) 1(Y)· Prove that (a) I is increasing on IR, where I (x) = 3x - 7. (b) g is decreasing on IR, where g(x) = 2 - 5x. (c) h is increasing on [0, 00), where hex) = x 2. (d)

I

is increasing on (- 3, 00), where i(x) = x - 31. x+

178

CHAPTER 4

Functions

(e)

g is decreasing on (-00, 3), where g(x)

= ~ ~ ;.

(f) h is decreasing on I, where hex) = -/(x) and I is increasing on 1. (g) lis increasing on I, wherel = go h, and g and h are increasing on I. (h) g is decreasing on I, where g = hoi, h is decreasing on I and I is increasing on I. (i) h is increasing on I, where h = I + g, and I and g are increasing on 1.

15.

*

16.

*

* 17.

Prove or give a counterexample: (a) If I is a linear function with positive slope, I is increasing on ~. (b) If I and g are decreasing functions on an interval I, then log is decreasing on I. (c) If I and g are decreasing functions on I, thenl 0 g is increasing on I. (d) If Dom (f) = ~ and I is increasing on the intervals [ - 2, -1] and [1, 2], then I is increasing on [-2,2]. (e) If I is decreasing on (-00, 0) and decreasing on [0, 00]' then I is decreasing on ~. (f) If I and g are increasing on ~, and h is decreasing on ~, then (f + g) + h is increasing on ~. Give two different examples of (a) a pair of functions I and g such that (f 0 g)(x) = (3x + 7)2. (b) a pair of functions I and g such that (f 0 g)(x) = ~2x2 - 5. (c) a pair of functions I and g such that (f 0 g)(x) = sin 12x + 41. (d) a function I that is neither increasing nor decreasing on [-1, 1]. (e) two quadratic functions I and g whose sum I + g is linear. Let I:

~ ---+ ~

and c E

~.

Define the scalar product cl by

cl = {(a, cd): (a, d) E/}. Prove c/: 18.

~ ....... ~

and find an explicit expression for (cf)(x).

Let I: A ---+ Band g: C ....... D. Define

I

Xg

= {«a,

c), (b, d»: (a, b) EI and (c, d) E

g}.

Provethatl X g: A X C---+ B X D. Find an explicit expression for (f X g)(a, c). 19.

Let Ii: ~ --+ ~ for i = 1,2'00" n. A linear combination of 11'/2,00''/n is a function of the form cdl + c2iz + 00. + cnln, where Ci E ~, i = 1,2, ... , n. Prove that a step function g: ~ ....... ~ with range {b l , b 2 , ••• , b n } is a linear combination of characteristic functions.

Proofs to Grade

20.

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Let I: A ....... B. If I-I is a function, then/- I 0 1= I A • "Proof." Suppose (x, y) E/- I 0 I. Then there is z such that (x, z) EI and (z, y) E/- I . But this means that (z, x) E/- I and (z, y) E/- I . Since I-I is a function, x = y. Hence (x, y) E/- I 0 I implies (x, y) E I A ; that

4.3

Functions That Are Onto; One-to-One Functions

179

is, f- I 0 f ~ fA' Now suppose (x, y) E fA' Since A = Dom (f), there is wEB such that (x, w) Ef. Hence (w, x) Ef- I. But (x, y) E fA implies x = yand so (w, y) Ef-I. But from (x, w) Efand(w, y) Ef- I, we have (x, y) Ef- I 0 f. This shows fA ~f-I 0 f. Therefore, fA = f- I 0 f. • (b) Claim. Iffandf- I are functions on A, andf 0 f = f, thenf = fA"Proof." Suppose f: A ---+ A and f- I: A ---+ A. Since f = f 0 f, f-I 0 f= f-' 0 (fo f). By associativity, we have f- I 0 f =(f-I 0 f) 0 f. This gives fA = fA 0 f. Since fA 0 f = f, we have fA = f· • (c) Claim. If f, g, and f- I are functions on A, then g = f- I 0 (g 0 f). "Proof." Using associativity and Theorems 4.4 and 4.5,

f- I (d)

0

(g

Claim.

0

f) = f- I

0

(f

0

g) = (f-I

0

f)

0

g = fA

0

g = g.

•

If f'(x) > 0 on an open interval (a, b), then f is increasing on

(a, b).

"Proof." Assume thatf'(x) > 0 on the interval (a, b). Suppose XI and X2 are in (a, b) andxI < X2' We must show thatf(xI) X2)' By the Mean Value Theorem, there exists c in (x I, x2) such that f(X2) - f(x,) = f'(c). X2 -XI

Therefore, f(X2) - f(xI) = f'(C)(X2 - XI)' By hypothesis f'(c) > 0, and X2 - XI > 0 since XI < X2' Therefore f(X2) - f(xI) > O. We conclude thatf(xI)
4.3

Functions That Are Onto; One-to-One Functions For every function f: A ---+ B, Rng (f) ~ B. It must not be assumed that Rng (f) = B. In the case when the codomain and range are equal, we say the function maps onto the codomain.

DEFINITION A function f: A ---+ B is onto B iff Rng (f) = B. We write f: A onto,B. A function that maps onto its codomain is also called a surjection.

Since Rng (f) ~ B is always true, f: A ---+ B is a surjection if and only if B ~ Rng (f). Thus f: A~B iff (Vb E B)(3 a E A)(f(a)

= b).

Whether a functionfmaps onto its codomain or not, it is still correct to say f maps to the codomain. It is impossible to tell whether a function "is onto" unless we

180

CHAPTER 4

Functions

know the codomain, just as we can't tell whether {-2, 2} is a subset unless we know the set to which {-2, 2} is being compared. When the codomain is already clear from the context, or when we want to stress that the codomain we have in mind is Rng (I), it is common practice, even if not perfectly grammatically correct, to say that "f is onto." Up to this point, there has been little need to emphasize the codomain in the definition of a function. After all, it is understood (see the discussion preceding Theorem 4.1) that the functions f: N --+ N, where fen) = 2n and g: N --+ Z, where g(n) = 2n are equal. Here Rng (I) = Rng (g) = E, the set of even natural numbers, and any set that includes E is a perfectly acceptable codomain. When we discuss the function h: N --+ E given by hen) = 2n, we can say that every element of the codomain is an image. That is, h is a surjection. As an immediate consequence of the definition, every function maps onto its range. That is, if we have a function given as a set of ordered pairs and we wish to be sure the function "is onto," we can do so by choosing the range as the codomain. To prove that a given functionf: A --+ B is onto B, we choose an arbitrary y E B. We then show y E Rng (f) by showing there is x E A such that f (x) = y. This shows B <::;:: Rng (I), which we have noted earlier is sufficient to satisfy the definition.

Example. Let F: IR --+ IR be defined by F(x) = x + 2. We will show that F is onto IR by showing IR <::;:: Rng (F). To do this, we show that for every wEIR, there is x E IR such that F(x) = w. (This x must be chosen so that w = x + 2.) We choose x = w - 2, for with this choice, F(x) = F(w - 2) = (w - 2) + 2 = w. Hence F is onto IR. Let G: IR --+ IR be defined by G(x) = x 2 + 1. Then G is not onto IR. To show this, we find an element y in the codomain IR that has no pre-image in the domain IR. Let y be -2. Since x 2 + 1 ~ 1 for every real number x, there is no x E IR such that G(x) = -2. Hence G is not onto IR.

Example.

Example. Let f: IR-+ IR be the polynomial function f(x) = x 3 + 3x 2 - 24x (figure 4.8). To prove f is onto IR, let wEIR. We must show the existence of a E IR such thatf(a) = w. The equation f(x) - w = 0 is a third-degree polynomial equation in one variable x. Since its degree is odd and since complex, nonreal roots occur in conjugate pairs, the polynomial f(x) - w has an odd number of real roots. Let a be a real rootto f (x) - w = O. Then f (a) -;- W = 0 and hence f(a) = w. Thus f is onto IR. . We will show that f: Z X Z --+ Z given by f(x, y) = xy is a surjection. Let z E Z. We choose (z, 1) E Z X Z. Then

Example.

fez, 1) = z . 1 = z.

4.3

Functions That Are Onto; One-to-One Functions

181

-.~--------~-----+---L----~x

Figure 4.8 Thus f is a surjection. (Even though some integers have many pre-images (for example f(3, 8) = f( -2, -12) = f(6, 4) = 24), to prove thatfmaps onto 7L we need only one pre-image for each integer.) be defined by F(m, n) = 2m- 1 (2n - 1). To show that F is onto N, let sEN. We must show that there is (m, n) E N X N such that F(m, n) = s. If s is even, then s may be written as 2kt, where k::::: 1 and t is odd. Since t is odd, t = 2n - 1 for some n E N. Choosing m = k + 1, we have F(m, n) = 2m-l(2n - 1) = 2kt = s. If s is odd, then s = 2n - 1 for some n E N. Forthis nand m = 1, we find F(m, n) = 2o(2n - 1) = s. Therefore, Fis onto N.

Example. Let F: N

X N--+ N

The next two theorems relate composition and the property of being a surjection.

Theorem 4.7

If f: A onlO'B, and g: B~C, then g 0 f: A ~ C. That is, the composite of surjective functions is a surjection.

Proof.

Theorem 4.8

Exercise 4.

•

If f: A ----+ B, g: B ----+ C, and g 0 f: A ~ C, then g is onto C. That is, when the composite of two functions maps onto a set C, then the second function applied must map onto C.

Proof. (We must show C!: Rng (g).) Suppose c E C. Since go f maps onto C, there is a E A such that (g 0 /)(a) = c. Let b = f(a), which is in B. Then (g o/)(a) = g(j(a)) = g(b) = c. Thus there"is b E B such that g(b) = c, and g maps onto C. • For a relation f to be a function from A to B, every element of A must appear exactly once as a first coordinate. No restrictions were made for second coordinates, except that they be elements of B. When every element of B appears at least once as a second coordinate, we have called the function onto B. Functions for which every element of B appears at most once as a second coordinate are called one-to-one. DEFINITION A function f: A --+ B is said to be one-to-one, written as f: A..!..=.lB, iff (x, y) Ef and (z, y) Ef imply x = z. That is, f is one-to-one iff f(x) = fez) =:} x = z. A one-to-one function is also called an injection.

182

CHAPTER 4 Functions

To prove directly that a given function f: A --+ B is one-to-one, we assume x and z are elements of A such that f(x) = fez). We then show that x = z. An alternative, using the contrapositive, is to assume that x =1= z and to show that f(x) =1= fez). To show thatfis not one-to-one, it suffices to exhibit two different elements of A with the same image. The function F: IR --+ IR defined by F(x) = 2x + 1 is one-to-one. To show this, assume F(x) = F(z). Then 2x + 1 = 2z + 1. Therefore, 2x = 2z, sox = z.

Example.

Example. Let G(x) = G(x) = G(y). Then

+1' We attempt to show G is an injection by assuming that +

x

x2 + 1-

y2

+

l'

Therefore, x 2 + 1 = y2 + 1, so x 2 = y2. It does not follow from this that x = y. In fact, this failed "proof' suggests a way to find distinct real numbers with equal images. Indeed, G(3) = G( - 3) = Therefore, G is not injective.

10.

Example. Let f: [0, (0) --+ [0, (0) be the function given by f(x) = x 2 . To show that f is one-to-one, suppose f(x) = f(y). Then x 2 = y2, so X = ±y. Since both x, y E Dom (f), x 2: and y 2: 0. Thus from x 2 = y2 we can conclude x = y. There-

°

fore, f is one-to-one. This example shows that the domain may be quite important in determining whether a function is one-to-one. Define F: N X N --+ N by F(m, n) = 2m- I(2n - 1). We will show F is one-to-one. Assume that F(m, n) = F(r, s). We first prove that m = r. We may assume thatm 2: r. (/fm < r, we could relabel the arguments.) From F(m, n) = F(r, s), we have 2m- I(2n - 1) = 2 r- I (2s - 1), which implies (;::~11)(2n - 1) = 2s - 1. Therefore, 2(m-I)-(r-I)(2n - 1) = 2s - 1; that is, 2m- r(2n - 1) = 2s - 1. Since the right side of the equality is odd, the left side is odd. Thus 2m - r = 1. Therefore, m - r = 0, and we conclude that m = r. Dividing both sides of the equation 2m- I (2n - 1) = 2'-1(2s - 1) by 2m - 1 (2 m - I = 2 r- I ), we have 2n - 1 = 2s - 1, which implies 2n = 2s, or n = s. Thus m = rand n = s, which gives (m, n) = (r, s). Hence the function F is one-to-one.

Example.

It was observed in the previous section that the inverse of a function is not always a function. The situation is clarified when we understand the connection between inverses and one-to-one functions.

Theorem 4.9

Let F: A --+ B. F- 1 is a function from Rng (P) to A iff F is one-to-one. Furthermore, if F- 1 is a function, then F- 1 is one-to-one.

Proof. Assume that F: A..!..=.!.B. To show that F- 1 is a function, assume (x, y) E F- 1 and (x, z) E F- 1. Then (y, x) E F and (z, x) E F. Since F is one-toone, y = z. Therefore if F is one-to-one, F- 1 is a function. By Theorem 3.2(b), the domain of F- 1 is Rng (P).

4.3 Functions That Are Onto; One-to-One Functions

183

Assume now that p-I is a function. To show that P is one-to-one, assume that (x, y) E P and (z, y) E P. Then (y, x) E p- I and (y, z) E p-I. Since p-I is a function, x = z. Therefore, if p-I is a function, then P is one-to-one. The proof that if P and p-I are functions, then p-I is one-to-one, is left as exercise 5.

•

We must be careful not to conclude that if P: A.!..=..!.B, then p-I: B I-I,A, since p may not be onto B. Recall that the domain and range of a relation and its inverse are interchanged. Therefore if P: A.!..=..!.B, then p-I: Rng (P)'!"="!'A.

Corollary 4.10

If P: A ~:d B, then p-I: B ~:d A. That is, the inverse of a one-to-one and onto function is a one-to-one and onto function.

Theorem 4.11

If I: A.!..=..!.B and g: B.!..=..!.C, then g functions is an injection.

0

I: A.!..=..!.C. That is, the composite of injective

Proof. Assume that (g 0 f)(x) = (g 0 f)(z); that is, g(f(x)) = g(j (z)). Then I(x) = I(z) since g is one-to-one. Then x = z since lis one-to-one. Therefore, g 0 I

•

is one-to-one.

There exist functions that are one-to-one but not onto, onto but not one-to-one, neither, and both (see the exercises). A function that is both one-to-one and maps onto its codomain is called a one-to-one correspondence or a bijection.

Example. LetA = {a, b, c} and B = {p, q, r}. The function I = {(a, p), (b, r), (c, q)} is a bijection from A onto B. Example. In previous examples we have seen that F: N X N --+ N given by P(rn, n) = 2m - I (2n - 1) is both one-to-one and onto. Thus Pis a one-to-one correspondence between N

X

Nand N-a fact we will find useful in chapter 5.

Combining Theorems 4.7,4.10, and 4.11, we have the following theorem.

Theorem 4.12

If I: A --+ Band g: B --+ C, and each is a one-to-one correspondence, then (a) (b)

go I: A --+ C is a one-to-one correspondence. I-I: B --+ A is a one-to-one correspondence.

Analogous to Theorem 4.8 for surjections, we have the following theorem for injections.

Theorem 4.13

If I: A --+ B, g: B --+ C, and g 0 I: A.!..=..!.C, then I: A.!..=..!.B. That is, if the composite of two functions is one-to-one, then the first function applied must be one-to-one. Proof. Exercise 6.

•

184

CHAPTER 4

Functions

We consider now a result that relates the concepts of injection, surjection, composition, and inversion. It gives a simple, practical method using composition to determine whether a given function is the inverse of a function F, thus indirectly verifying that F is one-to-one and maps onto its codomain.

Theorem 4.14

Let F: A ---+ Band G: B---+ A. Then G = F- I iff Go F = IA and FoG = lB' If G = F- I , then G 0 F = IA and FoG = IB' by Theorem 4.5. (We use the fact that Rng (F) = Dom (F-I) = B.) Assume now that G 0 F = IA and FoG = lB' Then F is one-to-one by Theorem 4.13 and F maps onto B by Theorem 4.8. Thus F- I is a function on Band

Proof.

F- I

= F- I

IB

0

= F- I

0

(F

0

G) = (F-I

0

F)

0

G = IA

0

G = G.

•

Theorem 4.14 is most useful in cases where we wish to check whether a function G is the inverse of a given function F, and we already know that F is a one-toone correspondence from A to B. (Often, F: JR ~:d IR). In this case, the check is even easier, because we need to verify only one of Go F = lA' or FoG = IB to show that G = F- I . (See exercise 12(b).) 2x + 1, and let G(x) = We calculate the two composites

Example. Let F(x)

(G

0

=

F)(x)

x; .Then F: JROiii(jJRand G: JROiii(jJR. I

I-I

I-I

= G(F(x» = G(2x + 1) = [(2x + 1) - 1] 2

_ 2x_

-"2- x . (F

0

G)(x)

= F(G(x» = F[(X; 1)] = 2[(X; 1)] + 1 = (x - 1)

+ 1 = x.

Therefore, G 0 F = I IR and FoG = I IR' Either computation implies that G=F- I •

The function H (x) = {x22 + ~ -I - X We show that H = K, where

Example.

K(x)

={

~ff x ::; 00 maps JR one-to-one and onto JR. 1 x>

-~X-2 ifx2:2 ~2 -x

if x

<2

There are two cases to consider. If x::; 0, (K 0 H)(x) = K(H(x» = K(x 2 + 2). Since x 2 + 22:2, the value of K is _~(x2 + 2) - 2 = = -Ixl = x when x::; 0. If x> 0, (K 0 H)(x) = K(H(x» = K(2 - x 2 ). Since 2 - x 2 < 2, the value of K is (2 - x 2) = = x if x> 0. In either case, (K 0 H)(x) = x, so K = H- I .

-p

P-

P

Constructions of functions by restrictions and unions can also be related to injective and surjective properties of functions. These results will be used in the study of cardinality in chapter 5.

4.3 Functions That Are Onto; One-to-One Functions

Theorem 4.15

185

A restriction of a one-to-one function is one-to-one. If h: A~C, g: B~D, and A n B = 0, then hUg: AU Bonto,C U D. ]-] ]-] Ifh: A--C,g: B--D,A nB=0,and C nD = 0, then hUg: A U B2.::..2.C U D.

(a) (b) (e)

Proof.

Parts (a) and (b) are left as exercise 7. ]-]

]-]

Suppose h: A--C, g: B--D, A n B = 0, and C n D = 0. Then by Theorem 4.6, hUg is a function with domain A U B. Suppose x, yEA U B. Assume (h U g)(x) = (h U g)(y).

(e)

(i) (ii) (iii) (iv)

If x, yEA, then hex) = (h U g) (x) = (h U g)(y) = hey). Since h is one-to-one, x = y. If x, y E B, then g(x) = g(y), and g is one-to-one; so x = y. Suppose x E A and y E B. Then hex) = g(y) and hex) E C and g(y) ED. But C n D = 0. This case is impossible. Similarly, x E Band yEA is impossible.

•

In every possible case, x = y. Therefore, hUg is one-to-one.

Exercises 4.3 1. Which of the following functions map onto their indicated codomains? Prove

* *

*

each of your answers. (a) f: IR ....... 1R,f(x) = !x + 6 (e) f: N ....... N X N,f(x) = (x, x) (e) f: 1R ....... 1R,f(x) = ~x2 + 5 (g) f: IR ....... 1R,f(x) = sin x (i) f: IR ....... [-l, I]'f(x)=cosx

2.

3.

*

(f)

(h) (j)

f: f: f: f: f:

7l.. ....... 7l..,f(x) = -x + 1,000 1R ....... 1R,f(x) = x 3 1R ....... 1R,f(x) = 2x IR X 1R ....... 1R,f(x, y) = x - y IR ....... [I, oo),f(x) = x 2 + I

x-2

(k) f: [2, 3) ....... [0, (0), f(x) = 3 _ x (I)

"*

(b) (d)

x f: [I, (0) ....... [I, oo),f(x) = x _

r

Which of the functions in exercise I are one-to-one? Prove each of your answers. Let A = {I, 2, 3, 4}. Describe a codomain B and a functionf: A ....... B such that is (a) onto B but not one-to-one. (b) one-to-one but not onto B. (e) both one-to-one and onto B. (d) neither one-to-one nor onto B.

f

•

onto

onto

onto

4.

Prove that If f: A--B, and g: B--C, then go f: A--C. (Theorem 4.7)

5.

Prove the remaining part of Theorem 4.9. That is, prove that if F and F-] are functions, then F -] is one-to-one.

6.

Prove that if f: A ....... B, g: B ....... C, and g rem 4.13)

]-]

0

]-]

f: A--C, then f: A--B. (Theo-

186

CHAPTER 4 Functions

7.

Prove parts (a) and (b) of Theorem 4.15.

8.

Give an example of functions I: A ....... Band g: B ....... C such that (a) I is onto B, but g 0 I is not onto C. (b) g is onto C, but g 0 I is not onto C. (c) go I is onto C, but I is not onto B. (d) I is one-to-one, but g 0 I is not one-to-one. (e) g is one-to-one, but go I is not one-to-one. (f) go I is one-to-one, but g is not one-to-one.

*

* 9.

Prove that 2- x = { ~

(a)

I(x)

(b)

I(x) =

{~~ 4 x-4

*

(e)

f(x)

~ {1 =~

(d) I(x) =

10.

* 11.~

* * 12.

{Ixl x-3

if x ::5 1 if x> 1 is one-to-one but not onto ~. if x ::5 -2 if -2 < x < 2 if x;::: 2 if x

*4

is onto

~

but not one-to-one.

is one-to-one and onto

~.

if x =4 if x ::5 2 if x> 2

is neither one-to-one nor onto

~.

Let A and B be sets and S ~A X B. Let nl be the projection function on S to A and n2 be the projection function on S to B. Give an example to show that (a) nl need not be one-to-one. (b) nl need not be onto A. (c) n2 need not be one-to-one. (d) n2 need not be onto B. Suppose S: A ....... B is a function. Then S ~ A X B. Let nl and n2 be as in exercise 10. (a) Is nl: S ....... A onto A? (b) Is nl: S ....... A one-to-one? (c) Is n2: S ....... B one-to-one? (d) Is n2: S ....... B onto B? (a) (b)

Give an example of sets A and B and functions F and G such that F: A ....... B, G: B ....... A, G 0 F = lA, and G F- 1• (See Theorem 4.14.) Suppose further that F: A~:dB. Prove that if either Go F = IA or FoG = I B , then F- 1 = G.

*

13.

Prove that if the real-valued function I is increasing (or decreasing), then I is one-to-one.

14.

Prove that

*

I: 1"4 ....... 1"8 given by 1(x/=4) = (2X)/=8 is an injection, but not a bijection. (b) I: 1"4 ....... 1"2 given by 1(x/=4) = (3X)/=2 is a surjection, but not a bijection. (c) I: 1"6 ....... 1"6 given by I(X/=6) = (x + 1)/=6 is a bijection. (d) I: 1"4 ....... 1"4 given by I(X/=4) = (2X)/=4 is neither an injection nor a surjection. (a)

4.3 Functions That Are Onto; One-to-One Functions

187

15. Give two examples of a sequence x of natural numbers (i.e., a function with domain N and range that is a subset of N) such that (a) x is neither one-to-one nor onto N. (b) x is one-to-one and onto N. (c) x is one-to-one and not onto N. (d) x is onto N and not one-to-one.

*

16. Suppose #A = m and #B = n. Then there are 2 mn relations from A to Band nn! functions from A to B (see exercise 17 in section 4.1). Find the number of one-to-one functions from A to B, assuming that (a) m < n. (b) m = n. (c) m > n. Find the number of functions from A onto B, assuming that (d) m = n. (e) m> n. (f) m = n + I. (g) Find the number of functions from A to B, assuming that m < n. (h) Find the number of bijections from A to A. Check that your answer is consistent with your answers for parts (b) and (d).

*

*

Proofs to Grade

17.

*

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. 1-1 [-I [-[ (a) Claim. Iff: A---->B and g: B---C, then go f: A---C (Theorem4.11). "Proof." We must show that if (x, y) and (z, y) are elements of go f, then x = z. If (x, y) Ego f, then there is u E B such that (x, u) Ef and (u, y) E g. If (z, y) Ego f, then there is v E B such that (z, v) Ef and (v, y) E g. However, (u, y) E g and (v, y) E g imply u = v since g is oneto-one. Then (x, u) Ef and (z, v) Ef and u = v; therefore, x = z, since f is one-to-one. Hence (x, y) and (z, y) in g 0 f imply x = z. Therefore, g 0 f is one-to-one. • (b) Claim. The functionf: JR--+ JR given by f(x) = 2x + 7 is one-to-one. "Proof." Supposex[ andx2 are real numbers withf(x[) *- f(X2)' Then 2x[ + 7 *- 2x2 + 7 and thus 2x[ *- 2X2' Hence x[ *- X2, which shows thatf is one-to-one. • (c) Claim. The function f: JR --+ JR given by f(x) = 2x + 7 is onto JR. "Proof." Suppose f is not onto JR. Then there exists b E JR with b Et: Rng (f). Thus b *- 2x + 7 for all real numbers x. But a = !(b - 7) is a real number and f(a) = b. This is a contradiction. Thus f is onto JR. • • onto onto (d) CI3Im. If f: A---->B and g: B---->C, then g 0 f: A --+ C maps onto C. (Theorem 4.8) "Proof." Suppose a EA.>Thenf(a) E B. Sincef(a) E B, g(j(a» E C. Therefore, (g 0 f)(a) = g(j(a» E C, so go f is onto C. • (e) Claim. Let I be the interval (0, I). The function f: I X I --+ I given by f(x, y) = x Y is a surjection. "Proof." Suppose tEl. Then 0 < t < I. If s is also in I, then 0< s < I. From t > 0, we have t S > O. From t < I, we have t S < 1'= l.Therefore,O
188

CHAPTER 4 Functions

Claim. Assume that the function I in part (e) is a mapping from I X I to 1. Then I maps onto 1. "Proof." Let tEl. Then O
(f)

(g)

•

(h)

(!, !)

!)

I(~,~) = (~y/4 = (Grr/4= GY/2 = IG, D. 4.4

•

Induced Set Functions Up to this point, a functionlfrom setA to set B has always been considered "pointwise." That is, we have considered the mapping of individual elements in A to their images in B or else considered pre-images of individual elements in B. The next step is to ask about collections of points in A or in B and what corresponds to them in the other set. Every function I from A to B induces a function that maps subsets of A to subsets of B. Also, regardless of whether I-I is a function, there is a well-behaved induced function in the other direction mapping subsets of B to subsets of A.

DEFINITIONS set of Xis

Let I: A ---+ B. If X ~ A, then the image of X or image

I(X) = {y E B: y = I(x) for some x EX}. If Y

~

B, then the inverse image of Y is

I-I(y) = {x EA:/(x) E Y}.

There need not be any confusion about the fact that the induced functions I: 9J>(A)---+9J>(B) and I-I: 9J>(B)--->9J>(A) have the same names as the function I from A to B and the relation I-I (which may not be a function). The induced functions I and 1-1 have as domain elements subsets of A and B, respectively. The image set I(X) is just the set of all images of elements of X, and the inverse image of I-I(y) is the set of all pre-images of elements of Y.

Example. Let A = {o, 1,2,3, -1, -2, -3}, B = {O, 1,2,4,6, 9}, and I: A ---+ B be given by I(x) = x 2 . Figure 4.9 shows that I({ -1, 3}) = {I, 9} and/-I({4, 6}) = {2, -2}. Also,f(A) = {o, 1,4, 9},f-I(B) = A,f-I({6}) = 0, and/({3, -3}) = {9}. Note that I-I is not a function from B to A, so it would not make sense to consider I-I (1). However, I-I({I}) is meaningful and equal to {I, -I}.

4.4

Induced Set Functions

A

189

B

Figure 4.9 Example. Let F: IR ---+ IR be given by F(x) = 2x + 1. For any subset X of the domain IR, F(X) = {F(x): x E X}. (The image set of X is the set of all images of elements of x.) For example, F({I, 2, 3}) = {3, 5, 7} and F(H, 6}) = {2, 13}. Furthermore, F(N) = {3, 5, 7, 9, .... }, F({n}) = {2n + I}, F([O, 5]) = [1, 11] and (because F is onto IR) F(IR) = IR. Figure 4.10 shows that for D = [1, 2], F(D) = [3,5]. For any subset Y of the codomain IR, F- 1 (Y) is the set of all pre-images of elements of Y. For example, F-l({13}) = {6}, F- 1({I, 2, 3}) = {O,!, I}, and F- 1([3, 5]) = [1, 2]. Example. Let f: IR ---+ IR be given by f(x) = x 2 . Then f({ -2, 2}) = f({2}) = 4, f([ -1,0]) = [0, 1] and f([I, 2]) = [1, 4]. Is f([ -1,2]) = [1, 4]? It is tempting to imagine that f ([ -1, 2]) is [( _1)2, 22] = [1, 4], but this is incorrect. By definition, f ([ -1, 2]) is the set of all images of elements of [-1, 2], and since 0, and. 7 are

-!,

y

F

5 F(D)

4

3

----F-+---~--L---L---~x

2 ~

D

Figure 4.10

3

4

190

CHAPTER 4

Functions

in [-1, 2], their images

t, 0, and .49 must be in f[ -1, 2]. Figure 4.11 shows that

1([ -1,2]) = [0,4]. Since all images of I are nonnegative, 1- 1([-4, -3]) =0. Even though 1([1,2]) =[0,4], 1- 1([0,4]) = [-2, -2]. (Look again at figure 4.11.) Also, 1([1,2]) = [1,4], but/- 1 ([I, 4]) =1= [1, 2]. Figure 4.12 shows why 1- 1([1,4]) = [-2, -1] U [1, 2]. Proofs involving induced set functions are likely to be more troublesome than others we have seen thus far. Before tackling such proofs, carefully study the definitions to see that each of these facts follows immediately from the definitions. If I: A ---+ B, D !: A, E !: B, and a E A, then: a ED==> I(a) E/(D) a E/-I(E) ==> I(a) E E I(a) E E ==> a E/- I (E)

Another useful fact requires some explanation: I(a) E/(D) ==> a ED,

provided that I is one-to-one. y

-4 -3 -2 -1

1

2

3

4

~

[-1,2]

Figure 4.11

f([ -1,2]) = [0,4]

y

-4

-3 -2 -1

2

[-2,-1]

Figure 4.12

r

l

([1,4])

3

4

[1,2] =

[-2, -1] U [1,2]

4.4

Induced Set Functions

191

This implication is false when f is not one-to-one. If in the preceding example we let D be the interval [0, 2], then f( -1) Ef(D), but -1 ff. D. In the case that f is one-to-one, then fromf(a) Ef(D) we know thatf(a) is the image of some element of D. Sincef(a) is the image of a and f(a) has only one pre-image, a ED. By the algebraic properties of the induced set functions, we mean their behavior as they interact with the operations of union, intersection, and set difference. We consider some examples before stating general results.

Examples. Let f: ~ ---> ~ be the function given by f(x) = x 2 , as in the preceding example. Let Ll = {n E N: 2:::; n :::; 1O} and for each nEll, let Dn = (-n, n). Then for each n E Ll,f(Dn) = [0, n 2 ]. Therefore, fen Dn) = f( -2, 2)) = [0,4] and

=

nf(Dn) nEt.

n [0, n

= [0,4], sOf(

2]

nEt.

U Dn) = f(

AlSO,f(

n

Dn)

~

nEt.

-10, 10))

nf(Dn). nEt.

=

[0, 100]

nEt.

and

Ek

Uf(D n) =

U [0, n

nEt.

nEt.

2] =

[0,100], sOf( U'Dn) nEt.

= Uf(D n). nEt.

For inverse images, consider r = {k E N: 5:::; k:::; 49} and for each k E [-1, k]. For each k E r, f-I(Ek) = Therefore,

[-.Jk,.Jk].

=

f- I (

n

Ek)

= f- I ([-I, 5]) =

r, let

[-)5,)5]

kEl

and nf-I(Ek') kEl

=

n [-.Jk,.Jk] = [-)5, )5], kEl

so f- I (

n

Ek) = nf-I(Ek)' AlsOf-l(

kEl

kEl

U Ek) = f- I [-I, 49] =

[-7, 7]

kEl

and

Uf-I(Ek) = U [-.Jk, .Jk] = [-7, 7], so f- I (U Ek) = Uf-I(Ek)' kEl

Theorem 4.16

kEl

kEl

kEl

Letf: A ---> B, and let {Da: a E Ll} and {Ep: PEr} be families of subsets of A and B, respectively. Then

(a)

f(

n

Du)

~

aEt.

(b)

f(

U Du) = uEt.

n

f(Du)·

uEt.

Uf(Du)'\ uEt.

192

CHAPTER 4

Fundions

(c)

f-I(n E p)

=

pEr

f- I(

(d)

U E p) = pEr

nf-I(Ep). pEr Uf-I(E p). pEr

Proof. (a)

Suppose b Ef( n Du). Then b = f(a) for some a E n Du· Thus a E Du uE~

uE~

for every a E d. Since b

=

f(a), b Ef(Da) for every a E d. Therefore,

bEn f(Da)· This proves thatf( n Da) ~~

~~

~

n

f(Da)·

~~

•

Parts (b), (c), and (d) are left as exercise 10.

Ifwe letf = {(a, x), (b, x)}, d = {I, 2}, DI = {a} and D2 = {b}, then f ( nDa) = uE~

f(D I n D 2) = f(0) = 0, while nf(Da) = f(D I) nf(D 2) = {x}

n {x}= {x}.

This

aE~

explains why we could not prove equality in part (a) of Theorem 4.16; see exercise 9.

Theorem 4.17

Letf: A-+ Band E ~ B. Thenf-I(B - E)

= A - f-I(E).

Proof. Suppose aEf-I(B-E). Then f(a)EB-E. That is, f(a)EB and f(a) t!. E. Therefore, a E A and a t!.f- I(E). (This is the contrapositive of a Ef-I(E) ==> f(a) E E.) Thus a EA - f-I(E). Therefore,f-I(B - E) ~A - f-I(E). The opposite inclusion is left as exercise II. •

Theorem 4.18

Letf: A-+ B, D (a)

(b) (c) (d)

~A,

and E

~

B. Then

f(f-I(E» ~ E. E = f(f-I(E» iff E ~ Rng (f). D ~f-I(f(D». f-I(f(D» = D ifff(A - D) ~ B - feD).

Proof. (a) Suppose b Ef(f-I(E». Then there is a Ef-I(E) such thatf(a) = b. Since a Ef-I(E),f(a) E E. Butf(a) = b, so bEE. Therefore,f(f-I(E» ~ E. (b) First, suppose E = f(f-I(E». Suppose bEE. Then b Ef(f-I(E». Thus there is a Ef-I(E) such thatb = f(a), so bE Rng (f). Therefore, E ~ Rng (f). Now assume E ~ Rng (f). We know by part (a) that f(f-I (E» ~ E, so to prove equality, we need only E ~f (f-I (E». Suppose bEE. Then b E Rng (f), so b = f(a) for some a EA. Since b = f(a) E E, a Ef- I(E). Thus b = f(a) and a Ef-I(E), so b Ef(f-I(E». Therefore, E ~f(f-I(E». Parts (c) and (d) constitute exercise 12.

•

As examples of the results in Theorem 4.18, consider again the functionf: IR-+ IR given by f(x) =x 2 and the sets D= [-1,2] and E= [-4,4]. Thenf(f-I(E» = f ([ - 2, 2]) = [0, 4], a proper subset of E, as in part (a) of the theorem. If we had chosen

4.4

Induced Set Functions

193

E=[1,4], then we would have f(f-I(E))=f([-2, -1]U[I,2])=E, as in part (b). Alsof- I(f(D)) = f- I[0,4] = [-2, 2], so thatD ~f-I (f(D)), as required by part (c). If we had chosen D = [-10, 10], thenf- I (f(D)) = f- I [0, 100] = D, as in part (d). For this choice of D, f(lR - D) = f« -00, -10) U (10, (0)) = (100, (0), and JR - feD) = JR - [0, 100] = (100, 00), so f(JR - D) ~ JR - feD).

Exercises 4.4 1. LetA = {I, 2, 3}, B = {4, 5, 6}, and h = (O, 4), (2,4), (3, 5)}.

*

2.

* *

3.

(a) (b)

List the eight ordered pairs in the induced function on QJ>(A) to QJ>(B). List the eight ordered pairs in the induced function on QJ>(B) to QJ>(A).

Let f(x) = x 2 + 1. Find (a) f([1,3]) (c) f- I ([-I,I]) (e) f- I ([5, 10])

(b) (d) (f)

Let f(x) = 1 - 2x. Find (a) f(A) where A = {-I,D, 1,2, 3} (c) (e)

f-I(JR) f((l,4])

(b) (d) (f)

f([ -1,0] U [2, 4])

f- I ([-2,3]) f- I ([-I, 5] U [17,

26])

f(l~J)

f- I ([2,5]) f(f-I(f([3,4])))

4.

Letf: N X N--+ N be given by fern, n) = 2m(2n + 1). Find (a) f-I(A) where A = {I, 2, 3, 4, 5, 6}. (b) f-I(B) where B = {4, 6, 8, 10, 12, 14}. (c) f(C) where C = (O, 1), (3, 3),A3, 1), (1, 3)}.

5.

Let f: A--+ B where A = {I, 2, 3, 4, 5, 6}, B= {p, q, r, s, t, z}, and f = (2, p), (3, s), (4, t), (5, z), (6, t)}. Find (a) f-I({P, q, s}) (b) f({I, 3,4, 6}) (c) f({3,5}) (d) f-I({P, r, s, z}) (e) f(f-I({p, r, s, z})) (f) f-I(f({I, 4, 5}))

*

{(l, p),

* *

6.

Let f: JR - {O} --+ JR be given by f(x) = x

*

(a) (c)

f«0,2)) f- I «3,4])

(e)

f(f-I(JR))

7.

* *

8.

'*

9.

*

+ .!.. Find

(b) (d)

x f([1,5]) f-I([O, 1))

(f)

f(r-I ([ _130 , 10.1]))

Let f: JR --+ JR be given by f(x) = lOx - x 2 . Find (a) f([1,6)) (b) f- I«0,21]) (c) f- I«3,4]) (d) f- I ([24,50]) (e) f([4,7]) (f) f- I (f([5,6])) Letf: N X N --+ N be given by fern, n) = 2m3n . Find (a) f(A X B) where A = {I, 2, 3}, B = {3, 4}. (b) f- I ({5, 6,7,8,9, 1O}). Give an example of a functionf: JR --+ JR, and a family {D(j: <5 E of JR such that

f(

n

D(j)

(jEll

'* nf(D

J).

JEll

il} of subsets

194

CHAPTER 4

Functions

"*

10.

Prove parts (b), (c), and (d) of Theorem 4.16.

11.

Prove the remaining inclusion of Theorem 4.17. That is, prove that if f: A -+ B andE r::;;,B, then A - f-I(E) r::;;,f-I(B - E).

12.

Prove parts (c) and (d) of Theorem 4.18.

13.

Let f: A -+ B and let X, Y r::;;, A and V, V r::;;, B. Prove (b) f(X) - fey) r::;;,f(X - Y). (a) f(X) r::;;, V iff X r::;;,f-1(V). (c) f-I(V) - f-I(V) = f-I(V - V).

14.

Let f: A -+ B. Prove that, iffis one-to-one, then f(X) nf(Y) = f(X all X, Y r::;;, A. Is the converse true? Explain.

15.

Let f: A -+ B. Prove that, if X r::;;, A and f is one-to-one, then f(A - X) f(A) - f(X).

=

16.

Let f: A -+ B. Prove that if X r::;;, A, Y r::;;, B, and f is a bijection, then f(X) ifff-I(y) =X.

Y

17.

Letf: A -+ B. Consider the function on QJ>(A) to QJ>(B) induced by f (a) What condition on f will make the induced function one-to-one? (b) What condition on f will make the induced function onto QJ>(B)?

18.

Letf: A -+ Band K r::;;, B. Prove thatfU- 1(K)) = K

19.

Let f: A -+ B. Let R be the relation on A defined by x R y iff f(x) = fCy). (a) Show that R is an equivalence relation. (b) Describe the partition of A associated with R.

20.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Iff: A -+ B and X r::;;, A, thenf- IU(X)) r::;;, X. uProoJ." If x Ef-IU(X», then by definition of f-I, f(x) Ef(X). Therefore x EX. Thus f-I U(X)) r::;;, x. (b) Claim. Iff: A-+BandXr::;;,A,thenX r::;;,f-1U(X)). uProoJ." Supposez EX. Thenf(z) Ef(X). Thereforez Ef-IU(X», which proves the set inclusion. (c) Claim. Iff: A -+ Band {Da: a E ~} is a family of subsets of A, then

*

Proofs to Grade

*

*

UProoJ."

n Y) for

=

n Rng U).

Suppose yEn f(Da). Then y Ef(Da) for all a. Thus there aELl

exists x EDasuch thatf(x) = y, foralla. Then x E n Daandf(x) = y, aELl

so y Ef( n Da). Therefore, n f(Da) r::;;,f( aELl

aELl

n Da).

aELl

-

CHAPTER

5

Cardinality

How many elements are in the set A=

in, 28, ji,!, -3, fl, cr, o}?

After a short pause, you said "eight." Right? Consider for a moment how you arrived at that answer. You probably looked at n and thought "I," then looked at 28 and thought "2," and so on up through 0, which is "8." What you have done is set up a one-to-one correspondence between the set A and the "known" set of eight elements {I, 2, 3, 4, 5, 6, 7, 8}. Counting the number of elements in sets is essentially a matter of one-to-one correspondences. This process will be extended when we "count" the number of elements in infinite sets in this chapter. Here is another counting problem. A certain shepherd has more than 400 sheep in his flock, but he cannot count beyond 10. Each day he takes his sheep out to graze, and each night he brings them back into the fold. How can he be sure all the sheep have returned? The answer is that he can count them with a one-to-one correspondence. He needs two containers and a pile of pebbles, one pebble for each sheep. When the sheep return in the evening, he transfers pebbles from one container to the other, one at a time for each returning sheep. Whenever there are pebbles left over, he knows there are lost sheep. The solution to the shepherd's problem illustrates the point that even though we have not counted the sheep, we know that the set of missing sheep and the set of leftover pebbles have the same number of elements-because there is a one-to-one correspondence between them.

5.1

Equivalent Sets; Finite Sets To determine whether two sets have the same number of elements, we see whether it is possible to match the elements of the sets in a one-to-one fashion. This idea may be conveniently described in terms of a one-to-one correspondence (a bijection) from one set to another. 195

196

CHAPTER 5

Cardinality

DEFINITION Two sets A and B are equivalent iff there exists a oneto-one function from A onto B. A and B are also said to be in one-to-one correspondence, and we write A = B.

If A and B are not equivalent, we write A*' B.

Example. The sets A = {5, 8, ep} and B = {r, p, m} are equivalent. The function f: A --+ B given by f(5) = r, f(8) = p, and f(ep) = m is one of six such functions that verify this. Example. The sets C = {x, y} and D = {q, r, s} are not equivalent. There are nine different functions from C to D. An examination of all nine will show that none of them is onto D and so none of them is a bijection. Thus C is not equivalent to D. Example. The set E of even integers is equivalent to D, the set of odd integers. To prove this, we employ the function f: E --+ D given by f(x) = x + 1. The function is one-to-one, because f(x) = fey) implies x + I = y + 1, which yields x = y. Also, f is onto D because if z is any odd integer, then w = z - 1 is even andf(w) = w + I = (z - 1) + I = z. Example. For a, b, c, dE IR, with a < band c < d, the open intervals (a, b) and (c, d) are equivalent. Proof. Let f: (a, b)--+ (c, d) be the linear function pictured in figure 5.1. The function is given by:

d-c

f(x) = b _ a (x - a)

+c

This is simply the linear function through the points (a, c) and (b, d) restricted to the domain (a, b). We leave the proof that f is a bijection to exercise 4. • y

d

c

/

--r-----~------~--~x

a

Figure 5.1

b

5.1

Equivalent Sets; Finite Sets

197

We now know that any two open intervals are equivalent, even when the intervals have different lengths. Also, the interval (5, 6) is equivalent to (1, 9), even though it is a proper subset of (1,9).

Example.

Let gji be the set of all functions from N to {O, I}. This set is sometimes denoted {a, I }N. We will show gji = QJI(N), the power set of N.

Proof.

(This proof will show that gji is precisely the set of all characteristic functions with domain N, which is in a one-to-one correspondence with the subsets of N.) To show gji QJI(N) , we define H: gji -+ QJI(N) as follows:

=

for g E gji,

H(g)

= {x E

N: g(x}

= I}.

(Note that under the function H, every function in gji has an image in QJI(N).) To show H is one-to-one, let g" g2 E gji. Suppose gl g2' Then there exists n E N such that g 1(n) g2 (n). Both gland g2 have codomain {O, I}, so we may assume gl (n) = 1 and g2(n) = 0. (The case gl (n) = and g2(n) = 1 is similar.) But then n E {x E N: gl (x) = I} and n ~ {x E N: g2(X} = I}. Thus H(gl} H(g2}' To show that H is onto QJI(N), let A E QJI(N). Then A k N. We note that XA: N --+ {a, I} and thus XA E gji. Furthermore,

*"

*"

°

HCxA) =

{x E N: XA(X)

=

I}

*"

=

A (.

by definition of XA- Thus H is onto. Because H is a bijection, gji = QJI(N).

Theorem 5.1

The relation = is reflexive, symmetric, and transitive. Thus relation on the class of all sets.

Proof.

• = is an equivalence

Exercise I.

•

The next theorem will be particularly useful for showing equivalences of sets.

Theorem 5.2

Suppose A, B, C, and D are sets with A = C and B = D. (a)

(b)

A X B = C X D. If A and B are disjoint and C and D are disjoint, then A U B = CUD.

Since A = C and B = D, there exist one-to-one correspondences h: A -+ C andg: B-+D.

Proof. (a) (b)

Let f: A X B-+ C X D be given by f(a, b) = (h(a), g(b)). We leave it as exercise 2 to show thatfis a one-to-one correspondence. Therefore, A X B = C X D. By Theorem 4.15, hUg: AU B-+ CUD is a one-to-one correspondence. Therefore, A U B = CUD. •

198

CHAPTER 5 Cardinality

We shall use the symbol Nk to denote the set {I, 2, 3, ... , k}. Think of Nk as the standard set with k elements against which the sizes of other sets may be compared.

DEFINITIONS A set S is finite iff S = 0 or S is equivalent to Nk for some natural number k. A set is infinite iff it is not finite. We say that the empty set has cardinal number 0 (or cardinality 0), and if S = Nk we say that S has cardinal number k (or cardinality k).

The set X = {98.6, c, n} is finite and has cardinal number 3. Exhibiting any one-to-one function from X onto N3 will prove this. One such correspondence f: X -+ N3 is given by f{J8.6) = 1, f(c) = 2, andf(n) = 3. We use the symbol S to denote the cardinal number ~ a set S. At this time S is defined only for a tinite set S. In the previous example, X = 3. For the empty set, 0= 0 and clearly, Nk = k for every natural number k. The setA = {8, 7, 3, 7, 2} is finite and has cardinality 4, since it is equal to the set {8, 7, 3, 2}. Since A = N4 ,

A=4. The symbol A is used for the cardinality of A so as to distinguish what we prove here from the more informal results in section 2~, where we used the symbol # A. We shall show that for finite sets the cardinality A of A corresponds to our intuitive notion of the number of elements in A. The empty set is finite by definition. Also, each set Nk is finite and has cardinal number k because the identity function Ir'Jk is a one-to-one function from Nk onto N k • In addition, any set equivalent to a finite set must also be finite. Suppose A is a finite set and A = B. If A = 0, then B = 0 (see exercise 3). Otherwise, A = Nk for some k and thus B = Nk by transitivity of = . In either case B is finite. The remaining theorems of this section give other properties of finite sets. One goal is to make sure that the notion of cardinality for a finite setA corresponds to our notion of the number of elements in A in the following important way: The cardinality ofafinite set should be unique. That is, if A has cardinality m (that is, A = N m ) and A has cardinality n (A = N n ), then m = n. You are asked to show this in exercise 14, using a property of finite sets called the pigeonhole principle, discussed later in this section. Our immediate goal is the key theorem that every subset of a finite set is finite. Our proof of this theorem uses two lemmas.

Lemma 5.3

If S is finite with cardinality has cardinality k + I.

k and x is any object not in S, then S U {x} is finite and

Proof. If S = 0, then S U {x} = {x}, which is equivalent to NI and thus finite. In this case S has cardinality 0 and S U {x} has cardinality 0 + 1. If S =1= 0, then S = Nk for some k. Also, {x} = {k + I}. Therefore, by Theorem 5.2, S U {x} = Nk U {k + I} = Nk+ I' This proves S U {x} is finite and has cardinality k + 1. •

5.1

Equivalent Sets; Finite Sets

199

f 1-1, onto

fiT 1-1, onto

Figure 5.2

Lemma 5.4

Every subset of Nk is finite.

Proof.

Let A be a subset of N k • (We prove A is finite by induction on the number k.) Let k = 1. Then either A = 0 or A = N I, both of which are finite. Assume all subsets of Nk are fini te and let A !: Nk+ I' Then A - {k + I} is a subset of Nk and is finite by the hypothesis of induction. If A = A - {k + I}, then A is finite. Otherwise, A = (A - {k + I}) U {k + I}, which is finite by Lemma 5.3. -

Theorem 5.5

Every subset of a finite set is finite. Assume S is a finite and T !: S. If T = 0, then T is finite. Thus we may assume T =1= 0 and hence S =1= 0. Since S = Nk for some kEN, there is a one-to-one function J from S onto Nk . Then the restriction off, J T, is a one-to-one function from Tonto J(T). Therefore, T is equivalent to J(T) (see figure 5.2). But J(T) is a subset of the finite set Nk and is finite by Lemma 5.4. Therefore, since T is equivalent to a finite set, T is finite. -

Proof.

I

At this point you may think that Lemmas 5.3 and 5.4 and Theorem 5.5 are a lot of hard work to prove the very obvious result that subsets of finite sets are finite. You may be right. The value of these results lies in the reasoning and in the use of functions to establish facts about cardinalities. This work will be helpful when we deal with infinite sets because there our intuition often fails us. The next result of this section is that the union of a finite number of finite sets is finite. To this end the next theorem is a special case: the union of two disjoint finite sets is finite. Its proof is a rigorous development of the sum rule (Theorem 2.17), which states that if A has m elements, B has n elements, and A n B = 0, then A U B has m + n elements.

Theorem 5.6

If A and B are finite disjoint sets, then A U B is finite and A U B = A

+ B.

Suppose A and B are finite sets and A n B = 0. If A = 0, then A U B = B; if B = 0-,- th~n. A U B = A. In either case A U B is finite, and since 0 = 0, A U B = A + B. Now suppose that A =1= 0 and B =1= 0. Let A = Nm and B = N n ,

Proof.

200

CHAPTER 5 Cardinality

and suppose that f: A --+ N m and g: B --+ Nn are one-to-one correspondences. Let = {m + 1, m + 2, ... , m + n}. Then h: N n --+ H given by hex) = m + xis aone-toone correspondence, and thus N n = H. Therefore, B = H by transitivity. Finally, by Theorem 5.2, A U B = N m U H = N m + m which proves that A U B is finite and that H

A UB=m+n.

Corollary 5.7

-

(a)

If A and B are finite sets, then A U B is finite.

(b)

If AI> A 2 , ... ,An are finite sets, then

n

U Ai is finite. i=1

Proof. We prove part (a) and leave part (b) as an exercise in mathematical induction (exercise 6). Assume that both A and B are finite. Since B - A ~ B, B - A is finite. Thus by Theorem 5.6, A U B = A U (B - A) is a finite set. Lemma 5.3 shows that adding one element to a finite set increases its cardinality by one. It is also true that removing one element from a finite set reduces the cardinality by one. The proof of Lemma 5.8 is left as exercise 13.

Lemma 5.8

Let r E N with r > 1. For all x E N" N r

-

{x} = N r -

I•

The final property of finite sets we consider is popularly known as the pigeonhole principle. In its informal version it says: "If a flock of n pigeons comes to roost in a house with r pigeonholes and n > r, then at least one hole contains more than one pigeon." If we think of the set of pigeons as Nn and the set of pigeonholes as N" then the pigeonhole principle says any assignment of pigeons to pigeonholes (function from Nn to Nr ) is not one-to-one.

Theorem 5.9

(The Pigeonhole Principle) Let n, r E N. Iff: Nn --+ N r and n > r, then f is not one-to-one.

Proof. The proof proceeds by induction on the number n. Since n> r, we begin with n = 2. If n = 2, then r = 1. In this casefis the constant function f(x) = 1, which is not one-to-one. Suppose the pigeonhole principle holds for some n; that is, suppose for all r < n, if f: Nn --+ N" then f is not one-to-one. Let r < n + 1. The case r = 1 is treated just as in the case n = 2 above, so we may assume r > 1. Suppose there is a one-to-one function f: Nn + 1 --+ N" Then fiNn is a one-to-one function. The range of this function may not contain fen + 1), but by Lemma 5.8 there is a one-to-one correspondence g: N r - {J(n + I)} --+ N r - I • Therefore, the composite g 0 fiNn: N n --+ N r - I is one-to-one. This contradicts the induction hypothesis. By the PMI, there is no one-to-one function f: Nn --+ ~r when n > r. The pigeonhole principle can be used to prove several important results about finite sets (see, for example, exercise 13). We use it here to give a characterization offinite sets.

5.1

Corollary 5.10

Equivalent Sets; Finite Sets

201

A finite set is not equivalent to any of its proper subsets.

Proof. We will show that Nk is not equivalent to any proper subset and leave the general case as exercise 15. The case k = 1 is trivial, so let k > 1. Suppose A is a proper subset of Nk and I: Nk --+ A is one-to-one and onto A. Case 1.

Case 2.

Suppose k tE A. Then A ~ N k - 1 and the inclusion function i: A --+ N k - 1 is one-to-one. But then i 0 I: Nk --+ N k- 1 is one-to-one, which contradicts the pigeonhole principle. Suppose k EA. Choose an element y E Nk - A and let A' = (A - {k}) U {y}. Then A = A' (because the function IA-(k} U {(k, y)} is a one-to-one correspondence). Thus A' = Nb A' is a proper subset of Nb and k tE A'. This is the situation of Case 1 with Nk and A' and again yields a contradictioo. •

Corollary 5.10 says: "If A is a finite set, then A is not equivalent to any of its proper subsets." The contrapositive is: "If A is equivalent to one of its proper subsets, then A is infinite." Earlier we observed that the open interval (0, 2) is equivalent to its proper subset (0, 1). Thus we know that (0, 2) is infinite. There will be many more interesting infinite sets in the next section.

Exercises 5.1 1. Prove Theorem 5.1. That is, show that the relation = is reflexive, symmetric, and transitive on the class of all sets. 2. Complete the proof of Theorem 5.2(a) by showing that if h: A --+ C and g: B--+ D are one-to-one correspondences, then I: A X B--+ ex D given by I(a, b) = (h(a), g(b)) is a one-to-one correspondence. 3. (a) Show that if A = 0, then A = 0. [See also exercise 12(b), section 4.1.] (b) Show that A = A X {x}, for any object x. 4. Complete the proof that any two open intervals (a, b) and (e, d) are equivalent by showing that I(x) = (%=~)(x - a) + e is one-to-one and onto (e, d). 5. Which of the following sets are finite? (a) the set of all grains of sand on Earth (b) the set of all positive integer powers of 2 (c) the set of four-letter words in English (d) the set of rational numbers (e) the set of rationals in (0, 1) with denominator 2k for some kEN (f) {x E IR: x 2 + 1 = o} (g) the set of all turkeys eaten in the year 1620 (h) {I, 3, 5} X {2, 4, 6, 8} (i) {x E N: x is a prime} (j) {x E N: x is composite} (k) {x EN: x 2 + x is prime}

"*

* * *

202

CHAPTER 5 Cardinality

{x E IR: x is a solution to 4x 8 - 5x6 + 12x4 - 18x3 + X 2 - X + 10 = o} (m) the set of all complex numbers a + bi such that a 2 + b2 = I

(I)

6.

Prove part (b) of Corollary 5.7.

7.

Let A and B be sets. Prove that (a) if A is finite, then A n B is finite. (b) if A is infinite and A ~ B, then B is infinite.

8.

(a) (b)

9.

Define BA to be the set of all functions from A to B. Show that if A and Bare finite, then BA is finite.

*

"*

10.

* "*

Prove that for all k, mEN, Nk X N m is finite. Suppose A and B are finite. Prove that A X B is finite.

If possible, give an example of each of the following: (a) an infinite subset of a finite set (b) a collection {Ai: i E N} of finite sets whose union is finite (c) a finite collection of finite sets whoseJInio..Q is infinite (d) finite sets A and B such that A U B =1= A + B

11. Using A UB

th~ meQlods

of this section, prove that if A and B are finite sets, then

= A + B - A n B. This fact is a restatement of Theorem 2.17.

12.' Prove that if A is finite and B is infinite, then B - A is infinite.

"*

13. Prove Lemma 5.8. 14. Show that if a finite set S has cardinal number m and cardinal number n, then m=n. 15. Complete the proof of Corollary 5.10 by showing that if A is finite and B is a proper subset of A, then B A. 16. Prove by induction on n that if r < nand f: N r ---+ N m then f is not onto Nn17. Let A and B be finite sets with A = B. Suppose f: A ---+ B. (a) If f is one-to-one, show that f is onto B. (b) If f is onto B, prove that f is one-to-one. 18. Prove that if the domain of a function is finite, then the range is finite.

'*'

"* "*

"*

19. Use Corollary 5.10 to show that each of the following sets is infinite. (a) {1O, II, 12, 13, 14, 15, ... } (b) {... , -3, -2, -I, o} (c) {3 k : kEN} (d) (0, (0)

*

Proofs to Grade

20.

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If A and B are finite, then A U B is finite. "Proof." If A and B are finite, then there exist m, n E N such that A = N m , I-I I-I I-I B = Nn- Let f: AOi1i6Nm and h: BOi1i6 Nn- Then f U h: A U BOi1i6 Nm+m which shows that A U B = Nm+n- Thus A U B is finite. (b) Claim. If S is a finite, nonempty set, then S U {x} is finite. "Proof." Suppose S is finite and nonempty. Then S = Nk for some integer k. Case 1. xES. Then S U {x} = S, so S U {x} has k elements and is finite. Case 2. x Et: S. Then S U {x} = Nk U {x} = Nk U NI = N k + l . Thus S U {x} is finite. -

5.2

(c)

5.2

Infinite Sets

203

Claim. If AX B is finite, then A is finite. "Proof." Choose any b* E B. Then A = A X {b*}. But A X {b*} = {(a, b*): a E A} C A X B. Since A X B is finite, A X {b*} is finite. Since A is equivalent to a finite set, A is finite. -

Infinite Sets This section will describe several infinite sets and their cardinal numbers. We will see that there are many infinite sets and that it is not true that all infinite sets are equivalent. Just as we use Nk as our standard set of k elements, we will specify other sets as standard sets for certain infinite cardinal numbers. In the previous section an infinite set was defined as a nonempty set that cannot be put into a one-to-one correspondence with any Nk . According to the next theorem, the set N is one such set. Since infinite means not finite, our proof, as might be expected, is by contradiction. /

Theorem 5.11

The set N of natural numbers is infinite.

Proof. Suppose N is finite. Clearly, N =I- 0. Therefore for some natural number k, there exists a one-to-one function f from Nk onto N. (We will contradict that f is onto N.) Let n = f(l) + f(2) + ... + f(k) + 1. Since each f(i) > 0, n is a natural number larger than each f(i). Thus n =I- f(i) for any i E Nk . Hence n t!. Rng if). Therefore, f is not onto N, a contradiction. Here is a second proof that N is an infinite set: Let E+ be the set of even positive integers. The function f: N --+ E+ defined by setting f(x) = 2x is clearly a one-to-one function from N onto E+. (Of course, there are many other one-to-one correspondences; the one given is the simplest.) Thus N is equivalent to one of its proper subsets. Therefore, by Corollary 5.10, the set of natural numbers cannot be finite. There are many other infinite sets, some-but not all-of which are equivalent to N. Infinite cardinal numbers are written using the symbol X, aleph, the first letter of the Hebrew alphabet. The cardinality of N is denoted Xo. The subscript 0 (variously read "zero," "naught," or "null") indicates that Xo is the smallest infinite cardinal number, akin to the integer 0 being the smallest finite cardinal. The set N is our "standard" set for the cardinal number Xo-a set has cardinal number Xo iff it is equivalent to N.

DEFINITION write S = Xo.

A set S is denumerable iff S is equivalent to N; we

204

CHAPTER 5 Cardinality

We showed above that E+, the set of positive even integers, is equivalent to N. Therefore, E+ is denumerable. Even though E+ is a proper subset of N, E+ has the same number of elements (~o) as N. Although our intuition might tell us that only half of the natural numbers are even, it would be misleading to say that N has twice as many elements as E + , or even to say that N has more elements than E +. We must rely on one-to-one correspondences to determine cardinality. In section 5.4 we will show that every infinite set is equivalent to one of its proper subsets. Together with Corollary 5.10, this will characterize infinite sets: a set is infinite iff it is equivalent to one of its proper subsets. The next theorem will show that ll, the set of all integers, is also equivalent to N.

Theorem 5.12

The set II of integers is denumerable.

Proof.

We define f: N ---+ II by

[(xl

=

{~ ;x

if x is even if x is odd.

Thus f(l) = 0, f(2) = 1, f(3) = -1, f(4) = 2, f(5) = -2, and so on. Pictorially,

f is represented as a one-to-one matching: N = {I,

2,

3,

4,

5,

6,

7, ... }

1 II = {O,

1

1

1

1

1

1

1, -1,

2, -2,

3, -3, ... }

We will first show that f is one-to-one. Suppose f(x) = f(y). If x and yare both i-x i-y even, then x2 = Y2' and thus x = y. If x and yare both 0 dd , then -2= -2-' so 1 - x = 1 - y and x = y. If one of x and y is even and the other is odd, then only one of fix) or f(y) is positive, so f(x) *- f(y). Thus whenever f(x) = f(y), x = y. It remains to show that the function maps onto ll. If wEll and w > 0, then 2w is even and f(2w) = = w. If wEll and w ::5 0, then 1 - 2w is an odd natural i-(i-2w) 2w 7J number and f(l - 2w) = 2 = 2"" = w. In either case, if wE fL, then wE Rng if). •

2;

Example. The set P of reciprocals of positive integer powers of 2 is denumerable. Writing the set P as P

=

{dk: kEN}

exhibits the one-to-one correspondence f: N ---+ P given by f(k) = 1/2k.

Example. The set N X N is denumerable. To see this, we must show there is a bijection f: N X N ---+ N. The function F in section 4.3 given by fern, n) = 2m - 1 (2n - 1) is one such function. Thus N X N has cardinality ~o. This example leads directly to the next theorem.

5.2

Theorem 5.13

Infinite Sets

205

If sets A and B are denumerable, then A X B is denumerable.

Proof. Since A = Nand B = N, A X B = N X N by Theorem 5.2(a). Since N = N, A X B = N. Therefore, A X B is denumerable.

X

N -

Theorem 5.13 is a surprising result if you rely on properties of finite cardinal numbers for insight into infinite cardinals. In chapter 2, you were asked to show that if A is a finite set with m elements and B is a finite set with n elements, then A X B has mn elements. There is no direct corresponding result for sets with cardinality Xo. In fact, Theorem 5.13 can b~ construed as saying that "Xo times Xo is Xo." We will not pursue the arithmetic of cardinal numbers, but simply remind you that extension of results for finite cardinals may not be straightforward.

DEFINITIONS A set is countable iff it is finite or denumerable; otherwise the set is uncountable.

The set N3 = {I, 2, 3} is countable because it is finite. The set of integers, Z, is a countable set because Z is denumerable. The concepts of countable, uncountable, and denumerable along with the ideas of finite and infinite are related in figure 5.3. Because denumerable sets are those that are countable and infinite, denumerable sets are sometimes referred to as "countably infinite" sets. We have seen examples of countable sets (both finite sets and denumerable sets), but no example, as yet, of a set that is uncountable. Before considering the next theorem, which states that the set of real numbers in the interval (0, 1) is such an example, we need to review decimal expressions for real numbers. In its decimal form, any real number in (0, 1) may be written as 0.ala2a3a4"" where each ai is an integer, 0:::; ai:::; 9. In this form, = 0.583333 ... , which is abbreviated to 0.583 to indicate that the 3 is repeated. A block of digits may also be repeated, as in ~ = 0.82142857. The number x = 0.ala2a3'" is said to be in normalized form iff there is no k such that for all n > k, an = 9. For example, 0.82142857 and ~ = 0.40 are in normalized form, but 0.49 is not. Every real number can be expressed uniquely in normalized/orm. Both 0.49 and 0.50 represent the same real number~, but only 0.50 is normalized. The importance of normalizing decimals is that two decimal numbers in normalized/orm are equal iff they have identical digits in each decimal position.

i2

Finite Sets

Uncountable Sets

Countable Sets

Infinite Sets

Figure 5.3

206

CHAPTER 5

Theorem 5.14

Cardinality

The interval (0, I) is uncountable.

Proof.

We must show that (0, I) is neither finite nor denumerable. The interval (0, I) is not finite since it contains the infinite subset (See Theorem 5.5.) Suppose there is a function f: N ---+ (0, I) that is one-to-one. We will show that f does not map onto (0, I). Writing the images of the elements of N in normalized form, we have

g, t, *, ... }.

f(l) =

0.a11 a I2a 13a I4a I5···

=

0.a21 a 22a 23a 24a 25···

f(3) = f(4) ~

0.a31a32a33a34a35···

f(2)

if au =1= 5 if au = 5.

0.a41 a 42 a 43 a 44a 45···

(The choice of 3 and 5 is arbitrary.)

Then b is not the image of any n E N, because it differs fromfi,n) in the nth decimal place. We conclude there is no one-to-one and onto function from N to (0, I) and a hence (0, I) is not denumerable. The open interval (0, I) is our first example of an uncountable set. The cardinal number of (0, I) is defined to be c (which stands for continuum) and is the only infinite cardinal other than Xo that will be mentioned by name. We add the interval (0, I) to our list (N k and N) of standard sets and will use (0, I) as the standard set of cardinality c.

DEFINITION A set S_has cardinality c iff S is equivalent to the open interval (0, I); we write S = c.

There is nothing special about using the interval (0, I) to define the cardinal number c; every open interval (a, b), where a, bE IR and a < b has cardinality c. In section 5.1, we showed that any two open intervals are equivalent. In particular, (0, 1) = (a, b). Thus, (a, b) has cardinal number c.

Example. The set A = (3, 4) U [5, 6) has cardinal number c. To see this, we note that the functionf: (0,1) ---+ A, given by 2X

+ 3,

f(x) = {

2x +4, is a one-to-one correspondence.

ifO <x


I ' if -::::; x < I 2

5.2

Infinite Sets

207

y

---+--~~--+-~x

Figure 5.4 The set of all real numbers also has cardinality c, as the next theorem shows.

Theorem 5.15

The set

~

is uncountable and has cardinal number c.

Proof. Define/: (0, I) -+ ~ by lex) = tan (nx -1)' See figure 5.4. The function / is a contraction and translation of one branch of the tangent function and is one-to• one and onto R Thus (0, I) = R Because we know that N X N has the same cardinality as N, you may suspect that ~ X ~ has the same cardinality as R Indeed, it is true that ~ X ~ has cardinality c. We omit the proof because the specialized methods needed will not be useful in our current study. Since N is a proper subset of ~, and N = Xo, ~ = c, and Xo c, it is natural to suspect (correctly) that Xo < c. We consider the ordering of cardinal numbers in section 5.4. In the meantime, we delve more deeply into the study of countable sets in section 5.3.

*"

Exercises 5.2 1. Prove that th~ following sets are denumerable.

*

"* 2.

"* "*

(a) (b) (c)

D+, the odd positive integers T+, the positive integer multiples of 3 T, the integer multiples of 3

(d)

{n: n E Nand n > 6} {x:xE£:andx<-12}

(e) (f) (g) (h)

N - {5, 6} {(x, y) E N X ~: xy = I} {x E £:: x == 1 (mod 5)}

Prove that the following sets have cardinality c. (a) (1,00) (b) (a, 00), for any real number a

208

CHAPTER 5

Cardinality

"*

*

(c) (d) (e) (f)

IR -

{O}

3.

State whether each of the following is true or false. (a) If a set A is countable, then A is infinite. (b) If a set A is denumerable, then A is countable. (c) If a set A is finite, then A is denumerable. (d) If a set A is uncountable, then A is not denumerable. (e) If a set A is uncountable, then A is not finite. (f) If a set A is not denumerable, then A is uncountable.

4.

(a) (b)

5.

* * *

Give an example of a bijection g from N to the set E+ of positive even integers such that g(l) = 20. Give an example of a bijection h from N to E+ such that h(l) = 16, h(2) = 12, and h(3) = 2.

Which sets have cardinal number Xo? c? (a) IR - [0, 1) (b) (c) (d) (e) (f) (g)

(S, (0)

{k: n E N} {2x: x E N} {(P,q)EIRXIR:P+~=I} {(p,q)ElRxlR:q= l-p2} {(x,y)EIRXIR:x,yEZ}

6.

Give an example of two denumerable sets A and B such that (a) A U B is denumerable and A n B is denumerable. (b) AU B is denumerable and A - B is denumerable. (c) AU B is denumerable and A - B is finite and nonempty.

7.

Give an example of two sets A and B each with cardinal number c such that (a) A - B is empty. (b) A - B is finite and nonempty. (c) A - B is uncountable.

8.

LetA" A 2 , A 3, ... ,An be denumerable sets. (a) Prove thatA I X (A 2 X A 3) and (AI X A 2 ) X A3 are denumerable. (Recall from exercise 16 of section 3.1 that these sets are not equal.) (b) Let B = {(a" a2, a3): ai E Ai, i = 1,2, 3}. Prove that B is denumerable. (c) Let Cn = {(ai, az, ... , an): ai EA i, i = 1, ... ,n}. Prove that Cn is denumerable, for every natural number n.

"* "*

(-00, b), for any reai number b [1,2) U (S, 6) (3, 6) U [10, 20)

9.

Prove Theorem S.13 directly. That is, if A and B are denumerable sets, construct a function from A X B to N that is one-to-one and onto N.

10.

GiveanotherproofofTheoremS.lSby showingthat!(x) = (x - !)/[x(x - I)] is a one-to-one correspondence from (0, I) onto IR.

11.

IR X IR has cardinality c. Use this fact to prove that the set C of complex numbers, has cardinality c.

5.3

Proofs to Grade

12.

*

209

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. The sets E+ of even natural numbers and D+ of odd natural numbers are equivalent. "Proof." E + is an infinite subset of N. Thus E + is denumerable. Likewise D+ is an infinite subset of N and is denumerable. Therefore, E+ =D+.

(b)

X

*

(c)

*

(d)

(e)

-

Claim. If A is infinite and x ~ A, then A U {x} is infinite. "Proof." Let A be infinite. Then A"'= N. Let I: N --+ A be a one-to-one correspondence. Then g: N --+ A U {x}, defined by g(t)= { l(t-l)

5.3

Countable Sets

if t = 1 ift>1

is one-to-one and onto A U {x}. Thus N = A U {x}, so A U {x} is infinite. Claim. If AU B is infinite, then A and B are infinite. "Proof." This is a proof by contraposition so assume the denial of the consequent. Thus assume A and B are finite. Then by Theorem 5.6, A U B is finite, which is a denial of the antecedent. Therefore, the result is proved. Claim. If a set A is infinite, then A is equivalent to a proper subset of A. "Proof." Let A = {Xl> X2,' .. }. Choose B = {X2, X3""}' Then B is a proper subset of A. The function I: A --+ B defined by I(Xk) = Xk+ 1 is clearly one-to-one and onto B. Thus A = B. Claim. The set N is infinite. "Proof." The function given by I(n) = n + 1 is a one-to-one correspondence between Nand N - {I}, so N is equivalent to a proper subset of N. Therefore, by Corollary 5.10, N is infinite. -

Countable Sets Countable sets have been defined as those sets that are finite or denumerable. Such sets are countable in the sense that they can be "counted" by using subsets of N: A finite non empty set is counted by using exactly the elements of some N k , while a denumerable set may be counted by using exactly the elements of N. This section will survey some of the important results about denumerable and countable sets. First we will deal with the cardinality of the rational numbers Q. Since N ~ Q ~ ~, you should suspect that the cardinality of Q is Xo, or c, or some infinite cardinal in between (the ordering of cardinal numbers is discussed in the next section). You might also suspect that, since there are infinitely many rationals between any two rationals, Q is not denumerable, but this is not the case. Georg Cantor (1845-1918) first showed that Q+ (the positive rationals) is indeed denumerable through a clever rearrangement of Q+. Every element in Q+ may be expressed as ~ for some p, q E N. Thus the elements of this set can be presented as in figure 5.5, where the nth row contains all the positive fractions with denominator n.

210

CHAPTER 5 Cardinality

'

//~/~

-I

-2

~ ll~~ -3

7 I

4 5 6 2 -

7 2

5 3

6 3

7 3

5 4

6 4

7 4

5 5

6 5

7 5

4 5

Figure 5.5 To show that Q+ is denumerable, Cantor gave a "diagonalization argument" in which the elements of Q+ are listed in the order indicated in figure 5.5. First are all then fractions in which the sum of the numerator and denominator is 2 (only those whose sum is 3 ~), then those whose sum is 4, and so on. We omit from the (= r), (= ~ = ~ (= ~), and all other fractions not in lowest terms. list ~ (= The remaining numbers are circled in the array. The result is this one-to-one correspondence from N to Q+:

i

t),1

(r,

t),

t),

N = {I, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ... }

1 1 1 1 1 1 1 1 1 1 1 1 1 1

Q+ =

{Il'

213143215165 3' l' 2' 3' 4' l' 5' l' 2'

l' 2' l'

~, ... }

This correspondence can be used to establish the following theorem (we omit the details). See also the discussion following Theorem 5.29.

Theorem 5.16

The set Q+ of positive rational numbers is denumerable. Adding one or any finite number of elements to a finite set yields a finite set. The next three theorems are analogous results for denumerable sets. An important distinction may be made between finite sets and denumerable sets: adding a finite number of elements or a denumerable number of elements does not change the cardinality of a denumerable set. Before we proceed with the theorems, we give an example.

Example. The setA = {... , -10, -8, -6, -4, -2} of negative even integers is a denumerable set. (The function/: N -+ A given by lex) = -2x is a bijection, so N = A.) Adding one additional integer, say 6, to A will result in a denumerable set: {... , -10, -8, -6, -4, -2,6}. Adding a finite number of additional integers, for example 2, 4, 6, and 8, yields a denumerable set: {... , -10, -8, -6, -4, -2,2,4, 6, 8}.

5.3

Countable Sets

211

Even adding a denumerable number of elements, say the nonnegative even integers, still yields a denumerable set: {... , -10, -8, -6, -4, -2,0,2,4,6,8, 1O, ... }.

Theorem 5.17

If A is denumerable, then A U {x} is denumerable. If x E A, then A U {x} = A, which is denumerable. Suppose that x fl. A. Since N = A, there is a one-to-one function f: N ---+ A that is onto A. Define g: N ---+ A U {x} by

Proof.

g(n) = {fen - I)

if n = I if n > I

It is now straightforward to verify that g is a one-to-one correspondence between N and A U {x}, which proves that A U {x} is denumerable. Theorem 5.16 may be loosely restated as Xo + I = Xo. Its proof is illustrated by the story of the Infinite Hotel, I' whose rooms are numbered I, 2, 3, 4, .... The Infinite Hotel has Xo rooms and is full to capacity with one person in each room. You approach the desk clerk and ask for a room. When the clerk explains that each room is already occupied, you say, "There is room for me! For each n, let the person in room n move to room n + I. Then I will move into room I, ·and everyone will have a room as before." There are Xo + I people and they fit exactly into the Xo rooms. See figure 5.6.

I Lobby I

Figure 5.6 t The Infinite Hotel is one of the topics discussed in Aha! Gotcha: Paradoxes to Puzzle and Delight by Martin Gardner (Freeman, New York, 1981).

212

CHAPTER 5 Cardinality

Rooms in the Infinite Hotel can also be found for any finite number of additional people (Theorem 5.18) or any denumerable number of additional people (Theorem 5.19). In fact, the clerk could find rooms if a denumerable number of additional people arrive a finite number of times (Theorem 5.21) or even a denumerable number of times (Theorem 5.22).

Theorem 5.18

If A is denumerable and B is finite, then A U B is denumerable.

Proof.

Theorem 5.19

•

Exercise 3.

If A and B are disjoint denumerable sets, then A U B is denumerable. 1-1

Proof.

I-I

.

Letf: N"0iit6A and g: N"0iit6B. Define h: N --+ AU B via

_{f(n;l) hen) -

g(~)

if n is odd if n is even.

It is left as exercise 4 to show that h is a one-to-one correspondence from N onto

AUR

•

Before finishing the theorems about denumerable sets, we note that Theorems 5.16, 5.17, and 5.19 provide a simple proof that the set of all rational numbers is denumerable.

Theorem 5.20

The set Q of all rational numbers is denumerable.

Proof. By Theorem 5.16, Q+ is denumerable. Likewise, the set of negative rationals, Q-, is denumerable. (The function f: Q+ -> ((J)- that assigns f(x) = -x is a bijection.) Therefore, Q+ U ((J)- is denumerable by Theorem 5.19. Finally, Q = (Q+ U ((J)-) U {a} is denumerable by Theorem 5.17. • The next theorem extends Theorem 5.19 to a finite union of pairwise disjoint denumerable sets. Its proof, as you might expect, is by induction on the number of sets in the family.

Theorem 5.21

Let {A;: i = 1, 2, ... , n} be a finite pairwise disjoint family of denumerable sets. n

Then

U Ai is denumerable. i=1

Proof.

See exercise 5.

•

We can extend Theorem 5.21 to the union of a denumerable number of pairwise disjoint denumerable sets, but this result (Theorem 5.22) cannot be proved from the properties of sets we have used thus far. The axioms of set theory allow us

5.3

Countable Sets

213

to form unions, intersections, power sets, and infinite sets such as N, Z, G, and even IR. However, a new property (The Axiom of Choice) to be introduced in section 5.5 is needed to prove Theorem 5.22. This does not make Theorem 5.22 suspect, for the Axiom of Choice is a widely accepted and useful tool. A complete proof of Theorem 5.22 is given in section 5.5.

Theorem 5.22

Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerable sets. That is, if i E N, then Ai is denumerable, and if i, j E Nand i =1= j, then Ai =1= Aj and Ai n Aj = 0. Then Ai is denumerable.

U

i=N

We continue with a sequence of important results about denumerable and countable sets, leading to the fact that the union of any countable collection of coutable sets is countable.

Theorem 5.23

Every subset of a countable set is countable.

Proof. Let A be a countable set and let B ~ A. If B is finite, then B is countable. Otherwise, B is infinite, so A is infinite and therefore denumerable. Let f be a bijection from A to N. Then fiB is also a bijection from B to f (B). We now define a function g: N ---+ feB) by induction. Let g(l) be the smallest integer in feB). For each n E N,f{B} - {g{I}, g{2}, ... , g{n)} is nonempty because feB) is infinite. Let g(n + 1) be the smallest element in feB) - {g{I}, g{2}, ... , g{n)}. If r, sEN and r < s, then g(r) E {g{I}, g{2}, ... , g{s - I)} but g(s) is not; thus g(r) =1= g(s). Thus g is an injection. Also, if t Ef(B), and there are k natural numbers less than t inf(B), then t = g(k + I). Thus g is a bijection from N tof(B). Therefore, N = feB) = B and so B is denumerable. We have seen that the set G of all rational numbers is denumerable and therefore countable. Thus such sets as {k n EN}, G n (0, 1), {~,~, ~}, and Z are countable sets.

Corollary 5.24

A set A is countable iff A is equivalent to some subset of N.

Proof. If A is countable, then A is either finite or denumerable. Thus A = 0, or A = Nk for some kEN, or A = N. In every case, A is equivalent to some subset of N. If A is equivalent to some subset of N, then A is equivalent to a countable set, since all subsets of (countable) N are countable. Therefore, A is countable. -

Theorem 5.25

If A and B are countable sets, then A U B is countable.

Proof. This theorem has been proved in the cases where A and B are finite (Corollary 5.7), where one set is denumerable and the other is finite (Theorem 5.18), and where A and B are denumerable and disjoint (Theorem 5.19). The only remaining case is where A and B are denumerable and not disjoint. Write A U B asA U (B - A), a union of disjoint sets. Since B - A ~ B, B - A is either finite or denumerable by Theorem 5.23. If B - A is finite, then A U B is denumerable by Theorem 5.18, and if B - A is denumerable, then A U B is denumerable by Theorem 5.19. -

214

CHAPTER 5 Cardinality

Theorem 5.25 may be extended (by induction) to any finite union of countable sets.

Theorem 5.26

Let .sil be a finite collection of countable sets. Then .sil is countable.

Proof.

•

See exercise 7.

Theorem 5.26 may, in tum, also be extended to any countable union of countable sets. For example, let.sil = {An: n EN}, where An = {x E Z: x :::; n}. Then

1... '

Al = -2, -I, a, I}, A 2 = ... ,-2,-I,a,I,2}, A3 = ... , -2, -I, a, 1,2, 3}, and so on. Each An is countable and

U An = N is countable. Even though .sil is a nEN

denumerable collection of denumerable sets, we cannot use Theorem 5.22 to conclude that the union is denumerable because the family .sil is not pairwise disjoint. However, the family .sil does determine another family with the same union. Let CJ3 be the family {Bn: n EN}, where B, =AI ={ ... , -2, -I,a, I}, B2 = A2 - A, = {2}, B3 = A3 - (A 2 UA,) = {3}, B4 = A4 - (A3 UA 2 UA,) = {4},

and so on. This family is a pairwise disjoint collection. Furthermore,

U Bn = nEN

U AI!" Such a family CJ3 may be defined for any denumerable family of sets (in nEN

Lemma 5.27) and used to show that the union of any denumerable family of denumerable sets is denumerable (Theorem 5.28).

Lemma 5.27

Let {Ai: i E N} be a denumerable family of sets. For each i E N, let B;

A; -

(CJ A

k ).

=

Then {B;: i E N} is a denumerable family of pairwise disjoint sets and

k='

UA;=

UBi.

iEN

iEN

Proof.

See exercise 8.

•

5.3

Theorem 5.28

Countable Sets

Let {Ai: i E N} be a denumerable family of denumerable sets. Then denumerable.

Proof. Al ~

215

U Ai

IS

iEN

UAi, so the union is infinite by exercise 7(b) of section 5.1. Ifwe let iEN

Bi

=

Ai -

(0

A k ), then by Lemma 5.27 it is sufficient to show that UBi is iEN

k=!

denumerable.

If all the Bi, i E N, are denumerable, then by Theorem 5.22,

U Bi is denumeriEN

able. Likewise, if all the Bi , i E N, are finite, then

U B; is denumerable (exercise 9). ;EN

Let ~I = {Bi: Bi is finite} and ~2 = {B;: B; is denumerable}. Both ~I and ~2 are nonempty and UBi = ( ;EN

U

B) U (

BE!'Jl 1

U

B).

If~1 is finite, then U

BE!'Jl2

B is fi-

BE!'Jl 1

U B is denumerable by Theorem 5.22. If~1 is infinite, then U B is denumerable (exercise 9 again). If ~2 is finite, then U B is denumerable (by Theorem 5.21); otherwise U B is denumerable by Theorem 5.22. Thus U B; is either the union of a finite set and a denumerable set or the union of two denumerable sets. In either case, U B; is denumerable. • nite (by Theorem 5.7) and

BE!'Jl2

BE!'Jll

BE!'Jl2

BE!'Jl2

;EN

iEN

We can pull all these results together in one simply stated theorem:

Theorem 5.29

Let .il be a countable collection of countable sets. Then

U.il is countable.

Proof. You should check to see that every case is considered, either as a theorem or an exercise.

•

Theorem 5.29 provides another means to prove that the positive rational Q+ is denumerable. Q+ is certainly infinite, and Q+ may be written as Am where

U

nEN

An = {~: a EN}. For each n E N, An is denumerable. By Theorem 5.29, Q+ is denumerable.

216

CHAPTER 5 Cardinality

Example. Each set Nb kEN is a finite set and therefore countable. Their union, Nk = N, is of course countable.

U

kEN

Example. For each n E N, let Bn = {-n, n}. The countable union of these countable sets is 71 - {O}, which is countable. We will give one more example of a denumerable set, one with particular significance for computer programmers. A computer program is written in a given programming language and consists of a finite sequence of symbols. These symbols are selected from a finite set called an "alphabet" (typically all 26 upper and lowercase letters, the 10 digits, a blank space, certain punctuation marks, arithmetic operations, etc.). For any particular language, let PI be all programs of length I, let P2 be all programs of length 2, P3 be all programs of length 3, and so on. We note that in most programming languages, the first few P; are empty sets. An example of an element of Pso in the language BASIC is

Program Xl Let X = 1 Let Y = 2 Print "X + Y is End

II

•

I

X + Y

(10 symbols) (9 symbols) (9 symbols) (19 symbols) (3 symbols)

Since any computer program is finite in length, every program is an element of Pn for some n. Thus, the set of all possible programs is

By Theorem 5.29, this countable union of countable sets is countable. Hence only a countable number of programs could ever be written in a given language. However, we have seen that the set of all functions from N to {O, I} is uncountable. This means that there are many functions from N to {O, I} for which there can be no computer programs to compute them! Put a different way, there are not enough solutions (programs) for all the possible problems (functions).

Exercises 5.3 1.

What is the 28th term in the sequence of positive rationals given in the discussion of Theorem 5.16?

2.

Explain how a diagonalization argument can be used to show that the set of rationals {2X /3 Y: x, yEN} is denumerable. Omit details of the proof.

3.

Prove Theorem 5.18 by induction on the number of elements in the finite set B.

4.

Complete the proof of Theorem 5.19 by showing that the function h as defined is one-to-one and onto A U B.

5.3 *:

Countable Sets

217

5.

Prove Theorem 5.21: The union of a finite pairwise disjoint family of denumerable sets {Ai: i = I, 2, ... , n} is denumerable.

6.

Use the theorems of this section to prove that (a) an infinite subset of a denumerable set is denumerable. (b) Q n (1,2) is denumerable. 20

(c)

U ((j) n (n, n + I») is denumerable. n=l

(d)

U ((j) n (n, n + I») is denumerable. nEN

(e)

U {nk: kEN} is denumerable. nEN

*:

2

7.

Prove Theorem 5.26 by induction on the number of sets in the family si.

8.

Complete the proof of the set equality in Lemma 5.27.

9.

Prove that if {Bi: i E N} is a denumerable family of pairwise disjoint distinct Bi is denumerable. finite sets, then

U

iEN

10. The Infinite Hotel is undergoing some remodeling, and consequently some of the rooms will be taken out of service. Show that, in a sense, this does not matter as long as only a "few" rooms are removed. That is, (a) prove that if A is denumerable and x E A, then A - {x} is denumerable. (b) prove that if A is denumerable and B is a finite subset of A, then A - B is denumerable. *: 11. Use an argument similar to the diagonalization argument for Theorem 5.16 to prove Theorem 5.13 (if A and B are denumerable sets, then A X B is denumerable). 12. Give an example, if possible, of (a) a denumerable collection of finite sets whose union is denumerable. (b) a denumerable collection of finite sets whose union is finite. (c) a denumerable collection of pairwise disjoint finite sets whose union is finite. 13. (a) Let S be the set of all sequences of O's and I's. For example, 1 0 1 0 1 0 1... , 0 I I I I 1 I I I ... and 0 1 0 I I 0 I I 1... are in S. Prove that S is uncountable. *: (b) Let Tbe the set of all sequences of 0' sand l' s where all but a finite number of terms are O. Use a diagonalization argument, like that in Theorem 5.16, to show that Tis a denumerable subset of S.

*

14. Let A be a denumerable set. Prove that (a) the set of all I-element subsets of A, {B: B ~ A and B = I} is denumerable. (b) the set of all 2-element subsets of A, {B: B ~ A and II = 2} is denumerable. (c) for any kEN, {B: B ~A and II = k} is denumerable. (d) thnet of all finite subsets of A, {B: B ~ A and B is finite} is denumerable. (Hint: Use Theorem 5.22.)

218

CHAPTER 5 Cardinality

Proofs to Grade

15.

Prove that if A and B are countable, then A X B is countable.

16.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If A is denumerable, then A - {x} is denumerable. "Proof." Assume A is denumerable. Case 1. Ifx E,lA, then A - {x} =A, which is denumerable by hypothesis. Case 2. Assume x EA. Since A is denumerable, there exists I: N ~~~lA. Define g by setting g(n) = I(n + 1). Then g: N ~:td (A - {x}), so N = A - {x}. Therefore, A - {x} is denumerable. _ (b) Claim. If A and B are denumerable, then A X B is denumerable. "Proof." Assume A and B are denumerable, but that A X B is not denumerable. Then A X B is finite. Since A and B are denumerable, they are not empty; therefore, choose a E A and bE B. By exercise 3(b) of section 5.1, A = A X {b}, and by an obvious modification of that exercise, B = {a} X B. Since A X B is finite, the subsets A X {b} and {a} X B are finite. Therefore, A and B are finite. This contradicts the statement that A and B are denumerable. We conclude that A X B is denumerable.

*

(c)

(d)

-

Claim. (Theorem 5.16) The set Q+ of positive rationals is denumerable. "Proof." Consider the positive rationals in the array in figure 5.5 (Theorem 5.16). Consider the order formed by listing all the rationals in the first row, then the second row, and so forth. Omitting fractions that are not in lowest terms, we have an ordering of Q+ in which every positive _ rational appears. Therefore, Q+ is denumerable. Claim. If A and B are infinite, then A = B. "Proof." Suppose A andB are infinite sets. LetA = {ai' az' a3, ... } and B = {b J, b2 , b3 , ... }. Define I: A ---+ B as in the picture

{al> a2, a3, a4, ... }

! 1 1 1 {bl> b2 , b3 , b4 , ..• } Then, since we never run out of elements in either set, and onto B, so A = B.

5.4

I

is one-to-one _

The Ordering of Cardinal Numbers The theory of infinite sets was developed by Georg Cantor over a 20-year span, culminating primarily in papers that appeared during 1895 and 1897. He described a cardinal number of a set M as "the general concept which, with the aid of our intelligence, results from M when we abstract from the nature of its various elements and from the order of their being given." This definition was criticized as being less precise and more mystical than a definition in mathematics ought to be. Other definitions were given, and eventually the concept of cardinal number was made precise. One way this may be done is to determine a fixed set from each equivalence class

5.4

The Ordering of Cardinal Numbers

219

of sets under the relation = , and then to call this set the cardinal number of each set in the class. Under such a procedure we would think of the number 0 as being the empty set and the number I as being the set whose only element is the number O. That is, 1 = {o}; 2 = {o, I}; 3 = {o, I, 2}, and so on. We will not be concerned with a precise definition of a ~rdinal number. For our purposes the essential point is that the cardinal number A of a set A is an object ass~iated with all sets equivalent to A and with no other set. The double overbar on A is suggestive of the double abstraction referred to by Cantor. Another common notation for the cardinal number of set A is IA I. Cardinal numbers may be ordered (compared) in the following manner:

DEFINITIONS

= B; otherwise A oF B. 1-1 B if and only if there exists f: A--+B. A< Bif and only if A:S Band AoF B. A = B if and only if A =

A

:S

=

A < B is read "the cardinality of A is strictly less tha!!. th~ cardi.!!ality of B" while :S is read "Ie~ th~n or ~uaIJo." In addition, we use A 1:. B and A "$ B to denote the denials of A < ~ anQ.A :S B, respectively. Usually a proof of1::s §. will involve constru£.tin~a one-to-one function from A to B, while a proof of A < §. wi!! have a proof of A :S B together with a proof, generally by contradiction, that A oF B. Once we have developed some properties Qf ca!:::, dinal inequalities those facts can be used to prove statements of the form A :S B without resorting to the construction of functions. Since 1, 2, 3, ... are cardinal numbers, the natural numbers may be viewed as a subset of the collection of all cardinal numbers. In this sense the properties of :S and < that we will prove for cardinal numbers in the next theorem may be viewed as extensions of those same properties of :S and < that hold for N. Parts (a), (b), (d), and (f) are left as exercise 6. Theorem 5.30

For sets A, B, and C, (a) (b) (c) (d) (e) (f)

A :§A._ (RefkxivJ!y) If 1: = §. and §. = ~, then 1: = ~. !f A :§ B a,!!d B...? C..1. the.!! A :S C.

(Transitivity of =) (Transitivity of:S)

A :S B iff A < B or A = B. If A C B, then A :S B.

A :S 13 iff there is a subset W of B such that W = A.

Proof. (c) Suppose A:s Band B:s C. Then there exist functions f: A..!..=.!.B and 1-1 [[ B-=---C. Since the composite go f: A -+ C is one-to-one, we conclude

A :sC. (e)

Let A C B. We noteJhatJhe inclusion map i: A -+ B, given by i(a) = a, is one-to-one, whence A :S B. •

220

CHAPTER 5 Cardinality

Example. k < Xo for every finite cardinal number k. From Nk ~ N we have k :::; Xo, and since Nk is not equivalent to N, k < Xo, from Theorem 5.30(d). Example. We have implied that Xo < c in previous sections. This is true because (i) Xo :::; c follows from N ~ ~ and (ii) N is not equivalent to ~. Although Cantor developed many aspects of the theory of infinite sets, his name remains attached particularly to the next theorem, which states that the cardinality of a set A is strictly less than the cardinality of its power set. We already know the result to be true if A is a finite set with n elements, since QJ>(A) has 2n elements (by Theorem 2.4) and n < 2 n for all n E N. The proof given here holds for all sets A.

Theorem 5.31

(Cantor's Theo~m)_ A < QJ>(A).

For every set A,

Proof. To show A < QJ>(A), we must show that (i) A:::; QJ>(A), and (ii) A *- QJ>(A). Part (i) follows from the facLthat F: A -+ QJ>(A) defined by F(x) = {x} is one-to-one. To prove (ii), suppose A = QJ>(A); that is, assume A = QJ>(A). Then there exists g: A ~~td QJ>(A). Let B = {y E A: y t!. g(y)}. Since B ~ A, BE QJ>(A), and since g is onto QJ>(A), B = g(z) for some z EA. Now either z E B or z t!. B. If z E B, then z t!. g(z) = B, a contradiction. Similarly, z t!. B implies z E giz), which implies z E B, again a contradiction. We conclude A :f;QJ>(A) and hence A < QJ>(A). • Cantor's Theorem has some interesting consequences. First, there are infinitely many infinite cardinal numbers. We know one, Xo, which corresponds to N. By Cantor's Theorem, Xo < QJ>(N). Since QJ>(N) is a set, its power set QJ>(QJ>(N)) has a strictly greater cardinality than that of QJ>(N). In this fashion we may generate a denumerable set of cardinal numbers: Xo < QJ>(N) < QJ>(QJ>(N)) < QJ>(QJ>(QJ>(N))) < ... Exactly where c, continuum, fits within this string of inequalities will be taken up later in this section. It is also an immediate consequence of Cantor's Theorem that there can be no largest cardinal number (see exercise 7). In section 5.1 we showed that the set cg; of all functions from N to {O, I} is equivalent to QJ>(N). Since N < QJ>(N), we know there are uncountably many functions from N to {O, I}. Since a function from N to {O, I} is a sequence of 0' sand 1's, Cantor's Theorem provides another proof for exercise 13(a) of section ~3. _ It appears to be obvious that if B has at ~ast..ils many elements as A (A :::; Ii), a.Qd A_has at least as many elements as B (Ii:::; A), then A and B are equivalent (A = Ii). The prooL ho~ever,ls n~ obvious. The situation mar_~e represente9 ~~ in figure 5.7. From A :::; Band B:::; A, there are functions F: A--B and G: B----+A. The problem is to construct H: A -+ B, which is both one-to-one and onto B. Cantor solved this problem in 1895, but his proof used the controversial Axiom of Choice (section 5.5). Proofs not depending on the Axiom of Choice were given independently by Ernest SchrOder in 1896 and two years later by Felix Bernstein.

5.4

The Ordering of Cardinal Numbers

A

221

B

F

C = Rng (G)

G

Figure 5.7

Theorem 5.32

((!nto!-S(h!iid~-Berl!Ste~

If A ::5 Band B ::5 A, then A =

Theorem)

B.

Proof. We may assume that A and B are disjoint, for otherwise we could replace A and B with the equivalent disjoint sets A X {a} and B X {I}, respectively. Let 1-1 1-1 F: A---+B, with D = Rng (F), and let G: B---+A, with C = Rng (G). If B = D we already have A = B, so assume B - D =1= 0. Define a string to be a function f: N ---> A U B such that f(1) EB - D, fen) EBimpliesf(n fen) E A implies fen

+ 1) = G(j(n», and + 1) = F(j(n».

We think of a string as a sequence of elements of A U B with first term in B - D, and such that thereafter the terms are alternately in C and in D. Each element of B - D is the first term of some string. See figure 5.8.

G

F G

F G

Figure 5.8

String I: 1(1), 1(2), 1(3), 1(4), 1(5), 1(6), ...

222

CHAPTER 5 Cardinality H

Figure 5.9 Let W = {x EA: x is a term of some string}. We note that We Rng (G) and that x E W iff x = f(2n) for some string f and natural number n. Let H: A -+ B be given by _ {F(X) H(x) - G-1(x)

if x ff. W ifxEW.

See figure 5.9. We will show that H is a one-to-oe correspondence from A onto B. Suppose x, yEA and H(x) = H(y). We will first show that x and y must both be in W or both in A - W. (We will use a proof by contradiction to show this.) Suppose x E A - Wand yEW. (The symmetric situation where yEA - Wand x E W produces a similar contradiction.) In this case, H(x) = H(y) is F(x) = G-1(y), so y = G(F(x)). Since yEW, Y = f(2n) for some string f and some natural number n. Therefore,f(2n - I) = F(x) (because G(j(2n - I)) = f(2n) = y = G(F(x)) and G is one-to-one). If n = 1, then f(1) = F(x), which implies f(I) E Rng (F), a contradiction to the definition of string f Thus n:::=:: 2. But then f(2n - 2) = x, since f(2n - 1) = F(x). This implies that x is a term in the string!, a contradiction to x E A - W. Therefore, we know that x and yare either both in W or both in A - W. If x, yare both in W, then H(x) = H(y) implies that G-1(x) = G-1(y). Therefore, x = y since G- 1 is one-to-one. Likewise, if x, yare both in A - W, then F(x) = F(y) and x = y because F is one-to-one. In either case, we conclude x = y and H is one-to-one. Next we show that H is onto B. Let bE B. We must show that b = H(x) for some x EA. There are two cases: Case 1. Case 2.

If G(b) E W, let x = G(b). Then H(x) = H(G(b)) = G- 1 (G(b)) = b. If G(b) ff. W, then bE Rng (j). (If b ff. Rng (j), then bE B - D. Therefore, b is the first element of some string and G(b) is the second element of that string, a contradiction to G(b) ff. W.) Since bE Rng (f), there exists x E A such that F(x) = b. Furthermore, x ff. w. (If x E W, then x is

5.4

The Ordering of Cardinal Numbers

223

a term in some string and therefore F(x) and G(F(x)) are the next two terms of the same string. But this is a contradiction, since G(F(x)) = G(b) and we are in the case where G(b) is not on any string.) From x EA - W we conclude H(x) = F(x) = b.

•

In both cases, H(x) = b, so H is onto B.

The Cantor-Schrader-Bernstein Theorem may be used to prove equivalence between sets in cases where it would be difficult to exhibit a one-to-one correspondence.

Example. We will show that (0, I) = [0, 1]. First, note that (0, I) <: [0, I], so by Theorem 5.30(e), (0,1)::5[0,1]. Likewise, since [0,1]<:(-1,2), we have [0, 1] <: (-I, 2). But we know (0, I) = (-I, 2) and thus (0, I) = (-1,2). Therefore, we may write [0, I] ::5 (0, I ). We conclude (0, I) = [0, I] by the Cantor-SchraderBernstein Theorem and thus (0, I) = [0, I]. Example. The Cantor-Schrader-Bernstein Theorem can be used to determine the relationship of c, the cardinal number of the open interval (0, I), to the increasing sequence of cardinal numbers N

< rJ>(N) < rJ>(rJ>(N)) < rJ>(rJ>(rJ>(N))) < ....

First, recall that any real number in the interval (0, I) may be expressed in a base 2 (binary) expansion 0.b)b 2b 3b4... , where each b i is either or 1. Ifwe exclude repeating 1's, such as 0.01011111 I 1.. . (which has the same value as 0.01 100000 ... ), then the representation is unique. Thus we may define a function f: (0, I) ---+ rJ>(N) such that for each x E (0, 1),

°

f(x) =

{n E N: bn

=

1 in the binary representation ofx}.

The uniqueness of binary representations assures that the function is defined and is one-to-one. We note that f is not onto rJ>(N) since inductive sets such as {5, 6, 7, 8, 9, ... } are not images of any x E (0, I). Sincefis one-to-one, (0, 1)::5 rJ>(N). Next, define g:rJ>(N) ---+ (0, I) by g(A) = 0.a)a2a3a4"" where

_{25

an -

ifnEA if n I$A.

For any set A <: N, g(A) is a real number in (0, I) with decimal expansion consisting of2's and 5's. (Any pair ofdigits not 9 will do.) The function g is one-to-one but certainly not onto (0, 1). Therefore, rJ>(N) ::5 (0, 1). _ _By the Cantor-Schrader-Bernstein Theorem, rJ>(N) = (0, IL Therefore, ~ = c. We can now identify the first two terms of the sequence N < rJ>(N) < rJ>(rJ>(N)) < rJ>(rJ>(rJ>(N))) < ... as being Xo and c. The Cantor-Schrader-Bernstein Theorem is another result in the extension of the familiar ordering properties of N to properties for all cardinal numbers. It, in tum, leads to others. In the following, parts (a) and (c) are proved; (b) and (d) are given as exercise 13.

224

CHAPTER 5 Cardinality

Corollary 5.33

For sets A, B, and C, if;i ::5 ~, the~E 1::_A. if;i::5 ~ and ~ < f, then;i < f· if;i < ~ and ~ ::5 f, then;i < f. if A < E and E < C, then A < C.

(a)

(b) (c) (d)

Proof. (a)

(c)

"*

Supp.Qse ~ < A. Then A E and E ::5 A. Combining this with the hypothesis Qlat ;i ::5 E, we conclude by the Cantor-Sc.hrod~r-Bernstein Theorem that A = E, wQich.is a c~tra~ction. !.her~ore, E 1:: A. SuppQ§eA_< E andj1::5 Then A ::5 E; sQJJy '"Q1eorel!!..5.21(c), A..? C~Sup­ pose 1:: (, Then A = C, which imp.lies ~ ::5 A. But E ::5 C and C ::5 A implies E::5 A. Combining this with... A:§. E, we conclude by th~ C~tor­ Schro-.ger-~ernstein Theorem that A = E. Since this contradicts A < E, we _ have A
JL

S.

It is tempti~ to ~xtend .Qur r~sults even further to include the converse of Corollary 5.33ia): JIE 1::j1, t~nA ::5 E." (As far as we know now, for two given sets A and B, both A ::5 E and E::5 A may be false.) The Cantor-Schroder-Bernstein Theorem turned out to be more difficult to prove than one would have guessed from its simple statement, but the situation regarding the converse of Coroll~ 513(a) is ~en '!pore remarkable. This is _disc.!!sseQ. in --.?ectism ~5, where "If E 1:: A, then A::5 E" is rephrased as "Either A < E or A = E or E < A."

Exercises 5.4

"*

1.

Prove that for any natural number n, n < c.

2.

Prove that QP(N) < QP(IR).

3.

Prove that if A ::5 E and E = C, then A ::5 C.

4.

Prove that if A ::5 B and A = C, then C ::5 B.

5.

State ~hefuer each of the following is true or false. (a) A::5 E im}21ies that A <;::; B. (b) 1n!!.::5E. (c) ;i::5 ~ implies ~(AL::5 QP(B). (d) A = E implies A ::5 E.

*

"*

6. Prove the remaining parts of Theorem 5.30. 7.

Prove that there is no largest cardinal number.

8.

Arrange the following cardinal numbers i.!!.-order: ~;;;;=;;~ (a) (0,1), [0, 1], {O,T}, 101, ~, Q, 0, IR - g, QP(QP(IR)), 1R (b) Q U {n}, IR - {n}, QP({O, I}), [0,2], (0, 00), E, IR - Il, QP(IR)

9.

Apply the proof of the Cantor-Schroder-Bemstein Theorem to this situation: A = {2, 3, 4, 5, ... },B = t,···},F: A-+BwhereF(x) = x!6,andG: B-+A

*

H, t,

5.4

The Ordering of Cardinal Numbers

t

225

where G(x) = i + 5. Note that and! are in B - Rng (F). Let J be the string that begins at and let g be the string that begins at (a) Find J(l ),f(2),f(3),f(4). (b) Find g(1), g(2), g(3), g(4). (c) Define Has in the proof of the Cantor-SchrOder-Bernstein Theorem and find H(2), H(8), H(l3), and H(20).

t,

!.

I-I

I-I

I-I

10. Suppose there exist three functions J: A--B, g: B--C, and h: C--A. Prove A = B = C. Do not assume that the functions map onto their codomains. 11. If possible, give an example of (a) functions J and g such that J: ~N, g: N1.=..!.Q, but neither J nor g is an onto map. I-I (b) a functionJ: IR--N. 1-1 (c) a function J: 0J>\~~--N. (d) a functionJ: IR--Q. 12. Prove that if there is a functionJ: A -+ N that is one-to-one, then A is countable. 13. Prove parts (b) and (d) of Corollary 5.33. 14. Use a cardinality argument to prove that there is no universal set V of all sets. 15. Use the Cantor-SchrOder-Bernstein Theorem to prove the following. (a) The set of all integers whose digits are 6, 7, or 8 is denumerable. (b) IR X IR = IR. Note that this means there are just as many points on the real line as there are in the Cartesian plane. (c) .!f A <: IR and there exists an open interval (a, b) such that (a, b) <: A, then

*

A=c. Proofs to Grade

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades ot~r t~n A. _ 16. (a) Claim. If A ::5 B a'.!...d A-== C, t1:!en ~ ::5 B. "Proof." Assume A ::5 B and A = C. Then there exists a functionJsuch I-I I-I = = thatJ: A--B. Since A-== CJ: C--B. Therefore, C::5 B. (b) Claim. If B <: C and B = C, then B = C. "Proof." Suppose B oF C. Then B is a proper subset of C. Thus C - B oF 0. This implie~C -_B> O. But C = B U (C - B)~nd,~ince B and C - B are disjoint, C = B + (C - B). By hypothesis, B = C. Thus (C - B) = 0.1... a c2..ntrad~tio~ (c) Claim. If A ::5 B a~ B =' C, then A ::5~. "Proof." Assume A ::5 B. Then, since B = C, we have A ::5 C by substitution. (d) Claim. If A oF 0 and A::5 B, then there exists a function J: Bonto,A. = = I-I "Proof." From A ::5 B, we know there exists g: A--B. Fix a particular a* EA. Define J: B -+ A as follows:

*

*

(i) For bE B, if bE Rng (g), then g-I({b}) consists of a single element of A, since g is one-to-one. Definej{b) to be equal to that element ofA. (ii) If bEt: Rng (g), defineJ(b) = a*.

The function J is onto A, for if a E A, let g(a) = b. Then bE Band g-I({b}) = a, soJ(b) = a. -

226

CHAPTER 5 Cardinality

5.5

Comparability of Cardinal Numbers and the Axiom of Choice The goal of section 5.4 was to establish results for the relations:::; and < on the class of all cardinal numbers while thinking of these results as extensions of properties of 1\1. One of the most useful ordering properties of 1\1 is the trichotomy property: if m and n are any two natural numbers, then m > n, m = n, or m < n. The analog for cardinal numbers is stated in the Comparability Theorem.

Theorem 5.34

(The Comparability Theorem)_ If A and B are any two sets, then A <

B, A = B, or B < A.

Surprisingly, it is impossible to prove the Comparability Theorem from the axioms and other theorems of Zermelo-Fraenkel set theory (see section 2.1). In a formal study of set theory one can build up, starting with a few axioms specifying that certain collections are sets, to the study of the natural, rational, real, and complex numbers, polynomial, transcendental, and differentiable functions and all the rest of mathematics. Still, comparability cannot be proved. On the other hand, it is impossible to prove in Zermelo-Fraenkel set theory that comparability is false. Theorem 5.34 is undecidable in our set theory; no proof of it and no proof of its negation could ever be constructed in our theory. At this point we could choose either to assume that Theorem 5.34 is true (or assume true some other statement from which comparability can be proved) or else assume the truth of some statement from which we can show comparability is false. Of course, we have revealed the fact that we want comparability to be true by labeling the statement as a theorem. It has become standard practice by most mathematicians to assume the Comparability Theorem is true by assuming the truth of the following statement:

The Axiom of Choice

If .il is any collection of non~mpty sets, then there exists a function F (called a choice function) from .il to U A such that for every A E .il, F(A) EA. AEd

The Axiom of Choice at first appears to have little significance: From a collection of nonempty sets, we can choose an element from each set. If the collection is finite, then this axiom is not needed to prove the existence of a choice function. It is only for infinite collections of sets that the result is not obvious and for which the Axiom of Choice is independent of other axioms of set theory. Many examples and uses of the Axiom of Choice require more advanced knowledge of mathematics. The example we present is not mathematical in content yet has become part of mathematical folklore. A shoe store has in the stockroom an infinite number of pairs of shoes and an infinite number of pairs of socks. A customer asks to see one shoe of each pair and one sock of each pair. The clerk does not need the Axiom of Choice to select a shoe from each pair of shoes. His choice rule might be "From each pair of shoes, choose the left shoe." However, since the socks of any pair of socks in stock are indistin-

5.5

Comparability of Cardinal Numbers and the Axiom of Choice

227

guishable, and since there are an infinite number of pairs of socks, the clerk must employ the Axiom of Choice to show the customer one sock from each pair. Let.sa = {A: A ~ IR and A =1= 0}. If we are to select one element from each set A in .sa, then we will need to use the Axiom of Choice. However, if we let 'ZA = {A: A ~ IR, A =1= 0, and A is finite}, then we do not need the Axiom of Choice to select one element from each set in 'ZA. Our choice rule might be: for each B E 'ZA, choose the greatest element in B. Since B is finite, such an element exists for each BE'ZA.

Example.

With the Axiom of Choice, we could, but will not, prove Theorem 5.34. For a proof, see R. L. Wilder, Introduction to the Foundation of Mathematics, 2nd ed. (Krieger, 1980). In section 5.3, we delayed the proof of Theorem 5.22 because it uses the Axiom of Choice. We restate the theorem and give its proof now. Let {Ai: i E N} be a denumerable pairwise disjoint family of distinct denumerable sets. That is, if i E N, then Ai is denumerable and if i, j E Nand i =1= j then Ai =1= Aj and Ai n Aj = 0. Then Ai is denumerable.

Theorem 5.22 (restated)

U

iEN

Proof.

Since each Ai is denumerable, there exist functions /;: N ---+ Ai' i E N, each of which is one-to-one and onto Ai' We define the function h: N X N ---+ Ai by

U

iEN

hem, n) = fm(n). (To see that h is a bijection, refer tofigure 5.10. Infigure 5./O(a), we have an arrangement ofN X N. The union of the Ai, i E N, is presented in figure 5./O(b), where the nth row contains the elements of Ai as the range offi' The one-to-one correspondence h is easily seen.) Since all the functions fi are bijections, Ai' Since it is an easy exercise to show that h is a bijection. Therefore, N X N = N X N = N, A-I is denumerable. iEN •

U

U

iEN

(1, I)

(1,2)

(1, 3)

(1,4)

(1, 5)

AI:

fl(l)

f,(2)

fl(3)

f,(4)

f,(5)

(2, I)

(2,2)

(2,3)

(2,4)

(2,5)

A 2:

f2(1)

f2(2)

f2(3)

f2(4)

fz(5)

(3, 1)

(3,2)

(3, 3)

(3,4)

(3, 5)

A3:

f3(1)

13(2)

f3(3)

13(4)

13(5)

(4, I)

(4,2)

(4,3)

(4,4)

(4,5)

A4:

f4(1)

f4(2)

f4(3)

f4(4)

f4(5)

(5, 1)

(5,2)

(5, 3)

(5,4)

(5,5)

A5:

f5(1)

f5(2)

f5(3)

f5(4)

f5(5)

(a)

N

X

(b)

N

UAi iEN

Figure 5.10

228

CHAPTER 5 Cardinality

Why was the Axiom of Choice necessary in the proof? For each i E N, A; is denumerable, so there are (many) bijections from A; to N. In our proof, we select one bijection (and call itj;) from the set of all bijections from A; to N. We do this an infinite number of times, once for each i E N. Our collection consisted of sets of bijections, and we needed one bijection from each set. There is no way to select the!; without the Axiom of Choice. Many other important theorems, in many areas of mathematics, cannot be proved without the use of the Axiom of Choice. In fact, several crucial results are equivalent to it.t Some of the consequences of the axiom are not as natural as the Comparability Theorem, however, and some of them are extremely difficult to believe. One of these is that the real numbers can be rearranged in such a way that every nonempty subset of R has a smallest element-in other words, that the reals can be well ordered. Another, called the Banach-Tarski paradox, states that a ball can be cut into a finite number of pieces that can be rearranged to form two balls the same size as the original ball. Actually, this "paradox" is hardly more surprising than exercise l5(b) of section 5.4-that IR X IR = IR, i.e., there are exactly as many points on the x-axis as there are in the entire Cartesian plane, a result that can be proved without the Axiom of Choice. The Axiom of Choice has been objected to because of such consequences, and also because of a lack of precision in the statement of the axiom, which does not provide any hint of a rule for constructing the choice function F. Because of these objections, it is common practice to call attention to the fact that the Axiom of Choice has been used in a proof, so that anyone who is interested can attempt to find an alternate proof that does not use the axiom. We present two more theorems whose proofs require the Axiom of Choice. The first is that if there is a functionjfrom A onto B, then A must have at least as many elements as B. The proof chooses for each b E B, an a E A such that j(a) = b. The second theorem is that every infinite set has a denumerable subset. The axiom is used to define the denumerable subset inductively.

Theorem 5.35

If there exists a function from A onto B, then B :::; A.

Proof._ I(§ = 0, then B k A, so B :::; A. Let j: A ---> B be onto B where B =1= 0. To show B :::; If, we must construct a function h: B ---> A that is one-to-one. Since j is onto B, for each bE B, bE Rng (f) or, equivalently, j-l({b}) =1= 0. Thus .ii = U-1({b}): bE B} is a nonempty collection of nonempty sets. By the Axiom of Choice, there exists a functiong from.ii to j-l({b}) = A such thatg(j-l({b})) E

U

j-l({b})forallbEB. hEB Define h: B--->A by h(b) = g(j-l({b})). It remains to prove h is one-to-one. Suppose h(b l ) = h(b2 ). Then g(j-l({b l})) = g(j-I({b2 })). By definition of g, this element is in bothj-l({b l}) and j-l({b2 }). Thus j-l({b l}) nj-I({b2}) =1= 0. For t See H. Rubin and J. E. Rubin, Equivalents a/the Axiom a/Choice (North-Holland, New Amsterdam, 1963).

5.5

Comparability of Cardinal Numbers and the Axiom of Choice

229

any x Ef-I({b l }) nf- I({b 2}),f(xL= bl1. and f(x) = b2. Therefore, b l = b 2. This • proves that h is one-to-one and that B :5 A.

Theorem 5.36

Every infinite set A has a denumerable subset.

Proof.

We shall inductively define a denumerable subset of A. First, since A is infinite, A*' 0. Choose al EA. Then A - {al} is infinite, hence nonempty. Choose a2 E A - {al}' Note that a2 al and a2 EA. Continuing in this fashion, suppose al"'" ak have been defined. Then A - {al"'" ak} 0, so select any ak+l from this set. By the Axiom of Choice, an is defined for all n E N. The an have been constructed so that each an E A and ai aj for i j. Thus B = {an: n E N} is a subset of A, and the functionfgiven by fen) = an is a one-to-one correspondence from N to B. Thus B is denumerable. •

*'

*'

*'

*'

Theorem 5.36 can be used to prove that every infinite set is equivalent to one of its proper subsets. (See exercise 8.) This characterizes infinite sets because, as we saw in section 5.1, no finite set is equivalent to any of its proper subsets. Theorem 5.36 also confirms that Xo is the smallest infinite cardinal number. For any set~ wUh infinite cardinality, there is a denumerable subset B of A. Therefore, Xo = B :5 A.

Corollary 5.37

A nonempty ~et A is countable iff there exists a function f: N onlo'A.

Proof.

Exercise 9.

•

We have seen that Xo < c and that the first two terms of the sequence N < Q]>(N) < Q]>(Q]>(N)) < Q]>(Q]>(Q]>(N))) < ... are the cardinal numbers Xo and c. This does not necessarily mean that c is the next largest cardinal numb~r after Xo. Cantor conjectured that this is so: that is, no set X exists such that Xo < X < c. This conjecture, called the continuum hypothesis, is one of the most famous problems in modern mathematics. The combined work of the logicians Kurt Godel in the 1930s and Paul Cohen in 1963 shows that the continuum hypothesis can neither be proved nor disproved in Zermelo-Fraenkel set theory. Like the Axiom of Choice, the continuum hypothesis is undecidable.

Exercises 5.5 1.

*

Indicate whether the Axiom of Choice must be employed to select one element from each set in the following collections. (a) an infinite collection of sets, each set containing one odd and one even integer. (b) a finite collection of sets with each set uncountable.

230

CHAPTER 5 Cardinality

*

(c) (d) (e) (f) (g) (h)

2.

Prove this partial converse of Theorem 5.14 without using the Axiom of Choice. Let A and B be sets with B =1= 0. If B ::5 A then there exists g: A -+ B that is onto B.

3.

Let A and B be any two nonempty sets. Prove that there existsf: A -+ B that has at least one of these properties: (i)

*

an infinite collection of sets, each set containing exactly one integer. a denumerable collection of uncountable sets. {(a, co): a E JR} {A: A ~ N and both A and N - A and infinite} {A: A ~ JR and both A and JR - A are infinite} {A: A ~ JR and A is denumerable}

fis one-to-one

or

(ii)

fis onto B.

A.

4.

Prove that if f: A -+ B, then Rng if)

5.

Suppose A is a denumerable set and B is an infinite subset of A. Prove A

::5

= B.

6.

Suppose B<

7.

Let {Ai: i E N} be a collection of distinct pairwise disjoint nonempty sets. That is, if i andj are in Nand i =1= j, then Ai =1= Aj and Ai n Aj = 0. Prove that Ai includes a denumerable subset.

Cand B$ A. Prove A< C.

U

"* Proofs to Grade

iEN

8.

Let A be an infinite set. Prove that A is equivalent to a proper subset of A.

9.

Prove Corollary 5.37: A nonempty set A is countable iff there is a function f: N -+ A that is onto A.

10. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Jus-

*

*

tify assignments of grades other than A. (a) Claim. There exists an infinite set of irrational numbers, no two of which differ by a rational number. "Proof." For x, y E JR, define the relation S by x S y iff x - y E Q. It is easy to show that S is an equivalence relation. For the family of equivalence classes {xis: x E JR}, choose one element from each equivalence class except the equivalence class Q. The set of all such chosen elements is an infinite set of irrational numbers, no two of which differ by a rational. (You may accept, without proof, the claim that this set is infinite.) (b) Claim. Every infinite setA has a denumerable subset. "Proof." Suppose no subset of A is denumerable. Then all subsets of A must be finite. In particular A ~ A. Thus A is finite, contradicting the assumption. (c) Claim. Every infinite set A has a denumerable subset B. "Proof." If A is denumerable, let B = A, and we are done. Otherwise, A is uncountable. Choose XI EA. If A - {XI} is denumerable, let B= A - {XI}. Otherwise, choose X2 EA - {XI}. If A - {XI' X2} is denumerable, let B = A - {x I' X2}. Continuing in this manner, using the Axiom of Choice, we obtain a subset C = {x I, Xz, . .. } such that B = A - C is denumerable. -

5.5 Comparability of Cardinal Numbers and the Axiom of Choice (d)

(e)

(f)

231

Claim. Every infinite set has two disjoint denumerable subsets. "Proof." Let A be an infinite set. By Theorem 5.36, A has a denumerable subset B. Then A - B is infinite, because A is infinite, and is disjoint from B. By Theorem 5.36, A - B has a denumerable subset C. Then B and C are disjoint denumerable subsets of A. Claim. Every infinite set has two disjoint denumerable subsets. "Proof." LetA be an infinite set. By Theorem 5.36, A has a denumerable subset B. Since B is denumerable, there is a function f: N ~:dBLet C={J(2n):nEN} and D={J(2n-I):nEN}. Then C={J(2),f(4), f(6), ... } and D = {J(1),f(3),f(5), ... } are disjoint denumerable subsets

~A. Claim. Every subset of a countable set is countable. "Proof." Let A be a countable set and let B ~ A. If B is finite, then B is countable by definition. If B is infinite, since B ~ A, A is infinite. Thus A is denumerable. By Theorem 5.36,1l has a ~enu!!1er~le subset C. Thus {; ~!! ~ A, which implies Xo = C and C:::; B :::; A = Xo. Therefore A = B = Xo. Thus B is denumerable and hence countable. -

CHAPTER

6

concepts of Algebra: Groups

Much of modem mathematics is algebraic in nature. We have in mind a broad meaning of algebra as a system of computation and the study of properties of such a system. In this chapter we make precise the idea of an algebraic structure; study an especially important structure, the group; and investigate the notions of substructure and quotient structure.

6.1

Algebraic Strudures Let A be a nonempty set. A binary operation on A is a function from A X A to A. We will usually denote an operation by one of the symbols +, ., 0, or *. If the operation is 0 and the image 0 (x, y) of the ordered pair (x, y) is z, we usually write x 0 y = z. This notion is familiar from the operations of addition and multiplication on the set of real numbers, where we usually write 4 + 7 = 11 instead of + (4, 7) = 11. Often we omit completely the operation symbol and write xy = z, as is done with multiplication. The images xy, x 0 y, and x * yare usually called products, regardless of whether the operations involved have anything to do with multiplication. Similarly, s + t is referred to as the sum of sand t, even when the function + has nothing to do with addition. Besides the usual arithmetic operations on sets of numbers, you are already familiar with several binary operations. Composition is an operation on the set of all functions that map a set A onto itself. The operation U, union, is a binary operation on the power set of A. Intersection, set difference, and symmetric difference are other operations on (lJ>(A). There are operations more complicated than binary operations. Ternary operations, for example, map A X A X A to A. An algebraic structure or system is a

233

234

CHAPTER 6 Concepts of Algebra: Groups nonempty setA together with a collection of (at least one) operations onA and a collection (possibly empty) of relations on A. For, example, the system of real numbers with addition and multiplication and the relation "less than" is an algebraic structure. We could as well consider the system of rea Is with just addition, or the system of reals with multiplication and relation "less than." In this chapter when we say "operation," we mean a binary operation, and when we say "algebraic structure," we mean a structure with one binary operation.

DEFINITION Let A be a set with a binary operation *. Let B be a subset of A. We say that B is closed under the binary operation * iff for all x, y E B, x * Y E B.

For an algebraic system (A, *) the set A is of course closed under * (because the operation * is a function that maps to A). For any subset B of A that is closed under *, we will use the same operation symbol, *, to denote the restriction of the operation to B X B. The three statements "B is closed under *," "* is an operation on B," and "(B, *) is an algebraic system" are all equivalent. The algebraic system of real numbers with the operation' of multiplication has many subsets that are closed under'. The set of rational numbers is closed under multiplication because the product of any two rational numbers is rational. Likewise, the set of even integers, the open interval (0, I), and the set {-I, I} are each closed under multiplication. Similarly, (0, +) is an algebraic system because the sum of two rationals is rational. The set of even integers is closed under addition. The interval (0, 1) is not closed under addition because .59 + .43 is not in (0, 1). The set {O} is closed under addition but the set {I, -I} is not. If A is a finite set, the order of the algebraic system (A, *) is the number of elements in A. When A is infinite, we simply say (A, *) has infinite order. A convenient way to display information about a binary operation, at least for a system of small finite order, is by means of its operation table, or Cayley table. An operation table for a system (A, *) of order n is an n X n array of products such that x * y appears in row x and column y. Table 6.1 represents a system (A, *) with A = {I, 2, 3} in which, for example, 2 * 1 = 3. As an example of computation in this system of order 3, notice that (3 * 2) * (1 * 3) = 3 * 1 = 2.

a

Table 6.1 2

* 1

2 3

3 3 2

3

2

1

1

3 3

3

6.1

DEFINITIONS (i) (ii) (iii)

(iv)

Algebraic Structures

235

Let (A, *) be an algebraic system. Then

* is commutative on A iff for all x, yEA, x * y = y * x. * is associative on A iff for all x, y, z E A, (x * y) * z = x * (y * z). an element e of A is an identity element for * iff for all x E A, x * e = e*x =x. if A has an identity element e, and a and b are in A, then b is an inverse of a iff a * b = b * a = e. In this case a would also be an inverse of b.

You are familiar with the fact that the system (1', .), with the usual multiplication of integers, is commutative and associative. In this system, I is the identity element, and only the elements I and -I have inverses. For the system consisting of the real numbers with addition, the operation is commutative and associative, 0 is the identity, and every element has an inverse (its negative). Our study of functions provides the information needed to consider what can be regarded as the most important kind of algebraic system. The set of all functions mapping a set one-to-one onto itself is closed under the operation of function composition. This operation is associative, but generally not commutative. The identity function is the identity element, and every element I has an inverse I-I (see Theorem 4.9). This system will be studied in greater detail in section 6.3. The operation * of table 6.1 is not commutative because, for example, I * 3 =1= 3 * 1. It is not associative because (1 * I) * 2 =1= I * (1 * 2). There is no identity element, so the question of inverses does not even arise. Three different operations on A = {I, 2, 3} are shown in tables 6.2, 6.3, and 6.4. Tables 6.2 and 6.4 are not commutative. The fact that the operation of table 6.3 is commutative on A can be seen by noting that the table is symmetric about its main diagonal, from the upper left to lower right. It is not easy to tell by looking at a table whether an operation is associative. For a system of order n, verification of associativity may require checking n 3 products of three elements, each grouped two ways. The operations in tables 6.2 and 6.3 are associative, but + (table 6.4) is not associative onA. (Why?) The associative property is a great convenience in computing products. First, it means that so long as factors appear in the same order, we need no parentheses. For both x(yz) and (xy)z we can write xyz. This can be extended inductively to products

Table 6.3

Table 6.2 o

2

3

1

2 2 2

3 3 3

2 3

Table 6.4 2

3

2 3

1

3

1

2 3

1

2 3

2

1

+

2

3

1

3

3

1

2 3

1 1

1

2 3

2

236

CHAPTER 6 Concepts of Algebra: Groups

of four or more factors: (xy)(zw) = (x(yz)w) = (x(y(zw)), and so forth. Second, for an associative operation, we can define powers. Without associativity, (xx)x might be different from x(xx), but with associativity they are equal, and both can be denoted by x 3 . The element 2 is an identity for· of table 6.3 and in this system the inverses of 1,2, and 3 are respectively, 3, 2, and 1. Table 6.2 has no identity element. In table 6.4, where 3 is the identity element, only 1 and 3 have inverses. An algebraic structure can have at most one identity. In an algebraic structure with an identity and associative operation (such a structure is called a monoid), an element can have at most one inverse. The facts are gathered in the first theorem of this section. The proofs are left as exercises 8 and 9.

Theorem 6.1

Let (A, 0) be an algebraic structure. (a) (b)

If e and f are both identities for 0, then e = f. If 0 is associative, e is the identity for 0, a E A, and x and yare both inverses for a, then x = y.

One of the most interesting concepts in algebra involves mappings between systems and, in particular, operation preserving mappings. As we shall see, such mappings actually preserve the structure of an algebraic system.

DEFINITION Let (A, 0) and (B, *) be algebraic systems. The mapping f: A --+ B is called operation preserving (or OP) iff for all x, yEA, f(x

0

y) = f(x) *f(y).

A simplified statement of how an OP map works is "the image of a product is the product of the images." That is, the result is the same whether operating or mapping is done first. That "the image of the product is the product of the images" must be carefully understood. To compute f(x 0 y), one must use the operation of (A, 0), while in f(x) * fey), the operation is the operation of (B, *). Because f(x 0 y) and f(x) * fey) are elements of B, the definition of operation preserving is a statement about equality of elements of B. Examples ofOP maps abound and mathematicians delight in finding them. The familiar equation log (x . y) = logx

+ logy

tells us that the logarithm function from (~+, .) to (~, +) is operation preserving. As another example, let (P, +) be the set of polynomial functions on ~ with the operation of function addition. Then the differentiation map D: (P, +) --+ (P, +) defined by D(F) = F', the first derivative of F, is operation preserving. This mapping preserves addition because we know from calculus that D(F + G) = (F + G)' = F' + G' = D(F) + D(G). In other words, the derivative of the sum is the sum of the

6.1

Algebraic Structures

237

derivatives. This same mapping from (P, .) onto itself, where' is the operation of function multiplication, is not operation preserving, because D(F . G) is not usually equal to D(F) . D(G). These examples involve functions with properties you already know. To prove that a function f from (A, 0) to (B, *) is OP, we begin with two arbitrary elements x andy of A and either evaluate the two termsf(x 0 y) andf(x) 0 fey) or rewrite one until it has the form of the other. Let the operation 0 on JR X JR be defined by setting (a, b) 0 (c, d) = (a + c, b + d) and let· be the usual multiplication on JR. Let f be the function from (JR X JR, 0) to (JR, .) given by f(a, b) = 2a • 3h . To prove f is operation preserving, let (x, y) and (u, v) be in JR X JR. Then f((x, y)

0

(u, v)) = f(x

+ u, y + v) =

2 x +u .

3 y+v.

Also f(x, y) ·f(u, v) = (2 x , 3 Y). (2u, 3 V ) = 2 x +u, 3 y+v.

Therefore, f is operation preserving. The next series of theorems is a part of the explanation of what we mean by saying that an OP map preserves the structure of an algebra. We will see that under an OP map the image set (range) of an algebraic system is an algebraic system and that, if the original operation is commutative or has an identity element, then so does the image set. More information about OP maps appears in sections 6.2, 6.4, and 6.7.

Theorem 6.2

Let f be an OP map from (A, 0) to (B, *). Then (Rng (f), *) is an algebraic structure. (What we must show is that Rng (f) is closed under *; that is, if u, v E Rng (f), then u * v E Rng (f).) First, note that because A is nonempty, Rng (f) is nonempty. Assume u, v E Rng (f). Then there exist elements x and y of A such that f(x) = u and fey) = v. Then u * v = f(x) *fey) = f(x 0 y), so u * v is the image of x 0 y, which is in A. Therefore, u * v E Rng (f). •

Proof.

Now that we know that the range of an OP map is an algebraic system, we can simplify things by ignoring the part of the codomain that is not in the range. This amounts to assuming that f maps onto B. It does not mean that every OP map is onto every possible codomain.

Theorem 6.3

Let f be an OP map from (A, 0) onto (B, *). If 0 is commutative on A, then * is commutative on B. Assume that f is OP and 0 is commutative on A. Let u and v be elements of B. (We must show u * v = v * u.)

Proof.

238

CHAPTER 6

Concepts of Algebra: Groups

From the fact that f maps onto B, we know that there are x and y in A such that u = f(x) and v = fey). Then u*v = f(x) *fey) = f(x 0 y) = fey 0 x) = fey) * f(x) = v * u. (This equation uses the fact that 0 is commutative and (twice) that f is OP.) Therefore u * v = v * u.

•

The properties of associativity, existence of identities, and existence of inverses are aU preserved by an OP mapping.

Theorem 6.4

Let f be an OP map from (A, 0) onto (B, *). If 0 is associative on A, then * is associative on B. If e is an identity for A, then fee) is an identity for B. If X -I is an inverse for x in A, then f(x -I) is an inverse for f (x) in B.

(a) (b) (c) Proof.

•

Exercise 10.

Exercises 6.1 1. Which of the foUowing are algebraic structures? (The symbols and so on, have their usual meanings.)

*

"* "*

*

2.

(a) (d) (g)

(E, -) (IR, +)

(j)

(rJ>(A) -

*

(0 - {a}, +) {0}, -)

(b) (e) (h)

(k)

(c) (f) (i) (I)

(E, +) (N,-) (rJ>(A) ,

n) ({o, I}, .)

+, - , U , n ,

(IR, -)

(0, +) (rJ>(A) , U)

({o, I}, +)

Which of the operations in exercise 1 are commutative?

3.

Which of the operations in exercise 1 are associative?

4.

Consider the set A =

{a,

b, c, d} with operation

a

b

c

d

a b

a b

b a

c d

c d

c d

d c

a b

d c b

(a) (b) (c) (d) (e) (f)

(g)

0

given by the Cayley table:

a

Is (A, 0) an algebraic structure? N arne the identi ty element of this system. Is the operation 0 associative onA? Is the operation 0 commutative on A? For each element of A that has an inverse, name the inverse. Is BI = {a, b, c} closed under o? Is B2 = {a, c} closed under o?

Algebraic Structures

6.1

(h) (i)

Name all subsets of A that are closed under o. True or False? For all x, yEA, x 0 x = yo y.

given by the table on the right.

6. The Cayley tables for operations

tH a b

a

b

*

a

b

c

a b

b a

a b c

c a b

a b c

c c b

(a) (b) (c) (d)

"*

0,

*,

+, and

*

a

b

c

d

a

c d

d

a

a

b c d

b c d a

b c d

5. Repeat exercise 4 with the operation * X

a

b c

b

are listed below.

rl-t-i a b

239

a

b

x

a

b

c

a

a

a

a

a b c

a c b

c b a

b a c

Which of the operations are commutative? Which of the operations are associative? Which systems have an identity? What is the identity element? For those systems that have an identity, which elements have inverses?

7.

Let M = {A: A is an m X n matrix with real number entries}. (a) If· is matrix multiplication, is (M, .) an algebraic system? What if m = n? (b) If + is matrix addition, is (M, +) an algebraic system? What if m = n?

8.

Let (A, 0) be an algebraic structure. Prove that if e andfare both identities for 0, then e = f. .

9.

Let (A, 0) be an algebraic structure, a E A, and e the identity for o. (a) Prove that if 0 is associative, and x and yare inverses of a, then x = y. (b) Give an example of a nonassociative structure in which inverses are not unique.

"*

10.

Prove Theorem 6.4.

11.

Suppose (A, *) is an algebraic system and * is associative on A. (a) Prove that if ai' a20 a3, a4 are in A, then

"*

12.

(al

(b)

* a2) * (a3 * a4) =

al

* «a2 * a3) * a4)'

Use complete induction to prove that any product of n factors ai' a2' a3,oo" an in that order is equal to the left-associated product (00 . «al * a2) * a3) 00 .) * an- Thus the product of n factors is always the same, no matter how they are grouped by parentheses, as long as the order of the factors is not changed.

Let (A, 0) be an algebra structure. An element 1 E A is a left identity for 0 iff loa = a for every a EA.

(a) (b) (c)

Give an example of a structure of order 3 with exactly two left identities. Define a right identity for (A, 0). Prove that if (A, 0) has a right identity r and left identity I, then r = I, and that r = 1 is an identity for o.

240

CHAPTER 6 Concepts of Algebra: Groups

*

13.

Define SQRT: (~+, +) ---+ (~+, +) by SQRT(x) = {X. Is SQRT operation preserving?

14.

Is SQRT: (~+, .) ---+ (~+, .) operation preserving?

15.

Is SQR: (~, +) ---+ (~, +) defined by SQR(x) = x 2 operation preserving? (~,

Is SQR:

17.

Define 181 on ~ X ~ by setting (a, b) 181 (c, d) = (ac - bd, ad + bc). (a) Show that (~ X ~, 181) is an algebraic system. (b) Show that the function h from the system (C, .) to (~X ~, 181) given by h(a + bi) = (a, b) is a one-to-one function from the set of complex numbers that is onto ~ X ~ and is operation preserving.

18.

Let '!F be the set of all real-valued integrable functions defined on the interval [a, b]. Then ('!F, +) is an algebraic structure, where + is the addition of functions. Define I: ('!F, +) ---+ (~, +) by I(f) = I!/(x) dx. Use your knowledge of calculus to verify that I is an OP map.

19.

Let '!F be as in exercise 18. Consider the algebraic structures ('!F, .) and (~, .) where . represents both function product and real number multiplication, respectively. Verify that!: ('!F, .) ---+ (~, .), defined by I(f) = I!/(x) dx, is not operation preserving.

20.

Let/: (A, 0) ---+ (B, *) and g: (B, *) ---+ (C, X) be OP maps. (a) Prove that g 0 I is an OP map. (b) Prove that if I is a one-to-one correspondence, then I-I is an OP map.

21.

Let M be the set of all 2 X 2 matrices with real entries. Define Det: M ---+ Det [ ~ (a)

(b)

22.

(a)

(b)

.) ---+

(~,

16.

~ ] = ad -

.) operation preserving?

~

by

bc.

Let (M, .) represent M with matrix multiplication and (~, .) represent the reals with ordinary multiplication. Prove that Det: (M, .) ---+ (~, .) is operation preserving. Let (M, +) represent M with matrix addition. Prove that Det: (M, +)---+ (~, +) is not operation preserving. Let C denote the complex numbers with ordinary addition. Define Conj: (C, +) ---+ (C, +) by Conj (a + bi) = a - bi. Prove that Conj is an OP map. Prove that Conj: (C, .) ---+ (C, .) is an OP map.

Let t E ~. Define St: tion preserving.

24.

Let I: A ---+ B. (a) Prove that the induced function I: (QP(A), U) ---+ (QP(B), U) is an OP map. (b) Prove that the induced function I-I: (QP(B) , n)---+(QP(A), n) is an OP map. (c) Prove that the induced function I-I: (QP(B), U)---+ (QP(A), U) is an OP map.

*

(~,

+) ---+

(~,

+) by St(x) = tx. Prove that St is opera-

23.

6.2

Proofs to Grade

25.

6.2

241

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. Let (A, 0) be an algebraic structure. If e is an identity for 0, and if x and yare both inverses of a, then x = y. "Proof." x = x 0 e = x 0 (a 0 y) = (x 0 a) 0 y = e 0 y = y. Thus x = y and inverses are unique. _ (b) Claim. Let 0 be the operation on ~ X ~ defined by setting (a, b) 0 (c, d) = (a + c, b + d) and let - be the usual subtraction on ~. Then the function f given by f(a, b) = a - 3b is an OP map from (~ X ~, 0) to (~,

*

Groups

-).

"Proof." (4,2) and (3, 1) are in ~ X ~. Then f((4, 2) 0 (3, 1» = f(7, 3) = 7 - 3 . 3 = -2, whereasf(4, 2) - f(3, 1) = -2 - 0 = -2, so f is operation preserving. (c) Claim. If every element of a structure (a, 0) has an inverse, then 0 is commutative. "Proof." Let x and y be in A. The element y has an inverse, which we will call y'. Then y 0 y' = e, so y is the inverse of y'. Now x = x, and multiplying both sides of the equation by the inverse of y', we have y 0 x = x 0 y. Therefore, 0 is commutative. _

Groups We have considered algebraic structures (A, 0) with one binary operation. For the rest of this chapter we focus our attention on one particular type of such an algebraic structure, the group. The properties of associativity and existence of identities and inverses examined in section 6.1 are just the properties we need when we define a group. Our approach is axiomatic. That is, we shall list the desired properties (axioms) of a structure, and any system having these properties is called a group. There are some important observations to be made about such a method. Although stated for groups, the following comments apply generally to axiomatic studies. First, a small set of axioms is advantageous (and may be challenging to produce) because a small set means that few properties need be checked to be sure a structure is a group. It may be best to leave a desired property out of the axioms if it can be deduced from the remaining axioms. This is so because every consequence of the axioms must be true of all structures satisfying the axioms. Finally, the fact that axiom systems may be altered by adding or deleting specific axioms does not mean that axioms are chosen at random, or that all axioms are equally worthy of study. The group axioms are chosen because the structures they describe are so important to modern mathematics and its applications. It was the work of Evariste Galois (1811-1832) on solving polynomial equations that lead to the study of groups. The concept of a group has influenced and enriched other areas of mathematics. In geometry, for example, the ideas of Euclidean and non-Euclidean geometries were unified by the notion of the group. Group the-

242

CHAPTER 6 Concepts of Algebra: Groups ory has been applied outside mathematics, too, in fields such as nuclear physics and crystallography.

DEFINITION

A group is an algebraic structure (G, 0) such that

(1) the operation 0 is associative. (2) there is an identity element e E G for (3) every x E G has an inverse x -I in G.

o.

To prove that a structure is a group, we prove that axioms I, 2, and 3 hold for the structure. Implicitly, we must also verify that (G, 0) is actually an algebraic structure. That is, we must verify that G is nonempty and that 0 satisfies the closure property: if x, y E G, then x 0 y E G. A fourth axiom may be considered. A group is called abelian iff (4) the operation

0

is commutative.

The fourth axiom is an independent axiom because it cannot be proved from the other group axioms. The systems (JR, +), (Q, +), and (Ji:, +) are all abelian groups. The system (JR, .) is not a group because 0 has no inverse, but (JR - {O}, .) and (Q - {O}, .) are (abelian) groups. The systems (N, +), (Ji:, -), and (Ji: - {o}, .) are not groups. (Why?) However, {{O}, +) is a group. Table 6.5 is the Cayley table for the abelian group {{I, -I}, .). The hours of the day shown on a clock face make an everyday example of a group (H, +). The set for this group is H = {I, 2, 3, ... , II, 12}, the twelve hours. To add x + y, we start at x o'clock and find the time that is y hours later. For example, 7 + 2 = 9, 5 + 5 = 10 and 7 + 8 = 3. (Notice that A.M. and P.M. do not matter on a clock dial.) This operation is associative and commutative, the identity is 12, and each hour has an inverse. For example, the inverse of 5 is 7 (because 5 + 7 = 12) and the inverse of 9 is 3. (H, +) is an abelian group. By Theorem 6.1, the identity and the inverses of elements in a group are unique. Thus in the definition of a group we could have said "there is a unique identity element e E G for 0 " and "every x E G has a unique inverse x -1 in G." The reason for stating the definition as we did is to make it easier to verify that a structure is a group. Frequently we will simply say "the group G" meaning the set G where the product of elements x and y is denoted xy. Two elementary consequences of the group axioms facilitate calculations involving elements of a group.

Table 6.S -1

o

1

-1

1 -1

-1 1

6.2

Groups

243

Theorem 6.5

Proof. Because a - I is the inverse of a, a - I a = aa -I = e. Therefore, a acts as the (unique) inverse of a-I, so (a-I)-I = a. We know (ab) -I is the unique element x of G such that (ab)x = x(ab) = e. We see that b -I a -I meets this criterion by computing (ab)(b-Ia- I )

= a(bb-I)a- I = a(e)a- I = aa- I =

e.

Similarly, (b-Ia-I)(ab) = e, so b-Ia- I is the inverse of abo

Theorem 6.6

•

Let G be a group. Then left and right cancellation both hold in G. That is, for elements x, y, z of G, (a) (b)

if xy = xz, then y = z. if yx = ZX, then y = Z.

Proof. (a)

Suppose xy

=

= xz in the group G. Then, using the fact that x -I E G, we have

(xz). Therefore using the associative, inverse, and identity properties, we see that

x -I (xy)

X -I

x-I(xy)

=

(x-Ix)y

=

ey

=y

x-I(xz)

=

(x-Ix)z

=

ez

= z.

and Therefore, y

=

z. This proves (a), and the proof of (b) is similar (exercise 7).

•

For a group (G, .) with a finite number of elements, it follows from (a) of Theorem 6.6 that no element can occur twice in any row of the operation table for the group. In fact, if a and b are in G, then a(a -I b) = b, so every element b occurs in row a, for every a. Thus if G is a group, then every element of G occurs exactly once in every row and [using part (b) of the theorem] once in every column of the operation table. The converse of this statement is false: It is possible to have each element occur exactly once in every row and once in every column of an operation table for a structure that is not a group. (See exercise 5.) If G is a group, it is convenient to have notation for powers of elements of G. Let a E G and n E N. We define aO = e and a n+ 1 = ana. Thus an is defined inductively for all n 2: O. Define an for n < 0 by an

=

(a-I)-n.

Then we can prove that for n > 0, (a")-I = a-no Now we have the usual laws of exponents. For m, n E Z:

We also have

if and only if the group is abelian.

244

CHAPTER 6 Concepts of Algebra: Groups

When the group operation is addition, we usually write the inverse of an element a as -a and call it the negative of a. Rather than powers of a, we refer to multiples of a, but the only difference between these ideas is notation. Thus Oa = e and (n + l)a = na + a for n > 0, and na = -n( -a) for n < O.

Example. For each tEN, the set M t = {kt: k E Z} of all integer multiples of t is a group under the operation of addition. To verify this, we proceed as follows: closure: If x, y E M t , then x = kit and y = k2t for some klo k2 in Z. Therefore, x+y=(k[ +k 2)t.

associativity: (x + y) + z = x + (y + z) is true for all integers. identity: 0 is the identity element in M t • inverse: If x E M to then x = kt for some k E Z. Then - k E Z, and -x = - kt is the inverse of x, because x + -x = O.

DEFINITION An operation preserving map from the group (G, 0) to the group (R, *) is called a homomorphism from (G, 0) to (R, *).

A homomorphism is just an OP map for groups. The range of a homomorphism is called a homomorphic image. The function f mapping (Z, +) to (M3 , +) given by f(x) = 3x is an example of a homomorphism. To verify this, suppose x, y E Z. Then f(x + y) = 3(x + y) = 3x + 3y = f(x) + fey). This OP function maps onto M 3 , because for each w = 3k E M 3 , w = f(k). Thus the homomorphic image of the group (Z, +) under f is the group (M3 , +).

Theorem 6.7

(a) (b)

The homomorphic image of a group is a group. The homomorphic image of an abelian group is an abelian group.

Proof. The proof is nothing more than an observation that the image of a group under a homomorphism is an algebraic system (Theorem 6.2) that is associative, has an identity element, and has an inverse for every element (Theorem 6.4). Furthermore, if (G, 0) is abelian, then the image is also abelian (Theorem 6.3). • Let (G, 0) be any group with identity e. Define a mapping F: G --+ G by setting F(x) = e, for every x E G. Then F is a homomorphism on G. We can verify this by observing that for x and y E G, both F(x 0 y) = e and F(x) 0 F(y) = e 0 e = e, so F(x 0 y) = F(x) 0 F(y). In this case the group that is the image of the homomorphismis ({e}, 0). Remember that Theorems 6.2-6.4 and 6.7 do not promise that every OP map is a surjection. If F is an operation preserving map from the group (G, +) to the algebraic structure (K, *), it may be that (K, *) is not a group. However (Rng (I), *) must be a group.

6.2

Groups

245

Exercises 6.2 1. Show that each of the following algebraic structures is a group. * (a) ({I, -I, i, -i},·) where i is chosen so that i2 = -1, and· is complex (b) (c) *: (d)

*

- I - i.J3 . number multiplication. - I + i.J3 ({I, a, P}, .) where a = 2 ' P= 2 ' and· IS complex number multiplication. ({ 1, -I}, .) where' is multiplication of integers. (r;!P(X), Ll) where X is a nonempty set and Ll is the symmetric difference operation All B = (A - B) U (B - A).

2.

Given that G = {e, u, v, w} is a group of order 4 with identity e and u 2 = v 2 = e, construct the operation table for G.

3.

Given that G = {e, u, v, w} is a group of order 4 with identity e and e2 = u 2 = v 2 = w 2 = e, construct the operation table for G.

4.

Which of the groups of exercise 1 are abelian?

S.

Give an example of an algebraic system (G, 0) that is not a group such that in the operation table for 0, every element of G appears exactly once in every row and once in every column. This can be done with as few as three elements in G.

6.

Let G be a group, and a, b, c and ai' for i E N be elements of G. (a) Prove that (abc)-I = c-1b-1a- l . (b) Prove a similar result for (ala2a3' "an) -1 for all n E N by induction.

7.

Prove part (b) of Theorem 6.6. That is, prove that if G is a group, x, y, and z are elements of G, and yx = zx, then y = z.

*:

8.

Let G be a group. Prove that if g2

*:

9.

Give an example of an algebraic structure of order 4 that has both right and left cancellation but that is not a group.

*:

V,

= e for all g E G, then G is abelian.

10. Let G be an abelian group and let a, bEG. (a)

Provethata 2b 2 = (ab)2.

(b)

Prove that for all n E N, anbn = (ab/.

11. Show that the structure (IR - {I}, a

+b-

0),

with operation

0

defined by a

ab, is an abelian group. [You should first show that (IR

0

b=

- {I}, 0) is in-

deed a structure.]

*

12.

(a)

*: (b)

In the group G of exercise 2, find x such that v 0 x = e; x such that v 0 x = u; x such that v 0 x = v; and x such that v 0 x = w. Let (G, *) be a group and a, bEG. Show that there exist unique elements x and y in G such that a * x = band y * a = b.

13. Show that (Z, #), with operation # defined by a # b Find x such that 50 # x = 100. 14.

(a)

=

a

+b+

I, is a group.

Let I be the function from (M 3 , +) to (M 6 , +) given by I(x) that I is a homomorphism.

= 4x. Prove

246

CHAPTER 6 Concepts of Algebra: Groups (b) (c)

Proofs to Grade

15.

What group is the homomorphic image of (M 3, +) under f? Let g be the function from (M3' +) onto (M3' +) given by g(x) = x + 3. Is g a homomorphism? Explain.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignment of grades other than A. (a) Claim. If G is a group, then G is commutative. "Proof." Let a and b be elements of G. Then ab = aeb = a(ab)(ab)-lb = a(ab)(b-1a-1)b = (aa)(bb-1)a-1b = (aa)a-1b = (aa)(b-1a)-1 = a(a(b-1a)-I) = a«b-1a)-la) = a«a-1b)a) = (aa-1)(ba) = e(ba) = ba.

*

6.3

(b)

Therefore, ab = ba, and G is commutative. Claim. If G is a group with elements x, y, and z, and if xz = yz, then x=y. "Proof." If z = e, then xz = yz implies that xe = ye, so x = y. If z 1= e, then the inverse of z exists, and xz = yz implies ¥ = :q. and x = y. Hence in all cases, if xz = yz, then x = y. -

Examples of Groups In this section we will see several examples where the concept of a group arises from another mathematical structure. The first example was mentioned briefly in section 6.1. A function on a set A that is one-to-one and onto A is called a permutation of A. For example, let A = {I, 2, 3}. We shall adopt one of the common notations for a permutation. We think of a functionf on A as follows:

{I ! {J(l)

2 ! f(2)

3} ! f(3)}

and write f as (f(l)f(2)f(3». Then (123) represents the identity function on A, (213) represents the function g where g(1) = 2, g(2) = I, and g(3) = 3, and (312) represents the function h where h(l) = 3, h(2) = I, and h(3) = 2. In section 2.5 we defined a permutation of a set as an arrangement of the elements of the set in some order. A permutation function has exactly the effect of arranging (or permuting) the elements of its domain.

6.3

Examples of Groups

247

As we have already observed in section 6.1, we know from our study of functions that the set of all permutations of a set A is closed under function composition, that composition of functions is associative, that the identity permutation is an identity element, and that the inverse of a permutation is a permutation. This group of functions is called the group of permutations of A.

DEFINITION For n E N, the group of all permutations of{l, 2, ... , n} is called the symmetric group on n symbols and is designated by Sn.

The functions g and h above are elements of S3. Recall that the operation 0 here is function composition, so to compute the product of g = (213) and h = (312), we note that (g (g (g

0 0 0

h)(1) h)(2) h)(3)

= g(h(l)) = g(3) = 3 = g(h(2)) = g(l) = 2 = g(h(3)) = g(2) = 1

Therefore, go h = (321). A similar computation will show that hog = (312) 0 (213) = (132). Thus S3 is the first example we have seen of a group that is not abelian. We list all six elements of S3 and, after extensive computation, fill in operation table 6.6. There are exactly n! arrangements of the elements of a setA with n elements, so the number of permutation functions of A is n! In particular the order of S3 is 3! = 6 and the order of Sn is n! Groups whose elements are some (but not necessarily all) permutations of a set are called permutation groups. The reason for the importance of permutation groups is that, for every group with elements of any kind (numbers, sets, functions), there is a corresponding group of permutations with the same structure as the original group. (See Theorem 6.23.) Our next example of a group involves using symmetric properties of familiar geometric figures. A symmetry of a geometric figure is a rigid motion (without bending or tearing) of the figure onto itself. More formally, a symmetry is a one-to-one distance-preserving transformation of the points of the figure. Every regular polygon and regular solid has an interesting group of symmetries. We consider the symmetries of a square by imagining a cardboard square with vertices 1,2,

Table 6.6

53

0

(123)

(213)

(321)

(132)

(231)

(312)

(123) (213) (321) (132) (231) (312)

(123) (213) (321) (132) (231) (312)

(213) (123) (231) (312) (321) (132)

(321) (312) (123) (231) (132) (213)

(132) (231) (312) (123) (213) (321)

(231) (132) (213) (321) (312) (123)

(312) (321) (132) (213) (123) (231)

248

CHAPTER 6 Concepts of Algebra: Groups

13'

TJ -c

-c

1

2

2

3[J -c

4

2

3

Figure 6.2

Figure 6.1

1

Figure 6.3

3,4 and center C (figure 6.1). The square is carried onto itself by the following rigid motions: a 90° rotation clockwise around center C; R transforms the square to the position shown in figure 6.2. a 180° rotation clockwise around C. a 270° rotation clockwise around C. a reflection about the horizontal axis through C. a reflection about the vertical axis through C; V transforms the square to the position shown in figure 6.3. a reflection about the lower-left to upper-right diagonal. a reflection about the upper-left to lower-right diagonal. the identity transformation

R: R2:

R3: H: V:

D: D':

I:

We compute the product of two symmetries by performing them in succession. Thus VR is the result of a vertical axis reflection followed by a 90° clockwise rotation. By experimenting with a small cardboard square, we find that VR = D oF RV = D'. This permutation group (Y, called the octic group, is shown in the table.

I

R

R2

R3

H

R R2 R3

R2 R3

R3 I

I

I

R

I

I

R R2 R3

R R2 R3

H V

H V

D' D

V H

R R2 D D'

D D'

D D'

H V

D' D

V H

V

D

D'

H

V

D

D'

D

D'

V

H

V

H

D'

D

D'

D R2

H

V

I

R R3 R2

I

R2 R R3

I

R3 R

R3 R R2

I

Other algebraic structures are based on the modulo m relation =m on 7l., which, you recall from section 3.2, is an equivalence relation on 7l.. We have seen that for each natural number m there are exactly m equivalence classes, O/=m, l/=m, ... , (m - l)/=m' and the set of all these classes is called 7l. m • When the modulus m is fixed throughout a discussion, we usually denote these classes by 0, T, ... , m - 1. We define two operations + m and, m as follows:

6.3

Examples of Groups

249

That is, the sum of two classes is the class of the sum, and the product of two classes is the class of the product. The operation + 12 on ::l12 is exactly the twelve-hourclock arithmetic mentioned in section 6.2. Of course, the identity in (::lI2' + dis O. This apparent difference is explained by the fact that 0 = 12 in ::l12' As an example of a computation in ::l12' we have

because 11 + 8 = 19 and 19 = 7. A difficulty in the definitions of + m and, m must be overcome, for we must be sure that + m and .m are operations on ::lm. The problem is that + m and .m must be functions from ::lm X 7l.m to ::lm. Take, for example, m = 5. Then :3 = -12 and T= 31, so we must have :3 +5 T= -12 +531 or else +5 would not have unique images. We must be certain that addition and multiplication as defined in::lm do not depend on which elements of the equivalence class are added. This is all taken care of by means of the following lemma.

Lemma 6.8

=m band C =m d, then (i) a + C =m b + d. (ii) a . C =m b· d.

If a

Proof.

(::lm'

Theorem 6.9

•

Exercise 3.

In light of the preceding discussion and Lemma 6.8, we now have that + m) and (::lm' .m) are algebraic systems for every mEN.

For every natural number m, (::lm'

+m) is an abelian group.

Proof. The proof that + m is associative is a routine verification using the definition of + m and the associativity of + for ::l. Commutativity is similar. The identity element is 0, and the negative ofa is the element (m - a). • When the meaning is clear from the context, we simplify even further the notation afor the equivalence class of a modulo m and simply write a. At times we even use just + for + m' The tables for + 2, + 6' and, 6 are shown. (1':2' +2) +2

0

0

0

1

1

(1':6' +6)

(1':6' '6)

+6

0

2

3

4

5

'6

0

1

0

0

0

1

1

1

2

3 4 5 0

3 4 5 0

4 5 0

5 0

0

2

1

2

1

2

1

2

2

3

3 4

3 4 5

0 0 0 0 0 0

2

2

3 4 5

3 4 5

3 4 5 0

1

1

2

3

4

5

0

0

1

2

0 4

2

4 0

0 3 0 3 0 3

0 5 4 3

3 4 5

2

4

2

0 4 2

2 1

250

CHAPTER 6 Concepts of Algebra: Groups Notice from its table that (::l6, '6) is not a group and that it is possible to have a =1= 0, b =1= 0, but a '6 b = O. In ::lm, if a =1= 0 is an element such that a . b = 0 for some b =1= 0 in ::lm, we say that a is a divisor of zero. It can be shown that (::lm - {o}, 'm) has no divisors of zero iff m is a prime.

Theorem 6.10

The structure (::lm - {O}, . m) is an abelian group iff m is prime. Proof.

•

Exercise 4.

Exercises 6.3 1.

Construct the operation table for Sz, the symmetric group on 2 elements. Is S2 abelian?

2.

(a) (b)

*

(c) (d) *:

3.

Prove Lemma 6.8.

4.

(a) (b)

5.

Let e = (123), a = (231), P= (132) be permutations in S3' (a) Find a 2. (b) Show thatpa = a 2p. (c) Show that {e, a, a 2, p, ap, a2p} is the set S3' (d) Construct an operation table for S3 using the symbols in part (c).

6.

Let y be the permutation on {I, 2, 3,4,5, 6} given by y = (641253). Find (a) l. (b) y3. (c) y6. (d) y-I. For each of the following permutations a, find a-I, a 2, a 3, a 4, a 50 , a 51 . (c) a=(3412)ES4 (a) a=(21)ES2 (b) a=(231)ES3

7.

*

What is the order of S4, the symmetric group on 4 elements? Compute these products in S4: (1243) 0 (4213), (4321) 0 (4321), and (2143) 0 (1324). Compute these products in S4: (3124) 0 (3214), (4321) 0 (3124), (1432) 0 (1432). Show that S4 is not abelian. Prove that (::lm - {o}, 'm) has no divisors of zero iff m is a prime. Prove Theorem 6.10.

*

*

8.

List the symmetries of an equilateral triangle and compute four typical products.

9.

List the symmetries of a rectangle and give the group table.

10.

How many elements are there in the group of symmetries of a regular pentagon? a regular hexagon? a regular n-sided polygon?

11.

Construct the operation table for each of the following groups. (b) (::l3 - {o}, .) (c) (::l7 (a) (::lg, +) (d) (::l9 - {o, 3, 6},·) (e) (::lI5 - {o, 3, 5, 6, 9, 10, 12}, .)

12. Find all zero divisors in the following algebraic structures. (a) (::lIZ,') (b) (::lI5") (c) (::lm,')

*

{o}, .)

6.4

*

13.

If p is prime, show that (p - I)-I = P - I in (Ilp

14.

Find all solutions in (1l20' .) for the following equations. (a) 5· x = 0 (b) 3· x = 0 (c) x . x = 0 (d) x 2 = 9

*

15.

16.

*

6.4

251

{O}, .).

In (Il s, +), find all solutions for the following equations. 4 +x = 6 (b) x + 7 = 3 (c) 3 + x = I (d) 2x = 4 (e) 2x = 3 (f) 2x + 3 = I

(a) Proofs to Grade

-

Subgroups

*

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If a and b are zero divisors in (Ilm' .), then ab is a zero divisor. "Prooj." If a and b are zero divisors, then ab = O. Thus (ab)(ab) = o . 0 = 0 and ab is a zero divisor. (b) Claim. If a and b are zero divisors in (Ilm' .) and ab =1= 0, then ab is a zero divisor. "Prooj." Since a is a zero divisor, ax = 0 for some x =1= 0 in Il m • Likewise, by = 0 for some y =1= 0 in Il m. Therefore, (ab)(xy) = (ax)(by) = o . 0 = O. Thus ab is a zero divisor. -

Subgroups In general, a substructure of an algebraic system (A, *) consists of a subset of A together with all the operations and relations in the original structure, provided that this is an algebraic structure. This proviso is necessary, for it may happen that a subset of A is not closed under an operation. For example, the subset of IR consisting of the irrationals is not closed under multiplication. Substructures are a natural idea, and they can be useful in describing a structure.

DEFINITION Let (G, 0) be a group and H a subset of G. Then (H, 0) is a subgroup of G iff (H, 0) is a group.

It is understood that the operation 0 on H agrees with the operation 0 on G. That is, the operation on H is the function 0 restricted to H X H. If H is a nonempty subset of G and (G, 0) is a group, then to prove that H is a subgroup of G, all three of the group properties and closure must be proved for (H, 0). At least, that will be the situation until we find some way to shorten the work. The set E of even numbers is a subgroup of (Il, +) because E is nonempty, E is closed under addition, the identity 0 E E, the negative of an even number is even, and + is associative on E. More generally, for each integer t, the set M t = {oo., -3t, -2t, -t, 0, t, 2t, 3t,00.} of all multiples of t is a subgroup of (Il, +).

252

CHAPTER 6 Concepts of Algebra: Groups

Two subsets of 1'.6 that are closed under +6 can be seen in the following tables. It is easy to check that both H = {o, 3} and K = {o, 2, 4} are subgroups of (1'.6, +).

+

o 1

2 3 4 5

o o 1 2 3 4

5

1 2 3 4 5

o

3

4

5

+

2 3 4 5

3 4

4 5

o

5

o

o o

1

1

o

1

1 2

1

2

2 3 4 5

2 3 4 5

o

5

1

o 2 3

3 4

1

2 3 4 5

o

2

3

4

5

2 3 4 5

3 4 5

4 5

o

o

1

2

o

1

o 1

2 3

5

1 2 3 4

(K, +)

(H, +)

o

o o

3

3

+

2

3

+

3

o

o o

2 4

2 4

o

2

4

2 4

o

o

4

2

For every group (G, 0) with identity e, ({e}, 0) is a group. This group is called the identity subgroup, or trivial subgroup of G. Also, every group is a subgroup of itself. All subgroups of G other than G and {e} are called proper subgroups. Important questions to be answered are whether the identity element in a subgroup can be different from the identity element of the original group, and whether the inverse of an element in H could be different from its inverse in G. The answers are "no" and "no."

Theorem 6.11

Let H be a subgroup of G. The identity of H is the identity e of G, and if x E H, the inverse of x in H is its inverse in G.

Proof. If i is the identity element of H, then ii = i. But in G, ie = i, so ii = ie and, by cancellation, i = e. The proof for inverses is exercise 5. • The next theorem is a labor-saving device for proving that a subset of a group is a subgroup. It is given in "iff' form for completeness, but the important result is that only two properties must be checked to show that H is a subgroup of G. The first is that His nonempty. This is usually done by showing that the identity e of G is in H. The other is to show that ab- I E H whenever a and b are in H. This is usually less work than showing both that H is closed under the group operation and that bE H implies b- I E H.

Theorem 6.12

Let G be a group. A subset H of G is a group iff H is non empty and for all a, b E H, ab- I EH.

6.4

Subgroups

253

Proof. First, suppose H is a subgroup of G. Then H is a group, so H contains the identity e. Therefore H =1= 0. Also, if a and b are in H, then b -I E H (by the inverse property) and ab- I E H (by the closure property). Now suppose H =1= 0 and for all a, bE H, ab -I E H. (We show that H is a subgroup of G by showing that the group axioms and closure hold for H. It is convenient to proceed in the order that follows.) (i) (ii) (iii)

(iv)

Let x, y, and z be in H. Then x, y, and z are in G, so by associativity for G, (xy)z = x(yz). H =1= 0, so there is some a E H. Then aa- I = e E H. Suppose x E H. Then e and x are in H, so by hypothesis, ex -I = X -I E H. Let x and y be in H. Then by (iii), y -I E H. Then x and y -I are in H, so by • hypothesis, x(y -I )-1 = xy E H.

There are subgroups associated with homomorphisms. We know that if f is a homomorphism from (A, 0) to (B, *), then (Rng (f), *) is a subgroup of (B, *). This is so because the homomorphic image of a group is a group, and Rng (f) ~ B. The next definition and theorem show that for every homomorphism there is a subgroup on the domain side, too.

DEFINITION Let f: (A, 0) --+ (B, *) be a homomorphism and i be the identity element of B. The kernel of f is ker (f) = {x E A: f(x) = i}.

Theorem 6.13

The kernel of a homomorphism is a subgroup of the domain group.

Proof. Letf: (A, 0) --+ (B, *) be a homomorphism and let e and i be the identity elements of A and B, respectively (figure 6.4). (We apply Theorem 6.12.) First, ker (f) is nonempty, because fee) = i, by Theorem 6.4; so e E ker (f). Suppose a and b are in ker (f). Thenf(a) = feb) = i. By Theorem 6.4,f(b- l ) = (f(b))-I. Therefore, f(a

0

b- I) = f(a)

This shows that a

0

0

f(b-I) = f(a)

0

(f(b))-I = i

0

i-I

=

i.

b - I E ker (f). Therefore, ker (f) is a subgroup of A.

Figure 6.4

•

254

CHAPTER 6 Concepts of Algebra: Groups

For each natural number m, there is a particularly nice homomorphism from (Z, +) onto (Zm, + m). This is the canonical map defined by f(a) = a for each a E Z. By definition of +m, f(a + b) = a + b = a +m b = f(a) + feb). The kernel of this homomorphism is 0 = {o, ± m, ± 2m, .. .}, which we have already seen is a subgroup of Z under addition.

Example. Consider the groups (Z6, +6) and ({a, b, c}, 0) with the operation table for

0

as shown.

o

a

b

c

a b c

a b

b c a

c a b

c

It can be verified (exercise 17) by checking all cases that g: Z6 ~ {a, b, c} given by g(O) = g(3) = a, g(1) = g(4) = b, and g(2) = g(5) = c is a homomorphism. The kernel of g is {O, 3}, which we know is a subgroup of Z6' • If a is a member of a group G, then all powers of a are in G by the closure property. This set {... , a -2, a -I, aD, aI, a 2 , . .. } of all powers of a produces a subgroup ofG.

DEFINITIONS Let a be an element of a group G. The cyclic subgroup generated by a is (a) = {an: n E Z}. If there is bEG such that (b) = G, then G is called a cyclic group and b is called a generator for G.

The name cyclic subgroup for (a) is justified by the fact that (a) is a subgroup of G. We verify this by observing that (a) is not empty since aD is the identity of G. Also, if x, y E (a), then x = am and y = an for some integers m and n, so xy-I = a m(a l1 )-1 = ama- n = a m- n E (a). Thus (a) is a subgroup of G. Every cyclic group is abelian, because ama n = a m+ n = ana m. There are many examples of cyclic groups. With the operation of multiplication, the set of all integer powers of 2 is a cyclic group with generator 2. The element 2 (really 2/=5) is also a generator of the group (Zs - {o}, '5)' This is because with operation' 5,2 1 = 2, 22 = 4, 2 3 = 8 = 3, and 24 = 16 = I, so every element of Zs - {O} is a power of 2. Using additive notation, a group is cyclic when it consists of all multiples of some element. For example, (Z, +) is cyclic with generator I and every (Zm, + m) is cyclic with generator A cyclic group may have more than one generator: for the group (Z4, +4), Z4 = (1) = (3). This is because

I/=m.

I .I = I 2'1=1+1=2 3'1=2+1=3 4'1=3+1=0

and

1· 3 = 3 2'3=3+3=2 3'3=2+3=1

4· 3 = 1 + 3 = O.

6.4

Subgroups

255

The element 2 does not generate .if4; it generates the cyclic subgroup {o, 2}. The group S3 is not cyclic because none of its elements generates the entire group. For example, (312)1 (312)2 (312)3 (312)4 (312)5 (312)6

= (312) = (231) =

(123),

the identity

= (312)1 = (312)

= (312)2 = (231) =

(312)3

=

(123),

and so forth.

All other powers of (312) are equal to one of these three elements, so the cyclic subgroup generated by (312) is {(l23), (312), (231)}. Similarly, the other elements of S3 do not generate S3 [exercises 11(a) and 12(a)]. Of course, the fact that S3 is not abelian is sufficient to conclude that S3 is not cyclic, because every cyclic group is abelian. The order of an element a EGis the order of the cyclic subgroup (a) generated by a. If (a) is an infinite set, we say a has infinite order.

Example. In the octic group, the element V has order 2 since VO = I, Vi = V, V2 = VV = I. Thus any power of V is either I or V; therefore, (V) = {I, V}. Similarly, the element R has order 4 because (R) = {I, R, R2, R3J, The orders of the elements of the octic group are 1 (for I), 2 (for R2, V, D, H, and D'), and 4 (for Rand R\ Thus the octic group is not cyclic.

Theorem 6.14

Let G be a group and a be an element of G with order r. Then r is the smallest positive integer such that a r = e, the identity, and (a) = ie, a, a 2, ... , a r - l }. Proof. Since the order of (a) is finite, the powers of a are not all distinct. Let am = an with 0 ::; m < n. Then a n- m = e with n - m > O. Therefore, the set of positive integers p such that aP = e is nonempty. Let k be the smallest such integer. (This k exists by the Well-Ordering Principle.) We prove that k = r by showing that the elements of (a) are exactly aO = e, ai, a 2 , .•. , a k- l . First, the elements e, ai, a 2 , ... , a k- l are distinct, for if as = at with 0::; s < t < k, then a t - s = e and 0 < t - s < k, contradicting the definition of k. Second, every element of (a) is one of e, ai, a 2 , ••• , a k- l . Consider at for t E.if. By the division algorithm, t = mk + s with 0 ::; s < k. Thus at = amk+s = amka s = (ak)ma s = ema s = ea s = as, so that at = as with 0 ::; s < k. We have shown that the elements as for 0::; s < k are all distinct and that every power of a is equal to one of these. Since (a) has exactly r elements, r = k and a r = e.

•

If a E G has infinite order, then by the reasoning used in the proof of Theorem 6.14, all the powers of a are distinct and ( a) -_ { .. "'

} a -2 ,a -I ,a 0_ - e, a 1, a 2'" ...

Because homomorphisms preserve structure, you should not be surprised that a homomorphic image of a cyclic group is cyclic. In fact, the image of a generator

256

CHAPTER 6 Concepts of Algebra: Groups of the group is a generator of the image group. Furthermore, every subgroup of a cyclic group is cyclic. This means that if g = (a) and H is a subgroup of G, then H = (a') for some sEN. The proofs of these facts are exercises 22 and 23.

Exercises 6.4 1.

*

By looking for subsets closed under the group operation, then checking the group axioms, find all subgroups of (b) (1"7 - {o}, .). (a) (1"8, +). (c) (1"5, +). (d) (J, *), with J = {a, b, c, d, e,f} and table shown below.

"*

*

a

b

c

d

e

f

a b c

b a

c

d e

e d b

f

f

d e

a b c d e

f

f

e

a

f

e d b

c d

f

a b c

c

f a

c d b a e

2.

In the group S3, (231) 0 (312) = (123). Is there a subgroup of S3 that contains (231) but not (312)? Explain.

3.

Find the smallest subgroup of S3 that contains (312) and (321). (Hint: Use the closure property.)

4.

For the octic group (symmetries of a square) in section 6.3, give the five different subgroups of order 2 and the three different subgroups of order 4.

5.

Prove that if G is a group and H is a subgroup of G, then the inverse of an element x E H is the same as its inverse in G (Theorem 6.11).

6.

Prove that if Hand K are subgroups of a group G, then H ofG.

7.

Prove that if {Ha: a E is a subgroup of G.

nK

is a subgroup

il} is a family of subgroups of a group G,

then

n

Ha

QEd

8.

Give an example of a group G and subgroups Hand K of G such that H U K is not a subgroup of G.

9.

Let G be a group and H be a subgroup of G. (a) If G is abelian, must H be abelian? Explain. (b) If H is abelian, must G be abelian? Explain.

*

10. Let G be a group. If H is a subgroup of G and K is a subgroup of H, prove that K is a subgroup of G.

11.

*

Find the order of each element of the group (a) S3. (b) (1"7' +). (c) (1"8' +). (d) (1"7 - {o},

12.

List all generators of each cyclic group in exercise II.

.).

6.4

*

*

Subgroups

13.

Let G be a group with identity e and let a E G. Prove that the set Na = {x E G: xa = ax}, called the normalizer of a in G, is a subgroup of G.

14.

Let G be a group and let C = {x E G: for all y E G, xy center of G, is a subgroup of G.

15.

Prove that if G is a group and a E G, then the center of G is a subgroup of the normalizer of a in G.

16.

Let G be a group and let H be a subgroup of G. Let a be a fixed element of G. Prove that K = {a-Iha: hE H} is a subgroup of G.

17.

Let ({a, b, c},

= yx}. Prove that C, the

0) be the group with the operation table shown here.

o

a

b

c

a b c

a b c

b c a

c a b

Verify that the mapping g: (1'6, +) ---+ ({a, b, c}, 0) defined by g(O) g(l) = g(4) = b, and g(2) = g(5) = c is a homomorphism.

*

257

18.

= g(3) = a,

Let {O, 1,. .. , TI} = 1'18 and Ho], [1], ... , [23]} = 1'24' (a) Prove that the function I: 1'18 ---+ 1'24 given by I(x) = [4x] is well defined and is a homomorphism from (1'18, +) to (1'24, +). (b) Find Rng (f) and give the operation table for the subgroup Rng (f) of 1'24'

(c) (d) 19.

Let

Find ker (f) and give the operation table for the subgroup ker (f) of 1'18' Find I({O, 9}) and give the operation table for this subgroup of 1'24'

{a, 1,. .. , 14} = 1'15 and

I(x) = (a) (b) (c) (d)

Ho],

[1], ... , [II]} = 1'12' Define I: 1'15 ---+ 1'12 by

[4x].

Prove that I is a well-defined function and a homomorphism from (1'15, +15)to(E I2 , +d· Find Rng (f) and give the operation table for this subgroup of 1'12' Find ker (f) and give the operation table for this subgroup of 1'15' Find I({O, 2, 4, 6, 8, 10, 12, 14}) and give the operation table for this subgroup of 1'12'

20.

Let (C - {O}, .) be the group of nonzero complex numbers with ordinary com. l'IcatlOn. . L et a = -I +iJ5 PIex num b er mu Itip 2-' (a) Find (a). (b) Find a generator of (a) other than a.

21.

Let x be an element of the group G with identity e. What are the possibilities for the order of x if (~) x l5 = e? (b) x 20 = e? (c) x" = e?

* 22.

Prove that every subgroup of a cyclic group is cyclic.

23.

Let h: G ---+ K be a homomorphism. (a) Prove that if x E G, then h(x k) = (h(x»k for all kEN. (b) Prove that the homomorphic image of a cyclic group is cyclic.

258

CHAPTER 6

Concepts of Algebra: Groups

24.

*

"* 6.5

Let G = (a) be a cyclic group of order 30. (a) What is the order of a 6 ? (b) (c) List all elements of order 3. (d)

List all elements of order 2. List all elements of order 10.

25.

Let G = (a) be a cyclic group of order 9. Let G' be the group generated by the permutation (231). If I is a homomorphism from G to G' such that I(a) = (231), find all other images.

26.

Let I: (G, *) -+ (H, 0) be a homomorphism. Let e be the identity of G. Prove that I is one-to-one iff ker (f) = {e}.

27.

Let I: G-+ H be a homomorphism. For a E G, prove that the order of I(a) divides the order of a.

eosels and Lagrange's Theorem The purpose of this section is to show that the order of a subgroup of a given finite group must be a divisor of the order of the group. Thus, given a group of order, say, 30, we will know it cannot have a subgroup with 8 elements because 8 does not divide 30. The way we prove that the order of a subgroup divides the order of the group is to use the subgroup to partition the group into disjoint subsets, one of which is the subgroup itself. Then because it turns out that all these subsets have the same number of elements as the subgroup, we know the group order is a multiple of the subgroup order. We first give an example. The group (Z12, +) has 12 elements, which we denote simply as 0, 1,2, ... ,11. One subgroup of Z12 is the set H of all multiples of 4. The set H is a subset of ZI2, H = {O, 4, 8}.

If we add 1 to each element of H, we obtain another subset, which we will denote 1 +H: 1 + H = {I, S, 9}. We note that 1 + H is not a subgroup of Z12 (closure fails), but 1 + H has the same number of elements as H. We call 1 + H a coset of fl. Continuing, we have

2 + H = {2, 6, 1O} 3 + H = {3, 7, II} We could go on and compute 4 + H, S + H or any coset x but these are not new sets of elements; S + H, for example, is

+H

where x E ZI2'

S + H:: {S, 9, 13} - {S, 9, I} = 1 +H because 13 and 1 are the same equivalence class. Finally we note that H, 1 + H, 2 + H, and 3 + H each have three elements, are disjoint, and their union is ZI2 (figure 6.S). The number of elements in ZI2 is the sum of the number of elements in the cosets H, 1 + H, 2 + H, and 3 + H, which is 4 times the order of H. Thus we have an explanation for the fact that the order of H divides the order of Z12'

6.5

(osels and Lagrange's Theorem

259

(J Figure 6.5 The result we seek, that the order of a subgroup must divide the order of the group, is Lagrange's Theorem, first proved by J oseph-Louis Lagrange (1736-1813). The first step is to give a formal definition of coset.

DEFINITION Let (G, 0) be a group, H a subgroup of G, and x E G. The set x 0 H = {x 0 h: h E H} is called the left coset of x and H, and H 0 x = {h 0 x: hE H} is called the right coset of x and H.

Table 6.7 shows a group T with identity a and subgroups H = {a, b} and {a, e,f}. The cosets of Hand K are also shown in tables. Notice that the left cosetfH and the right coset Hfare not equal. However, it does happen that for every xin T, xK = Kx. K =

Table 6.7

a b e d e I

Group T

a

b

e

d

e

I

a b e d e I

b a e I e d

e I a e d b

d e I a b e

e d b e I a

I e d b a e

Cosets of H

Cosets of K

aH= {a, b} =Ha = {a, b} bH = {b, a} =Hb= {b, a} eH = {c, e} =fo He = {e,/} dH = {d,f} =foHd = {d, e} eH= {e, e} =foHe= {e, d} IH= {t, d} =foHI= {t, e}

aK = {a, e,f} = Ka = {a, e,f} bK = {b, d, e} = Kb = {b, e, d} eK = {e, b, d} = Ke = {e, b, d} dK = {d, e, b} = Kd = {d, b, e} eK = {e,f, a} = Ke = {e,l, a} IK = {t, a, e} = KI = {t, a, e}

260

CHAPTER 6 Concepts of Algebra: Groups

As another example of cosets, consider the subgroup H = {O, ::!:: 5, ::!:: 10, ... } of (Z,+). Then 2+H=H+2={ ... ,-8,-3,2,7, ...}=2/=s and O+H= H + 0 = H. There are 5 cosets: H, 1 + H, 2 + H, 3 + H, and 4 + H.

Theorem 6.15

If H is a subgroup of G and x E G, then Hand xH are equivalent (have the same number of elements).

Proof. To show that Hand xH are equivalent, we show that the function F given by F(h) = xh maps H one-to-one and onto xH. If w E xH, then w = xh for some h E H. Then for that h, F(h) = w. Therefore, F maps onto xH. Suppose F(h l ) = F(h2) for hI> h2 E H. Then xhl = xh 2. Then by cancellation, hi = h 2. Therefore, F is one-toone.

-

Because every left coset of His equi valent to H, any two left cosets of H have the same number of elements. This result and others to follow also apply to right cosets.

Theorem 6.16

The set of left cosets of H in G is a partition of G.

Proof. Every element x of G is a member of some left coset because x = xe E xH. Therefore, G is a union of left cosets. It remains to show that left cosets are either identical or disjoint. Suppose xH n yH oF 0. Let bE xH n yH. Then b = xhl = yh2 for some -I -I hi> h2 E H. Therefore, x = yh2hl . Now let xh be in xH. Then xh = yh2hl hand h2h~ I h E H, so xh E yH. This shows that xh ~ yH. A similar argument shows yH ~ xH. Therefore, xH = yH. -

Theorem 6.17

(Lagrange) Let G be a finite group and H a subgroup of G. Then the order of H divides the order ofG.

Proof.

Let the order of G be n and the order of H be m. Let the number of left cosets of H in G be k. Denote the distinct cosets of H by xlH, x2H, ... ,XkH. By Theorem 6.l5, each coset contains exactly m elements. By Theorem 6.l6, G may be written as G = xlH U X2H U ... U XkH, and the cosets are pairwise disjoint. We have divided the elements of G into k classes, each class having m elements. Thus there are mk elements in G. Therefore, n = mk, and so m divides n. For a group G with order n and subgroup H of order m, the integer fi; is called the index of H in G and is the number of distinct left cosets of H.

Example.

Let (i) be the octic group of symmetries of a square (section 6.3). The subgroup m = {I, R, R2, R3} has two cosets: m and Vm = {V, D, H, D'}. Thus the index of min (i) is 2, which is (order of (i) / order of m).

The converse of Lagrange's Theorem is false. For example, there exists a group of order 12 that has no subgroup of order 6.

6.5

Corollary 6.18

(osets and Lagrange's Theorem

261

Let G be a finite group and a E G. Then the order of a divides the order of G.

Proof.

Corollary 6.19

If G is a finite group of order n and a E G, then an

Proof.

Corollary 6.20

•

Exercise 6. =

e, the identity.

Exercise 7.

•

A group of prime or~er is cyclic.

Proof.

Let G have prime order p. Let a be an element of G other than e. By Corollary 6.18, the order of a divides the order of G, and so must be either I or p. But (a) contains both e and a, so the order of a must be p. Therefore, G = (a) and G is cyclic. (What we have proved is that every element of G other than e is a generator.) • By Corollary 6.20, every group of order II must be cyclic. Of course there are cyclic groups whose order is not a prime, because the group (Em' + m) is cyclic for every m. By Corollary 6.19, x 8 is the identity, I, for every element x in the octic group ~. We have already seen in section 6.4 that the orders of the elements of ~ are either 4 (m and m3 have order 4) or 2 (all of V, H, D, D', m2 have order 2) or I (which is the order of J). All these orders divide 8, the order of~, as required by Corollary 6.18.

Exercises 6.5 *:

1. Let H be the subgroup of S3 generated by the permutation (213). Find all left and right cosets of H in S3'

*

2.

Let H be the subgroup of S3 generated by the permutation (231). Find all left and right cosets of H in S3'

3.

Find all the cosets of H = (a 3 ) in a cyclic group G = (a) of order 12.

4.

Show that a group Gof order 4 is either cyclic oris such thatx 2 = e for all x E G.

5.

Show that a group G of order 27 is either cyclic or is such that x 9 = e for all xEG.

6.

Prove Corollary 6.18.

7.

Prove Corollary 6.19.

8.

Suppose G is a group of prime order p, and a E G. What are the possible orders of a? If a is not the identity, what is the order of a? How many elements of G are in the subgroup of G generated by a? How many elements of G are generators of the cyclic group G?

9.

Let H be a subgroup of G. (a) Prove that aH = bH iff a-1b E H. (b) If aH = bH, must b-1a be in H? Explain. (c) If aH = bH, must ab - I be in H? Explain.

262

CHAPTER 6 Concepts of Algebra: Groups

"* Proofs to Grade

10.

Prove that if G is a cyclic group of order n and the natural number m divides n, then G has a subgroup of order m.

11.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If H is a subgroup of G and aH = bH, then a -I bE H. "Proof." Suppose aH = bH. Then ah = bh for some hE H, and so h = a-Ibh. Thus hh- I = a-Ib and since hh- I EH, we conclude that a-lbEH. • (b) Claim. If H is a subgroup of G and a-Ib E H, then aH C bH. "Proof." Suppose a - I b E H. Then a -I b = h for some h E H. Thus a-I = hb- I, so that a = (hb-I)-I = bh- I. Thus a E bH. Suppose that x E aH. Then x = ah' for some hi E H. Thus x = (bh-I)h ' = b(h-Ih ' ), so x E bH. Therefore, aH C bH. • (c) Claim. If H is a subgroup of G and aH = bH, then b-Ia E H. "Proof." Suppose aH = bH. Then b-laH = b-lbH = eH = H. Therefore, b-laH = H. Since b-Ia E b-laH, b-Ia E H. •

*

*

6.6

Quotient Groups Let G be a group and let H be a subgroup of G. The set of left cosets of H in G is a partition of G, by Theorem 6.16. Therefore, there is a corresponding equivalence relation on G with the property that a, bEG are related iff they belong to the same left coset of H. (See chapter 3, section 3.) The equivalence classes under this relation are exactly the left cosets of H. The set of all equivalence classes (cosets) is denoted G/H (read "G mod H').

Example.

In (1'.6, +), let H be the subgroup {O, 3}. The left cosets of Hare

o + H = 3 + H = {o, 3} I +H=4+H={1,4} 2

+H

=

5

+H

=

{2, 5}.

Here 1'.6/H = {O + H, 1+ H, 2 + H}. It is our intention to impose a structure-an operation we hope will satisfy the group axioms-on G/ H. If we try to define a "product" for cosets xH and y H, nothing could be more natural than xH . yH = xyH. In order to tryout this "structure" (G/ H, .), we consider as examples the sets T / Hand T / K where T, H, and K are the groups presented in table 6.7. The left cosets of H in T are {a, b}, {e, e}, and {d,f}. Applying the definition for· on T /H, we have {e, e} . {d,f} = eH . dH = edH = fH = {d,f} {e, e} . {d,f} = eH . fH = efH = aH = {a, b}.

The fact that· yields two different results means that it is not an operation on T /H. In other words, the operation is not well defined.

6.6

Quotient Groups

263

The left cosets of K in T are {a, e,f} = aK = eK = fK and {b, c, d} = bK = dK. In this case' is indeed an operation on T IK because the product xK . yK does not depend on the representatives for the cosets. For example, eK = fK, and bK = cK, and eK' bK = ebK = cK, and fK' cK = fcK = bK but cK = bK. Table 6.8 is the operation table for T IK. cK

=

Table 6.8

aK bK

aK

bK

aK bK

bK aK

It is more than a coincidence that (I) the proposed operation on left cosets works for T I K and xK = Kx for every x E T, and (2) the proposed operation is not well defined for left cosets in T IH and for some x E T, xH =1= Hx. We are led to the following definition.

DEFINITION Let G be a group and H a subgroup of G. Then H is called normal in G iff for all x E G, xH = Hx.

Thus K is a normal subgroup of T but H is not. If G is any group, then both G and {e} are normal in G (see exercise 1). If G is an abelian group, then every subgroup of G is normal. This is because xH = {xh: h E H} = (by commutativity) {hx: h EH} = Hx.

The condition xH = Hx for a normal subgroup does not require that xh = hx for every hE H. (That is, normality does not require commutativity.) For example, for the subgroup K of Tconsidered above,f E K and dK = Kd, but df =1= fd. (See table 6.7.)

Theorem 6.21

If H is a normal subgroup of G with xH = yH and wH = vH, then xwH = yvH.

Proof., Let xwh be an element of xwH. Then wh E wH = vH = Hv (because H is normal), so wh = h]v for some h] E H. Thus xwh = xh]v. Now xh] E xH = yH = Hy, so xh] = hi)! for some h2 E H. Thus xwh = hi)!v E Hyv = yvH. Since • cosets are either identical or disjoint, xwH = yvH. If H is a normal subgroup of G and if xH = yH and wH = vH, then xH . wH = xwH = yvH = yH' vH, by Theorem 6.21. This establishes that (GIH, .) is an algebraic structure.

Theorem 6.22

If H is a normal subgroup of G, then (GIH, .), with' as defined above, is a group called the quotient group of G modulo H (G mod H). The identity element is the coset H = eH, and the inverse of xH is x -] H.

Proof.

Exercise 9.

•

264

CHAPTER 6 Concepts of Algebra: Groups

Example. For tEN, the group M/ of all multiples of t is a subgroup (1':, +). M/ is a normal subgroup because (1':, +) is abelian. Elements of 1':/M/ are cosets of the form x + M/ = {x + kt: k E 1':}, which is just the equivalence class 4=/. The operation on cosets in 1':/M/ is the same as the operation + / on 1':/. That is, the sum of the cosets x + M/ and y + M t is the coset (x + y) + M/, just as x/=/ + y/=/ = (x + y)/=/. It turns out then, that the quotient group 1':/Mt of the integers modulo multiples of t is exactly the same group as the group (1':/, +/) of integers modulo t. Table 6.9 for the group (1':, + )/M3 = (1':3, +3) is shown to illustrate that these groups are the same. Table 6.9

The Quotient Group (1':, + )/M3 = (1':3' +3)

o={kE:£::k =30} =

0 + {3k: k E:£:}

1 = {kE:£:: k =3 I} = 1 + {3k: k E:£:}

2={kE:£::k=3 2} = 2 + {3k: k E:£:}

0= {kE:£:: k3 =3 o} = 0 + {3k: k E:£:}

2

1 = {kE:£:: k =3 I} = 1 + {3k:kE:£:} 2={kE:£::k=3 2}

1

= 2 + {3k: k E:£:}

What the quotient group concept does is allow us to treat whole blocks of elements of a group as objects that can be operated on, forming another group. The blocks of elements are just the cosets of the normal subgroup.

Example. The group H = ({I, 6}, '7) is a normal subgroup of the abelian group (1':7 - {o}, '7)' See table 6.10(a). Thecosets of H areH, 2H = {2, 5}, and 3H = {3, 4}. Listing the elements of 1':7 - {O} in a different order, so that elements in the same coset are adjacent, helps to visualize the coset blocks (table 6.10(b)). Notice (b)

Table 6.1 O(a) '7

1 2 3 4 5 6

1 2 3 4 5 6

2

3

4

5

6

2 4 6 1 3 5

3 6 2 5 1 4

4 1 5 2 6 3

5 3 1 6 4 2

0 5 4 3 2 1

1 6 1 6

2 5 3 4

2 5

3 4

[] [] [] [] [] []] [] []] [] 1

5 2

4 3

5 2

3 4

1 6

4 3

1 6

5 2

6

6.6

Quotient Groups

265

that the block {2, 5} can be multiplied by the block {3, 4} to produce the block {1, 6}. (Actually there are four products involved: 2 . 3 = 6, 2 . 4 = 1,5 . 3 = 1, and 5· 4 = 6.) The key idea is that when any two blocks are multiplied, the result is another block. This will work in any group G as long as one of the blocks is a normal subgroup of G. The group table for (1"7 - {O}/H, .) is shown in table 6.11.

((d:7 -{O}/H,·)

Table 6.11 H

2H

3H 3H

H

H

2H

2H

2H

3H

H

3H

3H

H

2H

What are the coset blocks formed by the normal subgroup {e} of a group G? For each x E G, the left coset x{e} = {x} has only one element. Thus there is one coset for each element of G, and the operation G/{e} works just like the product inG. At the other extreme, we have the normal subgroup G of G. For every x E G the left coset xG is equal to G, so all coset blocks are the same. Thus the quotient group GIG has only one element, G, which is the identity of GIG.

Exercises 6.6 1. Let G be a group with identity e. (a) Prove that {e} is a normal subgroup of G. (b) Prove that G is a normal subgroup of G. 2. The subgroup M = {o, 4} of (1"8, +8) is a normal subgroup. (Why?) (a) Find the left cosets of Min 1"8' (b) Construct an operation table for the quotient group E8/M. 3. For the octic group 0 of symmetries of a square (section 6.3), let J = {I, R, R2, R3}, K = {I, R2, H, V}, and L = {I, H}. (a) Construct the operation table for 01J. (b) Construct the operation table for OIK. (c) Is L normal in (d) Is L normal in K?

"*

m

"*

4.

If HI is a normal subgroup of G, and H2 is a normal subgroup of HI, must H2 be a normal subgroup of G? Explain.

S.

Consider the subgroup e = {I, R2} of the octic group O. (a) Prove that e is normal in the octic group. (b) Construct the operation table for Ole. (c) Is Ole abelian? (d) Is Ole cyclic?

266

CHAPTER 6 Concepts of Algebra: Groups

6.

"*

Let H be a normal subgroup of G. Prove that (a) if G is abelian, then G/H is abelian. (b) if G is cyclic, then G/H is cyclic.

*

7.

"*

Let G be a group and H be a subgroup of G. Prove that H is normal iff a-IHa ~ H for all a E G, where a-IHa = {a-Iha: hE H}.

8.

Prove that a subgroup H of G is normal in G iff Ha

9.

Prove Theorem 6.22.

10.

(a) (b)

~

aH for every a E G.

Let C = {x E G: for all y E G, xy = yx} be the center of G. (See exercise 14 of section 6.4.) Prove that the subgroup C is normal in G. The intersection over a family of subgroups of G is a subgroup of G. (See exercise 7 of section 6.4.) Prove that if Hu is normal in G for every a E Ll, then Hu is normal in G.

n

aEL1.

11. 12.

Let J be a homomorphism of G onto H. Let N be a normal subgroup of G. Prove thatJ(N) = {fen): n EN} is a normal subgroup of H. Let J be a homomorphism of G onto H. Let M be a subgroup of H. Prove that = {x: J(x) EM} is a subgroup of G that includes ker (f). Show that if Mis normal in H, then N is normal in G.

N Proofs to Grade

13.

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If H is a normal subgroup of G, and K is a normal subgroup of H, then K is a normal subgroup of G. "Proof." We must show xK = Kx for all x E G. Since H is normal in G, xH = Hx for all x E G. Since K is a normal subgroup of H, xK = Kx for all x E H and hence for all x E G. Therefore, K is a normal subgroup (b)

(c)

6.7

~a

-

If Hand K are normal subgroups of G, then H n K is a normal subgroup of G. "Proof." Let x E G. Then x(H n K) = xH n xK = Hx n Kx (since H and K are normal) = (H n K)x. Therefore H n K is normal. Claim. If H is a normal subgroup of G and G/H is abelian, then Gis abelian. "Proof." Let x, y E G. Then xH, yH E G/H. Since G/H is abelian, (xH)(yH) = (yH)(xH). Therefore,xyH = yxH. Therefore (xy)h = (yx)h for some hE H. By cancellation, xy = yx. Claim.

Isomorphism; The Fundamental Theorem of Group Homomorphisms A homomorphism that is one-to-one is called an isomorphism. If there is an isomorphismJ: (A, o)~(B, *), then (A, 0) is said to be isomorphic to (B, *). Inverses and composites of isomorphisms are also isomorphisms, so the relation of being isomorphic is an equivalence relation on the class of all groups.

6.7 Isomorphism; The Fundamental Theorem of Group Homomorphisms

267

The word isomorphic comes from the Greek words isos (equal) and morphe (form). Isomorphic groups are literally "equal form" because they differ only in the names or nature of their elements; all their algebraic properties are identical. For example, the groups (Ez, +2), ({I, 6}, '7), and ({0, A}, Ll) are isomorphic, where A is any non empty set and Ll is the symmetric difference operation defined by X Ll Y = (X - Y) U (Y - X). The three groups are shown in table 6.12. Table 6.12

o

o o

1

1

o

o o

A

A

6

1 6

1

1

o

6

6 1

A A

o

In fact, any two groups of order 2 are isomorphic. This observation can be extended: Every cyclic group of order m is isomorphic to (Em' + m). Every infinite cyclic group is isomorphic to (E, +) (see exercises 6-9). Because of these results, we can say that (Em' + m) and (E, +) are the only cyclic groups, up to isomorphism. The non cyclic group T of section 6.5 is isomorphic to the group (S3, 0) under the isomorphism F = {(a, (123)), (b, (213)), (c, (321)), (d, (132)), (e, (231)), (j, (312))}.

The operation tables for Tand S3 are shown in table 6.13(a) and (b). The tables clearly have the same structure. Table 6.13(a)

T

e

f

d

e

e

f

d b c

a

b

c

d

a b c d

a b c d

b a

c

f

e

e

f

f

(b)

e

a

f

e

c d

d b

f a b c

f a

c d b a e

S3

0

(123)

(213)

(321)

(132)

(231)

(312)

(123) (213) (321) (132) (231) (312)

(123) (213) (321) (132) (231) (312)

(213) (123) (231) (312) (321) (132)

(321) (312) (123) (231) (132) (213)

(132) (231) (312) (123) (213) (321)

(231) (132) (213) (321) (312) (123)

(312) (321) (132) (213) (123) (231)

268

CHAPTER 6 Concepts of Algebra: Groups

We claimed in section 6.3 that for every group there is a permutation group with the same structure. We now prove this result, due to Arthur Cayley (1821-1895).

Theorem 6.23

Every group G is isomorphic to a permutation group.

Proof.

As the set whose elements are to be permuted, we choose the set G. If a E G, then {ax: x E G} = G. Therefore, the function Oa: G ---+ G defined by Oa(x) = ax is a mapping from G onto G that (by the cancellation property 01 the group) is one-to-one. Therefore, Oa is a permutation of G associated with the element a of G. The function 1 sending a to Oa will be our isomorphism from G to the set H = {Oa: a E G}. Let x E G. Then 0ab(X) = (ab)x = a(bx) = 0a(Bb(x)) = (Ba 0 0b)(X), so Oab = 0" 0 0b. Therefore, 1 is a homomorphism. Also, 1 maps onto H by definition of H. Thus H (the homomorphic image) is a permutation group. Now suppose Oa = Bb . Then ax = bx for every x E G, so a = b. Therefore, 1 is one-to-one; G and Hare isomorphic. -

Example.

Let G be the group ({ I, --I, i, - i}, .), where' is the usual multiplication of complex numbers. As elements of the permutation group H, we take the functions that are left translations by elements of G (that is, Oa(x) = a . x). For example, O;(1)=i·l =i, O;(-I)=i·(-I)= -i,

= -1, = i· (-i) = 1,

O;(i) = i . i O;(-i)

so 0; = (i, -i, -I, I). Similar computations show that 0,

= (1,

-I, i, -i),

0_ 1 = (-I, I, -i, i), and

0_; = (-i, i, I, -1).

Thus H = {OJ, B-1' 0;, B-J and the operation on H is composition. The tables for G and H clearly have the same structure. (H,o) -i

0

81

-i

81 8_ 1

81 8_ 1

8i 8_ i

8i 8_ i

-1

8_ 1

8i

8_ 1 81 8_ i

8i 8_ i 8_ 1

8i

81

8_ i 8_ i 8i 81 8_ 1

The main purpose of this section is to establish the connection between normal subgroups and homomorphisms defined on a group. First, for every homomorphism, there is a corresponding normal subgroup. This subgroup determines a quotient group that is isomorphic to the image group.

Isomorphism; The Fundamental Theorem of Group Homomorphisms

6.7

Theorem 6.24a

269

The kernel K of a homomorphism f from a group (G, 0) onto (B, *) is a normal subgroup of G. Furthermore, (GIK, .) is isomorphic to (B, *).

Proof. We saw in Theorem 6.13 that K = {y E G:f(y) = eB} (where eB is the identity of B) is a subgroup of G. We first prove that for all x and a E G, x E a 0 K iff f(x) = f(a). Suppose x E a 0 K. Then x = a 0 k for some k E K. Then f(x) = f(a 0 k) = f(a) *f(k) = f(a) * eB = f(a), so f(x) = f(a). Now suppose f(x) = f(a). Then f(a- I 0 x) = f(a-I) * f(x) = (f(a))-I *f(a) = eB' Thus a-lox E K. Therefore, x = a 0 (a-lox) E a 0 K. Similarly, x E K 0 a iff f(x) = f(a). Hence x E a 0 K iff f(x) = f(a) iff x E K 0 a, so a 0 K = K 0 a for all a E G. Therefore, K is normal in G. This argument shows that for every left coset a 0 K of the kernel K, all the elements of a 0 K are mapped by f to the same elementf(a) in B. Therefore, we may define a mapping h: GIK --+ B by setting h(a 0 K) = f(a). Further, h is one-to-one.

(If heal

K) = h(a2 0 K) then f(a,) = f(a2), so a2 E al 0 K. As a2 = a2 0 e is K, and eosets form a partition of G, al 0 K = a2 0 K.) The mapping h is onto B: If bE B, then b = f(a) for some a E G, and b = f(a) = h(a 0 K). Finally, h is operation preserving: 0

also in a2

0

h((al

0

K)(a2

0

K)) = h(al 0 a2 0 K) = f(a, 0 a2) = f(al) * f(a2) = h(al 0 K) * h(a2

0

K).

Therefore, h is an isomorphism from (GIK, .) onto (B, *).

Example.

•

Let ({a, b, e}, .) be the group with the table shown.

o

a

b

c

a b c

a b c

b c

c a b

a

The function g: (Z6, +6) -+ ({a, b, e}, 0) given by g(O) = g(3) = a, g(l) = g(4) = b and g(2) = g(5) = e is a homomorphism (exercise 17 of section 6.4). The kernel of g is {o, 3}. By Theorem 6.24a, {o, 3} is a normal subgroup of Z6 and Z6/{o, 3} is isomorphic to ({a, b, e}, 0). The table for the quotient group is

{O,3}

{I, 4}

{2,5}

{O,3}

{O,3}

{I, 4} {2,5}

{I, 4} {2,5}

{l,4} {2,5}

{O,3}

{O,3}

{I, 4}

{2,5}

270

CHAPTER 6 Concepts of Algebra: Groups

Example. The function f: LEIs to LE24 that maps x in LEIs to 4x in LE24 is a homomorphism (exercise 18 of section 6.4). By Theorem 6.24a, K = ker (f) = {D, 6, TI} is a normal subgroup of LEIs and LEls/ker (I) is isomorphic to R = Rng (I) = {[o], [4], [8], [12], [16], [20n, with the operation +24' The group tables are shown. The isomorphism h is given by hex + K) = [4x).

K I+K 2+K 3+K 4+K 5+K

K

I+K

2+K

3+K

4+K

5+K

K I+K 2+K 3+K 4+K 5+K

1 +K 2+K 3+K 4+K 5+K K

2+K 3+K 4+K 5+K K I+K

3+K 4+K 5+K K I+K 2+K

4+K 5+K K 1 +K 2+K 3+K

5+K K 1 +K 2+K 3+K 4+K

[8]

[12]

[16]

[20]

[4]

[8]

[8]

[12] [16] [20] [4]

[12] [16] [20] [0]

[16] [20] [0] [4] [12] [16]

[20] [0] [4]

+24

[0]

[4]

[0] [4]

[0] [4]

[8]

[8]

[12] [16] [20]

[12] [20] [0]

[12] [16] [0] [4]

[8]

[8] [12]

[8] [16] [20]

The other aspect of the connection between homomorphism and normal subgroups is that it is possible to start with any normal subgroup and construct a homomorphic image of the group. This image is the quotient group.

Theorem 6.24b

Let H be a normal subgroup of (G, 0). Then the mapping g: G-+ G/H given by 0 H is a homomorphism from G onto G/H with kernel H.

g(a) = a

Proof. We have seen the mapping g before (section 4.1) and called it the canonical map from G to the set of cosets (equivalence classes). First, g maps onto G/H because, if a 0 H is a left coset, then a 0 H = g(a). To verify that g is operation preserving, we compute g(a

0

b)

=a

0

b

0

H

= a 0 H . b 0 H = g(a) . g(b).

It remains to show that H is the kernel of g. The identity of (G/H, . ) is the coset H, so x E ker (g) iff g(x) = x 0 H = H iff x E H. Therefore, ker (g) = H. •

Example. Since K = ker (f) in the previous example is a normal subgroup of LE IS' the function g: LEIs -> LE ls/ K given by g(x) = x + K must be a homomorphism with kernel K. To verify this, we examine g more closely. By definition g(D) = K = {D, 6, TI}. Also, g(l) = 1 + K = {T, 7, TI}, g(2) = 2 + K, and so forth. Thus g(x + y) = x + Y + K = x + K + Y + K = g(x) + g( y), as required. The two preceding theorems are summarized in the next one.

6.7 Isomorphism; The Fundamental Theorem of Group Homomorphisms

f

(G,o)

271

(homomorphism) I

(8,*)

g(',"oni"l~ ~mO",hi,m) (G/K.-)

Figure 6.6

Theorem 6.25

(The Fundamental Theorem of Group Homomorphisms) Every homomorphism of a group G determines a normal subgroup of G. Every normal subgroup of G determines a homomorphic image of G. Up to isomorphism, the only homomorphic images of G are the quotient groups of G modulo normal subgroups. Finally, the diagram in figure 6.6 "commutes"; that is, I = hog where h and g are the mappings of Theorem 6.24(a) and (b).

Proof. What remains to be proved is that I is the composite of h with g. Let a E G. Then g(a) = a 0 K and h(g(a)) = h(a 0 K) = I(a). Therefore, 1= hog. •

Example. In the example above, the isomorphism h maps E 18/K onto R = Rng (f), and the canonical function g maps E 18 ---+ E 18/K. For x E E 18, h 0 g(x) = hex

+ K) =

[4x] = I(x), soh

0

g = I.

The primary assertion of the Fundamental Theorem is that every homomorphic image can be identified with (is isomorphic to) the quotient group G/ker (f). Furthermore, with this identification, the homomorphism I acts like the canonical homomorphism. Thus we can think of homomorphic images as though they are quotient groups, and homomorphisms as though they are canonical maps. The Fundamental Theorem is used to conclude that two groups are isomorphic to each other without actually constructing the isomorphism. As an example of the use of the Fundamental Theorem, we describe all homomorphic images of (E6 , +). Every such image must be isomorphic to a quotient group of E6 . The subgroups of E6 have orders 1, 2, 3, and 6, and every subgroup is normal in E6 . The corresponding quotient groups have orders 6, 3, 2, and 1. These quotient groups are cyclic [see exercise 6(b) of section 6.6] and therefore isomorphic to E6 , E3 , Ez, and {OJ, respectively.

Exercises 6.7 1.

Show thatthe mapping I: (IR+, .) ....... (IR, +) given by I (x) = 10gIOx is an isomorphism.

2.

(a) (b)

*

Is (E 4 , +) isomorphic to ({ 1, -1, i, - i}, .)? Explain. Is (E 4 , +) isomorphic to the subgroup K = {I, R2, H, V} of the octic group?

272

CHAPTER 6 Concepts of Algebra: Groups

"*

3.

Let f: (G, .) ---+ (H, *) be an isomorphism. For a E G, prove that the order of a equals the order of f(a).

4. Is S3 isomorphic to (1'.6, +)? Explain. 5. Give an example of two groups of order 8 that are not isomorphic. 6.

Prove that any two groups of order 2 are isomorphic.

7. Prove that any two groups of order 3 are isomorphic. 8. Show that every cyclic group of order m is isomorphic to (Em' +). 9. Show that every infinite cyclic group is isomorphic to (1'., +). 10.

Use the method of proof of Cayley's Theorem to find a group of permutations isomorphic to (a) (1'.5 - {O}, '5)' (b) the subgroup {I, R2, D, D'} of the octic group. (c) (JR, +).

11.

(a) (b) (c)

(d)

*

Define a homomorphism from (E 2no +) onto (En' +). What is the kernel K of the homomorphism you defined in (a)? List the elements of E2n/ K and give an isomorphism h from E2n/ K onto En· Construct the operation tables for the groups described in (a), (b), and (c) in the case n = 4.

12. Describe all homomorphic images of Eg. 13. Describe all homomorphic images of Ell' 14. Describe all homomorphic images of S3'

CHAPTER

7

concepts of Analysis: Completeness of the Real Numbers

In this chapter we give an introduction to the analyst's point of view ofthe real numbers. Section 7.1 discusses the reals as afield (a set of numbers with operations of addition and multiplication) that is ordered (so that all the real numbers may be thought of as forming a line) and complete (in the sense that there are no gaps or missing numbers anywhere along the line). After the initial definitions of section 7.1, each of the next three sections presents a new aspect of the concept of completeness for the real number system. Section 7.5 ties together all the other sections and gives an application of completeness.

7.1

Ordered Field Properties of the Real Numbers In this section we assume a knowledge of rational and real number systems commonly obtained in elementary calculus. Real numbers may be defined as equivalence classes of Cauchy sequences (see exercise 12 of section 7.4) or as "Dedekind cuts"-collections of pairs of subsets that partition the rationals. t Since we are more concerned in this text with completeness of the reals than with the development of the real number system, we choose to define the reals as a complete ordered field. Although we shall give general definitions of the words field, ordered, and complete, you should think of the real numbers as your principal example of each concept and your only example of all three. To understand fully the formal notion of a field in this section requires some familiarity with the first concepts of algebraic structure presented in section 1 of chapter 6. The concepts of ordering in this section can be understood without studying section 3.4 on orderings, although students who have covered that section will find the concepts familiar. 7 For a full development of the reals from Cauchy sequences, see Robert Stoll's Set Theory and Logic (Freeman, 1963). Walter Rudin's Principles of Mathematical Analysis (McGraw-Hill, 1953) develops the reals using Dedekind cuts.

273

274

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

DEFINITION A field is an algebraic structure (F, are binary operations on F such that

+, .) where + and,

+) is an abelian group with identity denoted by 0; that is, + is an operation on F so that for all x, y, z E F,

(1) (F, (a)

(x

(b)

x

+ y) + z = x + (y + z) + 0 = 0 + x = x.

For every x E F, there exists an additive inverse -x E F such that x + (-x) = (-x) + x = O. (d) x + y = Y + x. (2) (F - {o}, .) is an abelian group with identity denoted by I; that is, . is an operation on F - {O} so that for all x, y, z E F - {O}, (c)

x' (y . z) = (x' y) . z. X' I = I . x = x. (c) For every x E F - {O}, there exists a multiplicative inverse X-I E F - {O} such that x . X-I = X-I . X = 1. (d) x·y=y·x. (3) For all x, y, z in F, x . (y + z) = x . y + x . z. (a) (b)

(4) 0

* 1.

The rationals, the reals, and the complex numbers, where + and, are the normal arithmetic operations of addition and multiplication, are all fields. Some of the modular arithmetic structures (Z, +, .) discussed in section 6.3 are also fields. If p is a prime, then (Zp' +, .) is a field with p elements. Every field must have an additive identity 0 and a multiplicative identity 1. Thus the smallest field is Z2 = {o, I} with operations shown here.

o

o o

1

1

+

1

o

o 1

o o o

o 1

We now develop properties that collectively will distinguish the real numbers from all other fields. To begin, we consider a property of the fields Q and ~ not shared by the other fields we have named.

DEFINITION A field (F, such that for all x, y, z E F,

+, .) is ordered iff there is a relation < on F

(1) x <. x (irreflexivity). (2) If x < y and y < z, then x < Z (transitivity). (3) Either x < y, x = y, or y < x (trichotomy). (4) If x < y, then x + z < Y + z. (5) If x < y and 0 < z, then xz < yz.

7.1

Ordered Field Properties of the Real Numbers

275

Properties 0), (2), and (3) are familiar if you studied orderings in section 3.4. Property (3) ensures that the field elements are linearly arranged (a total ordering). Properties (4) and (5) ensure the compatibility of the order relation < with the field operations + and '. Familiar properties (such as 0 < I, -I < 0, and a < b implies -b < -a) can be derived from the order axioms. Therefore, these properties must be true in every ordered field. It can also be shown that, in an ordered field, if x < y and z < 0, then xz > yz. The notation x < y means x < y or x = y, while x> y stands for y < x. Both the rational numbers and the real numbers are ordered fields under their usual orderings. It can be shown that the complex numbers do not have an ordering that satisfies properties (1)-(5). Consider, for example, the situation if we assume 0< i. Then 0 = 0 . i < i . i = i 2 = - I and thus 0 < -1. On the other hand, if i < 0, then i 2 > 0 . i, so again -1 > O. By trichotomy, either 0 < i or i < 0 must be true, but both us lead to the false statement 0 < - I. Therefore, the field of complex numbers cannot be an ordered field. The finite fields also cannot be ordered. To show, for example, that 712 is not ordered, suppose 0 < 1. Then I = 0 + 1 < I + I = 0, so 1 < O. Thus, by transitivity, o < 0, contradicting property (1). Developing a distinction between JR and iQ takes a bit more care. We first note that every ordered field has the property that between any two elements there is a third element. If a < b then a + a < a + b < b + b, or (1 + I)a < a + b < (1 + l)b. Using! as the symbol for the inverse of the element I + I, we have a < !(a + b) < b. Thus ordered fields do not have "gaps" between any two given elements. However, some ordered fields, like the rationals, seem to be missing some points. For instance, if we look at a sequence of rationals such as 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213,00', formed by approximating we see the sequence consists entirely of rationals, yet the value (ft) approached by the sequence is not a rational number. In this respect the rationals appear to be incomplete. We shall see that this situation does not happen for sequences of real numbers; if there is a limiting value for a sequence of reals, it must be a real number. To make precise the idea of being incomplete or having missing points, we need to use the idea of bounded subsets.

ft,

DEFINITIONS Let A be a subset of an ordered field F. We say u E F is an upper bound for A iff a:::; u for all a EA. If A has an upper bound, A is bounded above. Likewise, I E F is a lower bound for A iff I :::; a for all a E A, and A is bounded below iff any lower bound for A exists. The set A is bounded iff A is both bounded above and bounded above. In JR, the half-open interval [0, 3) has 3 as an upper bound. In fact, n, 18, and 206 are also upper bounds for [0, 3). Both -0.5 and 0 are lower bounds. We note that a bound for a set might or might not be an element of the set. Any finite nonempty subset A = {XI> Xz, X3,00" x n } of JR is both bounded above and below; u = max {lxJ i = 1,2,3'00" n} is an upper bound for A and -u is a lower bound for A. In JR, the subset N is bounded below but not above, while the sets iQ and 71 are neither bounded above nor bounded below.

276

CHAPTER 7

Concepts of Analysis: Completeness of the Real Numbers

n, i, i, n"

In 0, the set i,···} of negative integer powers of 2 is bounded above (by 1) and below (by 0). The set A = {x E 0: x < ji} has many upper bounds: 8, 1.4, 1.4149, and so on. However, A has no lower bounds. The set B = {x E 0: x 2 < 2} is bounded above by 3 and below by -3. The best possible upper bound for a set A is called the supremum of A.

DEFINITIONS Let A be a subset of an ordered field F. We say s E F is a least upper bound or supremum of A in F iff (i) (ii)

s is an upper bound for A. s :::; t for all upper bounds t of A.

Likewise, i E F is the greatest lower bound or infimum of A in F iff (i) (ii)

i is a lower bound for A. I :::; i for all lower bounds I of A.

While several numbers may serve as an upper bound for a given setA, when the supremum of A exists it is unique (see exercise 8). We shall denote the supremum and infimum of A by sup (A) and inf (A), respectively.

Example. In the field JR, sup ([0, 3» = 3 and inf ([0, 3» = O. If A = {Tk: kEN}, then sup (A) = and inf (A) = O. For B = {x E JR: x 2 < 2}, sup (B) = ji and inf (B) = - ji. Also, inf (N) = 1 and sup (N) does not exist.

!

=

=!

Example. In the field of rationals, if A {Tk: kEN}, then sup (A) and inf (A) = O. For B = {x E 0: x 2 < 2}, even though B is bounded in 0, B has no supremum or infimum in O. The following theorem provides a characterization of the supremum of a set. Its interpretation, which depends on taking the view that e may be very small, is that every element of A is strictly less than a little more than the supremum of A, but there is always an element of A that is greater than a little less than the supremum. For example, in the field JR, the supremum of the set (2, 4) is 4. Even if we take e to a very small positive number, say e = lO,boo, every element of (2, 4) is less than 4 + e. Furthermore, 4 - e = 3.9999 is not quite big enough to be the supremum, because 3.99995 is in the set and is greater than 4 - e.

Theorem 7.1

Let A be a subset of an ordered field F. Then s = sup (A) iff

< s + e. (2) for all e > 0, there exists yEA such that y > s - e.

(1) for all e> 0, if x E A, then x

Proof.

First, suppose s = sup (A). Let e > 0 be given. Then x :::; s < s

x E A, which establishes property (l).

+ e for all

7.1

Ordered Field Properties of the Real Numbers

277

To verify (2), suppose there were no y such that y > s - e. Then s - e is an upper bound for A less than the least upper bound of A, a contradiction. Suppose now that s is a number that satisfies conditions (I) and (2). To show that s = sup (A), we must first show that s is an upper bound for A. Suppose there is yEA such that y > s. If we let e = then y > s + e, which violates condition (I). (The idea here is that e is only half the distance from s to the larger number y, so s + e must still be less than y. To verify this algebraically, we have 2y > s + y, so s+y 2s+y-s y-s y > -2- = 2 = s + -2- = s + e.) We conclude that y:5 S for all yEA,

y;S,

so s is an upper bound for A. To show s is the least of all upper bounds, suppose that there is another upper bound t such that t < s. (We will show that t cannot be an upper bound.) If we let e = s - t, then by condition (2), there is a number yEA such that y > s - e. Thus y > s - e = s - (s - t) = t. Hence y > t and! is not an upper bound. Therefore, s is indeed the least upper bound of A. The next example makes use of a form of the Archimedean principle (see Theorem 2.11), which states that for any positive real number r there is always a large enough natural number K so that < r.

k

Example. The set A = supremum in Q.

{x E Q: x <

Ii} is bounded above in the field Q but has no

Proof. A is bounded above in Q by ~ (and many other rationals). Now suppose s = sup (A) does exist in Q. We will first show that both s < and s > are false. If s < then s is positive. Choose a natural number K such that < s. Then s + < and s + is rational. Thus s + is an element of A, which contradicts s being an upper bound for A. If s > then s is positive. Choose a natural number M such that <s Then s > and s is rational. Thus s is an upper bound for A, which contradicts s being the least upper bound for A. Since both s < and s > are false, we conclude that s = .j2. However, is not a rational number. So, within Q, sup (A) does not exist. -

k Ii -it

Ii,

Ii

I!- -

k Ii

Ii, -Ii·

-Ii

it Ii

Ii

Ii

k

k

it

it

Ii

Ii

The last example shows that the field of rational numbers includes bounded subsets with no supremums in Q, and therefore Q is not complete in this sense. On the other hand, every bounded subset of the reals does have a supremum in ~. A proof of this requires considerable preliminary study of the nature of real numbers and thus is beyond the goals of this text. The distinction we seek between ~ and Q is given in the next definition.

DEFINITION An ordered field F is complete iff every non empty subset of F that has an upper bound in F has a supremum that is in F.

278

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

The rationals are not complete. We state without proof that the reals are a complete ordered field. In fact, with much work we could show that if F and F' are any two complete ordered fields, then they are essentially the same both in number of elements and in structure: The fields are isomorphic. This means there is a one-toone correspondence from F to F' that preserves both operations, the order relations, and all supremums. Thus the real numbers form essentially the only complete ordered field, so you should think of the terms real number system and complete orde redfield as synonymous.

Exercises 7.1 1.

2.

Let F be an ordered field and x, y, z E F. Prove that (a) exactly one of x < y, x = y, or y < x is true. (b) ifxO. (c) 0 < 1. (d) -1 <0. (e) if x < y, then -y < -x. (f) if x < y and z < 0, then xz > yz. Find four upper bounds for each of the following setS'. {x E ~: x 2 < 1O}

(a)

*

*:

(b)

{;x: x EN}

(c)

{x E~: x+ ~ < 5}

(d)

{x E ~: 7x 2

+

14x ~

+2<

23}

3.

Find a lower bound in

4.

Find the supremum and infimum, if they exist, of each of the following reals.

* * * * 5.

(if one exists) for each of the sets in exercise 2.

(a)

U:

n EN}

(b)

{ n : 1: n EN}

(c)

{2x:xEZ}

(d)

{(-l)n(l

(e)

{n: 2: n E

(f)

{x E Q: x 2 < 1O}

(g)

{x: -l::5x ::51}

(h)

{x:

-1 ::5 x ::5 I} - {OJ

(i)

{;y: x, YEN}

(j)

{x:

Ixl > 2}

N} U {5}

+~): nEN}

Let x be an upper bound for A ~ ~. Prove that (a) if x < y, then y is an upper bound for A. (b) if x E A, then x = sup (A).

6. Let A ~ ~. Prove that *: (a) if A is bounded above, then A, the complement of A, is not bounded above. (b) if A is bounded below, then A is not bounded below.

7.1

Ordered Field Properties of the Real Numbers

279

7.

Give an example of a set A C JR for which both A and A are unbounded.

8.

LetA C JR. (a) Prove that sup (A) is unique. That is, prove that if x and yare both least upper bounds for A, then x = y. (b) Prove that inf (A) is unique.

9.

LetA CB C JR. (a) Prove that if sup (A) and sup (B) both exist, then sup (A) :S sup (B). (b) Prove that if inf (A) and inf (B) both exist, then inf (A) ;:::: inf (B).

*

10.

Formulate and prove a characterization of greatest lower bounds similar to that in Theorem 7.1 for least upper bounds.

11.

If possible, give an example of (a) a set A C JR such that sup (A) = 4 and 4 tt. A. (b) a set A C Q such that sup (A) = 4 and 4 tt. A. (c) a set A C N such that sup (A) = 4 and 4 tt. A. (d) a set A eN such that sup (A) > 4 and 4 tt. A.

12.

Give an example of a set of rational numbers that has a rational lower bound but no rational greatest lower bound.

13.

Let A C JR. Prove that (a) if sup (A) exists, then sup (A) = inf {u: u is an upper bound of A}. (b) if inf (A) exists, then inf (A) = sup {I: I is a lower bound of A}.

*

14.

Prove that if B CAe JR and A is bounded, then B is bounded.

15.

For A C JR, let A - = {x: -x E A}. Prove that if sup (A) exists, then inf (A -) exists and inf (A-) = -sup (A).

16.

Let A and B be subsets of JR. (a) Prove that if sup (A) and sup (B) exist, then sup (A U B) exists and sup (A U B) = max {sup (A), sup (B)}. (b) State and prove a similar result for inf (A U B). (c) Give an example where A nB *0, sup (A nB) <sup (A), and sup (A nB) < sup (B). (d) For An B 0, state and prove a relationship between sup (A), sup (B), and sup (A n B). (e) Give an example where A n B 0, inf (A n B) > inf (A), and inf (A n B)

*

*

>

(f)

Proofs to Grade

17.

*

*

inf(B). For A n B 0, state and prove a relationship between inf (A), inf (B), and inf (A n B).

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. LetA C JR. If i = inf (A) and G > 0, then there is yEA such that

< i + G. "Proof." Let y = i + l Then i < y so yEA. By construction of y, y < i + G. • Claim. Let (F, +, .) be an ordered field and x E F. If -x < 0, then y

(b)

x>O.

280

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Suppose -x < O. Then by property (4) of ordered fields - x + x < 0 + x. But -x + x = 0 and 0 + x = x, so 0 < x. Thus x> O. Claim. If I: JR--+ JR and A is a bounded subset of JR, then I(A) is bounded. "Proof." Let m be an upper bound for A. Then a $ m for all a EA. Therefore, I(a) $/(m) for all a EA. Thus I(m) is an upper bound for "Proof."

(c)

(d)

I(A). Claim. Let A!: JR. If A is bounded above, then A is bounded below. "Proof." If A is bounded above, then sup (A) exists (because JR is complete). Since sup (A) = inf (A) (see the figure), inf (A) exists. Thus A is bounded below. fl.

A

sup(A) = inf(A)

7.2

~

The Heine-Borel Theorem In the next three sections we shall limit our discussion to the field of real numbers. It is not necessary to do so, but this provides a familiar model of an ordered field in which to work. Many of the most significant properties of the real numbers are based on certain "topological" concepts. In this section we shall introduce some of these concepts and establish a key result, the Heine-Borel Theorem, that is based on completeness.

DEFINITION For a real number a, if J is a positive real number, the ()-neighborhood of a is the set N(a, <5) = {x E JR: Ix - al < J}.

Because Ix - al < J is equivalent to a - J < x < a + J, the J-neighborhood of a is the open interval centered at a with radius J:

N(a, J)

=

(a - J, a + <5).

Thus, for example, N(3, 0.5) = (2.5,3.5) and NO, 0.01) = (0.99, 1.01). Although the number J may be any positive real number, your intuition is often better served by thinking of J as a small positive number. Familiar concepts from calculus may be expressed in neighborhood terminology. For example, recall that a function I: JR --+ JR is continuous at a E JR means that whenever x is near a, then/(x) must be near I(a). The precise way to express this is that for all e > 0 there exists J > 0 such that if Ix - al < J, then Vex) - I(a) I < e. In terms of neighborhoods this may be written: for all e > 0 there exists J > 0 such that if x E N(a, J), then I(x) E N(f(a), e).

7.2

The Heine-Borel Theorem

281

This approach has the advantage of specifying "closeness to a" in terms of sets (neighborhoods) rather than as points within a certain distance of a. Since there are number systems other than the real numbers for which the neighborhood concept may be introduced, the version above allows us to define continuity for those systems as well. The study of such systems along with continuity and other functional properties is a branch of mathematics called topology.

DEFINITION For a set A k JR, a point x is an interior point of A iff there exists <5 > 0 such that .N'(x, <5) k A.

Thus, point x is an interior point of A means that not only is x an element of A, but all the elements of some neighborhood (perhaps one wi th a small radius 6) are also in A. For the interval [2,5), 3 is an interior point since .N'(3, 0.5) k [2, 5). Also, 4.9998 is an interior point because .N'(4.9998, 0.000l) k [2, 5). See figure 7.1. In fact, every point in (2, 5) is an interior point of [2, 5). The point 2 is not interior to [2, 5), since every <5-neighborhood of 2 contains points that are less than 2 and hence not in [2, 5).

DEFINITIONS The set A k JR is open in JR iff every point of A is an interior point of A. The set A is closed in JR iff its complement A is open in JR.

As we have seen above, the interval (2, 5) is open because every point in (2, 5) is an interior point. Thus its complement (-00, 2] U [5, (0) is closed. On the other hand, the interval [2, 5) is not open since 2 is not an interior point. Should we N(2, 8)

(,

N(3, 0.5)

.) (, 2

.) 3

Figure 7.1

282

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

conclude that [2, 5) is closed? In ordinary conversation, with reference to objects like doors and eyes, "open" and "closed" are opposites. This is not so when the terms open and closed are applied to subsets of the real numbers. The complement of [2, 5) = (-00, 2) U [5, 00), which contains 5 but not as an interior point. Thus [2:5) is not open, so [2, 5) is not closed.

Example. The set of all real numbers is open since for every x E ~,H(x, 1) ~ ~. The empty set 0 is also open since the statement "for all x E 0, x is an interior point of 0" is vacuously true. Since ~ and 0 are complements, they are also closed. There are no other subsets of the reals that are both open and closed. A set is open if about each point you can find aJ-neighborhood that lies entirely within the set. This means no point of the set can be on the "boundary" or outer edges of the set (see exercise 7). With experience you should come to recognize open subsets of ~. The next two theorems will help.

Theorem 7.2

Every open interval is an open set.

Proof. Let (a, b) be an open interval. To show (a, b) is open, we let x E (a, b) and show that x is an interior point of (a, b). (That is, show H(x, J) ~ (a, b) for some J> 0.) We chooseJ = min {x - a, b - x}. (This minimum is the largest possibleJwe can use. See figure 7.2.) Then J > O. To show that H(x, J) ~ (a, b), let y E H(x, 6). Then a = x - (x - a)::5 x - J < y < x + J::5 x + (b - x) = b; and so y E (a, b) . •

Theorem 7.3

(a)

If d is a collection of open sets, then

U A is open. AE.s1l

(b)

If d is a finite collection of open sets, then

n A is open. AE.s1l

Proof. (a) Suppose x E

U A. Then there exists BEd such that x E B. Since B is in AE.s1l

the collection d, B is open and thus x is an interior point of B. Therefore, A, H(x, J) ~ A. there exists J > 0 such that H(x, J) ~ B. Since B ~ Therefore, x is an interior point of

U

U

AE.s1l

AE.s1l

U A, which proves U A is open.

n A. Then x E A for all A E d, and so for each open set A E d AE.s1l

(b)

Suppose x E

AE.s1l

AE.s1l

there corresponds J A > 0 such that H (x, J A) ~ A. Let J = min {JA: A E d}. N(x, 0) I

(

1_ x -

)

a -I~.----- b -

X - - - - - - - + I I1

) a

b

x

N(x, 0) in the case 0

= min{x -

Figure 7.2

a, b - xl

=

x - a

7.2

The Heine-Borel Theorem

283

(It is always possible to find a minimum or maximum from a finite set of real numbers. Note especially that the minimum of a finite set of positive numbers must be positive.) Then J> 0 and H(x, J) C A for all A E.il. Thus H(x, J) C A, which proves A is open. -

n

n

AE~

AE~

If in part (b) the set of JA'S were infinite, we could not be sure of finding a minimum. The set {k: n EN}, for example, does not have a minimum element. Furthermore, the statement of part (b) would be false without the word "finite." Indeed, the infinite collection .il = {(2 - ft, 5): n E N} of open sets has intersection

n (2 - 1.,5)

nEN

=

[2, 5),

n

which we have seen is not open. For A = (1, 5), we might choose JA = 0.75, for = (I!, 5) we could choose J A = .41, and so forth, but there is no minimum for the set {JA : A E .il}. Theorems 7.2 and 7.3 can be combined to produce many examples of open subsets of~. For example, the following are open sets:

A

(5,7) U (-3, 4) U (10,20)

U A,

(2,00) =

where.il = {(2, x):x > 2}

AE~

(-00, 2)

=

U A,

where.il

= {(x,

2): x < 2}

AE~

(-5,0) U (2,00) and ~

- {2}

=

(-00, 2) U (2,00)

In exercise II you are invited to state and prove a theorem similar to Theorem 7.3 for closed sets. Some examples of closed sets are closed intervals and every finite subset of ~. (See exercise 7(b ).) The set 71. is closed in ~ because its complement, ~ - 71. = (a, a + 1), is a union of open sets and hence is open.

U

aEZ

The remainder of this section will concentrate on closed.and bounded sets. Every finite set is closed and bounded, but even closed and bounded sets that are infinite behave like finite sets in some ways. For example, every finite set contains its infimum and supremum, and so does every set that is closed and bounded. The closed and bounded sets [-1,0.83], [6, 10], and [-2,2] U [7, 9] U {12}, for example, all contain their infima and suprema. We know of sets (n, 4) and [2,00) that do not contain their suprema or infima or both, but neither of these sets is both closed and bounded.

Theorem 7.4

If A is a non empty closed and bounded subset of~, then sup (A) E A and inf (A) EA. Proof. Denote sup (A) by s and suppose s tt. A. Then sEA, which is open, since A is closed. Thus H(s, J) C A for some positive J. This implies s - J is an upper bound

284

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

for A, since the interval (s - J, s + J) is a subset of A. (Notice that no element of A is greater than or equal to s + J, because s is an upper boundfor A.) This contradicts Theorem 7.1. Therefore, sEA. The proof that inf (A) E A is similar.

Theorem 7.5

Let A be a closed set and x E IR. If A

Proof.

n .N'(x, J) =1= 0

for all J > 0, then x EA.

-

Exercise 12.

To further extend the properties of closed and bounded sets, we will need the concept of a cover for a set. A cover for a set A is simply a collection of sets whose union includes A.

DEFINITION Let A be a set. A collection C(6 of sets is a cover for A iff AC C. A subcover of C(6 for A is a subcollection C!A of C(6 that is also a

U

CE'€

cover for A.

If we let Ha = (-00, a) for each a E IR, then '3£ = {Ha: a E IR} is a cover Ha = IR. The collection OW' = {Ha: a E Z} is a subcover of '3£; for the reals, since

Example.

U

aE~

and:£ = {Ha: a E N} is a subcover ofOW' and of '3£. We note in passing that there is no finite subset of '3£ that covers IR.

Example.

For any set C, {C} is a cover for C. Also with no proper subcover at all.

Hc}: c E C} is a cover for C

Let An = (n - k, n + k) for each n E N. The collection.sil ={An: n EN} is a cover for N that has no sub cover other than itself (figure 7.3).

Example.

The collection C(6 = {(l, 1): x E (1, oo)} is a cover of the interval (0, 1) by open sets, and CZlJ = {(l, 1): x E N - {I}} is a subcover of C(6. The cover C(6 of (0, 1) has no finite subcover. It also happens that C(6 is a cover of the set [!, 1). Among the many finite subcovers of C(6 for [!, 1) is '3£ = {(!, 1), (n, I)}.

Example.

A cover C(6 of a set A may be imagined by thinking of the covering sets of C(6 as providing shade for the set A. In figure 7.4 we have the sun directly over the set A. The covering sets Co., Cp, Co,'" are drawn slightly above A. Think of the sun's rays

Al

Az I

(

j

0

2

A4

A3

As

A6

( I )

( I )

5

6

) )

( 3

Figure 7.3

( 4

7.2

The Heine-Borel Theorem

285

Sun

Sun's rays

A

Figure 7.4 as parallel beams of light. As the sun shines straight down, 't6 is a cover for A if A is c. within the region shaded by the sets in 't6-that is, if A <::;::

U

CE'ii:

DEFINITION A subset A of ~ is compact iff every cover of A by open sets has a finite subcover.

We have seen that 'Je = {( -00, a): a E ~} is an open cover of ~ with no finite subcover. Thus ~ is not compact. Likewise, (0, 1) is not compact, since 't6 = {(t, 1): x E (1, oo)} has no finite subcover. What kind of sets are compact?

Examples.

The empty set is compact; anyone set from the covering set will cover = {Xlo Xl> X3,"" xn} is compact. If F <::;:: Ou where each Ou is

0. Any finite set F

U

uELl

open, simply select one covering set 0Ui' such that Xi E 0Ui for each i = 1, 2, 3, ... , n. Then {Ou: i = 1, 2, 3,. '" n} is a finite cover for F. The concept of compactness may be difficult to imagine initially. You should think of compactness as a concept akin to finiteness: if A is compact, then any time A <::;:: Ou, a union of open sets, then there are a finite number of 0Ui: i = I, ... , n

U

uELl

286

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

(

8 )

I

• • • 6

5

•4

•3

5

4

3

"2

•

2

Figure 7.5 11

such that A C

U Our Thus any cover of A with open sets needs to use only finitely i='

many of its sets to cover A. It does not matter whether the 0Ui are large or small sets; what matters is that there are only a finite number of them. Here is another example.

Example. The set A

=

{(n : 1): nEN} U {I} is compact.

Proof. Let {Ou: a E~} be any cover of open sets for A. One of the covering sets, call it Ou~' contains the element I. Since Ou* is open, there is a J-neighborhood .NO, J) C 0u*. See figure 7.5. (We will show that all but afinite number of elements of A are in.NO, J) and therefore in 0u*.) Choose N such that N > If n > N, then ~ < n and so 1 < n6. This implies n + 1 < n + n6, which means n! ' < I + J. Therefore, if n > N, then n! ' E.NO, J) and n! 1 E 0u*. Now choose 0UI' 0U2' 0U3;· .. , 0UN such that 2 E OUI' ~ E 0U2' E OU3'···' and N~l E DUN. Then A C Ou* U OUI U 0U2 U 0U3 U ... U DUN. (The first N elements ofA are in 0UI' ... ' OUN' and everything else is in Ou*.) We have succeeded in finding a finite subcover of N + 1 sets for the open cover {Ou: a Ell}. Therefore, A is compact. •

l

1

We close this section with an elegant characterization of the compact subsets of

JR. The next theorem can be traced to the works of Edward Heine (1821-1881) and Emile Borel (1871-1938). Watch how heavily the proof depends on the completeness of JR.

Theorem 7.6

(The Heine-Borel Theorem) A subset A of JR is compact iff A is closed and bounded.

Proof. (i)

Suppose A is compact. We first show that A is bounded. We note that JR= (-n,n); thus AC (-n,n). Therefore, ~={(-n,n):nEN}

U

U

nEN

nEN

is an open cover of A. By compactness, ~ has a finite subcover ~' = {( -n, n): n E {n" nb ... , nd}. IfwechooseN = max {n" nb ... ' nk}' then A C (-N, N). Therefore, A is bounded. We next show that A is closed by proving A is open. Suppose yEA:. (We must show y is an interior point of A:.) For each x E A, x 1= y and thus

7.2

OX 1 (

I

I

I

)

X4

XI

287

Ox 3 )

OX 4 (

The Heine-Borel Theorem

(

( OXS I

I

) Ox, I

)

)

I Xs

X3

X2

yEA

A

Figure 7.6 6x = ~Ix - yl is a positive number. The collection {.N'(x, 6x): x E A} is a family of open sets that covers A. Hence by compactness, A c;: .N'(XIo 6xl ) U N(X2, 6X2 ) U ... U N(Xko 6Xk )

(ii)

for some x" Xz, ... , Xk EA. By choosing 6 = min {6xl , 6X2 '"'' 6Xk }' we have .N'(y,6) c;: A. (Ifz E A, then Iz - xii < 6xJor some i. Thus ifz E .N'(y, 6), then Iz - yl < 6:::; 6xj and IXi - yl :::; IXi - zl + Iz - yl < U5Xi = IXi - yl.) Thus A is open. Hence A is closed (figure 7.6). Conversely, suppose A is a closed and bounded set and Cf6 is an open cover for A. (We showCf6 has afinite subcover.) For each X E IR, letA x = {a E A: a :::; x}. Also, let D = {x E IR: Ax is included in a union of finitely many sets from Cf6}. Since A is bounded, inf (A) exists. Thus if x < inf (A), Ax = 0 and it follows that xED. Therefore, (-00, inf (A)) c;: D and so D is nonempty. We claim D has no upper bound. (Thefollowing proofofthisfact involves showing that if D is bounded above, then sup (D) is in A, and then using the nature of D to build a contradiction.) Suppose D is bounded above. Then Xo = sup (D) exists (by the completeness of IR). Let 6 > 0 and choose tED such that Xo - 6 < t :::; Xo (applying Theorem 7.1). If A n .N'(xo, 6) = 0, then At=!aEA:a:::;t}= {aEA:a:::;xo+~}= Axo +(J/2)' But t is in D, so Xo + '2 is in D. This is a contradiction to Xo = sup (D). Therefore, for all 6> 0, we have A n .N'(xo, 6) i= 0. By Theorem 7.5, Xo is in the closed set A. Let C * be an element of Cf6 such that Xo E C*. Since C* is open, there exists e> 0 such that .N'(xo, e) c;: C*. Choose Xl ED such that Xo - e < XI:::; Xo· Since XI ED, there are open sets C" C2, ... , Cn in Cf6 such that AXI c;: C1 U C2 U ... U Cn' Now letx2 = Xo +~. Thenx2 E C* andA x2 c;: C, U C2 U ... U Cn U C*. Thus X2 ED, a contradiction, since X2 > Xo and Xo = sup (D). We conclude that D has no upper bound. Finally, since D has no upper bound, choose xED such that x> sup (A) (sup (A) exists because IR is complete). Thus Ax = A and since xED, A is included in a union of finitely many sets from Cf6. Therefore, A is compact. _

Exercises 7.2 1. For x 10 X2 E 1R,61 > 0 and 62 > 0, describe (a) .N'(XIo 6 1) n N(XI, 6 2 ), (b) .N'(XI' 6 1) (c) .N'(XI, 6,) n .N'(Xz, 6 2),

"*

n N(xz, 6,).

288

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers 2.

*:

3.

Find x and 0 E IR such that (a) H(x,o) = (7, 12). (e) H(x,o) = (6.023, 6.024).

*

(b)

Write the calculus definition of limf(x)

H(x,o)

= (3.8,

3.8S).

= L in terms of neighborhoods.

x~a

4.

Find the set of interior points for each of these subsets of IR. (a)

(-I, I)

(b)

W

Q

~

(e) (g) (i)

{:&: kEN}

(f) (b)

IR - N

U (n + 0.1, n + 0.2)

(-I, I]

IR-Q

{k kEN} U {O} IR -

{:&: kEN}

nEN

5.

Classify each of the following subsets of IR as open, closed, both open and closed, or neither open nor closed. (a) (-00, -3) (b) H(a,o) - {a}, a E IR, 0> 0 (e) (S, 8) U {9} (d) Q (e) IR - N (f) {x: Ix - SI = 7} (g) {x: Ix - SI > 7} (b) {x: Ix - SI 7} (i) {x: Ix - SI :::; 7} (j) {x: 0 < Ix - SI :::; 7}

*

* * 6.

*

Prove that (a) every open ray, either (a, (0) or (-00, a), is an open set. (b) every closed ray, either [a, (0) or (-00, a], is a closed set. (e) every closed interval, [a, b], is a closed set.

Let A and B be subsets of IR and x E IR. Prove that (a) if A is open, then A - {x} is open. (b) if A is finite, then A is closed. (e) if A is open and B is closed, then A - B is open. (d) if A is open and B is closed, then B - A is closed. *: (e) if A and B are closed, then A U B is closed. (f) if A and B are closed, then A n B is closed.

7.

*

8.

Give an example of a family of sets .ii such that each A E .ii is closed but A is not closed.

U

AEsd

9.

LetA be a subset of IR. Prove that the set of all interior points of A is an open set. 10. A point x is a boundary point of the set A iff for all 0 > 0, H(x, 0) n A 0

*

*

andH(x,o) nA 0. (a) Find all boundary points of (2, S], (0, I), [3, S] U {6}, Q. (b) Prove that x is a boundary point of A iff x is not an interior point of A and not an interior point of A(e) Prove that A is open iff A contains none of its boundary points. (d) Prove that A is closed iff A contains all of its boundary points. 11.

State and prove a theorem similar to Theorem 7.3 for closed sets.

12.

Prove Theorem 7.S.

13. Prove that if A and B are both compact subsets of IR, then *: (a) AU B is compact. (b) An B is compact.

7.2

14.

The Heine-Borel Theorem

289

Give an example of (a) a bounded subset of IR and an open cover of that set that has no finite sub cover. (b) a closed subset of IR and an open cover of that set that has no finite subcover.

15. Which of the following subsets of IR are compact? (a) 71. (b) [0, 10] U [20, 30] (c) [n,JTO] (d) IR-A,whereAisanyfiniteset (e) {I, 2,3,4,9, 12, 18} (f) {OJ U {k: n E N} (g) (-3,5] (h) [0, I] n Q

*

*

*

16.

Use the Heine-Borel Theorem to prove that every finite set is compact.

17. Let S = (0, I] and let C(6 = {( n ; 2, 2 0In (a) (b) (c)

»): n E N

l-

Prove that C(6 is an open cover of S. Is there a finite subcover of C(6 that covers S? What does the Heine-Borel Theorem say about S?

Use the Heine-Borel Theorem to prove that if {Au: a E ~} is a collection of compact sets, then Au is compact. uELl 19. Give an example of a collection {Au: a E ~} of compact sets such that Au is not compact. uELl 20. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If A and B are compact, then A U B is compact. "Proof." If A and B are compact, then for any open cover {au: a E~} of A, there exists a finite subcover aU" OU2"'" OUn' and for any open cover {Up: {JE 1} of B, there exists a finite subcover Up" Up2 ,... , Up",. Thus A ~ au, U OU2 U ... U OUn and B ~ Up, U UP2 U ... U Up",. Therefore, A U B ~ au, U OU2 U ... U OUn U Up, U UP2 U ... U Up"" a union of a finite number of open sets. Thus A U B is compact. (b) Claim. If A is compact, B ~ A, and B is closed, then B is compact. "Proof." Let {au: aE~} be an open coverofB. If{Ou: aE~}isanopen cover of A, then there is a finite subcover of {au: a E ~} that covers A and hence covers B. If {au: a E~} is not an open cover of A, add one more open set 0* = IR - B to the collection to obtain an open cover of A. This open cover of A has a finite subcover of A that is a cover of B. In either case B is covered by a finite number of open sets. Therefore B is compact. (c) Claim. If A is compact, B ~ A, and B is closed, then B is compact. "Proof." B is closed by assumption. Since A is compact, A is bounded. Since B ~ A, B is also bounded. Thus B is closed and bounded. Therefore, B is compact. (d) Claim. An open ray (a, 00) is an open set. "Proof." Let x E (a, 00). Let 0 = a-x. If y E H(x, 0), then y > x - O. Therefore, y > a and so y E (a, 00). Thus H(x, 0) ~ (a, 00). This proves x is an interior point of (a, 00). Since every point of (a, 00) is an interior point, (a, 00) is open. 18.

n

U

Proofs to Grade

*

290

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

7.3

The Bolzano-Weierstrass Theorem In the previous section we used the completeness of IR to prove the Heine-Borel Theorem. In this section we use the Heine-Borel Theorem to prove another classical result of modern analysis, the Bolzano-Weierstrass Theorem. Bernard Bolzano (1781-1848) was a leader in developing highly rigorous standards of mathematics. Karl Weierstrass (1815-1897) has been called the father of modern analysis because of his many deep contributions and his absolute insistence upon rigor. The theorem that bears their names is easy to visualize if you think of the terms of a sequence "piling up" around some number. Consider the set A = {(_l)n+1/ n: n EN}, which is bounded above by I and below by Since A = {I, -~, is infinite and bounded, it would seem that its points must pile up or "accumulate" around some number or numbers between and I. The number 0 is a point about which points of A accumulate (figure 7.7). The fact that such a number must exist in IR is a consequence of the Bolzano-Weierstrass Theorem. Before we get to that result, we first define what it means for a set to accumulate around a point.

-!, t,

-!.

!, ... }

-!

DEFINITION The number x is an accumulation point for the setA iff for all J > 0, H(x, J) contains at least one point of A 4istinct from x. When x is an accumulation point for A, there are so many elements of A near x that every neighborhood of x, H(x, J), no matter how small J may be, must contain points of A other than x. The point x is not necessarily an element of A; the idea is that there are always points of A arbitrarily close to x. This also means that when x is an accumulation point for a set A, every neighborhood of x must contain an infinite number of points of A.

Theorem 7.7

A number x is an accumulation point for a set A iff for all J > 0, H(x, <5) contains an infinite number of points of A.

Proof. If every neighborhood of x contains an infini te number of points of A, then each neighborhood certainly contains at least one point of A distinct from x. Therefore, x is an accumulation point. Now suppose that x is an accumulation point for A. Suppose that H(x, J) n A is finite for some J> O. Let J 1 = min {Ix - yl: y E H(x, J) n A, x oF y}. (Our choice of J 1 is so small that H(x, ( 1 ) will have no points of A other than perhaps x itself.) Then H(x, ( 1) n A = {x}, which contradicts the initial assumption that x is an accumulation point for A. Therefore, every neighborhood of x must contain an infinite number of points of A. •

•

•

•

I

I

4

. . . . I ...

-6--

0

•I •I 7

•

5

Figure 7.7

•

7.3 The Bolzano-Weierstrass Theorem

291

From Theorem 7.7 it follows that no finite set A can have any accumulation points. It also implies that isolated points within sets (such as the number 9 in the set [1,8] U {9}) are not accumulation points for the set.

Example. Every point between 3 and 5, including 3 and 5, is an accumulation point for the half-open interval [3, 5). Using the point 3.1 as an example, every .N(3.1, J) contains points of [3,5) other than 3.1. In particular, if J 2: 1.9, then .N(3.1, J) includes all of A; if .1 < J < 1.9, then .N(3.1, J) includes [3, 3.1 + J); ifO<J:::;.I,then.N(3.1,J)includes(3.1-J,3.1 +J); thus 3.1 is an accumulation point of [3,5). Similar reasoning applies to every number between 3 and 5. The number 3 is also an accumulation point for [3, 5), since .N(3, J) n [3, 5) will include (3, 3 + J) for J < 2. Even though 5 tt. [3, 5), similar reasoning shows that 5 is an accumulation point for [3, 5).

Examples.

Every element of the closed interval [0, 1] is an accumulation point of [0, 1]. No other real numbers are accumulation points of [0, 1]. The number 0 is the only accumulation point of {( - 1)" + 1/n: n E N}, although ois not in the(-1)"11+1 set. The set { 11 : n E N} has exactly two accumulation points, -1 and 1.

Example.

Every real number is an accumulation point for Q. To see this we observe that for any real number r, .N(x, J) = (r - J, r + J) contains rationals. Since those same neighborhoods always contain irrationals, every real number is also an accumulation point for the set of irrationals. The set of accumulation points of A is called the derived set of A, denoted A'. As examples: (3, 5)' = [3, 5)' = (3, 5]' = [3, 5]' = [3, 5] ([1,8] U {9})' = [1, 8] (( - 1, 6] U (7, 8»' = [- 1, 6] U [7, 8] ((-1, 6) U (6, 8»'= [-1, 8] Q'= IR

N'=0 LetB={1.4, 1.41, 1.414, 1.4142, 1.4l421,oo.} be the set of successive decimal approximations to Then B' = {Pl. The following theorem relates derived sets and closed sets.

p.

Theorem 7.8

A set A is closed iff A' CA. Proof. Suppose A is closed andx E A'. We prove x E A by contradiction. If x tt. A, then x E Ii, an open set. Thus .N(x, J) C Ii for some positive J. But then .N(x, J) can contain no points of A. Thus x tt. A', a contradiction. We conclude x EA.

292

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Suppose now A' CA. To show that A is closed, we show A is open by contradiction. If A is not open, there is at least one x E A that is not an interior point of A. Therefore, no 6-neighborhood of x is a subset of A; that is, each 6-neighborhood of x contains a point of A. This point must be different from x, since x E A. Thus x EA'. ButA' CA, so x EA. This is a contradiction. We conclude that A is closed .

•

At the beginning of this section we suggested that the bounded infinite set A = {( _1)n+ n: n E N}, and by inference all other bounded infinite sets, must have at least one accumulation point in JR. We are now in a position to prove this is so, using the Heine-Borel Theorem.

1/

Theorem 7.9

(The Bolzano-Weierstrass Theorem) Every bounded infinite subset of real numbers has an accumulation point in JR.

Proof. Suppose the set A is bounded and infinite but has no accumulation points. Then A is closed by Theorem 7.8 (A' = 0, hence A' C A.) Thus by the Heine-Borel Theorem, A is compact. Since A has no accumulation points for each x E A there exists 6x > such that H(x,6x ) n A = {x}. Thus for x =1= y in A, H(x, 6x ) =1= H(y, 6y). But then the family .ii = {H(x, 6x ): x E A} is an infinite collection of open sets that covers A and has no subcover other than itself. Hence A has no finite subcover. This contradicts the fact that A is compact. Therefore, A must have an accumulation point.

°

•

Exercises 7.3 an example of an infinite subset of JR that has no accumulation points. exactly one accumulation point. exactly two accumulation points. denumerably many accumulation points. an uncountable number of accumulation points.

1.

Find (a) (b) (c) (d) (e)

2.

Find the derived set of each of the following sets.

*

(a)

{n ~ 1: n EN}

(b)

{2": n E N}

*

(c)

{6n: n E N}

(d)

{;: n,mEN}

(e)

(0, 1]

(f)

(3, 7) U {4, 6, 8}

(b)

e

*

(g)

{I + (-n+l 1)"n: n EN}

(i)

Q

n (0,1)

(j)

7l..

+ n2 (1 n+ (_1)n): n EN}

7.3 The Bolzano-Weierstrass Theorem (k)

*

"*

"*

x: xE (-;n,~)} {k + ~: k, n EN}

{sin

*

(m)

3.

Let S = (0, I]. Find S' n (S)'.

(I)

{Si:X: xE (0, n)}

(n)

{;,: x, y

293

EZ}

4.

Prove that if A <::: JR, Z = sup (A), and Z tt. A, then z is an accumulation point of A.

5.

(a) (b)

6.

Prove that (A U B)' = A' UB'. (The operation of finding the derived set preserves unions.)

7.

Prove that (A

S.

(a) (b) (c) (d)

9. 10.

* 11.

* *

12.

*

Prove that if A <::: B <::: JR, then A' <::: B'. Is the converse of part (a) true? Explain.

n B) <::: A' n B'. Show by example that equality need not hold. I

Prove that if x is an interior point of the set A, then x is an accumulation point for A. Is the converse of part (a) true? Explain. Prove that if S <::: JR is open, then every point of S is an accumulation point of S. Is the converse of part (c) true? Explain.

Prove that if B is closed and A <::: B, then A I <::: B. Define the closure of a set A to be c(A) = A U A'. (a) Prove that c(A) is a closed set. (b) Prove that A <::: c(A). (c) Prove that c(A) is the smallest closed set containing A. That is, if B is closed and A <::: B, show that c(A) <::: B. (d) Prove that the derived set of the closure of A is the same as the derived set of A. Which of the following must have at least one accumulation point? (a) An infinite subset of N (b) An infinite subset of (-10, 10) (c) An infinite subset of [0, 100] (d) An infinite subset of Q (e)

{;k:kEN}

(g)

An infinite subset of Q n [0, I]

(f)

{~:P,qEN,p
Suppose that A is an infinite subset of [-100, -5] U [5, 100] and for all x EA, -x EA. (a) Could A have exactly one accumulation point? If so, give an example. (b) Could A have exactly two accumulation points? If so, give an example. (c) Prove that ify is an accumulation point of A, then -y is an accumulation point of A. (d) Give an example of an infinite subset B of [-100, 100] such that for all x E B, -x E Band B has exactly three accumulation points.

13.

Prove that if a bountled subset of the real numbers has no accumulation points, the set is finite. -

14.

Let x be an accumulation point of the set A and let F be a finite set. Prove that x is an accumulation point of A-F.

294

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Proofs to Grade

15.

"*

*

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. For A, B C JR, (A U B)' = A' U B'. "Proof." (i) Since A C A U B, A' C (A U B)' by exercise 5(a). Likewise, B' C (A U B)'. Therefore, A' U B' C (A U B)'. (ii) To show that (A U B)' CA' U B', let x E (A U B)'. Then for all b > 0, H(x, b) contains a point of A U B distinct from x. Restating this, we have, for all b > 0, that H(x, b) contains a point of A distinct from x or a point of B distinct from x. Thus for all b > 0, H(x, b) contains a point of A distinct from x, or, for all b> 0, H(x, b) contains a point of B distinct from x. But this means x EA' or x E B'. Therefore, x EA' U B'. • (b) Claim. If A is a set with an accumulation point, B CA, and B is infinite, then B has an accumulation point. "Proof." First, A is infinite because B C A and B is infinite. Since A has an accumulation point, by the Bolzano-Weierstrass Theorem, A must be bounded. Since B C A, this means B is bounded. Hence by the Bolzano-Weierstrass Theorem again, B has an accumulation point. • (c) Claim. For A, B C JR, (A - B)' CA' - B'. "Proof." (A - B)' = (A n iJ)' (definition of A - B) C A' n (iJ)' (exercise 7) C A' n (B') (since (iJ)' C (B'») =A'-B'. • (d) Claim. For A C JR, (A)' = (A') "Proof." x E fA') ¢=} x $. A' ¢=} x is not an accumulation point for A ¢=} x is an accumulation point for A ¢=}x

(e)

7.4

E (A)'.

•

Claim. If A is closed, then A' CA. "Proof." Suppose A is closed. Then A is open. Let x E A. Then x is an interior point of A. Therefore, there exists b> 0 so that H(x, b) C A. Hence H(x, b) n A = 0. Thus x is not an accumulation point for A. From x EA =:}x $.A', we conclude A' CA. •

The Bounded Monotone Sequence Theorem The concept of a sequence should be familiar to you from calculus. For example, the sequence given by Xn = 2n has as its first few terms XI = 2, X2 = 4, X3 = 8, and so on. The sequence Ym where Yn = (_l)n, has the alternating terms -1 and 1, and . few terms- 2'I 3'I - 4'I 5'···· I the sequence Zn = (-1) n/ (n + 1) has flfSt A sequence is simply a function x whose domain is N. Usually the image of n E N under the sequence x is written as x n , rather than the functional notation x(n). We call x n the nth term of the sequence. A sequence of real numbers (or real sequence) has codomain JR, a rational sequence has codomain Q, and so on. In this section we shall use the Bolzano-Weierstrass Theorem to prove an important result about sequences of real numbers.

7.4 The Bounded Monotone Sequence Theorem

295

A sequence x is bounded above iff there exists a real number B such that B for all n E N. For example, the sequence whose nth term is Cn = e- n is bounded above by B = e -I. Of course, once one value of B is selected, any other larger value for B works just as well. It would be correct to say that Cn = e -n is bounded above by B = 945. In a similar way, we say that a sequence x is bounded below iff there exists a real number B such that Xn ;::::: B for all n E N. A bounded sequence is both bounded above and bounded below. Boundedness may be described in terms of absolute values of the terms of the sequence. Xn

Theorem 7.10

::=;

A sequence x of real numbers is bounded iff there exists a real number B such that Ixnl ::=; B for all n E N.

•

Proof. Exercise 7.

The sequence given by Xn = 2n is bounded below by 0 but not bounded above, whereas the sequence whose nth term is tn = e- n is bounded above by e- I and below by O. The sequence z, where Zn = (_I)n I(n + I) is bounded above by and We may select B = to satisfy Iznl ::=; B for all n EN. below by The bounded sequence x given by Xn = (n + I)ln has for its first few terms 2, ~, ~, ~, .... Furthermore, for large values of n, the values ofx n are close to the number I. This observation gives rise to the concept of the limit of x.

-i.

!

!

DEFINITIONS A sequence x has the limit L, or x converges to L, iff for every real e > 0 there exists a natural number N such that if n > N, then IXn - LI <e. If x converges to L, we write lim Xn = L or Xn --+ L. If no such L exn_ oc ists, the sequence diverges.

=( -It I(n + I) converges to O. We can see that as n . II II II h h I ( grows arger wntten as n --+ 00)h , t e terms - 2' 3' - 4' 5' - 6' 7"" approac zero. T e sequence xn = 2n diverges. As n --+ 00, the terms 2, 4, 8, 16,32, ... become progressively larger without bound; there is no number L about which the terms of Xn gather.

Examples. The sequence Zn

With some practice, you will be able to "eyeball" (easily discern) limits of some sequences involving rational expressions. For example, consider 12n4 + 5n

Xn

+

I

= 70n2 - 18n4 .

For large values of n, the term 5n + I is quite small compared to 12n4. Likewise, -18n4 will be much larger than 70n 2 for large values of n. Thus for large n, the sequence Xn behaves rather like the sequence Wn = 12n41(-18n 4 ), which we see converges to -,/. We guess (correctly) that Xn converges to -/

296

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Using the definition to prove that a given sequence converges usually involves these two steps: first, decide the value of the limit L, and then verify the definition of convergence. Often, as in the next example, one must do some preliminary scratchwork, beginning with the desired result IX n - LI < e and continuing until we find a relationship between nand e that will imply this result.

Example. The sequence x given by Xn

3n 2

= - 2 - - converges.

n +1 Scratchwork. We calculate that Xs is approximately 2.8846, XIO is approximately 2.9703, and Xso is approximately 2.9995, and make a guess that x converges to 3. Next we must show that the limit is 3 by showing that, for every e > 0, there is a natural number N such that n > N implies 13n2 /(n 2 + 1) - 31 < e. Since 13n2 /(n 2 + 1) - 31 = I -3/(n 2 + 1) 1= 3/(n 2 + 1), we require an integer N such that n > N implies 3/(n 2 + 1) < e or, equivalently, n 2 + 1 > i. We know that n 2 + 1 > n, so by selecting N to be any natural number greater than we have that n > N implies n2 + 1 > N > This scratchwork leads to the formal proof that follows.

i,

i.

+

= 3. Let e> O. Let N be a natural number greater than i. Suppose n > N. Then n > i, and since n 2 + 1 > n, n 2 + 1 > i. Therefore, if n > N, then 3/(n 2 + 1) < e. Thus 13n2 /(n 2 + 1) - 31 < e. Therefore, lim 3n 2 /(n 2 + 1) = 3. • n_ Proof.

We will show that lim 3n 2 /(n 2 n_oo

1)

OO

Intuitively, a sequence x converges if the terms Xn get closer to a single number L for larger values of n. It follows that the alternating sequence Yn = (_I)n must diverge. To prove this we use a denial of the definition of convergence: if lim Yn =1= L, then "there exists a real e> 0 such that for all N E N, there exists n > N such that IYn - LI 2:: e."

Example.

Prove that the sequence given by Yn = (_I)n has no limit.

= L for some number L. Let e = 1. We must show that for all N E N, there exists n > N such that IYn - LI 2:: 1. Let N E N. If L > 0, let n be any odd integer greater than N. Then Yn = -1. For L ::5 0, let n be any even integer greater than N; here Yn = 1. In each of these cases n was selected so that n > Nand Proof.

Suppose lim Yn

IYn - LI

2::

n_ oo

1. Therefore, lim Yn n_ oo

=1=

L.

•

The next theorem establishes that the terms of a convergent sequence cannot approach two different points.

Theorem 7.11

If a sequence x converges, then its limit is unique. Suppose x n ---+ L and x n ---+ M and L =1= M. Let e = tiL - M I. (The idea of the proof is to suppose there are two different limits and select e so small that the terms cannot simultaneously be written e of each limit.) See figure 7.8.

Proof.

7.4

• •

! •••

L

..

The Bounded Monotone Sequence Theorem

)

(

•

• ••• !

M

•

..

)

297

•x

- MI ----~·I e = ~IL - MI

'~'-----IL

Figure 7.8 Since Xn --+ Land xn --+ M, there are natural numbers NI and N2 such that IX2 - MI < e. Let N = NI + N 2.

n > NI implies IXn - LI ::5 e and n > N2 implies Then since N is greater than both NI and N 2 ,

IL -

MI =

I(L - xn) + (x n - M)I ::5 IL - x,,1 + IXn - MI = Ix" - LI + Ix" - MI <e+e = 2e 2

=3"IL-MI.

Thus the assumption L oF M leads to IL - MI < ~IL - MI, which is a contradiction since IL - MI > O. We conclude that the limit of x is unique. • The next theorem is useful for determining and verifying limits without resorting to the definition. Sometimes dubbed the "sandwich" or "squeeze" theorem, it states that if a sequence b is "sandwiched" between two sequences a and e, both of which converge to L, then b must also converge to L.

Theorem 7.12

Suppose a, b, and e are real sequences such that an ::5 b" ::5 e" for all n E N. If an --+ L and en --+ L, then bn --+ L.

Proof. Suppose an --+ L and en --+ L. Let e > O. There are natural numbers NI and N2 such that n > NI implies Ian - LI < e and n > N2 implies len - LI < e. Let N be the larger of N I , N 2. Since an ::5 bn ::5 em an - L::5 bn - L::5 en - L. Therefore, for n > N, -e < an - L ::5 bn - L ::5 en - L < e. Thus Ibn - LI < e for all n > N, so bn--+N. • Example. To illustrate Theorem 7.12 consider the sequence whose nth term is Xn = si~ n. Since sine is a function with range [ -I, 1], - k::5 si~ n ::5 k, for all n E N. Because both - k--+ 0 and k--+ 0, we conclude that si~ n --+ O. We have seen that the sequence Xn = 2 n diverges because the terms increase without bound and cannot stay "close" to anyone real number. In general, a sequence whose terms form an unbounded set must diverge. In other words, a convergent sequence must be bounded because all but the first few terms will be close to the limit. In this setting, closeness is determined by the choice of e, and by "the first few terms" we mean the first N terms, where N depends on e.

298

CHAPTER 7

Theorem 7.13

Concepts of Analysis: Completeness of the Real Numbers

Every sequence that converges is bounded.

Proof.

Suppose x is a sequence convergent to a number L. For e = I, there is a natural number N such that if n > N, then IXn - LI < 1. Since Ilxnl - ILl 1 :::::; IXn - LI, we have for all n > N, Ixnl - ILl < 1. Thus for all n > N, Ixnl < ILl + 1. (All but the first N terms are bounded by ILl + 1. We now take care of the first terms as well.) Let B = max {lxII, Ix21. ... , IxNI. ILl + I}. Then Ixnl:::::; B for all n E N, and x is bounded. ' •

Example.

Let x be the sequence

C-2)n { Xn= ~ n+1

I :::::; n:::::; 1,000

n>

1,000.

This sequence converges to 15 and therefore must be bounded. In this case the first "few" (1,000) terms hop around a bit before the terms settle in close to 15. A bound for the sequence is given by Ixnl : : :; 2 1000 for all n E N.

DEFINITIONS A sequence x is increasing iff for all n, mEN, Xn :::::; xm whenever n < m. A decreasing sequence y requires Yn 2:: Ym for n < m. A monotone sequence is one that is either increasing or decreasing. The sequences given by Xn = 2n and Sn = n~1 are increasing while Zn = e- n is decreasing. The sequence Y defined by Yn = C_l)n is not monotone. A constant sequence t, where tn = c for some fixed c E IR and for all n, is monotone since it is both increasing and decreasing. The next theorem relates all the concepts of this section: A sequence that is both bounded and monotone must converge. Its proof makes use of the BolzanoWeierstrass Theorem.

Theorem 7.14

(The Bounded Monotone Sequence Theorem) For every bounded monotone sequence x, there is a real number L such that xn ---+ L.

Proof.

Assume x is a bounded and increasing sequence. The proof in the case where x is decreasing is similar. If {xn: n E N} is finite, then let L = max {xn: n EN}. For some N E N, XN = L and, since x is increasing, Xn = L for all n > N. Therefore, xn ---+ L. Suppose {xn: n E N} is infinite. Then by the Bolzano-Weierstrass Theorem, the bounded infinite set {xn: n E N} must have at least one accumulation point. Let L be an accumulation point. We claim Xn :::::; L for all n E N. If there exists N such that XN > L, then xn> L for all n 2:: N. Since L is an accumulation point of {xn: n EN}, and {xn: n :::::; N} is finite, L is an accumulation point of {xn: n > N}, by exercise 14 of section 7.3. But for J = IXN - LI. NCL, J) contains no points of {xn: n 2:: N}. This is a contradiction. Thus xn :::::; L for all n.

The Bounded Monotone Sequence Theorem

7.4

299

We claim that the sequence x converges to L. Let 8 > O. Since L is an accumulation point of {xn: n EN}, there exists MEN such that XM E ,N(L, 8). Thus L - 8 < XM, and so, for n > M, L - 8 < XM :S Xn :S L < L + 8. Therefore, for n > M, IXn - LI < 8. Thus Xn --+ L. • The Bounded Monotone Sequence Theorem can be used to prove the existence of several important real numbers. Recall that the constant e, the base of the natural logarithm function, may be defined as n

e

= n_OO lim (1 +.!.) . n

Its existence is guaranteed by the Bounded Monotone Sequence Theorem. We will show that the sequence whose nth term is Xn = (1 + k)n is both bounded and increasing. By the binomial expansion theorem (Theorem 2.25), n

( 1 +.!.) = 1 +~.!.+n(n-l)-.L+n(n-l)(n-2) 1 + ... +-.L n I! n 2! n2 3! n3 nn = 1 + 1 + -.L n(n - 1) + -.L n(n - 1)(n - 2) + ... + -.L n! 2! n2 3! n3 n! nn 1 1 1 :S1+1+-+-+"'+2! 3! n! 1 1 ... + -1:S 1 + 1 +-+-+ 2 4 2n - 1 = 1 + 2n - 1

2n -

1

<1+2 = 3. Thus the sequencex n = (1 + k)n is bounded above by 3. We next show that x is an increasing sequence. We again use the Binomial Theorem to compare Xn andxn+l:

and xn+1 = ( 1 + n

1 )n+1

+1

= 1 + 1 + -.L (n +

l)n + -.L (n + 1)(n)(n - 1) + ... 2! (n + 1)2 3! (n + 1)3 + 1 (n + 1)(n)(n - 1)' ... ·3·2 + 1 (n + I)! . n! (n + l)n (n + I)! (n + l)n+1

300

CHAPTER 7

Concepts of Analysis: Completeness of the Real Numbers

We leave it as exercise 10 to show that each term in the expansion of Xn is less than or equal to the corresponding term in the expansion of x n + I· Additionally, x n + I has one more term, _ I- (n+ in its binomial expansion. Thus Xn :::; x n+I for all F>.I (n+I)! (n+1)

Ill"

nE,\lI.

Because Xn = (I + k)n is bounded and increasing, it must converge. The limit of this sequence is, by definition, the number e.

Exercises 7.4 1.

* * * * *

For each sequence x, determine whether x is bounded, bounded above, or bounded below. 10 (a) Xn = IOn (b) x = n n (c)

(e) (g)

(i) (k)

Xn

=

10- 11

(d)

10 n! x =tann -" n Xn = (-0.9)n (_1)n + I xn = n x

n

=

(h)

xn

=

(j)

Xn

=

(f)

=-

loglon n x =cos -n n Xn

(I)

(-I)nn

(-1.1)n ((_1)n - 1)((_1)n xn = n

+

2. For each sequence x, estimate n_ lim Xn or determine that it does not exist. oo (b)

10 xn =n-

* *

(a)

*

(e)

* *

(g)

xn

=

( I +;;Ir

(h)

x" = (I +

(i)

Xn

=

(-0.9)n

(j)

Xn

(k)

xn=((-I) 11 +1) ( 1+;;Iy

(I)

n x =2-+-3" n 5n

(c)

Xn

=

IOn

4n 2 + 7n + 12 II - n + 5n 2 8n2 + 4n + I x = n IIn 3 - n + 5 xn =

(d) (f)

n

6n 3 + 5n 2 + 3n + 8 IOn 2 + 7n + 5 6n 3 + 5n 2 + 3n + 8 x = n IOn 3 + 7n 2 + 5n - 8 xn =

~rn

= (-1.1)"

3. For each sequence x, prove that x converges or diverges.

*" *"

(-1)"

(a)

x

(c)

x"

=

(e)

x

=--

(g)

x"

"

n

=--

n

n2 cosn n

=

(~y

*"

(b)

x n =n+1 -n

(d)

Xn

=

(_1)n n 2n + I

(f)

Xn

=

[rl+l - {n

(h)

x

=-

"

6 2n

I)

7.4 (i) (k)

5,000 x n =--,n. n! x =n nn

(m) {XI = 1

xn=JI +Xn-I,

xn

(I)

x n = nlln

(n)

{XI Xn

= =

1

2JXn-I'

n::::::.2

Give an example of (a) a bounded sequence that is not convergent. (b) an increasing sequence that is not convergent. (c) a convergent sequence that is not monotone. (d) a divergent sequence X such that the sequence whose nth term is Ixnl converges. n (e) a sequence that converges to 2.

5.

Prove that if Xn ---+ Land Yn ---+ M and r E ~, then (a) xn+Yn---+L+M. (b) xn-Yn---+L-M. (c) rXn ---+ rL. (d) XnYn ---+ LM. (e) Ixnl-- ILl. Prove that if Xn ---+ 0 and Y is a bounded sequence, then X/Jill ---+ O. Prove Theorem 7.10.

"* 6. 7. 8.

"*

n::::::.2

. (nn) sm "2

(j)

=

301

4.

*

"*

The Bounded Monotone Sequence Theorem

9.

Give a proof of the Bounded Monotone Sequence Theorem for the case in which the sequence X is bounded and decreasing. (a) Prove that if Xn ---+ Land L 1= 0, then there is a number N such that if n::::::' N, then Ixnl > (b)

Prove that if xn---+L, xn1=O for all n, L1=O, and if Yn---+M, then

(~:) 10.

---+ (

1)

Complete the proof that xn = ( I that for all k :::; n,

1.- n(n k!

('~I). n

+~)

I)(n - 2)· .... [n - (k - I)] nk

<:

is an increasing sequence by showing

1.- (n + I)(n)(n -

- k!

I)· .... [n - (k - 2)] + I)k

(n

11.

Let X be a bounded increasing sequence. Use the completeness property of the reals and properties of suprema and limits to prove directly (without reference to the Bolzano-Weierstrass Theorem) that X converges. (Hint: Consider the supremum of the set of terms of x.)

12.

A sequence x of real numbers is a Cauchy sequence iff for every e > 0, there exists an integer M such that if m, n > M, then IXn - xml < e. That is, terms in the sequence are arbitrarily close together if the terms are chosen far enough along the sequence. (a) Prove that if x is a Cauchy sequence, then x is bounded. (It can also be shown that every Cauchy sequence converges, though the proof is rather complicated. ) (b) Prove that if x is a convergent sequence, then x is a Cauchy sequence.

302

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Proofs to Grade

13.

A sequence y" is a subsequence of x" means that there is an increasing function j: N --+ N such that y" = xf(")' For example, Yn = X2n is the sequence formed by every even-numbered term of the sequence x". _ (-I)n + I . (a) Let xn . Descnbe the subsequences X2" and X2,,-" n (b) Prove that if a sequence x is bounded, then every subsequence of x is bounded. (c) Prove that if a sequence x converges to L then for every real e > 0, there exists a subsequence Y of x such that IYn - LI < e for all n E N. (d) Prove that if x" converges to Land Yn is a subsequence of x, then Y converges to L. (e) Prove that if x contains two convergent subsequences Y and z, y" --+ M and Zn --+ L, and M =I- L, then x diverges.

14.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. lim (I + k)" = e. n-OO

n

Let M = lim (I + k) . Take the natural logarithm of both

"Proof."

= In (limT!''':+ k)") = lim (In (I + k)n) = lim (n In (n + k)) n_x In«n + I)/n» = lim (n + I) - n Inn) = lim I = 1. Since !1_X 11_00

sides: In M

11_00

= nlim (n -X InM

(b)

*

(c)

(d)

=

11_00

-

I,M = e.

Claim. Every bounded decreasing sequence converges. "Proof." Let x be a bounded decreasing sequence. Then Yn = -XI! defines a bounded increasing sequence. By the proof of Theorem 7.14, Yn--+ L for some L. Thus Xn --+ - L. Claim. If the sequence x converges and the sequence Y diverges, then x + Y diverges. "Proof." Suppose Xn + Yn --+ K for some real number K. Since X,,--+ L for some number L, (x n + Yn) - Xn --+ K - L; that is, Yn --+ K - L. This is a contradiction. Thus Xn + Yn diverges. Claim. If two sequences x and Y both diverge, then x + Y diverges. "Proof." Suppose lim (x n + Yn) = L. Since x diverges, there exists

> 0 such that for ~llN E N there exists n > N such that Ixn - @I ::::: e,. Since Y diverges, there exists 1:2 > 0 such that, for all N E N, there exists n > N such that IYn - @ I : : : e2' Let e = min {e" e2}' Then for all N E N, there exists n > N such that 1:,

!

I(xn + Yn) - LI

(~)] + [Yn - (~)] I

= I[Xn -

(~)I + IYn - (~)I

::::: IXn -

::::: e, + e2 I I ::::: -e +-e

2

2

= e.

Therefore, lim (xn + Yn) =I- L. 11_00

-

7.5 Equivalents of Completeness

7.5

303

Equivalents of Completeness In the first section of this chapter we stated that the ordered field of real numbers is complete. Completeness was then used to prove the Heine-Borel Theorem in section 7.2. Then in section 7.3, the Heine-Borel Theorem was the main result used to prove the Bolzano-Weierstrass Theorem, which in tum was used to prove the Bounded Monotone Sequence Theorem of section 7.4. In this section we complete a circle of implications by showing that if the ordered field of reals is assumed to have the property that every bounded monotone sequence converges, then the reals are complete (see figure 7.9). Completeness

~

of,

~

Heine-Borel

Bounded Monotone

Figure 7.9 Thus completeness and the three properties described by the main theorems of this chapter are equivalent for IR. What we have chosen to omit is a proof of the (correct) statement that IR is complete. Furthermore, the theorems stated for the reals could be generalized to prove that the properties are all equivalent for any ordered field. Our proof of completeness from the Bounded Monotone Sequence Theorem will use the following two lemmas about convergent sequences. The proofs are exercises 1 and 2.

Lemma 7.15

If x and yare two sequences such that lim Yn = s and lim (xn - Yn) n_oo n_OC lim Xn = S.

= 0, then

n_oo

Lemma 7.16

If x is a sequence with lim

n_oo

exists N E N such that xn

Theorem 7.17

Xn

= sand t is a real number such that t < s, then there

> t for all n ::::: N.

Suppose the real number system has the property that every bounded monotone sequence must converge. Then IR is complete. Proof. Let A be a nonempty subset of IR that is bounded above by a real number b. To prove completeness, we must show sup (A) exists and is a real number. Since A =I- 0, we may choose a EA. If a is an upper bound for A, then a = sup (A) (exercise 5, section 7.1 ), and we are done. Assume that a is not an upper bound for A. If (a + b)/2 is an upper bound for A, let Xl = a and YI = (a + b)/2; if not, let Xl = (a + b)/2 and YI = b. In either case YI - Xl = (b - a)/2, Xl is not an upper bound, and YI is an upper bound for A.

304

CHAPTER 7 Concepts of Analysis: Completeness of the Real Numbers

Now if (x, + y,)/2 is an upper bound for A, let X2 = x, and Y2 = (x, + y,)/2; otherwise, letx2 = (x, + y,)/2 andY2 = y,. In either case the result is thatY2 - X2 = (b - a)/4, X2:::=:: x" X2 is not an upper bound for A, while Y2::; y" and Y2 is an upper bound for A. Continuing in this manner, we inductively define an increasing sequence x such that no x" is an upper bound for A, and a decreasing sequence Y such that every y" is an upper bound for A. In addition, y" - x" = (b - a)/2 n and Y is bounded below by a. Therefore by hypothesis, y converges to a point s E JR. Furthermore, since lim (y" - x n ) = lim (b - a)/2n = 0, lim Xn = S by Lemma 7.15. 11--+00

Il--+X

n-X

We claim that s is an upper bound for A. If z > s for some z E A, then z > YN for some N (because Yn--+ s). This contradicts the fact that YN is an upper bound for A. Finally, if t is a real number and t < s, then t < XN for some N E N, by Lemma 7.16. Since XN is not an upper bound for A, t is not an upper bound. Thus s is a real number that is an upper bound for A, and no number less than s is an upper bound; that is, s = sup (A). Therefore, JR is complete. We saw in section 7.1 that the field of rationals is not complete. Since the major theorems of this chapter are equivalent to completeness in a field, all three must fail for the rationals. (1) The set A = {x E 0: x 2 ::; 2} = [-.ft, ~] no is a closed subset of O. A is also bounded but not compact because {( - x, x): x E A and x =1= O} is an open cover of A with no finite subcover. This example shows that the Heine-Borel Theorem fails for O. (2) The set B = {1.4, 1.41, 1.414, 1.4142, ... } of (rational) decimal approximations of.ft is a bounded and infinite subset of 0 whose only accumulation point in JR is not in O. Thus the Bolzano-Weierstrass Theorem fails in O. (3) A counterexample to the Bounded Monotone Sequence Theorem in 0 is the sequence whose terms are the successive decimal expansions of .ft. This bounded and increasing sequence fails to converge to any rational number. Why is completeness such a 9,rucial property of the real number system? The convergent sequence Xn = (I + k) provides a clue. In section 7.4 we saw that the sequence is bounded and increasing, so by the completeness of JR (via the Bounded Monotone Sequence Theorem), the terms Xn must approach a unique real number, which is the number e. The fact that important constants such as e must exist in JR is a consequence of completeness. Not only must the limit of a bounded monotone sequence of rational numbers be in JR, but the same is true for a bounded monotone sequence of irrational numbers, or of rationals and irrationals. The completeness property and its equivalents assure us that every number that can be approached by a sequence of reals is in fact a real number.

Exercises 7.5 1.

Prove Lemma 7.15.

2.

Prove Lemma 7.16.

7.5

3.

"*

Equivalents of Completeness

305

"*

Give an example of a closed subset A of 0 such that A ~ [7, 8] and A is not compact. (b) a bounded infinite subset of On [7,8] that has no accumulation point in O. (c) a bounded increasing sequence x of rationals such that {xn: n E N} ~ [7, 8] and x has no limit in O.

4.

For the set IR - 0 of irrational numbers, give an example of (a) a closed subset A of IR - 0 such that A ~ [3, 4] and A is not compact. (b) a bounded infinite subset of IR - 0 n [3, 4] that has no accumulation point in IR - O. (c) a bounded increasing sequence x of irrationals such that {xn: n E N} ~ [3,4] and x has no limit in IR - Q.

5.

Prove the converse of Theorem 7.17 directly; that is, prove that the completeness of IR implies that every bounded monotone sequence in IR must converge.

6.

(a)

(a)

Find (i) (iii)

(b)

n

An where An' n E N is defined as follows:

=[2-~n2' 4+~] n2

n:N =[_1.n' 1.n ]

(ii)

A

An = [n 3 , co)

(iv)

An = ( 0,

n

n

~)

(The Nested Interval Theorem) Show that if An = [a", bn] is a sequence of closed intervals such that An+ I ~ All for all n E N, then An =1= 0.

n

nEN

Proofs to Grade

7.

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. (a) Claim. If every bounded monotone sequence in the reals is convergent, then the reals are complete. "Proof." Suppose the reals are not complete. Then there is a bounded infinite subset A such that A has no supremum. Letxl EA. Then XI is not an upper bound for A, or else XI would be the least upper bound (since xI E A). Thus there is an X2 E A such that XI < X2' Likewise, X2 E A and X2 is not an upper bound, so there exists X3 E A with X2 < X3' Continuing in this fashion, we build an increasing sequence XI> Xz, X3, .... This sequence is bounded since it is a subset of A. Therefore, L = lim Xn exists. n--+ OO

*

(b)

Since L > Xn for all n E N, L is the supremum of A. Therefore, sup (A) exists, which is a contradiction. Thus IR is complete. a Claim. The Bolzano-Weierstrass Theorem implies the completeness of IR. "Proof." (Direct proof). Suppose that every bounded infinite subset of the reals has an accumulation point. Let A be an infinite subset of IR with an upper bound aD. Then B = [0, aD] n A is a bounded set. If B is finite, then B has a least upper bound, so assume that B is infinite. By the Bolzano-Weierstrass Theorem, B has an accumulation point al which, by construction, is the least upper bound of A. a

Answers to Selected Exercises

Exercises 1.1 1.

(c)

(e) (g)

2. (a)

(c)

(e)

(g)

proposition proposition Not a proposition. The pronoun "she" acts as a variable, so the truth value depends on the assignment of values to this variable. The statement becomes a proposition for each such assignment. P

-P

PA-P

T F

F T

F F

P

Q

R

QVR

PA(QV R)

T F T F T F T F

T T F F T T F F

T T T T F F F F

T T T T T T F F

T F T F T F F F

P

Q

-Q

PA-Q

T F T F

T T F F

F F T T

F F T F

P

Q

-Q

PAQ

(PAQ)V-Q

T F T F

T T F F

F F T T

T F F F

T F T T 307

308

Answers to Selected Exercises

3. (a) (c)

4.

5.

equivalent equivalent (e) equivalent (g) not equivalent (i) not equivalent (k) equivalent (m) equivalent (a) false (c) true (e) false (g) false (i) true (k) false (a) x is not a positive integer.

(c)

5<3

(e)

Roses are not red or violets are not blue. Tis green and Tis not yellow. Since P is equivalent to Q, P has the same truth table as Q. Therefore, Q has the same truth table as P, so Q is equivalent to P. B /\ ~C; true p(j)Q P Q

(g) 6. (a) 7. (c) 9. (a)

T F T F

10. (c)

T T F F

F T T F

tautology P

Q

PAQ

-pv-Q

(PAQ)V(-pv-Q)

T F T F

T T F F

T F F F

F T T T

T T T T

Exercises 1.2 1. (a) (d) (f)

2. (a)

(d)

Antecedent: squares have three sides. Consequent: triangles have four sides. Antecedent: f is differentiable. Consequent: f is continuous. Antecedent: f is integrable. Consequent: f is bounded. Converse: If triangles have four sides, then squares have three sides. Contrapositive: If triangles do not have four sides, then squares do not have three sides. Converse: If f is continuous, then f is differentiable. Contrapositive: If f is not continuous, then f is not differentiable.

Answers to Selected Exercises (f)

4.

5. 6.

(a) (c) (e) (a) (c) (b)

(c)

7.

309

Converse: If f is bounded, then f is integrable. Contrapositive: If f is not bounded, then f is not integrable. true true true true true P Q -P

T

T

F T

T F

F

F

F T F T

P

Q

-Q

T F T F

T T F F

F F T T

T T T T

T F F T

T T F T

(f)

P

Q

R

QVR

P A(QV R)

PAQ

PAR

(P AQ) v (P AR)

T F T F T F T F

T T F F T T F F

T T T T F F F F

T T T T T T F F

T F T F T F F F

T F F F T F F F

T

T

F

F

T

T

F

F

F

T

F F F

F F F

Since the fifth and eighth columns are the same, the propositions S.

(a) (d)

(e) 9.

P A (Q V R) and (P A Q) V (P A R) are equivalent. ((f has a relative minimum at xo) A (f is differentiable at xo)) => (f'(xo) = 0). ((x = 1) V (x = -1)) => (Ixl = 1). (xo is a critical point for f) {=} ((f'(xo) = 0) V (f'(xo) does not exist».

(b)

P

Q

R

PAQ

T F T F T F T F

T T F F T T F F

T T T T F F F F

T F F F T F F F

T T T T F T T T

-R

-Q

F F F F T T T T

F F T T F F T T

F F F

T T T

F

T

T F

F T

T

T

F

T

Since the fifth and ninth columns are the same, the propositions ~R) => ~Q are equivalent.

P A Q => Rand (P A

310

Answers to Selected Exercises

10.

14.

(a) (c) (a) (d)

If 6 is an even integer, then 7 is an odd integer. not possible tautology neither

(a)

~(Vx)(x is precious => x is beautiful). Or, C3x)(x is precious and x is not beautiful). This exercise is not the same as lea). (3x)(x is a positive integer and x is smaller than all other positive integers). Or, (3x)(x is a positive integer and (Vy)(y is a positive integer => x ::5 y». (3x)(x cares about me). (Vx)( ~(Vy)(x loves y» or ~(3x)(Vy)(x loves y).

Exercises 1.3 1.

(b) (g)

(h) (i) (I)

(Va)(Vb)(3c)(3d)[(a + bi)(c

+ di) = ac -

bd]

or (Va)(Vb)(3c)(3d)[bc + ad = 0]. 2. (a) (Vx)(x is precious => x is beautiful). All precious stones are beautiful. (c) (Vx)(x is a positive integer => (3y) (y is a positive integer) Ax > y». For every positive integer there is a smaller positive integer. Or, ~(3x)(x is a positive integer A (Vy)(y is a positive integer => x ::5 y». There is no smallest positive integer. 3. Hint: To use part (a), note that ~(3x)( ~A(x»is equivalentto (Vx)(~~A(x». 4. (b) true (e) false (h) true 5. (b) Only one number is both nonnegative and nonpositive. (d) Exactly one number has a natural log of 1. 6. (a) true (d) false (f) false (i) false 10. (d) This statement is not a denial. It implies the negation of (3! x)P(x), but if (Vx) ~ P(x), then both the statement and (3! x)P(x) are false. 12. For every backwards E, there exists an upside down A!

Exercises 1.4 2.

(a)

Suppose (G, *) is a cyclic group.

(c)

Thus, (G, *) is abelian. Therefore, if (G, *) is a cyclic group, then (G, *) is abelian. Suppose (G, *) is not abelian. Thus, (G, *) is not a cyclic group. Therefore, if (G, *) is a cyclic group, then (G, *) is abelian.

Answers to Selected Exercises

311

4. (d) Proof. Suppose x is even and y is odd. Then x = 2k for some integer k, and y = 2j + 1 for some integer j. Therefore, x + y = 2k + (2j + 1) = 2(k + j) + I, which is odd. Hint: The four cases to consider are: case I, in which a ::::: 0 and b ::::: 0; case 2, in which a < 0 and b < 0; case 3, in which a ::::: 0 and b < 0; and case 4, in which a < 0 and b ::::: O. In case 3 it is worthwhile to consider two subcases: in subcase (i), a + b ::::: 0, so that la + bl = a + b; in subcase (ii), a + b < 0, so that la + bl = -(a + b). Now in subcase (i) we have la + bl= a + b < a (fromb < 0) and a < a + (-b) (from 0 < -b). Thus la + bl< a + (-b) = lal + Ibl. Subcase (ii) is treated similarly. Case 4 has two subcases, similar to case 3. 6. (b) Proof. Suppose a and b are positive integers and a divides b. Then for some integer k, b = ka. (We must show that a:::::: b, which is the same as a:::::: ka. To show a:::::: ka, we could multiply both sides of 1 :::::: k by a, using the fact that a is positive. To do that, we must first show I :::::: k.) Since b and a are positive, k must also be positive. Since k is also an integer, I :::::: k. Therefore, a = a . 1 :::::: a . k = b, so a:::::: b. (d) Proof. Suppose a and b are positive integers and ab = I. Then a divides I and b divides 1. By part (b) a :::::: I and b :::::: 1. But a and b are positive integers, so a = I and b = I. 9. (a) Proof. Suppose A > C > B > O. Multiplying by the positive numbers C and B we have AC > C 2 > BC and BC > B2, so AC > B2. AC is positive, so 4AC > AC. Therefore, 4AC > B2, so B2 - 4AC < O. Thus the graph must be an ellipse. 10. (a) F. The converse rather than the statement is proved. (c) A. (e) C. The order in which the steps are written makes it look as if the author of this "proof' assumed that x + i : : : 2. The proof could be fixed by beginning with the (true) statement that (x - 1)2::::: 0 and ending with the conclusion that x + i : : : 2. (f) F. This "proof' appears to show that the sum of the two even numbers is even but uses this fact in the "proof." 5.

(d)

1.

(a) (c)

Exercises 1.5

(f)

2.

(b)

Choose m = -3 and n = 1. Then 2m + 7n = I. Suppose m and n are integers and 2m + 4n = 7. Then 2 divides 2m and 2 divides 4n, so 2 divides their sum 2m + 4n. But 2 does not divide 7, so this is impossible. Hint: See the statement of part (d). Can you prove that m and n are both negative whenever the antecedent is true? Proof. Assume a divides b - I and c - 1. (The proof involves writing bc - 1 as a sum of multiples of a, using the fact that if a divides a number, it divides any multiple of that number (exercise 6(c) of section 1.4) or, more generally, the fact that if a divides two numbers, it divides any sum ofmultiples of the two numbers (exercise 2(a)).) Then a divides the

312

Answers to Selected Exercises

product (b - I)(e - I) = be - b - e + I. By exercise 2(a), a divides (be - b - e + I) + (b - I) = be - e. Then also by exercise 2(a), a divides the sum (be - e) + (e - I) = be - I. 5. (b) Hint: For a counterexample, choose x = I. Explain. (i) Hint: For a proof, choose y = x. For a different proof, choose y = I. 7. (c) Proof. Assume e divides a and e divides b. Since d = GCD(a, b), e divides d by the definition of GCD. Thus %is a natural number. Also ~ and ~ are natural numbers. To show that %is GCD(~, ~), we must show (i) % divides both ~ and ~ and (ii) if k divides both ~ and t then k divides %. (i) Since d = GCD(a, b), d divides a. Since d = (%)e and a = (~)e, (%)e divides (~)e. Thus %divides ~, by exercise 6(h) of section 1.4. Simd d"d b I'1 arI y, c IVI es c. (ii) Suppose k divides ~ and k divides ~. Then by exercise 6(h) of section lA, ke divides a and ke divides b. Therefore ke divides d, so k divides %. We conclude that %= GCD(~, ~). (f) Hint: Write I in the form ax + by and multiply both sides bye. Then show that a divides eax + eby. (g) Hint: One-half of the biconditional is immediate (exercise 6(g) of section 104). For the other direction, assume that GCD(a, b) - I, a divides m, and b divides m. Write m as be for some e and apply Euclid's Lemma. 8. (c) Proof. Let n be a natural number. Then both 2n and 2n + I are natural numbers. Let M = 2n + I. Then M is a natural number greater than 2n. (g) Proof. Let e > 0 be a real number. Then t is a positive real number and so has a decimal expression as an integer part plus a decimal part. Let M be the integer part of t plus I. Then M is an integer and M > To prove for all natural numbers n > M that k< e, let n be a natural number and n > M. Since M > we have n > Thus k< e. Therefore, for every real number e> 0, there is a natural number M such that for all natural numbers n > M,

t.

t,

t.

k< e. 9.

(a) (b) (d)

F. The false statement referred to is not a denial of the claim. C. Uniqueness has not been shown.

A.

Exercises 1.6 1. (e)

Consider x ; y.

7.

(a)

8.

(a)

Proof. Suppose (x, y) is inside the circle. Then from the distance formula, Ix-312+ ly-21 2 <4. Therefore, Ix-31 2 <4 and ly-21 2 <4. It follows that Ix - 31 < 2 and Iy - 21 < 2, so -2 < x - 3 < 2 and -2
Answers to Selected Exercises

(c)

10. (b) (f)

(g)

11.

(c) (g)

If GCD(p, a) = p, then p divides a by definition of GCD. If p divides a, then since p divides p, p divides GCD(p, a) by definition of GCD, and so p:::; GCD(p, a). But GCD(p, a) :::; p, so GCD(p, a) = p. Note: A shorter proof is to observe that the statement to be proved is just a special case of exercise 7 (b) of section 1.5. Hint: Assume that p divides ab and p does not divide a. Applyexercise 8(b) and Euclid's Lemma (exercise 7(f) of section 1.5). Hint: To show that m = b if a divides b, use part (a) (or the definition of LCM) to show m :::; b. Show also that b :::; m. Hint: Write m = LCM(a, b) as multiples an and bv of a and b respectively. Use part (c) and cancellation to show v :::; a. Use the fact that a divides bv and Euclid's Lemma (exercise 7(f) of section 1.5) to show that a divides v. Derive a = v and conclude that m = abo Hint: 'Let GCD(a, b) ~ d. Apply part (d) to find LCM(~, ~). Use exercise 7(d) of section 1.5 and part (f) to find another expression for LCM(~, ~). Equate the two values for LCM(~, ~). (Or apply part (e)). F. The claim is false. A.

Exercises 2.1 1. (a) (c) 2. (a) (c) 3. (a) (c) (e) (g) (i)

4. (a) (c) (e) (g) (i) (k)

5. (a) (c) 6. (a) (c) 7. (a) (c) 9. (a) (c)

313

{x E N: x < 6} or {x: x E N and x < 6}. {x E ~: 2 :::; x:::; 6} or {x: x E ~ and 2 :::; x:::; 6}.

{I, 2, 3,4, 5} not possible true true false true false true true false false false true A = {I, 2}, B = {I, 2, 4}, C = {I, 2, 5} A = {I, 2, 3}, B = {I, 4}, C = {I, 2, 3, 5}

Ha}, {L:.}, {D}, {a, L:.}, {a, D}, {L:., D}, X, 0} H0}, Han, Hbn, Ha, bn, {0, {an, {0, ibn, {0, {a, bn, Ha}, ibn, Ha}, {a, bn, Hb}, {a, bn, {0, {a}, ibn, {0, {a}, {a, bn, {0, {b}, {a, bn, Ha}, {b}, {a, bn, x, 0}. no proper subsets {I}, {2} true true

314

Answers to Selected Exercises

(e) (g) (i)

10. 12.

14.

18.

19.

(k) (a)

true true true true

Proof. Let x be any object. Suppose x EA. Then P(x) is true. Since (Vx)(P(x) => Q(x», Q(x) is true. Thus x E B. Therefore, A ~ B. Proof. To prove that if A ~ Band B ~ C, then A ~ C, begin by assuming that A ~ Band B ~ C. To show that A ~ C, we recall that A ~ C means (Vx)(x EA => x E C). The proof begins with: Let x be any object. Suppose x EA. Now use the fact that A ~ Band B ~ C. Hint: To prove A = B, use the hypothesis A ~ B and show B ~ A by using Theorem 2.3. Suppose that X is a set. If X E X then X is not an ordinary set, so X tt. X. On the other hand, if X tt. X, then X is an ordinary set, so X EX. Both X E X and X tt. X lead to a contradiction. We conclude that the collection of ordinary sets is not a set. (c) C. The "proof' asserts that x E C, but fails to justify this assertion with a definite statement that x E A or that x E B. This problem could be corrected by inserting a second sentence "Suppose x E A" and a fourth sentence "Then x E B." (g) A. Every statement in the proof is correct. No reasons are given, but no explanation of the steps is required for correctness. (h) F. The claim is false. (For example, A = {I, 2}, B = {I, 2, 3, 4, 5}, C = {I, 2, 5, 6, 7}.) The statement "Since x E B, x EA ... ," would be correct if we knew that B ~ A.

Exercises 2.2 1. (a) (c) (e) (g) (i)

2. (a)

3.

(c) (e) (g) (i) (k) (b) (f) (k)

{~I, ~ 3,4, 5,6,7, 8, 9} {I, 3, 5, 7, 9} {3,9} {I, 5, 7} {I, 5, 7} {O, -2, -4, -6, -8, -IO, ... }

D N U {O} D

{O, 2, 4, 6, 8, ... }

0 (1, 8)

[2,4) (-00, 3) U [8, oc)

4. (a) {{I}, {2}, {I, 2}, 0} (d) {0, {I}, {3}}

s.

(f) {{3}} A and B are disjoint.

Answers to Selected Exercises 10.

(a)

315

To prove that A C B implies A - B = 0, assume that A C Band show that x E A - B is false for every object x. To prove the converse, begin by assuming A - B = 0 and that some object x EA.

Hint:

(c)

Proof. (i) Assume C CA n B. Suppose x E C. Then x EA n B, so x EA and x E B. Since x E C implies x E A, C CA. Similarly, C C B. (ii) Assume C C A and C C B. Suppose x E C. Then from C C A we have x E A and from C C B we have x E B. Therefore, x E A n B. We conclude that C CAn B. 11. (c) Proof. We show that if C and D are not disjoint, then A and B are not disjoint. If C and D are not disjoint then there is an object x E C n D. But then x E C and xED. Since C C A and DC B, x E A and x E B. Therefore, x E A n B, so A and B are not disjoint. 13. (a) Proof. S E g>(A n B) iff S CAn B iff (by exercise lO(c)) S CA and S C B iff S E g>(A) and S E g>(B) iff S E g>(A) n g>(B). (d) Proof. Let A and B be any sets. Since 0 is a subset of every set we have E g>(A - B). Also 0 E g>(A) and 0 E g>(B), so 0 tt. g>(A) - g>(B). This shows that g>(A - B) cJ g>(A) - g>(B). 14. (a) A = {I, 2}, B = {I, 3}, C = {2, 3, 4}

o

( c) (e)

17.

(a)

(b) (c) (d) (h) (i) (j)

A

= {I, 2}, B = {I, 3}, C = {2, 3}

A={1,2},B={1,3},C={1}

C. The serious error is the assertion that x tt. c, which has no justification. The author of this "proof' was mislead by supposing x E A, which is an acceptable step but not useful in proving A - C C B - C. After assuming that A C B, the natural first step for proving that A - C C B - C is to suppose that x E A - C. Since it is grammatically incorrect to "suppose a set," consider what the author of this "proof" might have intended by the sentence "Suppose A - C" C. The proof that A n B = A is incomplete. F. The claim is false. The statement "x E A and x E 0 iff x E A" is false. C. The idea is correct, but the discussion is probably too brief. Proofs should be understandable to those reading this text. A. C, or possibly F. Although a picture may help by suggesting ideas around which a correct proof can be made, a picture alone is rarely sufficient for a proof. Thus a proof that consists only of Venn diagrams will usually have a grade of F. This "proof' is made better because of the explanation that is included.

Exercises 2.3 1.

(a)

U A = {I, 2, 3, 4, 5, 6, 7, 8}; AE~

(c)

U An = N; nEN

n

AE~

nAn nEN

= {I}

A

= {4, 5}

316

Answers to Selected Exercises

(e)

UA=Z; nA={lO} AE.sil

(g)

(i)

7. 5.

n n

AE.sil

nEN

U An = (0, 1); U Ar = [0, 00);

nEN

rEIR

rEIR

Ar = 0

The family in exercise I (a) is not pairwise disjoint. The family in I (b) is pairwise disjoint. (a) Let fJ E L\. Suppose x E Au. Then x E Au for each a E L\. Since fJ E L\,

x EAp. Therefore

6.

An = 0

(a)

xEBn

U Au

n

n

uEL1.

A CAp.

uEL1.

iff x E B and x E

uEL1.

U

Au uEL1. Au for some

iff x E B and x E a E L\ iff x E B n Au for some a E L\ iff x E (B n Au).

U

uEL1.

7.

(a)

8. 10.

(c) (a)

true

U Au. Then x E Au for some a E r. Since r C L\, a E L\. Thus x E Au for some a E L\. Therefore, x E U Au. Suppose x E

uEr

uEL1.

11. (a)

n

Proof. Let x E B. For each A E s'l, B CA. Thus for each A E s'l, x EA. Therefore x E A. AE.sil

(b)

X= nA. AE.sil

17.

(c)

m

U Ai. Then there exists j E N such that k ::; j ::; m and i=k Since j E N and x E A x E U Ai.

Proof. Let x E

x

x E Aj • 19. 20.

(a)

LetA k

j,

=[ _~, I +~).

i=l

(a) A. Note that if we allowed L\ = 0, the claim would be false. (b)

C. No connection is made between the first and second sentences.

(f)

F. The claim is false.

DC

U [n, n + I) = n=l

Exercises 2.4 1.

(b) (d)

not inductive not inductive

[I, 00).

Answers to Selected Exercises

(f)

317

not inductive

2. (b) true 6.

(e) (c)

false A = {n: n = 2k for some kEN} may be defined as (i) 2 EA. (ii) ifx E A, then 2x EA.

8.

(h)

Proof. (i) (ii)

For n = 1,4 1 - 1 = 3, which is divisible by 3. Suppose for some kEN that 4k - 1 is divisible by 3. Then

4k+I-l =4(4k )-1 =4(4k -l)-1 +4 = 4(4k - 1) + 3.

15.

Both 3 and 4k - 1 are divisible by 3, so 4k+ 1 - 1 is divisible by 3. (iii) By (i), (iii) and the PMI, 4" - 1 is divisible by 3 for all n E N. (a) F. Let S = {n E N: all horses in every set of n horses have the same color}. Then 1 E S. In fact, S = {I}. The statement n E S ==> n + 1 E S is correct for n:::=:: 2. The only counterexample to (Vn)(n E S ==> n + 1 E S) occurs when n = 1. The "proof' fails to consider that a special argument would be necessary when n = 1. In this case there would be no way to remove a different horse from a set of n horses. (d) F. The factorization of xy + 1 is wrong, and there is no reason to believe x + 1 or y + 1 is prime.

Exercises 2.5 1.

S.

Hint: Show that the statement is true for n = 4 and n = 5, then apply the PCI induction hypothesis to n - 2. (a) Hint: The induction hypothesis is "Suppose 13k is even and both hk+l and hk+2 are odd for some natural number k." From this use the definition of Fibonacci numbers to show thath(k+I) is even and bothh(k+l)+l and 13(k+ 1)+2 are odd. (d) Proof. In the case of n = 1, the formula is II = II +2 - 1, which is 1 = 2 - 1. Suppose for some k that II + 12 + 13 + ... + h = 1k+2 - 1. Then

II + I2 + I3 + ... + h + Ik+ I = UI + 12 + 13 + ... + h) + Ik+ I = Uk+2 - 1) + Ik+1 = Uk+2 + 1k+I) - 1 = 1k+3 - l. Thus, by the PMI, II + I2 + I3 + ... + I" = I,,+2 - 1 for all natural 6.

(d)

numbers n. Consider the cases n = 1 and n = 2 separately. For n > 2, you will find it useful to multiply the equation a 2 = a + 1 by a,,-2 and p2 = P+ 1 by

pn-2. 7.

Hint: Let A = {b E N: there exists a E N such that a 2 = 2b 2 .}, assume A is nonempty, and use the WOP to reach a contradiction.

318

Answers to Selected Exercises

14.

(b)

Proof. Let S be a subset of ~ such that 1 E Sand S is inductive. We wish to show that S = ~. Assume that S"* ~ and let T = ~ - S. By the WOP, the nonempty set T has a least element. This least element is not 1, because 1 E S.1f the least element is n, then nET and n - 1 E S. But by the inductive property of S, n - 1 E S implies that n E S. This is a contradiction. Therefore, S = ~.

Exercises 2.6 2. 3.

(b) 16 Hint: Since 1,000,000 = (10 3)2 = (10 2)3 = 106 , there are 103 squares less than or equal to 1,000,000; 102 cubes less than or equal to 1,000,000; and 10 natural numbers that are both squares and cubes (sixth powers) less than or equal to 1,000,000. 11. (a) 3!' 4! (b) Hint: Think of arranging 5 objects, where each boy and the group of girls are the objects. Then consider that the girls can be arranged in 3! ways. 16. (a) The algebra in the inductive step is:

%(n; l)ar n

b + 1-

r

(n; l)arbn+1-r + an+1 = bn+1+ ~ ((;) + C~ 1) )a rbn+1-r+ an+1 = bn+1 + ~

C~ l)a rbn+1-r+ an+1 ~ + ~ (;)a'b"-') + a(~ (, ~ ,)r'b"+'-' +a") ~ b(~ (;)a'b"-') + a(%, (; )a'b"-' + a") = bn+1+ ~ (;)a rbn+1-r+ ~

+"

b(~ (;)ar~-r) + a(~ (;)arb n-r) = (a + b)(~ (;)arbn-r). =

(d)

Proof. Choose one particular element x from a set A of n elements. The number of subsets of A with r elements is (~). The collection of r-element subsets may be divided into two disjoint collections: those subsets containing x and those subsets not containing x. We count the number of subsets in each collection and add the results. First, there are (n-; I) r-element subsets of A that do not contain x, since each is a subset

Answers to Selected Exercises

319

of A - {x}. Second, there are (~= Dr-element subsets of A that do contain x, because each of these corresponds to the (r - 1)-element subset of A - {x} obtained by removing x from the subset. Thus the sum of the number of subsets in the two collections is

18. (b) Hint:

Consider two disjoint sets containing nand m elements.

Exercises 3.1 x B = {(1, q), (I, {t}),

(1, n), (2, q), (2, {t}), (2, n), ({I, 2}, q), ({I, 2}, {t}), ({I, 2}, n)} B x A = {(q, I), ({t}, I), (n, I), (q, 2), ({t}, 2), (n,2), (q, {I, 2}), ({t}, {I, 2}), (n, {I, 2})}

1. (b) A

3. (b)

6.

(a) (c) (e)

7.

(a)

Proof. (a,b)EAx(BnC) iffaEAandbEBnC iff a E A and b E Band b E C iff a E A and b E B and a E A and b E C iff (a, b) EA X Band (a, b) EA X C iff (a, b) E (A X B) n (A X C). domain~, range ~ domain [I, 00), range [0, 00) domain R range ~ Y

(c)

Y

-~+----x

(e)

8. (a)

-I

RI

Y

= R)

{(x, y) E ~ X ~: y = ~(X + 1O)}

(c)

R;I =

(e)

R;)={(X'Y)E~X~:y=±~5~X}

(g)

x+4} R7-) = { (x, y) E ~ X ~: y < -3-

(i)

R;I

= {(x, y) E

P X P: y is a child of x, and x is male}.

320

Answers to Selected Exercises

9.

(b) (d)

RoT = {(3, 2), (4, 5)}

(d) (g)

= {(l, 2), (2,2), (5, 2)} = {(x, z): x = z} = RI R2 0 R3 = {(x, z) E ~ X ~: z = -35x + 52} R4 0 R5 = {(x, z) E ~ X R z = 16x 4 - 40x 2 + 27}

(j)

R6oR6={(X,z)E~X~:z<x+2}

10. (a)

R

RI

(m) R3 (0)

20.

(a) (b) (d)

R

0

RI

0

0

Rs =

{(X, z) E ~ X IR: z = xl~2 -

1O}

R9 0 R9 is not {(x, z): z is a grandfather of x}. F. The statements "x E A X B" and "x E A and x E B" are not equivalent. C. The only correction required is that (a, c) Et: B X D implies a Et: B or c Et: D. A.

Exercises 3.2 1.

(a)

(e) (k)

2.

3.

4.

(a) (d) (f)

(a)

(d)

S. 6.

(a) (c) (a)

10. (a)

not reflexive, not symmetric, transitive reflexive, not symmetric, transitive not reflexive, symmetric, not transitive (Note: Sibling means "a brother or sister. ") {(l, I), (2, 2), (2, 3), (3, I)} {(l, I), (2, 2), (3, 3), (1, 3), (2, 3), (3, I), (3, 2)} Y

~x

This is tho gmph ofth, "lation {(x. y). y

-ox).

To show that R is reflexive, let a be a natural number. All prime factorizations of a have the same number of 2's. Thus a R a. It must also be shown that R is symmetric and transitive. Three elements of 4/R are 4 = 2· 2,28 = 2·2· 7, and 300 = 2·2·3·5·5. To show transitivity, begin by supposing (x, y) R (z, w) and (z, w) R (u, v). Then xw = yz and zv = wu. Multiply the first equation by v and the second by y. Comparison of the resulting equations will show that (x, y) R (u, v). It must also be shown that R is reflexive and symmetric. The equivalence class of (2, 3) contains pairs (x, y) such that y= ~. transitive, but not reflexive and not symmetric reflexive, symmetric, and transitive 0/=5 = {... , -15, -10, -5,0,5, 1O,00.} 1/=5 = {oo., -9, -4, 1,6, ll,oo.} 2/=5 = {oo., -8, -3,2, 7, 12,00.} 3/=5 = {oo., -7, -2,3,8, 13,00.} 4/=5 = {oo., -6, -1,4,9, 14,00.} Proof. Let (x,y)ERUR- I . Then (x,y)ER or (x,y)ER- I . If (x, y) E R, then (y, x) E R -I. Likewise, if (x, y) E R -I, then (y, x) E R. In either case, (y, x) E R U R- 1• Thus, R U R- 1 is symmetric.

Answers to Selected Exercises

12.

15. 16.

321

(b)

Proof. Assume R is symmetric. Then (x, y) E Riff (y, x) E Riff (x, y) ER- 1. Thus R = R-]. Now, suppose R = R-]. Then (x, y) ER implies (x, y)E R -], which implies (y, x) E R. Thus R is symmetric. One part of the proof is to show that R is symmetric. Suppose x R y. Then x L y and y L x, so y L x and x L y. Therefore, y R x. (d) F. The last sentence confuses R n S with R 0 S. A correct proof requires a more complete second sentence.

Exercises 3.3 2. 4. 5.

7. 10.

13. 14.

(b) {E, O}, where E is the set of even integers and 0 is the set of odd integers. Hint: The partition has four elements. One of them is the set {(l, - 1), (- 1, 1), (i, i), (- i, - i) }. (b) x R y iff x = y or both x > 2 and y > 2. if x E E or y E E (d) R ·ff{X = Y X yl [xff] __ [yff] otherwise. (recall [x~ denotes the greatest integer function) {(l, 1), (1,2), (2, 1), (2, 2), (3, 3), (3, 4), (3, 5), (4, 3), (4,4), (4,5), (5, 3), (5,4), (5,5)} No. Let R be the relation {(l, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2,1), (3, I)} on the set A = {I, 2, 3}. Then R(1) = {I, 2, 3}, R(2) = {I, 2}, and R(3) = {I, 3}. The set.sil = HI, 2, 3}, {I, 2}, {I, 3}} is not a partition of A. (b) Yes, {S], S2} is a partition of A, because {S], S2} = {B2, B]}. If B] = B2 then B] = B2 = A and S] = S2 = 0, so {S]' S2} is not a partition. (b) C. This is a tough one because the ideas are all there and every statement is true. We give it C because the ideas are not well connected. (a)

Exercises 3.4 1. (a)

11.

(c) (e) (a)

No, since (2, 4) and (4, 2) are in R. No, since (2, -2) and (-2, 2) are in R. No, since (1,3) and (3, 1) are in R. There are multiple correct answers for this question, depending upon preferences. One answer is: t9

I t6

I

t2

tF

I

t4

12. 13.

(a) (a)

{(a, a), (b, b), (c, c), (c, a), (c, b)} Proof. Suppose B - {x} <::;:: C <::;:: B. For any y E B - {x}, we have y E C. If x E C, then for all y E B we have y E C. Therefore, B <::;:: C, which

322

Answers to Selected Exercises

14.

(b)

23.

(c) (c)

25.

(b)

shows that C = B. On the other hand, if x tE C, then for all y E C we have y E B andy oF x. Thus, C ~ B - {x}, which shows that C = B - {x}. Therefore, there is no C different from Band B - {x} such that B - {x} ~ C ~ B. Hence B - {x} is an immediate predecessor of B. No; for example, consider a set of two squares where the squares are side by side within the rectangle. A set containing two disjoint squares does not have a lower bound. Proof. Let x E z and yEw. Then x P z, z P x, y P w, and w P y. Suppose x p y. Then x P y. But this with z P x and y P w let us conclude that z P w, hence z p w. A similar argument shows that z p w implies x p y. Thus x p y iff zp w. C. This proof looks as if it deserves an A grade, but it does not show that sup (B) exists (all it shows is that if sup (B) exists, then u = sup (B)). A correct proof would show that u = sup (B) by showing u has the two supremum properties (u is an upper bound and u R v for all other upper bounds v.)

Exercises 3.5 9.

16.

Suppose the graph G has order n ;::: 2, and all vertices have different degrees. Consider what these degrees must be. Can one vertex have degree n - 1 and another have degree o? (a) F. The claim is false. To discover the error in the proof, consider exercise 10.

Exercises 4.1 1.

(a) (g)

2. (a) 4. 5.

(d) (c)

(a) 7. (a) (c)

8. (b) (e)

10. 14. 15.

R[ is a function. Dom (R[) = {O, 6, D, n, U}. A possible codomain is Rng (R]) = {O, 6, D, n, U}. R7 is a function. Dom (R7) = IR. Possible codomains are IR and [0, (0).

Domain = IR - {I}. Range = {y E IR: y oF O}. A possible codomain is IR. Domain = IR - {~ + kn: k E Z}. A possible codomain is IR.

5, -5 Dom(f) = IR - {3}, Rng(f) = IR - {-I} Letx n = -n.Thenx=(-I, -2, -3, -4, ... ). Hint: Find a rule so that X2 = 3~ and X3 = 3~. This rule is a function. Not a function: for example, the modulo 3 congruence classes of 0, 3, 6, and 9 are all equal to 0/== 3, so the rule assigns this equivalence class all these answers: 0/== 4,3/== 4,2/== 4' and 1/== 4·

(a) A. (a) 1(3) = 3/== 6 = {... , -9, -3,3, 9, ... }. (a) Dom(S)

Answers to Selected Exercises

16.

(a)

323

Proof. Let x, y, zEN. (i)

By definition of absolute value, d(x, y) = Ix - yl 2': 0 for all x, yEN.

d(x, y) = Ix - yl = 0 iff x - y = 0 iff x = y. d(x, y) = Ix - yl = Iy - xl = dey, x). (iv) By the triangle property of absolute value, Ix - yl + Iy - zl2': Ix - zl. Thus d(x, y) + dey, z) 2': d(x, z). (~)n2. We choose 2 elements from A for first coordinates; each may then be assigned any element of B as image. (ii) (iii)

17.

(c)

1.

(a) (c) (e)

(f 0 g)(x) = 17 - 14x, (g 0 f)(x) = -29 - 14x (f 0 g)(x) = sin (2x 2 + 1), (g 0 f)(x) = 2 sin 2 x log = {(k, r), (t, r), (s, l)}. go 1= ([J.

(I')

x :5 -1 g () x -- {/(2X) 1(-x) if if x> -1. We must consider cases. If x:5 -1,f(g(x)) = 1(2x). Since in this case 2x:5 0,f(2x) = 2x + 1. If x> -1,f(g(x)) = I( -x). There are two subcases: If x 2': 0, then -x :5 0, so I( - x) = -x + 1. If -1 < x < 0, then I > -x> 0, so I( -x) = -2x. Therefore,

Exercises 4.2

Ob

h I serve t at

1

g(x) = {

0

+1

0

2X + 1 -2x

-x + I

if x:5 -1 if -1 <x < 0 if x 2': O.

Similarly,

go/(x)

2.

(a) (c) (e) (i)

=

{g(X + 1) g(2x)

if x:5 0 if x> O.

If x:5 0, g(f(x)) = g(x + 1). There are two subcases: If x :5 - 2, then x + I :5 - 1, so g(x + 1) = 2x + 2. If x > - 2, then x + 1 > - I, so g(x + 1) = -x - 1. If x> 0, g(f(x)) = g(2x). Since x> 0, 2x> 0, so 2x> -1. Thus g(2x) = -2x. if x :5 - 2 2x + 2 Therefore, g o I (x) = { - x - I if -2 < x :5 0 -2x if x> O. Dom (f 0 g) = JR = Rng (f 0 g) = Dom (g 0 I) = Rng (g 0 I) Dom (f 0 g) = JR, Rng (f 0 g) = [-1, I]. Dom (g 0 I) = JR, Rng (g 0 f) = [1, 3]. Dom (f 0 g) = {k, t, s}, Rng (f 0 g) = {r, I}. Dom (g 0 f) = 0 = Rng (g 0 f). Dom (f 0 g) = JR, Rng (f 0 g) = (-00, 2). Dom (g 0 f) = JR, Rng (g 0 f) = (-00, 1).

324

Answers to Selected Exercises

3.

(a)

f-I(x)

=x~2

(c)

f-I(X)

=

I - 2x

x-I

11.

f-I(X) = -3 + Inx (a) {(x, y) E JR X JR: y = 0 if x < 6 and y = x 2 if x;::: o} {(x, y) E JR X JR: y = x 2 }. Hint: Write A U C as A U (C - E). Then show hUg = h U (gl C-E) and

12.

use Theorem 4.6. (a) hUg is a function.

(e)

7.

13. (a)

14.

(a)

Y

Proof. We show thatfl + f2 is a function with a domain JR. Firstfl + fz is by definition a relation. For all x E JR there is some u E JR such that (x, u) Efl because fl: JR ---+ JR and there exists v E JR such that (x, v) E f2 because fz: JR ---+ IR. Then (x, u + v) Efl + fz, so ·VE Dom (fl +12)' It is clear from the definition of fl + f2 that x E Dom (fl + f2) implies x E JR, so Dom (fl + f2) = JR. Let x E JR. Suppose (x, c) and (x, d) are in fl + f2' Then c = fl (x) + f2(X) = d. Therefore, fl + fz is a function. Proof. Suppose x
(d)

Proof. Suppose x and yare in (-3,00) and x < y. Then 4x < 4y, 3x - y <3y - x. Thus xy + 3x - y - 3 < xy + 3y - 3y - x - 3 (x- l)(y+ 3) «y - I)(x + 3). Using the fact that x and yare (-3,00), we know x + 3 ang:y + 3 are positive. Dividing both sides the last inequality by x + 3 and y + 3, we conclude that (x-I)«y-l). (x

+ 3)

(y

so or in of

+ 3)

Thusf(x)
15.

(d)

(Note: This proofwasfound by working backwardfrom the conclusion.) The function f given by

+ I if x:::; 0 I- x if 0 < x < I x - I if x;::: I is a counterexample. Another counterexample is f given by f(x) (x + I)(x - 1)2. Example I:f(x) = x 2, g(x) = 3x + 7. Example 2:f(x) = (x + 7)2, g(x) = 3x. Example I:f(x) = Ixl. Example 2:f(x) = 3. A X

f(x) =

16.

(a) (d)

20.

(a)

{

=

Answers to Selected Exercises

325

Exercises 4.3 1. (a) (c)

(k)

Onto IR. Proof. Let wEIR. Then for x = 2(w - 6), we have f(x) = ![2(w - 6)]+ 6 = w. Thus w E Rng (f). Therefore,fmaps onto IR. Not onto N X N. Since (5, 8) E N X Nand (5, 8) t!. Rng (f), f does not map onto N X N. Proof. First, if x E [2, 3), then x - 2 :::=:: 0 and 3 - x > 0, so f(x) =

~-

2 :::=:: O. This shows that Rng (f) C [0, 00). To show 3w + 2 let w E[O, 00). Choose x = - - I. Then w+

f

is onto [0, 00),

-x

f(x) = [

e::n- e::n] 2 ] -;- [ 3 -

= [3w + 2 - 2(w + I)] -;- [3(w + I) - (3w + 2)] =

2.

3. 8.

(a) (a)

(e)

9.

W"

(This value for x was found by working backward from the desired result.) Therefore, f maps onto [0, 00). (a) One-to-one. Proof. Suppose f(x) = fey). Then!x + 6 =!y + 6. Then !x=!y,sox=y. (g) Not one-to-one, because sin (~) = sin c~n) = !. x-2 (k) f is one-to-one. Proof. Suppose x E [2,3) andf(x) = fez). Then= 3z-2 -x -3- , so 3x - xz - 6 + 2z = 3z - xz - 6 + 2x. Thus x = z.

(c)

-z = {O,

3}, f = {(l, 0), (2,3), (3,0), (4,0)} Let f: IR -+ IR be given by f(x) = 2x and g: IR -+ IR be given by g(x) = x 2 . Then f maps onto IR but go f is not onto IR. LetA = {a, b, c}, B = {I, 2, 3}, C = {x, y, z},f = {(a, 2), (b, 2), (c, 3)}, g = (O, x), (2, y), (3, z)}. Then g is one-to-one, but go f = {(a, y), (b, y), (c, z)} is not. Proof. We verify that f maps onto IR as follows: I E Rng (f), because f(4) = 1. 4w+ 2 For w 1= I, choose x = - - I . Then B

w-

f(x)

=

[(4t_+;) - 2] -;- [(4t_+;) + 4] = 6; =

w.

Therefore, f maps onto IR. To show that f is one-to-one, suppose f(x) = x-2 w-2 few). Then for x 1= -4, w 1= -4. --4 = --4' Therefore, xw - 2w + x+ w+ 4x - 8 = xw - 2x + 4w - 8, so 6x = 6w and x = w. We must also consider whether f(x) might be I, for x 1= -4. But if x 1= -4 and f(x) =

10.

11.

x - 42 = I, then x - 2 = x + 4, so - 2 = 4. This is impossible. Therefore, x+ f is one-to-one. (a) LetA = B = lRandS = {(x, y) E IR X IR: x 2 + y2 = 25}. Then (3, 4) ES and (3, -4) E Sand 1[] (3,4) = 3 = 1[] (3, -4), so 1[] is not one-to-one. (a) Yes. (c) Not necessarily. The projection 1[2 is one-to-one iff S is one-to-one.

326

Answers to Selected Exercises

14.

(a)

15.

(b)

Proof. f is not a surjection because 1/=- 8 has no pre-image in Z4. This is because if f{x/=- 4) = 1/=- 8, then 2x/=- 8 = 1/=- 8, so 2x =-8 1. But then 8 divides the odd number 2x - I, which is impossible. To show f is an injection, suppose f{x/=- 4) = f{z/=- 4)· Then 2x/=- 8 = 2z/=- 8, so 2x =-8 2z. Therefore, 8 divides 2x - 2z, so 4 divides x - z, and thus X/=-4 = Z/=-4. Example I: x(n) = n for all n E N (the identity function) Example 2: x(n)

=

{so n

Example 3: x(n) = {n

+I

!f n < SO If n;::: SO

n

!f n !s odd

n - I If n IS even

16. (c) (e)

17.

(c)

(e)

The sequence of example 3 is (2, 1,4,3,6, S, 8, 7, ... ). None. (':i)n! nm-n. We choose n elements of A to be the pre-images of elements of B and then assign images to arguments in n! ways. Finally, each of the remaining m - n elements of A may be assigned any element of B as its image. A. F. The proof shows only that Rng (f) ~ I. To show that must prove that I ~ Rng (f).

f

is onto we

Exercises 4.4 1.

(a)

(0,0), {{I}, {4}), {{2}, {4}), {{2, 3}, {4, S}), {A, {4, S})

2.

(a) (c) (a)

[2, 10]

4. S. 6.

(b) (d) (b) (c)

{{3},

{S}), {{I,

2}, {4}),

{{I, 3}, {4, S}),

to} {(l, I)} {p, s, t} {1,2,3,S} [2, S.2]

[2 - ~, 3~ vIS) u C: vIS, 2+ ~ ]

[9, 2S] (S -lfi, S U [S + S + Ifi) 9. Hint: There must be at least two sets D J and D2 in the family, and b Ef(D(j) for all 0 E Ll, but no element a in D(j such that f(a) = b. The function f cannot be one-to-one. (jEt'. 13. (b) Proof. Let t Ef(X) - feY). Then t Ef(X), so there exists x E X such thatf(x) = t. We note x 1$ Y since t = f(x) I$f(Y). Thus x EX - Yand, therefore, t = f(x) Ef(X - Y). 14. The converse is true. To prove f is one-to-one, let x, yEA and x =1= y. Then {x} n {y}= 0 and thus f{{x} n {y}) = 0. By hypothesis, f{{x} n {y}) = f{{x}) nf{{y}) = {I(x)} n {I(y)}. Thusf(x) =1= fey). 7.

(a) (c)

fo]

fo,

n

Answers to Seleded Exercises

17. 20.

(a)

3.

(b)

(a)

327

If fis one-to-one, then the induced function is one-to-one. F. The claim is not true. We cannot conclude x EX fromf(x) Ef(X).

Exercises 5.1 Define f: A -+ A U {x} by f(a) = (a, x), for each a EA. Now show that

f is one-to-one and onto A U {x}. 5.

7. 8.

10. 11. 14. 17.

18. 19.

20.

(a) (c) (e) (a) (a)

finite finite infinite Suppose A is finite. Since A n B is a subset of A, A n B is finite. Hint: Consider f: Nk X N m-+ N km given by f(a, b) = (a - I)m + b. Use the Division Algorithm to show f is a one-to-one correspondence. This is a formal version of the Product Rule (Theorem 2.20) that tells us #(Nk X Nm) = km. (c) not possible Hint: Write A U B = (A - B) U B and A = (A - B) U (A n B) and apply Theorem 5.6. Hint: Use the pigeonhole principle. (a) Hint: Suppose fis not onto B and consider the range off (b) Hint: Suppose f is not one-to-one. Then A is not empty and since A and B are finite and A = B, there is some n E N such that N n "'" A and B = Nn . Use these facts to construct a function F from Nn onto Nn that is not one-to-one. Then for some x, y, Z E N m f(x) = fey) = z. Removing (y, z) from F produces a function from a proper subset of Nn onto Nn . Now apply exercise 15. Use induction on the number of elements in the domain. (a) Proof. Let f: N -+ {1O, 11, I2, ... } be given by f(x) = x + 9. If f(x) = fey), then x + 9 = y + 9 and therefore, x = y. Thus f is one-toone. To show that fis onto {1O, 11, I2, ... }, lett E {1O, 11, I2, ... }. Then choose z = t - 9. Since t > 9, t - 9 E N. For this z, fez) = f(t - 9) = (t - 9) + 9 = t. (b) c. In Case 2, it is not correct that Nk U NI "'" Nk + l . In addition, if x E Nk> then S U {x} + Nk U {x} = N k •

Exercises 5.2 1.

Proofs. Let f: N -+ D+ be given by fen) = 2n - 1 for each n E N. We show that f is one-to-one and maps onto D +. first, to show f is one-toone, suppose f(x) = fey). Thus 2x - 1 = 2y - 1, which implies x = y. Also, f maps onto D + since if d is an odd positive integer, then d has the . form d = 2r - 1 for some r E N. But then fer) = d. (e) Hint: Consider f(x) = -(x + 12) with domain N. (a)

328

Answers to Selected Exercises

2. (a) (b) (d)

Hint: Hint: Hint:

Let 1(0, I) ---+ (1, (0) be given by I (x) = k. Let/(O, 1)---+ (a, (0) be given by I (x) = (a - I) Let/(O, 1)---+ [1, 2) U (5,6) be given by

I(x) =

{2X + I 2x

4.

(a)

2x (a)

8. 9.

12.

2

Define g: N ---+ E+ by 20

(c) (e) (b)

::5.!

if ~ < x < 1.

+4

g(x) = { 2

5.

if 0 < x

+ k.

if x if x if x

=I =

10

"* I, x "* 10.

c

Xo c Hint: Show that B is equivalent to Al X (A 2 X A 3 ) and then use part (a). Hint: Since A is denumerable, there is a bijection I: A ---+ N. Since B is denumerable, there is a bijection g: B ---+ N. Let h: A X B ---+ N be given by h(a, b) = y(a)-1(2g(b) - I). (a) F. The main idea of the "proof' is that infinite subsets of N are denumerable. No justification for this is given. (c) F. The claim is false. Also "A and B are finite" is not a denial of "A and B are infinite." (d) F. Writing an infinite set A as {Xl> X2,"'} is the same as assuming A is denumerable.

Exercises 5.3 5.

Hint:

Use a proof by induction on n, the number of sets in the family.

9.

Hint:

For each mEN, let Bm

Defineh: 11.

= k m. Then there is a bijection 1m: Bm --+ N km •

i~ Bi--+ Nbyh(x) = ('~l k

i)

+ Im(x),forx EBm.

Hint: Since A is denumerable, there is a bijection I: A ---+ N. Since B is denumerable, there is a bijection g: B ---+ N. Apply the diagonalization process to A X B where the elements of A X B are arranged: (/(1), (/(2), (/(3), (/(4),

12. (a)

g(1» g(I» g(l) g(1»

Each set in {{n}:

(/(1), (/(2), (/(3), (/(4),

g(2» g(2» g(2» g(2»

(/(1), (/(2), (/(3), (/(4),

g(3» g(3» g(3» g(3»

(/(1), (/(2), (/(3), (/(4),

g(4» g(4» g(4» g(4»

n E N} is finite and the union of this family is N.

Answers to Selected Exercises

13.

(b)

For each n E 1\1, let Tn be the set of all sequences where all but n terms are 0. Show that Tn is denumerable [using exercise 8(d)] and that

Hint: T=

U Tn·

nEN

16. (a)

329

*"

C. The proof is valid only when f(l) = x. In the case when f(1) x, we must first redefine g. Let t be the unique element of 1\1 such that f(t) = x and define f* = (J - {(l,f(l», (t, x)}) U {(l, x), (t,f(l»}. Now let g(n) = f*(n + 1) for all n E 1\1.

Exercises 5.4 2. Note that QJ>(I\I) = ~. Now use Cantor's Theorem. (b) True 8. (a) 0 < {O} < {O, I} < Q < (0, 1) = [0, 1] = ~ - Q S.

11.

(b)

16. (b)

(d)

= ~ < QJ>(~) < ~QJ>(~QJ>=;;;;(~~».

not possible F. We have not defined or discusseQ..pr02erties of operations such as addition for cardinal numbers. Thus C = Ii + (C - B) cannot be used unless C and B are finite sets. The claim is false. A.

Exercises 5.5 1. (b)

Axiom of Choice is not necessary since there are only a finite number of sets in the collection. (h) Yes. S. Proof. Let B k A with Ii infinite and A denumerable. Since B k A, Ii :5 A. Since A is denumerable, A = N. Since B is infinite, B has a denumerable subset D by TheoreQl5.~6. Thus A= j\j = D:5 B. By the Cantor-SchrOder-Bemstein Theorem, Ii = A. Thus B = A. 8. Hint: Let x EA. By Theorem 5.27, A - {x} has a denumerable subset {an: n E Construct a one-to-one correspondence between A and A - {x}. 10. (a) A (c) A

1\If.

Exercises 6.1 1. (a) (e)

2.

(a) (e)

yes no not commutative not an operation

330

Answers to Selected Exercises

3. (a)

4.

(e) (a)

not associative not an operation Yes, 0 is a binary operation because for all x, yEA the product x always one of the elements of A, as shown by the table.

(b)

(a)

(c)

0

y is

Yes. This is tedious to verify because one must verify that 64 equations of the form (x 0 y) 0 z = x 0 (y 0 z) are all true. It helps to observe that if x = a (the identity) then (x 0 y) 0 z = y 0 z = x 0 (y 0 z). Similarly, if y = a or z = a, the equation is easily seen to be true. This leaves only 27 cases to verify when none of x, y, or z is a. For example, (b 0 c) 0 b = d 0 b = c, while b 0 (c 0 b) = (b 0 d) = c, so the equation is true when x = b, y = c, and z = b. (d) Yes, because the table is symmetric about its main diagonal. To verify by cases that the equation x 0 y = y 0 x is true for every choice of x and y, consider first that case that one of x or y is a, then the case that x = y, and finally the other 3 cases. (e) The inverses of a, b, c, d, are a, b, c, d, respectively. (f) No. The product b 0 c = d is not in B I , so BI is not closed under o. (g) Yes. a 0 a = a, a 0 c = c, c 0 a = c, co c = a. (h) {a}, {a, b}, {a, c}, {a, d}, and {a, b, c, d}. (i) True. In fact for all x E A, x 0 x = a. 8. Hint: Compute eo f. 9. (a) Hint: Compute x 0 (a 0 y) and (x 0 a) 0 y. 11. (b) Hint: Assume that for some natural number n, every product of t elements at> az, ... , at of A is equal to ( ... ((al * a2) * a3) ... ) * at for every t::5 n. Now consider a product of n + I factors at> az, . .. , an+1 in that order. This product has the form b l * b2 , where b l is a product of some k factors (k ::5 n) at> a2, ... , ak in that order and b2 is a product of the remaining factors ak+ I' ... , a n + I in that order. First use the induction hypothesis to write b l and b2 in left-associated form. Now consider two cases. If k > I, then there are at least two factors al and a2 in the product b l . Denote the product al * a2 by c, which is an element of A. Replace al * a2 by c and use the hypothesis of induction to write b l * b2 as a left-associated product. If k = I and b2 has only one factor, then the product b l * b2 is al * a2, which is already in left-associated form. Otherwise, b2 has at least 2 factors, and we may denote by d the product of the first two factors a2 and a3 of b2. Replace a2 * a3 by d in the product b l * b2 and apply the hypothesis of induction. Finally, apply the associative property to al * (a2 * a3) to write the entire product in left-associated form. 18. ~(f + g) = (f + g)(x) dx = f(x) dx + g(x) dx = l(f) + leg). 24. (a) Let C, D E 0J>(A). Then fCC U D) = f(C) Uf(D) by Theorem 4.16(b). Therefore, f is an OP mapping. 25. (c) F. The claim is false. One may premultiply (multiply on the left) or postmultiply both sides of an equation by equal quantities. Multiplying one side on the left and the other on the right does not always preserve equality.

f:

f:

f:

Answers to Selected Exercises

331

Exercises 6.2 1. (a)

(d)

2.

4. 8. 9. 12. 15.

-1 1 -1

1 -1

-i

-i

Hint:

-i

-1 1 -i

-i -1 1

-i i 1 -1

We see from the table that the set is closed under', I is the identity, and each element has an inverse. Also, . is associative.

o is the identity.

e

u

e u v

e u v

w

w

u v w e

v

w

v

w

w

e u v

e u

(a) The group is abelian. Hint: For a, bEG, compute a 2b2 and (ab)2 two ways. Hint: In order to have both cancellation properties, every element must occur in every row and in every column of the table. (a) v; w; e; u. (b) Hint: Let a, bEG. For an element x such that a * x = b, try a -I * b. (b) F. A minor criticism is that no special case is needed for e. The fatal flaw is the use of the undefined division notation.

Exercises 6.3 2. 3.

6. 7•

10. 12. 13. 14. 15. 16.

(b)

(3214), (1234), (2413). Suppose a =m band c =m d. Then m divides a - band c - d. Thus there exist integers k and I such that a - b = km and c - d = 1m. Hence (a + c) - (b + d) = (a - b) + (c - d) = km + 1m = (k + I)m. Then k + I is an integer so m divides (a + c) - (b + d). Therefore, a + c =m b + d. (b) (143256) (b) a 2 = (312) and a 3 = (123), the identity. Therefore, a-I = a2 ; a4 = a 3a = a; a So = a48 a 2 = (a3) 12 a 2 = a 2; a SI = asoa = a 2a = a 3 = (123). 10, 12, 2n (a) The zero divisors are 2, 3,4,6, 8, 9, 10. Since (p - 1)(p - I) = p2 - 2p + 1 = pep - 2) + 1, (p _ 1)2 =p I. Therefore (p - I )(p - I) = I in lLp, and hence (p - I)-I = P - 1. (a) x = 0,4,8, 12, 16. (c) x = 2, x = 6 (e) No solution. (b) C. There is no justification that xy =1= 0. One must begin differently by considering prime factorizations of a, b, and m. (i)

332

Answers to Selected Exercises

Exercises 6.4 {O}, 1:8 , {o, 4}, {o, 2,4, 6} There are six subgroups. 9. Yes. Assume G is abelian and H is a subgroup of G. Suppose x, y E H. Then x, y E G. Therefore, xy = yx. 11. (c) The order of 0 is 1. The elements 1, 3, 5, and 7 have order 8. The elements 2 and 6 have order 4. The order of 4 is 2. 13. Proof. Na is not empty, because ea = a = ae, so e E N a. Let x, y E N a. Then xa = ax and ya = ay. Multiplying both sides of the last equation by y -I, we have y-I(ya)y-I = y-I(ay)y-I. Thus (y-Iy)(ay-I) = (y-Ia)(yy-I), or ay-I = y-I a . Therefore, (xy-I)a = X(y-I a ) = x(ay-I) = (xa)y-I = (ax)y-I = a(xy-I). This shows xy-I E N a. Therefore, Na is a subgroup of G, by Theorem 6.12. 16. Proof. The identity e E H because H is a group, and thus a-lea E K. Thus K is not empty. Suppose b, e E K. Then b = a-Ihla and e = a- I h 2a for some hI> h2 E H. Thus be- I = (a- Ih l a)(a- Ih 2a)-1 = (a-Ihla)(a-lh:;la ) = a-Ihl(aa-l)h:;la = a-Ihlh:;la . But H is a group, so hlh:;1 E H. Thus be - I E K. Therefore, K is a subgroup of G. 18. (a) Suppose y = x in 1: 18 , Then 18 divides x - y. Therefore, 6 divides x - y, and so 24 divides 4(x - y) = 4x - 4y. Thusf(x) = [4x] = [4y] = fCY) in 1:24 , so f is well defined. Now let x, y E 1: 18 , Then 1.

(a) (d) (a)

f(X (b)

(c)

(d)

+ 18

y)

Rng (f) =

= [4(x + y)] = [4x + 4y] = [4x] +24 [4y] =

(a)

+24

fCY)·

{[o], [4], [8], [12], [16], [20n. The table is: [0]

[4] [4] [8]

[0]

[0]

[4] [8]

[4] [8]

[12] [16] [20]

[12] [16] [20]

ker (f) = {o, The table is:

6, TI}

[12] [16] [20] [0]

[12]

[16]

[20]

[8]

[12] [16] [20] [0]

[16] [20] [0]

[20] [0]

[12] [16] [20] [0]

[4] [8]

[4]

[4] [8]

[4] [8]

[12] [16]

[12]

6

12

0 6

0 6

6

12

12

0

12

12

0

6

[0]

[12]

[0] [12]

[12] [0]

ker (f) = {[a], [TIn, with table

1,3,5,orI5

[8]

0

[0] [12]

21.

f(X)

Answers to Selected Exercises

24. 27.

(c)

Hint:

333

a lO ,a 20 Use exercise 23(a).

Exercises 6.5 1.

4.

10. 11.

H = {(213),(l23)}. The left cosets of Hare (l23)H = H, (132)H = ((32), (312)}, and (321)H = {(321), (213)}. Let G be a group of order 4. By Corollary 6.18, the order of an element of G must"be 1,2, or 4. The only element of order 1 is the identity. If G has an element of order 4, then G is cyclic. Otherwise, every element (except the identity) has order 2. Hint: Let G = (a), and suppose n = mk. Consider the element a k • (a) F. From aH = bH we can conclude only that for some hI and h2 E H, ahl = bh 2· (c) A.

Exercises 6.6 1. (b)

4. 6.

Hint: For each x E G, show that the left cosetxG = G. To do this, you will need to show that G ~ xG. For this, use the fact that for y E G, y = x(x-Iy), and x-Iy E G. After establishing that every left coset is G, consider right cosets. Hint: See exercise 3. (b) Hint: If G = (a), the group generated by a E G, consider (aH), the group generated by the element aH in GIH. As a lemma prove that if x E G, then [xH( = xnH for all n E 71.

7. Proof. (i) Suppose H is normal. Let a E G. To show a-IHa ~ H, let t E a-IHa. Then t = a-Iha for some hE H. Thus at = ha. Since ha E Ha, at E Ha. By normality Ha = aH. Therefore, at E aH, so there exists k E H such that at = ak. By cancellation, t = k. Thus t E H. (ii) We letx E G and show thatxH = Hx. Lety ExH. Theny = xh for some hE H. But thenyx- I = xhx- I. Sincexhx- I ExHx- 1 ~ H,yx- I E H. Therefore, yx -I = k for some k E H. Thus y = kx, which proves y E Hx. Therefore, xH ~ Hx. A proof that Hx ~ xH is similar. 8. Hint: Use exercise 7. 13. (b) C. The proof is correct but a verification that x(H n K) = xH n xK is not provided. Such a verification might not be required in some classes.

Exercises 6.7 2.

(b)

No. Suppose f: K --+ 714 were an isomorphism. Then f(l) = O. Thus x = f-I(1) is one of R2, H or V. Since f is OP,f(x 2) = f(x) + f(x) = I + I = 2. But x 2 = /, so f(x2) = O. This is impossible. Note: We

334

Answers to Selected Exercises

3. 12.

should not expect the cyclic group Z4 to be isomorphic to the noncyclic group K. Isomorphism preserves the order of elements (i.e., the order of a is the same as the order of the image of a under an isomorphism). This fact is established in exercise 3. Hint: See exercise 27 of section 6.4. By Theorem 6.24 the homomorphic images of.ifs are .ifs/{O}, which is isomorphic to.if s .ifs/.ifs, which is isomorphic to {o} .ifs/{o, 4}, which is isomorphic to.if4 .ifs/{o, 2,4, 6}, which is isomorphic to .if2

Exercises 7.1 2. 3. 4.

6.

8.

(b) (b)

(a) (c) (e) (g)

1, 1,2,3 0

supremum: 1; infimum: 0 supremum: does not exist; infimum: 0 supremum: 1; infimum: supremum: 5; infimum: -1 Hint: Suppose b is an upper bound for A. Then for all x, if x E A then x ::5 b. This means that for all x, if x > b, then x $. A. Explain why Ii is not bounded above. (a) Proof. Let x and y be least upper bounds for A. Then x and yare upper bounds for A. Since y is an upper bound and x is a least upper bound, x ::5 y. Since x is an upper bound and y is a least upper bound, y ::5 x. Thus

13.

(a)

16.

(a)

17.

(a)

1

x=y. Proof. Let s = sup (A), B = {u: u is an upper bound for A} and t = inf (B). We must show s = t. (i) To show t ::5 s we note that since s = sup (A), s is an upper bound for A. Thus s E B. Therefore, t ::5 s. (ii) To show s ::5 t we will show t is an upper bound for A. If t is not an upper bound for A, then there exists a E A with a > t. Let e = alt. Since t = inf (B) and t < t + e, there exists u E B such that u < t + e. But t + e < a. Therefore, u < a, a contradiction, since u E B and a EA. Proof. Let m = max {sup (A), sup (B)}. (i) Since A ~A U B, sup (A)::5 sup (A U B). Also, B ~A U B implies sup (B)::5 sup (A U B). Thus m = max {sup (A), sup (B)}::5 sup (A U B). (ii) It suffices to show m is an upper bound for A U B. Let x E A U B. If x E A, then x ::5 sup (A) ::5 m. If x E B, then x::5 sup (B) ::5 m. Thus m is an upper bound for A U B. Hence sup (A U B) ::5 m. F. The claim is true but y = i + ~ might not be in A.

Answers to Selected Exercises

335

Exercises 7.2 1. (b)

In the case where IX2 - xIi < 261,r

J~ (x"

2. 3.

0,)

n J~I\r (xl> 0,)

_ 'r(X' + X2 -2-,0,

-J~

- IX2 -2 xIi) .

(b) x = 3.825,0 = 0.025 lim/ex) = L iff for all e > 0 there exists 0> 0 such that if 0
al

< 0,

x~a

then I/(x) - LI < e. (b) open (e) open (i) closed 7. (b) Proof. If A = {x" Xl>oo"x n} with x, <x2 <x3 < 00. <x'" then A = (-00, x,) U (XI- X2) U 00. U (xn-I- x n ) U (x", (0). Thus A is open since it is a union of open sets. Therefore, A is closed. (e) Hint: Use De Morgan's Laws. 137 (a) Hint: First show that an open cover of A U B is an open cover of A and an open cover of B. A different proof uses the Heine-Borel Theorem. 15. (a) not compact (not bounded) (d) not compact (neither closed nor bounded) (g) not compact (not closed) 20. (b) C. With the addition of 0* to the cover {au: a E L\} we are assured that there is a finite subcover of {O*} U {au: a E L\}, but not necessarily a subcover of {au: a E L\}. Since 0* = A - B is useless in a cover of B, it can be deleted from the subcover after it is used. 5.

Exercises 7.3] 2.

m

(a) (c) 0 (g) {O, (m) N

2}

3. {O, I} Hint: First show that for all a E A, Z > a. Then show z is an accumulation point of A by using Theorem 7.1. 7. Hint: Show (A n B)' ~A' n B' by using exercise 5(a). 10. (c) By parts (a) and (b), c(A) is closed and includes A. Let A ~ B, and let B be closed. Then, by exercise 9, A' ~ B. Thus c(A) = A U A' ~ B. 11. (a) has no accumulation points (c) has at least one accumulation point 12. (b) Yes. A could be the set {-100 + k: n E N} U {IOO - k: n EN}. The two accumulation points of A are -100 and 100. 15. (a) F. Hint: See the end of section 1.5 on the misuse of quantifiers. (c) F. (B) I need not be a subset of (ij). 4.

336

Answers to Selected Exercises

Exercises 7.4 1.

2.

3.

(a) (c) (e) (g) (i) (a) (c) (e) (g) (i) (a) (c)

(f)

bounded below by 10, not bounded above bounded below by 0 bounded; bounded above by bounded; bounded above by 10, bounded below by 0 not bounded above nor bounded below bounded; bounded above by 0.81, bounded below by -0.9 does not exist

10,

4

5'

o e2

o

~ 0; for 8 > 0, use N > i. x diverges; use 8 = 1. Xn ~ 0; for 8> 0, use N. > (28)-2 and Xn

+ [n' [n) [rl+l - [n = ([rl+l - [n) ([rl+l [rl+l n+ 1+ n 5.

6.

9. 14.

Proof. Let 8> O. Then ~ > O. Since Xn ~ L, there exists N) E N such that if n > NI> then IXn - LI <~. Likewise, there exists N2 E N such that n > N2 implies IYn - MI < l Let N3 = max {N), N 2}, and assume n > N 3 . Then I(x n + Yn) - (L + M)I = I(x n- L) + (Yn - M)I ::5 IXn - LI + IYn - MI < ~ + ~ = 8. Therefore, Xn + Yn ~ L + M. (e) Hint: Ilxnl - ILl 1 ::5 IXn - LI. Hint. If Y is bounded by a positive number B, use the definition of Xn ~ 0 with ~. (a) Hint: Since Xn ~ L, IXnl---+ ILl, by exercise 6(a). Now apply the definition of Ixnl---+ ILl with 8 = I~I. (c) A. The proof uses exercise 4(b).

(a)

Exercises 7.5 (a)

J56.

3. 5.

Consider a set of rational numbers in [7, 8] that approach Suppose x is a bounded increasing sequence (the proof is similar if x is decreasing). Let A = {xn: n EN}. A is bounded and infinite and so by completeness, sup (A) exists. Now show that lim Xn = sup (A).

7.

(b)

Hint:

Hint:

n~X

F. The claim is correct but there is little that is correct in this proof. For instance, the upper bound ao for A may be negative, in which case B

would have to be defined differently. More seriously, there is no connection between being an accumulation point and being an upper bound for a set.

N D E X

Abelian group, 242 Accessible vertex, 155 Accumulation point, 290 Adjacent vertices, 152 Algebraic structure (system), 233-241 Antecedent, 9 Antisymmetric relation, 141-142 Appel, Kenneth, 36 Argument, 163 Associative operation, 235 Atomic proposition, 2 Axiom consistent, 37 defined, 26 independent, 242 Axiomatic approach, 241 Axiom of Choice, 226-228 Bernstein, Felix, 220 Biconditional sentence, 11-12 proof of, 34-35 translation of, 12, 14 Bijection (one-to-one correspondence), 183, 196 Binary operation defined, 233 set closed under, 234 on sets, 67 Binet's formula, 103 Binomial coefficient, 108 Binomial expansion, 110-111 Bolzano, Bernard, 290 Bolzano-Weierstrass Theorem, 290, 292, 294, 303,304 Borel, Emile, 286 Boundary point, 288 Bounded Monotone Sequence Theorem, 298-300,303,304 Bounded sequence, 295 Bounded set, 275 Bridge, 158

Canonical map, 165,254 Cantor, Georg, 60, 209, 210, 218, 220, 221, 229 Cantor-Schrader-Bernstein Theorem, 221-224 Cantor's Theorem, 220-221 Cardinality, 198,203,206 Cardinal numbers, 198,203,219 comparability of, 226 ordering of, 218-225 Cartesian product, 116 Cauchy sequence, 273, 301 Cayley, Arthur, 268 Cayley's Theorem, 268 Cayley (operation) table, 234 Center, of group, 257 Characteristic function, 164 Chartrand, G., 152 Choice function, 226 Closed interval, 60 Closed path, 154 Closed ray, 60 Closed set, 281-282 Closure of set, 293 symmetric, 135 transitive, 135 Codomain, 162, 163 Cohen, Paul, 229 Combination Rule, 109-110 Commutative operation, 235 Compact, 285 Comparability Theorem, 226 Complement of digraph, 135 of graph, 135, 158 of set, 70 Complete digraph, 131 Complete field, 277-278 Complete graph, 152 Complete induction, 97-100 Completeness, equivalents of, 303-305

337

338

Index Component of graph, 156 Composition offunction, 171-173 of relation, 122, 124 Compound proposition, 2, 3 Conditional sentence contrapositive of, II con verse of, II defined,9 inverse of, 17 translation of, 14-15 Congruence modulo m, 130, 132, 248-249 Conjunction, 2 Connected graph, 155 Consequent, 9 Consistent axiom systems, 37 Constant function, 164 Constructions of functions, 171-179 Constructive proof, 40 Continuum, 206 Contradiction, 7 proof by, 33-34, 36,41,43 Contrapositive, II Contrapositive proof, 33, 36 Convergent sequence, 295 Converse, II Coordinate, lIS Coset, 258-262 Countable set, 205, 209-218 Counterexample, 42 Counting principles, 104-114 Cover of set, 284 Cross (Cartesian) product, 116 Cycle, 155 Cyclic group, 254 Decreasing sequence, 298 Deductions, incorrect, 45-46 Deductive reasoning, I Degree of vertex, 152 DeMoivre's Theorem, 92 De Morgan's Laws, 27, 29, 70, 79 Denial,5-6 Denumerable set, 203-205 Derived set, 291 Descartes, Rene, 116 Difference of sets, 68 Digraph complement of, 135 complete, 131 defined, 121 subdigraph, 131 Directed graph. See Digraph Direct proof, 29-33, 42, 61 Disconnected graph, ISS Disjoint sets, 68 Disjunction, 2, 9 Divergent sequence, 295 Division Algorithm for integers, 103 for natural numbers, 101-102

Di visor of zero, 250 Domain, 118 Edge, lSI, 152 Element-chasing proofs, 65 Elkies, Noam, 40 Empty set, 61 Equality of sets, 62 Equivalence class, 130 Equivalence relation, 128-136 Equivalent forms of induction, 97-104 Equivalent open sentences, 18 Equivalent proposition,S Equivalent propositional forms,S Equivalent quantified sentences, 21 Equivalent sets, 196-197 Equivalents of completeness, 303-305 Equivalents of the Axiom of Choice (Rubin and Rubin), 228n Euclid, 34, 37 Euler, Leonhard, 40-41, 161 Exclusive or, 9 Exhaustion, proof by, 36 Existence theorems, 40 Existential quantifier, 18 Extension of function, 173 Family of sets, 74-84 indexed,76-77 intersection over, 75 nested,83 union over, 75 Fibonacci, Leonardo, 102 Fibonacci numbers, 102, 103 Field complete, 277-278 defined,274 ordered, 274 Finite set, 104, 198-201 Formulas Binet's, 103 wen-formed,3 Four-Color Theorem, 36 Fraenkel, Abraham, 60, 226 Function(s),161-194 argument of, 163 canonical, 165, 254 characteristic, 164 choice, 226 codomain, 162, 163 composition of, 171-173 constant, 164 as constructions, 171-179 defined, 161 extension of, 173 greatest integer, 164 identity, 166 image of element, 163 induced set, 188-194 inverse of, 171, 182-184 one-to-one (injection), 181-183

Index onto (surjection), 179 operation preserving, 236 pre-image under, 163 projection, 166 real-valued, 166 as relations, 161-170 restriction of, 173 step, 164 union of, 174 Fundamental Theorem of Group Homomorphisms, 271 Galois, Evariste, 241-242 Generator, 254 Godel, Kurt, 229 Graph(s), lSI bridge, 158 complement of, 135, 158 complete, 152 component of, 156 connected, ISS defined, 152 directed. See Digraph disconnected, ISS isomorphic, 151-152 multi graph, 152 null, 152 order of, 152 path in, 154 of relations, 119, 151-159 simple, 152 size of, 152 subgraph, 154 walk, 154 Greatest (largest) element, 146 Greatest integer function, 164 Greatest lower bound (infimum), 145,276 Group(s),246-258 abelian, 242 center of, 257 cyclic, 254 defined, 242 isomorphic, 266-272 octic,248 of permutations, 247 quotient, 262-266 subgroups, 254-258 symmetric, 247 Haken, Wolfgang, 36 Handshaking Lemma, 153 Hasse diagram, 144-145, 148 Heine, Edward, 286 Heine-Borel Theorem, 286-287, 290, 303 Homomorphism, 244, 253-254, 271 Horizontal section, 127 Hypothesis of induction, 89 Identity element, 235 Identity function, 166 Identity relation, 120

339

Identity subgroup, 252 Image of element, 163 Image set, 188 Immediate predecessor, 143-144 Improper subset, 62 Incident edge, 152 Inclusion map, 166 Inclusive or (disjunction), 2, 9 Increasing sequence, 298 Independent axiom, 242 Index of set, 76-78 of subgroup, 260 Indexed family, 76-78 Induced set functions, 188-194 Induction, 85-104 Principle of Complete Induction (PCI) (strong), 97-100 equivalent forms of, 97-104 hypothesis of, 89 Principle of Mathematical Induction (PMI), 86-97 proof by, 87-91 Inductive definition, 86 Inducti ve reasoning, I Inducti ve set, 86-97 Infimum, 145,276 Infinite sequence, 165 Infinite set, 198-20 I, 203-209 Initial vertex, 154 Injection, 181-183 Integers, division algorithm for, 103 Interior point, 281 Intersection of two sets, 68 Intersection overfamily, 75 Interval closed, 60 nested, 305 open, 60 Interval notation, 60-61 Introduction to Mathematical Logic, An (Elliot), 3 Introductory Graph Theory (Chartrand), 152 Inverse of conditional sentence, 17 of function, 171, 182-184 multiplicative, 44 of relation, 122 Inverse element, 235 Inverse image of set, 188 Irreflexive relation, 148 Isolated vertex, 152 Isomorphism of graphs, 151-152 of groups, 266-272 Kernel, 253 Lagrange, Joseph-Louis, 259 Lagrange's Theorem, 259, 260-261 Lander, L.J., 40

340

Index Largest (greatest) element, 146 Least (smallest) element, 146 Least upper bound (supremum), 145,276 Left coset, 259 Left identity, 239 Leibniz, G. W., 161 Limit of sequence, 295 Linear combination, 178 Linear order, 146 Loop, 129, 152 Lower bound, 145,275,276 Main diagonal of table, 235 Map canonical, 165,254 inclusion, 166 Mapping, 162 operation preserving, 236 See also Function(s) Mendelson, Elliot, 3 Modus ponens, II, 27 Monoid,236 Monotone sequence, 298 Moschovakis, Y.N., 60n Multigraph, 152 Multiple of element, 244 Multiplicative inverse, unique, 44 Negation, 2, 6 Negative of element, 244 Neighborhood,280 Nested family of sets, 83 Nested Interval Theorem, 305 Normalized form of decimal, 205 Normalizer, 257 Normal subgroup, 263 Notes on Set Theory (Moschovakis), 60n Null graph, 152 Octic group, 248 One-to-one correspondence (bijection), 183, 196 One-to-one function (injection), 181-183 Onto function (surjection), 179 Open interval, 60 Open ray, 60 Open sentence, 18 Open set, 281-282 Operation preserving (OP) mappings, 236 Operation (Cayley) table, 234 Opposite of conditional sentence, 17 Order of algebraic system, 234 of element, 255 of graph, 152 linear, 146 partial, 142 total,146 Ordered field, 274 Ordered n-tuple, 115 Ordered pair, 115

Ordered triple, 115 Ordering of cardinal numbers, 218-225 of relations, 141-150 well, 100-101, 147 Pairwise disjoint, 80 Paradox, 2, 60 Parkin, Thomas, 40 Partial order, 142 Partition, 136-141 Pascal, Blaise, 110-111 Pascal's Triangle, 110-111 Path,154 Peano, Guiseppe, 85 Permutation, 107, 246 Permutation group, 247 Permutation Rule, 108-109 Pigeonhole principle, 200-201 Pointwise product, 177 Pointwise sum, 177 Poset, 142 Postulates, 26 Power set, 63 Predicate, 18 Pre-image under function, 163 Principle of Complete Induction (PCI), 97-100 Principle of Inclusion and Exclusion, 106 Principle of Mathematical Induction (PMI), 86-97 Product Cartesian, 116 pointwise, 177 scalar, 178 Product Rule, 64, 106-107 Projection function, 166 Proof of biconditional sentence, 34-35 by complete induction, 97-100 constructive, 40 by contradiction, 33-34, 36, 41, 43 by contraposition, 33, 36 defined,26 direct, 29-33, 42, 61 element-chasing, 65 by exhaustion, 36 by induction, 87-91 by Well Ordering Principle, 100-101 Proofs to Grade instructions, 39 Proper subgroup, 252 Proper subset, 62 Proposition biconditional,II-12 compound, 2, 3 conditional, 9 defined,2 denial,5-6 equivalent, 5 simple (atomic), 2 Propositional form, 3-4

Index Quantifier existential, 18 unique existential, 24 universal, 18 Quotient group, 262-266 Range, 118 Reachable vertex, 155 Real-valued function, 166 Recursive definition, 86 Reflexive relation, 128 Relation(s) antisymmetric, 141-142 composite of, 122, 124 defined, 117 equivalence, 128-136 functions as, 161-170 graphs of, 119, 151-159 identity, 120 inverse of, 122 irreflexive, 148 ordering, 141-150 reflexive, 128 symmetric, 128 transitive, 128 Replacement, 28 Restriction of function, 173 Right coset, 259 Right identity, 239 Rolle's Theorem, 53-54 Rubin, H., 228n Rubin, J. E., 228n Russell, Bertrand, 60 Russell paradox, 60 Scalar product, 178 SchrOder, Ernest, 221 Sequence, 165,294 bounded,295 Cauchy,273,301 convergent, 295 decreasing, 298 defined, 165,294 divergent, 295 increasing, 298 infinite, 165 limit of, 295 monotone, 298 nth term of, 294 Set(s) binary operation on, 67 bounded,275 c1osed,281-282 closure of, 293 complement of, 70 countable, 205, 209-218 cover of, 284 defined,59 denumerable, 203-205 derived,291

difference of, 68 disjoint, 68 empty, 61 equality of, 62 equivalent, 196-197 family of, 74-84 finite, 104, 198-20 I image, 188 indexing, 76-77 induced set functions, 188-194 inductive, 86-97 infinite, 198-201,203-209 intersection of, 68 inverse image of, 188 open, 281-282 ordinary, 66 partially ordered, 142 power, 63 uncountable, 205, 206-207 union of, 68 Set notation, 59 Set operations, 67-74 Set Theory and Logic (Stoll), 273n Simple graph, 152 Simple proposition, 2 Smallest (least) element, 146 Step function, 164 Stoll, Robert, 273n Strong (complete) induction, 97-100 Subcover, 284 Subdigraph, 131 Subgraph, 154 Subgroup, 254-258 cyclic, 254 defined, 251 identity (trivial), 252 index, 260 normal,263 proper, 252 Subset defined,61 improper, 62 proper, 62 Sum Rule, 104 Supremum, 145,276 Surjection, 179 Symmetric closure, 135 Symmetric group, 247 Symmetric relation, 128 Symmetry, 247 Tautology, 7 Terminal vertex, 154 Theorem defined,26 existence, 40 See also specific theorems Total order, 146 Towers of Hanoi, 95 Transitive closure, 135

341

342

Index Transitive relation, 128 Traverse, 154 Trichotomy property, 226 Trivial (identity) subgroup, 252 Truth set, 18 Unary operation, 70 Uncountable set, 205, 206-207 Undecidable statement, 37 Undefined terms, 26 Union overfamily, 75 of function, 174 of two sets, 68 Unique existential quantifier, 24 Unique multiplicative inverse, 44 Universal quantifier, 18 Universe of discourse, 18 Upper bound, 145,275,276

Vertex, 151 adjacent, 152 defined, 152 degree of, 152 distance between, 155 initial,154 isolated, 152 reachable (accessible), 155 terminal, 154 Vertical section, 127 Walk,154 Weierstrass, Karl, 290 Well-formed formulas, 3 Well ordering, 147 Well Ordering Principle (WOP), 100-101 Zermelo, Ernst, 60, 226 Zermelo-Fraenkel set theory, 60, 226, 229

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