Discrete Mathematics with Applications

Preface A journey of a thousand miles begins with a single step. - - A Chinese proverb p eople often ask: What is disc...

Author: Thomas Koshy

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Preface A journey of a thousand miles begins with a single step. - - A Chinese proverb

p

eople often ask: What is discrete mathematics? It's the mathematics of discrete (distinct and disconnected) objects. In other words, it is the study of discrete objects and relationships that bind them. The geometric representations of discrete objects have gaps in them. For example, integers are discrete objects, therefore (elementary) number theory, for instance, is part of discrete mathematics; so are linear algebra and abstract algebra. On the other hand, calculus deals with sets of connected (without any gaps) objects. The set of real numbers and the set of points on a plane are two such sets; they have continuous pictorial representations. Therefore, calculus does not belong to discrete mathematics, but to continuous mathematics. However, calculus is relevant in the study of discrete mathematics. The sets in discrete mathematics are often finite or countable, whereas those in continuous mathematics are often uncountable. Interestingly, an analogous situation exists in the field of computers. Just as mathematics can be divided into discrete and continuous mathematics, computers can be divided into digital and analog. Digital computers process the discrete objects 0 and 1, whereas analog computers process continuous d a t a ~ t h a t is, data obtained through measurement. Thus the terms discrete and continuous are analogous to the terms digital and analog, respectively. The advent of modern digital computers has increased the need for understanding discrete mathematics. The tools and techniques of discrete mathematics enable us to appreciate the power and beauty of mathematics in designing problem-solving strategies in everyday life, especially in computer science, and to communicate with ease in the language of discrete mathematics.

The Realization of a Dream

This book is the fruit of many years of many dreams; it is the end-product of my fascination for the myriad applications of discrete mathematics to a variety of courses, such as Data Structures, Analysis of Algorithms, Programming Languages, Theory of Compilers, and Databases. Data structures and Discrete Mathematics compliment each other. The information in this book is applicable to quite a few areas in mathematics; discrete xiii

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mathematics is also an excellent preparation for number theory and abstract

algebra. A logically conceived, self-contained, well-organized, and a user-friendly book, it is suitable for students and amateurs as well; so the language employed is, hopefully, fairly simple and accessible. Although the book features a well-balanced mix of conversational and formal writing style, mathematical rigor has not been sacrificed. Also great care has been taken to be attentive to even minute details.

Audience The book has been designed for students in computer science, electrical engineering, and mathematics as a one- or two-semester course in discrete mathematics at the sophomore/junior level. Several earlier versions of the text were class-tested at two different institutions, with positive responses from students.

Prerequisites No formal prerequisites are needed to enjoy the material or to employ its power, except a very strong background in college algebra. A good background in pre-calculus mathematics is desirable, but not essential. Perhaps the most important requirement is a bit of sophisticated mathematical maturity: a combination of patience, logical and analytical thinking, motivation, systematism, decision-making, and the willingness to persevere through failure until success is achieved. Although no programming background is required to enjoy the discrete mathematics, knowledge of a structured programming language, such as Java or C + +, can make the study of discrete mathematics more rewarding.

Coverage The text contains in-depth coverage of all major topics proposed by professional associations for a discrete mathematics course. It emphasizes problem-solving techniques, pattern recognition, conjecturing, induction, applications of varying nature, proof techniques, algorithm development, algorithm correctness, and numeric computations. Recursion, a powerful problem-solving strategy, is used heavily in both mathematics and computer science. Initially, for some students, it can be a bitter-sweet and demanding experience, so the strategy is presented with great care to help amateurs feel at home with this fascinating and frequently used technique for program development. This book also includes discussions on Fibonacci and Lucas numbers, Fermat numbers, and figurate numbers and their geometric representations, all excellent tools for exploring and understanding recursion.

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A sufficient amount of theory is included for those who enjoy the beauty in the development of the subject, and a wealth of applications as well for those who enjoy the power of problem-solving techniques. Hopefully, the student will benefit from the nice balance between theory and applications. Optional sections in the book are identified with an asterisk (.) in the left margin. Most of these sections deal with interesting applications or discussions. They can be omitted without negatively affecting the logical development of the topic. However, students are strongly encouraged to pursue the optional sections to maximize their learning.

Historical Anecdotes and Biographies Biographical sketches of about 60 mathematicians and computer scientists who have played a significant role in the development of the field are threaded into the text. Hopefully, they provide a h u m a n dimension and attach a h u m a n face to major discoveries. A biographical index, keyed to page, appears on the inside of the back cover for easy access.

Examples and Exercises Each section in the book contains a generous selection of carefully tailored examples to clarify and illuminate various concepts and facts. The backbone of the book is the 560 examples worked out in detail for easy understanding. Every section ends with a large collection of carefully prepared and wellgraded exercises (more than 3700 in total), including thought-provoking true-false questions. Some exercises enhance routine computational skills; some reinforce facts, formulas, and techniques; and some require mastery of various proof techniques coupled with algebraic manipulation. Often exercises of the latter category require a mathematically sophisticated mind and hence are meant to challenge the mathematically curious. Most of the exercise sets contain optional exercises, identified by the letter o in the left margin. These are intended for more mathematically sophisticated students. Exercises marked with one asterisk (.) are slightly more advanced than the ones that precede them. Double-starred (**) exercises are more challenging than the single-starred; they require a higher level of mathematical maturity. Exercises identified with the letter c in the left margin require a calculus background; they can be omitted by those with no or minimal calculus. Answers or partial solutions to all odd-numbered exercises are given at the end of the book.

Foundation Theorems are the backbones of mathematics. Consequently, this book contains the various proof techniques, explained and illustrated in detail.

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They provide a strong foundation in problem-solving techniques, algorithmic approach, verification and analysis of algorithms, as well as in every discrete mathematics topic needed to pursue computer science courses such as Data Structures, Analysis of Algorithms, Programming Languages, Theory of Compilers, Databases, and Theory of Computation.

Proofs Most of the concepts, definitions, and theorems in the book are illustrated with appropriate examples. Proofs shed additional light on the topic and enable students to sharpen their problem-solving skills. The various proof techniques appear throughout the text.

Applications Numerous current and relevant applications are woven into the text, taken from computer science, chemistry, genetics, sports, coding theory, banking, casino games, electronics, decision-making, and gambling. They enhance understanding and show the relevance of discrete mathematics to everyday life. A detailed index of applications, keyed to pages, is given at the end of the book. Algorithms Clearly written algorithms are presented throughout the text as problemsolving tools. Some standard algorithms used in computer science are developed in a straightforward fashion; they are analyzed and proved to enhance problem-solving techniques. The computational complexities of a number of standard algorithms are investigated for comparison. Algorithms are written in a simple-to-understand pseudocode that can easily be translated into any programming language. In this pseudocode: 9 Explanatory comments are enclosed within the delimeters (* and *). 9 The body of the algorithm begins with a B e g i n and ends in an E n d ; they serve as the outermost parentheses. 9 Every compound statement begins with a b e g i n and ends in an end; again, they serve as parentheses. In particular, for easy readability, a while (for) loop with a compound statement ends in e n d w h i l e (endfor).

Chapter Summaries Each chapter ends with a summary of important vocabulary, formulas, and properties developed in the chapter. All the terms are keyed to the text pages for easy reference and a quick review.

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Review and Supplementary Exercises Each chapter summary is followed by an extensive set of well-constructed review exercises. Used along with the summary, these provide a comprehensive review of the chapter. Chapter-end supplementary exercises provide additional challenging opportunities for the mathematically sophisticated and curious-minded for further experimentation and exploration. The book contains about 950 review and supplementary exercises.

Computer Assignments Over 150 relevant computer assignments are given at the end of chapters. They provide hands-on experience with concepts and an opportunity to enhance programming skills. A computer algebra system, such as Maple, Derive, or Mathematica, or a programming language of choice can be used.

Exploratory Writing Projects Each chapter contains a set of well-conceived writing projects, for a total of about 600. These expository projects allow students to explore areas not pursued in the book, as well as to enhance research techniques and to practice writing skills. They can lead to original research, and can be assigned as group projects in a real world environment. For convenience, a comprehensive list of references for the writing projects, compiled from various sources, is provided in the S t u d e n t ' s Solutions Manual.

Enrichment Readings Each chapter ends with a list of selected references for further exploration and enrichment. Most expand the themes studied in this book.

Numbering System A concise numbering system is used to label each item, where an item can be an algorithm, figure, example, exercises, section, table, or theorem. Item m . n refers to item n in Chapter "m". For example, Section 3.4 is Section 4 in Chapter 3.

Special Symbols Colored boxes are used to highlight items that may need special attention. The letter o in the left margin of an exercise indicates that it is optional, whereas a c indicates that it requires the knowledge of calculus. Besides, every theorem is easily identifiable, and the end of every proof and example

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is marked with a solid square (l l). An asterisk (.) next to an exercise indicates that it is challenging, whereas a double-star (**) indicates that it is even more challenging. While " - " stands for equality, the closely related symbol "~" means is approximately equal to: 0 C

II

optional exercises requires a knowledge of calculus end of a proof or a solution a challenging exercise a more challenging exercise is equal to is approximately equal to

Abbreviations For the sake of brevity, four useful abbreviations are used throughout the text: LHS, RHS, PMI, and IH: LHS RHS PMI IH

Left-Hand Side Right-Hand Side Principle of Mathematical Induction Inductive Hypothesis

Symbols Index An index of symbols used in the text and the page numbers where they occur can be found inside the front and back covers.

Web Links The World Wide Web can be a useful resource for collecting information about the various topics and algorithms. Web links also provide biographies and discuss the discoveries of major mathematical contributors. Some Web sites for specific topics are listed in the Appendix.

Student's Solutions Manual The Student's Solutions Manual contains detailed solutions of all oddnumbered exercises. It also includes suggestions for studying mathematics, and for preparing to take an math exam. The Manual also contains a comprehensive list of references for the various writing projects and assignments.

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Instructor's Manual

The Instructor's Manual contains detailed solutions to all even-numbered exercises, two sample tests and their keys for each chapter, and two sample final examinations and their keys. Acknowledgments

A number of people, including many students, have played a major role in substantially improving the quality of the manuscript through its development. I am truly grateful to every one of them for their unfailing encouragement, cooperation, and support. To begin with, I am sincerely indebted to the following reviewers for their unblemished enthusiasm and constructive suggestions: Gerald Alexanderson Stephen Brick Neil Calkin Andre Chapuis Luis E. Cuellar H. K. Dai Michael Daven Henry Etlinger Jerrold R. Griggs John Harding Nan Jiang Warren McGovern Tim O'Neil Michael O'Sullivan Stanley Selkow

Santa Clara University University of South Alabama Clemson University Indiana University McNeese State University Oklahoma State University Mt. St. Mary College Rochester Institute of Technology University of South Carolina New Mexico State University University of South Dakota Bowling Green State University University of Notre Dame San Diego State University Worcester Polytechnic Institute

Thanks also go to Henry Etlinger of Rochester Institute of Technology and Jerrold R. Griggs of the University of South Carolina for reading the entire manuscript for accuracy; to Michael Dillencourt of the University of California at Irvine, and Thomas E. Moore of Bridgewater State College for preparing the solutions to the exercises; and to Margarite Roumas for her excellent editorial assistance. My sincere thanks also go to Senior Editor, Barbara Holland, Production Editor, Marcy Barnes-Henrie, Copy Editor, Kristin Landon, and Associate Editor, Thomas Singer for their devotion, cooperation, promptness, and patience, and for their unwavering support for the project. Finally, I must accept responsibility for any errors that may still remain. I would certainly appreciate receiving comments about any unwelcome surprises, alternate or better solutions, and exercises, puzzles, and applications you have enjoyed.

Framingham, Massachusetts September 19, 2003

Thomas Koshy [email protected]

A W o r d to t h e S t u d e n t Tell me a n d I will forget. S h o w me a n d I will remember. Involve me a n d I will u n d e r s t a n d . n

Confucius

The SALT of Life Mathematics is a science; it is an art; it is a precise and concise language; and it is a great problem-solving tool. Thus mathematics is the SALT of life. To learn a language, such as Greek or Russian, first you have to learn its alphabet, grammar, and syntax; you also have to build up a decent vocabulary to speak, read, or write. Each takes a lot of time and practice.

The Language of Mathematics Because mathematics is a concise language with its own symbolism, vocabulary, and properties (or rules), to be successful in mathematics, you must know them well and be able to apply them. For example, it is important to know that there is a difference between perimeter and area, area and volume, factor and multiple, divisor and dividend, hypothesis and hypotenuse, algorithm and logarithm, reminder and remainder, computing and solving, disjunction and destruction, conjunction and construction, and negation and negative. So you must be fluent in the language of mathematics, just like you need to be fluent in any foreign language. So keep speaking the language of mathematics. Although mathematics is itself an unambiguous language, algebra is the language of mathematics. Studying algebra develops confidence, improves logical and critical thinking, and enhances what is called mathematical maturity, all needed for developing and establishing mathematical facts, and for solving problems. This book is written in a clear and concise language that is easy to understand and easy to build on. It presents the essential (discrete) mathematical tools needed to succeed in all undergraduate computer science courses.

Theory and Applications This book features a perfect blend of both theory and applications. Mathematics does not exist without its logically developed theory; in fact, theorems are like the steel beams of mathematics. So study the various xxi

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A Wordto the Student proof techniques, follow the various proofs presented, and try to reproduce them in your own words. Whenever possible, create your own proofs. Try to feel at home with the various methods and proofs. Besides developing a working vocabulary, pay close attention to facts, properties, and formulas, and enjoy the beautiful development of each topic. This book also draws on a vast array of interesting and practical applications to several disciplines, especially to computer science. These applications are spread throughout the book. Enjoy them, and appreciate the power of mathematics that can be applied to a variety of situations, many of which are found in business, industry, and scientific discovery in today's workplace.

Problem-Solving Strategies To master mathematics, you must practice it; that is, you must apply and do mathematics. You must be able to apply previously developed facts to solve problems. For this reason, this book emphasizes problem-solving techniques. You will encounter two types of exercises in the exercise sets: The first type is computational, and the second type is algebraic and theoretical. Being able to do computational exercises does not automatically imply that you are able to do algebraic and theoretical exercises. So do not get discouraged, but keep trying until you succeed. Of course, before you attempt the exercises in any section, you will need to first master the section; know the definitions, symbols, and facts, and redo the examples using your own steps. Since the exercises are graded in ascending order of difficulty, always do them in order; save the solutions and refine them as you become mathematically more sophisticated. The chapter-end review exercises give you a chance to re-visit the chapter. They can be used as a quick review of important concepts.

Recursion Recursion is an extremely powerful problem-solving strategy, used often in mathematics and computer science. Although some students may need a lot of practice to get used to it, once you know how to approach problems recursively, you will certainly appreciate its great power.

Stay Actively Involved Professional basketball players Magic Johnson, Larry Bird, and Michael Jordan didn't become superstars overnight by reading about basketball or by watching others play on television. Besides knowing the rules and the skills needed to play, they underwent countless hours of practice, hard work, a lot of patience and perseverance, willingness to meet failures, and determination to achieve their goal.

A Word to the Student

xxiii

Likewise, you cannot master mathematics by reading about it or by simply watching your professor do it in class; you have to get involved and stay involved by doing it every day, just as skill is acquired in a sport. You can learn mathematics only in small, progressive steps, building on skills you have already mastered. Remember the saying: Rome wasn't built in a day. Keep using the vocabulary and facts you have already studied. They must be fresh in your mind; review them every week.

A Few Suggestions for Learning Mathematics 9 Read a few sections before each class. You might not fully understand the material, but you'll follow it far better when your professor discusses it in class. In addition, you will be able to ask more questions in class and answer more questions. 9 Whenever you study the book, make sure you have a pencil and enough paper to write down definitions, theorems, and proofs, and to do the exercises. 9 Return to review the material taught in class later in the same day. Read actively; do not just read as if it was a novel or a newspaper. Write down the definitions, theorems, and properties in your own words, without looking in your notes or the book. Good note-taking and writing aid retention. Re-write the examples, proofs, and exercises done in class, all in your own words. If you find them too challenging, study them again and try again; continue until you succeed. 9 Always study the relevant section in the text and do the examples there; then do the exercises at the end of the section. Since the exercises are graded in order of difficulty, do them in order. Don't skip steps or write over previous steps; this way you'll progress logically, and you can locate and correct your errors. If you can't solve a problem because it involves a new term, formula, or some property, then re-study the relevant portion of the section and try again. Don't assume that you'll be able to do every problem the first time you try it. Remember, practice is the only way to Success.

Solutions Manual The Student's Solutions Manual contains additional helpful tips for studying mathematics, and preparing for and taking an examination in mathematics. It also gives detailed solutions to all odd-numbered exercises and a comprehensive list of references for the various exploratory writing projects.

A Final Word Mathematics is no more difficult than any other subject. If you have the motivation, and patience to learn and do the work, then you will enjoy

xxiv

A Word to the Student

the beauty and power of discrete mathematics; you will see that discrete mathematics is really fun. Keep in mind that learning mathematics is a step-by-step process. Practice regularly and systematically; review earlier chapters every week, since things must be fresh in your mind to apply and build on them. In this way, you will enjoy the subject, feel confident, and to explore more. The name of the game is practice, so practice, practice, practice. I look forward to hearing from you with your comments and suggestions. In the meantime, enjoy the beauty and power of mathematics. Thomas Koshy

Chapter 1

The L a n g u a g e of Logic Symbolic logic has been disowned by many logicians on the plea that its interest is mathematical and by many mathematicians on the plea that its interest is logical. - - A . N. W H I T E H E A D

L

ogic is the study of the principles and techniques of reasoning. It originated with the ancient Greeks, led by the philosopher Aristotle, who is often called the father of logic. However, it was not until the 17th century that symbols were used in the development of logic. German philosopher and mathematician Gottfried Leibniz introduced symbolism into logic. Nevertheless, no significant contributions in symbolic logic were made until those of George Boole, an English mathematician. At the age of 39, Boole published his outstanding work in symbolic logic, A n Investigation of the L a w s of Thought. Logic plays a central role in the development of every area of learning, especially in mathematics and computer science. Computer scientists, for example, employ logic to develop programming languages and to establish the correctness of programs. Electronics engineers apply logic in the design of computer chips. This chapter presents the fundamentals of logic, its symbols, and rules to help you to think systematically, to express yourself in precise and concise terms, and to make valid arguments. Here are a few interesting problems we shall pursue in this chapter: 9 Consider the following two sentences, both There are more residents in New York City head of any resident. No resident is totally sion: Is it true that at least two residents hairs? (R. M. Smullyan, 1978)

true: than there are hairs on the bald. What is your concluhave the same number of

9 There are two kinds of inhabitants, "knights" and "knaves," on an island. Knights always tell the truth, whereas knaves always lie. Every inhabitant is either a knight or a knave. Tom and Dick are two residents. Tom says, "At least one of us is a knave." What are Tom and Dick?

Chapter I The Language of Logic

A r i s t o t l e (384-322 B.C.), o n e of the greatest philosophers in Western culture, was born in Stagira, a small town in northern Greece. His father was the personal physician of the king of Macedonia. Orphaned young, Aristotle was . ~,; ~" i. ~'' % ~ ,,. "-":. ;'i ' . ' ~ ' raised by a guardian. At the age of 18, Aristotle entered Plato's Academy in Athens. He was the "brightest and most learned student" at the Academy which he left when Plato died in 34 7 B.C. About 342 B.C., the king of Macedonia invited him to supervise the education of his young son, Alexander, who later became Alexander the Great. Aristotle taught him until 336 B.C., when the youth became ruler following the assassination of his father. ...-*:4 ......}-"~:~i:~:.r Around 334 B.C., Aristotle returned to Athens and founded a school called the Lyceum. His philosophy and followers were called peripatetic, a Greek word meaning "walking around," since Aristotle taught his students while walking with them. The Athenians, perhaps resenting his relationship with Alexander the Great, who had conquered them, accused him of impiety soon after the Emperor's death in 323 B.C. Aristotle, knowing the fate of Socrates, who had been condemned to death on a similar charge, fled to Chalcis, so the Athenians would not "sin twice against philosophy." He died there the following year.

W h a t are t h e y if T o m says, " E i t h e r I ' m a k n a v e or Dick is a k n i g h t " ? (R. M. S m u l l y a n , 1978) 9 Are t h e r e positive i n t e g e r s t h a t can be e x p r e s s e d as t h e s u m of t w o d i f f e r e n t cubes in two d i f f e r e n t ways? 9 Does t h e f o r m u l a E ( n ) = n 2 - n + 41 yield a p r i m e n u m b e r for e v e r y positive i n t e g e r n?

A d e c l a r a t i v e s e n t e n c e t h a t is e i t h e r t r u e or false, b u t not both, is a p r o p o s i t i o n (or a s t a t e m e n t ) , w h i c h we will d e n o t e by t h e l o w e r c a s e l e t t e r p, q, r, s, or t. T h e v a r i a b l e s p, q, r, s, or t are b o o l e a n v a r i a b l e s (or l o g i c variables). T h e following s e n t e n c e s are propositions: (1) (2) (3) (4)

S o c r a t e s w a s a G r e e k philosopher. 3+4=5. 1 + 1 = 0 a n d t h e m o o n is m a d e of g r e e n cheese. If i = 2, t h e n roses are red.

T h e following s e n t e n c e s a r e not propositions: 9 Let m e go!

(exclamation)

9 x+3=5

(x is an u n k n o w n . )

1.1

Propositions

B a r o n Gottfried Wilhelm Leibniz (1646-1716), an outstanding German mathematician, philosopher, physicist, diplomat, and linguist, was born into a Lutheran family. The son of a professor of philosophy, he "grew up to be a genius with encylopedic knowledge." He had taught himself Latin, Greek, and philosophy before entering the University of Leipzig at age 15 as a law student. There he read the works of great scientists and philosophers such as Galileo, Francis Bacon, and Rend Descartes. Because of his youth, Leipzig refused to award him the degree of the doctor of laws, so he left his native city forever. During 1663-1666, he attended the universities of Jena and Altdorf, and receiving his doctorate from the latter in 1666, he began legal services for the 7 Elector of Mainz. After the Elector's death, Leibniz pursued scientific studies. In 1672, he built a calculating machine that could multiply and divide and presented it to the Royal Society in London the following year. In late 1675, Leibniz laid the foundations of calculus, an honor he shares with Sir Isaac Newton. He discovered the fundamental theorem of calculus, and invented the popular notations--d/dx for differentiation and f for integration. He also introduced such modern notations as dot for multiplication, the decimal point, the equal sign, and the colon for ratio. From 1676, until his death, Leibniz worked for the Duke of Brunswick at Hanover and his estate after the duke's death in 1680. He played a key role in the founding of the Berlin Academy of Sciences in 1700. Twelve years later, Leibniz was appointed councilor of the Russian Empire and was given the title of baron by Peter the Great. Suffering greatly from gout, Leibniz died in Hanover. He was never married. His works influenced such diverse disciplines as theology, philosophy, mathematics, the natural sciences, history, and technology. , ..........

Lb~,

9 Close the door!

(command)

9 Kennedy was a great president of the United States.

(opinion)

9 What is my line?

(interrogation)

1

Truth Value

The truthfulness or falsity of a proposition is called its t r u t h v a l u e , denoted by T(true) and F(false), respectively. (These values are often denoted by 1 and 0 by computer scientists.) For example, the t r u t h value of statement (1) in Example 1.1 is T and that of statement (2) is F. Consider the sentence, This sentence is false. It is certainly a valid declarative sentence, but is it a proposition? To answer this, assume the sentence is true. But the sentence says it is false. This contradicts our assumption. On the other hand, suppose the sentence is false. This implies the sentence

Chapter I The Language of Logic

George Boole (1815-1864), the son of a cobbler whose main interests were mathematics and the making of optical instruments, was born in Lincoln, England. Beyond attending a local elementary school and briefly a commercial school, Boole was self-taught in mathematics and the classics. When his father's business failed, he started working to support the family. At 16, he began his teaching career, opening a school of his own four years later in Lincoln. In his leisure time, Boole read mathematical journals at the Mechanics Institute. There he grappled with the works of English physicist and mathematician Sir Isaac Newton and French mathematicians Pierre-Simon Laplace and Joseph-Louis Lagrange. In 1839, Boole began contributing original papers on differential equations to The Cambridge Mathematics Journal and on analysis to the Royal Society. In 1844, he was awarded a Royal Medal by the Society for his contributions to analysis; he was elected a fellow of the Society in 1857. Developing novel ideas in logic and symbolic reasoning, he published his first contribution to symbolic logic, The Mathematical Analysis of Logic, in 184 7. His publications played a key role in his appointment as professor of mathematics at Queen's College, Cork, Ireland, in 1849, although he lacked a university education. In 1854, he published his most important work, An Investigation to the Laws of Thought, in which he presented the algebra of logic now known as boolean algebra (see Chapter 12). The next year he married Mary Everest, the niece of Sir George Everest, for whom the mountain is named. In addition to writing about 50 papers, Boole published two textbooks, Treatise on Differential Equations (1859) and Treatise on the Calculus of Finite Differences; both were used as texts in the United Kingdom for many years. A conscientious and devoted teacher, Boole died of pneumonia in Cork. .....

I

1

is true, which again contradicts our assumption. Thus, if we assume t h a t the sentence is true, it is false; and if we assume t h a t it is false, it is true. It is a meaningless and self-contradictory sentence, so it is not a proposition, but a p a r a d o x . The t r u t h value of a proposition may not be known for some reason, b u t t h a t does not prevent it from being a proposition. For example, around 1637, the F r e n c h m a t h e m a t i c a l genius Pierre-Simon de F e r m a t conjectured t h a t the equation x n + yn = z n has no positive integer solutions, where n >_ 3. His conjecture, known as F e r m a t ' s L a s t " T h e o r e m , " was one of the celebrated unsolved problems in n u m b e r theory, until it was proved in 1993 by the English m a t h e m a t i c i a n Andrew J. Wiles (1953-) of Princeton University. Although the t r u t h value of the conjecture eluded m a t h e m a t i c i a n s for over three centuries, it was still a proposition! Here is a n o t h e r example of such a proposition. In 1742 the P r u s s i a n m a t h e m a t i c i a n Christian Goldbach conjectured t h a t every even integer greater t h a n 2 is the sum of two primes, not necessarily distinct. For example, 4 - 2 + 2, 6 - 3 + 3, and 18 = 7 + 11. It has been shown true for every

1.1 Propositions

F e r m a t (1601-1665) was born near Toulouse as the son of a leather merchant. A lawyer by profession, he devoted his leisure time to mathematics. Although he published almost none of his discoveries, he did correspond with contemporary mathematicians. Fermat contributed to several branches of mathematics, but he is best known for his work in number theory. Many of his results appear in margins of his copy of the works of the Greek mathematician Diophantus (250 A.D. ?). He wrote the following about his famous conjecture: "I have discovered a truly wonderful proof, but the margin is too small to contain it."

C h r i s t i a n Goldbach (1690-1764) was born in K6nigsberg, Prussia. He studied medicine and mathematics at the University of K6nigsberg and became professor of mathematics at the Imperial Academy of Sciences in St. Petersburg in 1725. In 1728, he moved to Moscow to tutor Tsarevich Peter H and his cousin Anna of Courland. From 1729 to 1763, he corresponded with Euler on number theory. He returned to the Imperial Academy in 1732, when Peter's successor Anna moved the imperial court to St. Petersburg. In 1742, Goldbach joined the Russian Ministry of Foreign Affairs, and later became privy councilor and established guidelines for the education of royal children. Noted for his conjectures in number theory and work in analysis, Goldbach died in Moscow.

even integer less than 4 • 1014, but no one has been able to prove or disprove his conjecture. Nonetheless, the Goldbach conjecture is a proposition. Propositions (1) and (2) in Example 1.1 are s i m p l e p r o p o s i t i o n s . A compound proposition is formed by combining two or more simple propositions called c o m p o n e n t s . For instance, propositions (3) and (4) in Example 1.1 are compound. The components of proposition (4) are I = 2 and Roses are red. The truth value of a compound proposition depends on the truth values of its components. Compound propositions can be formed in several ways, and they are presented in the rest of this section.

Conjunction The conjunction of two arbitrary propositions p and q, denoted by p A q, is the proposition p a n d q. It is formed by combining the propositions using the word and, called a connective.

Chapter I The Language of Logic Consider the s t a t e m e n t s p: S o c r a t e s w a s a G r e e k p h i l o s o p h e r q: E u c l i d w a s a C h i n e s e m u s i c i a n .

and

Their conjunction is given by p A q: S o c r a t e s w a s a G r e e k p h i l o s o p h e r a n d E u c l i d w a s a Chinese musician.

m

To define the t r u t h value of p A q, where p and q are a r b i t r a r y propositions, we need to consider four possible cases: 9 p is true, q is true. 9 p is true, q is false. 9 p is false, q is true. 9 p is false, q is false. (See the t r e e d i a g r a m in Figure 1.1 and Table 1.1.) If both p and q are true, t h e n p A q is true; i f p is t r u e and q is false, t h e n p A q is false; i f p is fhlse and q is true, t h e n p A q is false; and if both p and q are false, t h e n p A q is also false.

F i g u r e 1.1

Truth value ofp

Truth value ofq T

T F F

T a b l e 1.1

P

q

T T F F

T F T F

P^q

This information can be s u m m a r i z e d in a table. In the third column of Table 1.1, enter the t r u t h value ofp A q c o r r e s p o n d i n g to each pair of t r u t h values of p and q. The resulting table, Table 1.2, is the t r u t h t a b l e for pAq.

1.1 Propositions Table 1.2

p

q

pAq

T r u t h table for p A q

T T F F

T F T F

T F F F

Expressions t h a t yield the value t r u e or false are b o o l e a n e x p r e s s i o n s , and they often occur in both m a t h e m a t i c s and computer science. For instance, 3 < 5 and 5 < 5 are boolean expressions. If-statements and whileloops in computer programs often use such expressions, and their values determine w h e t h e r or not if-statements and while-loops will be executed, as the next example illustrates. Determine w h e t h e r the assignment s t a t e m e n t , sum <-- sum + i + j , * will be executed in the following sequence of statements: i <--3 j <--5 sum ~ 0 i f ( i < 4) and ( j _< 5) t h e n sum ~- sum + i + j

SOLUTION: The assignment s t a t e m e n t will be executed if the t r u t h value of the boolean expression (i < 4) and (]" _< 5) is T. So, let us evaluate it. Since i ~-- 3, i < 4 is true; s i n c e j ~-- 5 , j _< 5 is also true. Therefore, (i < 4) and (j < 5) is t r u e (see row 1 of Table 1.2). Consequently, the given assignment s t a t e m e n t will be executed. I1

Disjunction A second way of combining two propositions p and q is by using the connective or. The resulting proposition p or q is the disjunction o f p and q and is denoted by p v q. Consider the s t a t e m e n t s

p: Harry likes pepperoni pizza for lunch and

q: Harry likes mushroom pizza for lunch. *The s t a t e m e n t x ~- y m e a n s the value of the expression y is assigned to x, w h e r e ~-- is the assignment operator. The general form of an a s s i g n m e n t s t a t e m e n t is variable <--expression.

Chapter I The Language of Logic Their disjunction is given by p v q: Harry likes pepperoni pizza for lunch or Harry likes m u s h r o o m pizza for lunch. This sentence, however, is often written as

m

p v q: Harry likes pepperoni or m u s h r o o m pizza for lunch.

A n interesting observation: In this example, H a r r y could like pepperoni pizza or m u s h r o o m pizza, or both, for lunch. In other words, the connective or is used in the inclusive sense and~or to mean at least one, maybe both. Such a disjunction is an i n c l u s i v e d i s j u n c t i o n . Table 1.3 gives the t r u t h table for an inclusive disjunction.

T a b l e 1.3

p

q

pvq

T r u t h table for p v q

T T F F

T F T F

T T T F

The disjunction of two propositions is true if at least one component is true; it is false only if both components are false. Consider the s t a t e m e n t s r: Bernie will play basketball at 3

P.M.

today

s: Bernie will go to a matinee at 3

P.M.

today.

and

Then r v s: Bernie will play basketball or go to a matinee at 3

P.M.

today,

m

In this example, Bernie cannot play basketball and go to a matinee at the same time, so the word or is used in the exclusive sense to m e a n at least one, but not both. Such a disjunction is an e x c l u s i v e d i s j u n c t i o n . (See Exercise 31.) T h r o u g h o u t our discussion, we will be concerned with only inclusive disjunction, so the word "disjunction" will mean "inclusive disjunction."

Negation The n e g a t i o n of a proposition p is It is not the case that p, denoted by ~p. You may read ~p as the negation o f p or simply not p.

1.1

Propositions

Let p. P a r i s is the capital of F r a n c e a n d q: Apollo is a H i n d u god. T h e n e g a t i o n of p is ~p: It is not the case t h a t P a r i s is the capital of France. This sentence, however, is often w r i t t e n as ~p: P a r i s is not the capital of France. Likewise, the n e g a t i o n of q is ~q: Apollo is not a H i n d u god.

m

If a proposition p is true, t h e n ~ p is false; i f p is false, t h e n ~ p is true. This definition is s u m m a r i z e d in Table 1.4.

Table 1.4

p

~p

T r u t h table for ~ p

T F

F T

E v a l u a t e each boolean expression, w h e r e a = 3, b - 5, and (1) [~(a > b)] A (b < c)

c

- 6.

(2) ~ [ ( a < b) v (b > c)]

SOLUTION: (1) Since a > b is false, ~ ( a > b) is true. Also, b < c is true. Therefore, !~ (a > b)] A (b < c) is true. (See row 1 of Table 1.2.) (2) a _ c is false. So (a _< b) v (b > c) is true. (See row 2 of Table 1.3.) Therefore, ~ [ ( a < b) v (b > c)] is false, m N e x t we p r e s e n t a n o t h e r way of c o n s t r u c t i n g new propositions.

Implication Two propositions p and q can be c o m b i n e d to form s t a t e m e n t s of t h e form: I f p, then q. Such a s t a t e m e n t is an i m p l i c a t i o n , denoted by p --~ q. Since it involves a condition, it is also called a c o n d i t i o n a l s t a t e m e n t . T h e c o m p o n e n t p is the h y p o t h e s i s (or p r e m i s e ) of the implication a n d q t h e

conclusion. Let

p: AABC is e q u i l a t e r a l

and q" AABC is isosceles. Then

p ~ q ' I f A A B C is equilateral, then it is isosceles.

10

Chapter I The Language of Logic

Likewise,

q ~ p: I f A A B C is isosceles, then it is equilateral. (Note: In the implication q conclusion.)

--~ p, q is the hypothesis a n d p is the m

Implications occur in a variety of ways. commonly k n o w n occurrences:

The following are some

9 I f p , t h e n q.

9 I f p , q.

9 p implies q.

9 p only if q.

9 q ifp.

9 p is sufficient for q.

9 q is necessary for p. Accordingly, the implication p ~ q can be read in one of these ways. F o r instance, consider the proposition p --. q: If AABC is equilateral, t h e n it is isosceles. It m e a n s exactly the same as any of the following propositions: 9 If AABC is equilateral, it is isosceles. 9 AABC is equilateral implies it is isosceles. 9 AABC is equilateral only if it is isosceles. 9 AABC is isosceles if it is equilateral. 9 T h a t AABC is equilateral, is a sufficient condition for it to be isosceles. 9 T h a t AABC is isosceles, is a necessary condition for it to be equilateral.

Warning: The s t a t e m e n t p only if q is often m i s u n d e r s t o o d as h a v i n g the same m e a n i n g as the s t a t e m e n t p if q. R e m e m b e r , p if q m e a n s I f q, then p. So be careful. T h i n k of only if as one phrase; do not split it. .

.

.

.

.

.

.

To construct the t r u t h table for an implication I f p, then q, we shall t h i n k of it as a conditional promise. If you do p, t h e n I promise to do q. If the promise is kept, we consider the implication true; if the promise is not kept, we consider it false. We can use this analogy to c o n s t r u c t the t r u t h table, as shown below. Consider the following implication: p --~ q: If you wax my car, t h e n I will pay you $25.

1.1

Propositions

11

I f y o u wax my car (p true) and ifI pay you $25 (q true), t h e n the implication is true. If you wax my car (p true) and ifI do not pay you $25 (q false), t h e n the promise is violated; hence the implication is false. W h a t if you do not wax my car (p false)? T h e n I may give you $25 (being generous!) or not. (So q may be t r u e or false). In either case, my promise has not been tested and hence has not been violated. Consequently, the implication has not been proved false. If it is not false, it m u s t be true. In other words, i f p is false, the implication p ~ q is t r u e by default. ( I f p is false, the implication is said to be v a c u o u s l y t r u e . ) This discussion is s u m m a r i z e d in Table 1.5.

Table 1.5 T r u t h table for p -~ q

P

q

T T F F

T F T F

p'-*q

T F T T

In the ordinary use of implications in the English language, there is a relationship between hypothesis and conclusion, as in the car waxing example. This relation, however, does not necessarily hold for formal implications. For instance, in the implication, I f the p o w e r is on, then 3 + 5 - 8, the conclusion 3 + 5 = 8 is not even related to the hypothesis; however, from a m a t h e m a t i c a l point of view, the implication is true. This is so because the conclusion is true regardless of w h e t h e r or not the power is on. F r o m an implication we can form three new i m p l i c a t i o n s - - i t s converse, inverse, and c o n t r a p o s i t i v e m as defined below.

Converse, Inverse, and Contrapositive The c o n v e r s e of the implication p -~ q is q -~ p (switch the premise and the conclusion in p -~ q). The i n v e r s e o f p ~ q is ~p ~ ~q (negate the premise and the conclusion). The c o n t r a p o s i t i v e of p ~ q is ~q -~ ~p (negate the premise and the conclusion, and then switch them). L e t -P~

q. If AABC is equilateral, then it is isosceles.

Its converse, inverse, and contrapositive are given by: Converse

q ~ p: If AABC is isosceles, t h e n it is equilateral.

In verse

~p ~ ~q: If AABC is not equilateral, t h e n it is not isosceles.

Contrapositive

~q ~ ~p: If AABC is not isosceles, t h e n it is not equilateral.

A word o f caution: Many people mistakenly t h i n k t h a t an implication and its converse m e a n the same thing; they usually say one to m e a n

m

Chapter I The Language of Logic

12

the other. In fact, they need not have the same t r u t h value. You will, however, learn in Example 1.18 t h a t an implication a n d its contrapositive have the same t r u t h value, and so do the converse a n d the inverse. T h u s far, we have p r e s e n t e d four b o o l e a n o p e r a t o r s : A, v, --+, and 4. The first t h r e e enable us to combine two propositions; accordingly, t h e y are b i n a r y o p e r a t o r s . On the o t h e r hand, we need only one proposition to perform negation, so ~ is a u n a r y o p e r a t o r . These operators can be employed to construct more complex s t a t e m e n t s , as the next example d e m o n s t r a t e s . C o n s t r u c t a t r u t h table for (p --+ q) A (q -+ p).

SOLUTION: We construct the proposition (p --+ q) A (q --+ p) step-by-step. F r o m the propositions p and q, we can form p -+ q and q -+ p; t h e n take t h e i r conjunction to yield the given s t a t e m e n t . Thus, the t r u t h table for (p --+ q) A (q -+ p) requires five columns: p, q, p -+ q, q --+ p, and (p --+ q) A (q -+ p) in t h a t order (see Table 1.6). As before, first enter the possible pairs of t r u t h values for p and q in columns 1 and 2. T h e n use the t r u t h tables for implication (Table 1.5) and conjunction (Table 1.2) to complete the r e m a i n i n g columns. The resulting table is displayed in Table 1.6. It follows from the table t h a t (p --+ q) A (q -+ p) is true if both p and q have the s a m e t r u t h values. II

T a b l e 1.6

P

q

T T F F

T F T F

p--,q

T F T T

q--,p

T T F T

(p--,q)^(q--,p)

T F F T

The Island of Knights and Knaves The next two examples* illustrate the power of t r u t h tables in decisionm a k i n g and in arriving at logical conclusions in the midst of seemingly confusing and contradictory s t a t e m e n t s . ~

Faced with engine problems, Ellen Wright made an e m e r g e n c y landing on the beach of the Island of Knights and Knaves. The island is inhabited by two distinct groups of people, k n i g h t s and knaves. Knights always tell the t r u t h and knaves always lie. Ellen decided t h a t her best move was to reach the capital and call for service. *Based on C. Baltus, "A Truth Table on the Island of Truthtellers and Liars," Mathematics Teacher, Vol. 94 (Dec. 2001), pp. 730-732.

1.1

Propositions

13

Walking from the beach, she came to an intersection, where she saw two men, A and B, working nearby. After hearing her story, A told Ellen, "The capital is in the mountains, or the road on the right goes to the capital." B then said, "The capital is in the mountains, and the road on the right goes to the capital." Then A looked up and said, " T h a t man is a liar." S h r u g g i n g his shoulders, B then said, "If the capital is in the mountains, t h e n the road to the right goes to the capital." Ellen then made a table on the back of her guidebook, t h a n k e d the two men, and walked down the road on the left. Did Ellen make the correct decision?

SOLUTION: Let

c: The capital is in the m o u n t a i n s

and r: The road on the right goes to the capital. Now we build a t r u t h table, as Table 1.7 shows. Since B could not give both false and true s t a t e m e n t s (see rows 3 and 4 in columns 4 and 5), the last two rows of the table do not fit the given scenario; so they can be ignored.

T a b l e 1.7

c T T F F

r T F T F

A:cvr T T T F

B:c^r

T F F F

B:c~r

T F T T

It now follows from the rest of column 3 t h a t A is a knight. So his statement that "B is a liar" is true; thus B is a knave. Consequently, we can ignore row 1 also. This leaves us with row 2. Therefore, the s t a t e m e n t r is false; that is, the road on the left goes to the capital. Thus Ellen made the correct decision, m The following example is a continuation of Ellen's saga. ~

Walking u p the road to the left, Ellen encountered a group of people gathered at what she t h o u g h t to be a bus stop. She approached three women, C, D, and E, and asked them w h e t h e r the road went to the capital and w h e t h e r the location was indeed a bus stop. She received three different responses: C: "The road goes to the capital, and the bus stop is not here." D: "The road does not go to the capital, and the bus stop is here." E: "The road does not go to the capital, and the bus stop is not here." Confused and somewhat perplexed, Ellen asked t h e m w h e t h e r they are knights or knaves. To this they all answered, "Two of us are knights, and one is a liar." How m a n y of the three women are knights? Does the road go to the capital? Is the location where Ellen met t h e m a bus stop?

Chapter I The Language of Logic

14

SOLUTION: Once again, we build a t r u t h table. To this end, we let g: T h e road goes to t h e capital and b: T h e bus stop is here. Table 1.8 shows the r e s u l t i n g table, w h e r e only some c o l u m n s are s h o w n for convenience.

T a b l e 1.8

g

b C:g^~b

T T F F

T F T F

F T F F

D:~gAbE:~gA~b F F T F

F F F T

Since all t h r e e w o m e n m a d e the s a m e s t a t e m e n t , "Two of us are k n i g h t s , and one is a liar," t h e y m u s t all be knaves. Consequently, we can discard rows 2-4 in Table 1.8. (It now follows from row i t h a t the t h r e e w o m e n are all knaves.) So the road does indeed go to the capital and the location is a bus stop. I Ellen's story is c o n t i n u e d f u r t h e r in the exercises. See Exercises 76-78. Next we p r e s e n t yet a n o t h e r m e t h o d of combining propositions.

Biconditional Statement Two propositions p and q can be combined using the connective if and only if. The r e s u l t i n g proposition, p if and only if q, is the conjunction of two implications: (1) p only if q, and (2) p if q, t h a t is, p --~ q a n d q --* p. Accordingly, it is called a b i c o n d i t i o n a l s t a t e m e n t , symbolized b y p o q. Let

p: AABC is equilateral

and

q: AABC is equiangular.

T h e n the biconditional s t a t e m e n t is given by p o q: AABC is equilateral if a n d only if it is e q u i a n g u l a r .

I

Since the biconditional p o q m e a n s exactly the s a m e as the s t a t e m e n t (p --, q) A (q --, p), t h e y have t h e s a m e t r u t h value in every case. We can use this fact, and columns 1, 2, and 5 of Table 1.6 to c o n s t r u c t the t r u t h table for p o q, as in Table 1.9. Notice t h a t the s t a t e m e n t p ~ q is t r u e if both p a n d q have t h e s a m e t r u t h value; conversely, i f p ~ q is true, t h e n p a n d q have t h e s a m e t r u t h value.

1.1 T a b l e 1.9 T r u t h table for p ~ q

Propositions

P

q

T T F F

T F T F

15

P~q

T F F T

Here is a simple application of this fact with which you are already familiar (see Section 7.8). Let S denote the s u m of the digits in 2034. If 3 is a factor of S, t h e n 3 is a factor of 2034 also. Conversely, if 3 is a factor of 2034, t h e n 3 is a factor of S also. T h u s the biconditional, 2034 is divisible by 3 if and only if S is divisible by 3, is a t r u e proposition. Consequently, if one c o m p o n e n t m say, S is divisible by 3 m is true, t h e n the o t h e r c o m p o n e n t is also true. m Order of Precedence

To evaluate complex logical expressions, you m u s t k n o w the order of precedence of the logical operators. The order of precedence from the highest to the lowest is: (1) ~ (2) A (3) v (4) --. (5) ~ . Note t h a t parenthesized subexpressions are always evaluated first; if two operators have equal precedence, the c o r r e s p o n d i n g expression is evaluated from left to right. F o r example, the expression (p ~ q) A ~ q ~ ~ p is evaluated as [(p --. q) A (~q)] ~ (~p), and p --. q ~ -~q ~ -~p is evaluated as (p --+ q) ~ [(~q) --. (~p)]. .

.

.

.

.

.

.

.

The next example involves c o n s t r u c t i n g a t r u t h table for a conditional s t a t e m e n t and we shall use it shortly to m a k e a few definitions. C o n s t r u c t a t r u t h table for (p ~ q) ~ (~p v q). SOLUTION: We need columns for p, q, p ~ q, ~p, ~ p v q, and (p ~ q) ~ (~p v q). First, fill in the first two columns with the four pairs of t r u t h values for p and q. T h e n use the t r u t h tables for implication, negation, disjunction, and biconditional to complete the r e m a i n i n g columns. Table 1.10 shows the resulting table. T a b l e 1.10

p

q

p --~ q

~p

~p v q

(p --. q) .-~ (~p v q)

T T F F

T F T F

T F T T

F F T T

T F T T

T T T T always true!

Chapter 1 The Languageof Logic

16

Tautology, Contradiction, and Contingency An interesting observation: It is clear from Table 1.10 that the compound statement (p -~ q) ~ (~p v q) is always true, regardless of the t r u t h values of its components. Such a compound proposition is a t a u t o l o g y ; it is an eternal truth. For example, p v ~p is a tautology. (Verify this.) On the other hand, a compound statement that is always false is a contradiction. For instance, p A ~p is a contradiction (Verify this). A compound proposition that is neither a tautology nor a contradiction is a c o n t i n g e n c y . For example, p v q is a contingency. Next we show that there is a close relationship between symbolic logic and switching networks.

Switching Network (optional) A switching network is an a r r a n g e m e n t of wires and switches connecting two terminals. A switch that permits the flow of current is said to be closed; otherwise, it is open. Likewise, a switching network is c l o s e d if current can flow from one end of the network to the other; otherwise, it is open. Two switches A and B can be connected either in s e r i e s (see Figure 1.2) or in parallel (see Figure 1.3). The switching network in Figure 1.2 is closed if and only if both A and B are closed; accordingly, it is symbolically denoted by A A B. The network in Figure 1.3 is closed if and only if at least one of the two switches is closed; consequently, it is denoted by A v B.

Figure 1.2

|

|

Switches connected in series, A A B.

Figure 1.3 Switches connected in parallel, A v B.

|

! | An electrical network may contain two switches A and A' (A prime) such that if one is closed, then the other is open, and vice versa. (The o p e r a t o r ' corresponds to the logical operator ~.) A switching network, in general, consists of series and parallel connections and hence can be described symbolically using the operators A, v, and ', as the following example illustrates. Find a symbolic representation of the switching network in Figure 1.4. SOLUTION: Switches A and B' are connected in series; the corresponding portion of the circuit is symbolized by A A B'. Switch B is in parallel with A A B'; so we have (A A B') v B. Since A' and C are connected in series, the corresponding portion of the network is described by A' A C. The circuits (A A B') v B and (A' A C) are connected in parallel. Therefore, the given network is

1.1

Propositions

17

|

F i g u r e 1.4

|

| |

9

symbolized by [(A A B') v B] v (A' A C). Since the operation v is associative (see Table 1.13), this expression can be r e w r i t t e n as (A A B') v B v (A' A C). m

Exercises I.I Which of the following are propositions? 1. The e a r t h is flat.

2. T o r o n t o is the capital of Canada.

3. W h a t a beautiful day!

4. Come in.

Find the t r u t h value of each c o m p o u n d s t a t e m e n t . 5. (5 < 8) and (2 + 3 = 4) 7. I f l = 2 ,

6. Paris is in F r a n c e or 2 + 3 = 4.

then3=3.

8. AABC is equilateral if and only if it is equiangular. Negate each proposition. 9. 1 + 1 = 0.

10. The chalkboard is black.

Let x, y, and z be any real n u m b e r s . R e p r e s e n t each sentence symbolically, w h e r e p : x < y, q : y < z, and r: x < z. 11. (x>_y) o r ( y < z )

12. (y>__z) o r ( x > _ z )

13. (x >_y) and [(y < z) or (z > x)]

14. (x < y) or [(y >_ z) and (z > x)]

Evaluate each boolean expression, where a = 2, b = 3, c = 5, and d = 7.

15. [~(a > b)] v [~(c < d)]

16. [~(b < c)] A [~(c < d)]

17. ~ [ ( a > b )

18. ~ { ( a d)]}

Let t be a tautology and p an a r b i t r a r y proposition. Give the t r u t h value of each proposition.

19. ~ p v t

20. " - p A ~ t

21. ~ t A p

22. ~ ( ~ p A ~ t )

C o n s t r u c t a t r u t h table for each proposition.

23. ~ p v ~ q

24. ~ ( ~ p v q )

25. ( p v q ) v ( ~ q )

26. p A ( q A r )

18

Chapter I The Language of Logic

Give t h e t r u t h value of e a c h proposition, u s i n g t h e given i n f o r m a t i o n . 27. p A q, w h e r e q is n o t t r u e .

28. p A q, w h e r e u q is n o t false.

29. p v q, w h e r e u p is false.

30. p v q, w h e r e u p is n o t t r u e .

31. T h e e x c l u s i v e d i s j u n c t i o n of two p r o p o s i t i o n s p a n d q is d e n o t e d by p X O R q. C o n s t r u c t a t r u t h t a b l e for p X O R q. W r i t e e a c h s e n t e n c e in if-then form. 32. An e q u i a n g u l a r t r i a n g l e is isosceles. 33. Lines p e r p e n d i c u l a r to t h e s a m e line a r e parallel. 34.

X 2 ----

16 is n e c e s s a r y for x = 4.

35. x = 1 is sufficient for x 2 = 1. W r i t e t h e converse, inverse, a n d c o n t r a p o s i t i v e of e a c h implication. 36. If t h e c a l c u l a t o r is w o r k i n g , t h e n t h e b a t t e r y is good. 37. If L o n d o n is in F r a n c e , t h e n P a r i s is in E n g l a n d . Let x,y, a n d z be a n y real n u m b e r s . R e p r e s e n t each s e n t e n c e symbolically, w h e r e p" x < y, q" y < z, a n d r" x < z. 39. I f z >__y a n d x < y, t h e n

38. I f x >__y a n d x < z, t h e n

y
z>x.

40. x < z if a n d only i f x < y a n d y < z.

41. x >__y a n d y > z if a n d only i f x > z.

D e t e r m i n e w h e t h e r or not t h e a s s i g n m e n t s t a t e m e n t x ~- x + 1 will be e x e c u t e d in each s e q u e n c e of s t a t e m e n t s , w h e r e i ~ 2, j ~ 3, k ~ 6, a n d x~0. 42. I f ( i < 3) A (j < 4) t h e n

43. I f ( i < j ) v (k > 4) t h e n

x~---x+l

x<----x-1

else

else

y+--y+1

x~x+l

44. While u ( i < j ) do begin

45. While u ( i + j >_ k) do x ~-- x + 1

x<----x+l i~---i+l endwhile Let t be a t a u t o l o g y a n d p an a r b i t r a r y proposition. F i n d t h e t r u t h value of each.

46. (ut) ~ p

47. p ~ t

48. (p v t) ~ t

49. (p v t)---~ (ut)

50. ( p A t ) - - + p

51. p ~

52. t o ( p V t )

53. p o ( p A t )

(pAt)

1.1

Propositions

19

Construct a t r u t h table for each proposition. 54. p -~ (p v q )

55. (p Aq) ~ p

56. ( p A q ) ~

57. (p V q) ~ (p A q)

(pvq)

Determine whether or not each is a tautology. 58. p v ( ~ p) 60. [ ( p ~ q )

A(~q)]~p

59.

[p A (p ~ q)] --+ q

61.

[(pVq) A(~q)]~p

Determine whether or not each is a contradiction. 62. p

A (~p)

63. p ~ ~ p

64. ~ ( p v ~ p )

65. ~ p ~ (p v ~ p )

Indicate the order in which each logical expression is evaluated by properly grouping the operands using parentheses. 66. p v q A r

67. p Aq --* ~ p V ~ q

68. p v q ~ ~ p A ~ q

69. p - - + q ~ ~ p v q

Represent each switching network symbolically. 70.

-@

(~

71.

@ @

@

9 @--@

Draw a switching network with each representation. 72. ( A v B ) A ( A v C )

73. ( A v B ' ) v ( A v B )

74. ( A A B ' ) v ( A ' A B )

75. (AAB) v ( A ' A B ) v ( B ' A C )

76. (Examples 1.11 and 1.12 continued) At the bus stop, Ellen noticed signs for three buses, B1, B2, and B3, and approached another trio of women, F, G, and H. A conversation ensued: Ellen: Where do the buses go? F: At least one of B1 and B2 goes to the capital. G: B1 goes to the capital. H: B2 and B3 go to the capital. F: B3 goes to the beach. G: B2 and B3 go to the beach. H: B1 goes to the beach. Which bus did Ellen take?

20

Chapter I The Language of Logic 77. After reaching the bus terminal at the capital, Ellen saw three personal computers. She asked a y o u n g woman, I, w h e t h e r the computers had I n t e r n e t connections. She replied, " C o m p u t e r 1 is not connected to the Internet. Ask t h a t man, J; he is a knight." When Ellen approached the man, he told her, " C o m p u t e r 2 has an I n t e r n e t connection, but computer 3 does not." A second man, K, who overheard the conversation, t h e n said, "If computer 2 has an I n t e r n e t connection, t h e n so does computer 1. Computer 3 is not connected to the I n t e r n e t . " Which computer had an I n t e r n e t connection? 78. At the bus terminal, Ellen overheard the following conversation between two baseball fans, L and M: L: I like the Yankees. M: You do not like the Yankees. You like the Dodgers. L: We both like the Dodgers. Does fan L like the Yankees? Who likes the Dodgers?

Two compound propositions p and q, although they may look different, can have identical t r u t h values for all possible pairs of t r u t h values of their components. Such s t a t e m e n t s are l o g i c a l l y e q u i v a l e n t , symbolized by p -- q; otherwise, we write p ~ q. I f p -- q, the columns headed by t h e m in a t r u t h table are identical. The next two examples illustrate this definition. Verify t h a t -P- *

q - u p v q.

SOLUTION: Construct a t r u t h table containing columns headed by p --* q and u p v q, as in Table 1.11. Use the t r u t h tables for implication, negation, and disjunction to fill in the last three columns. Since the columns headed by p --, q and u p v q are identical, the two propositions have identical t r u t h values. In other words, p --* q - u p v q.

T a b l e 1.11

p

q

p--,q

~p

,,~pvq

T T F F

T F T F

T F T T

F F T T

T F T T

L identical columns --~

1.2

Logical Equivalences

21

(Note" This example shows t h a t an implication can be expressed in t e r m s I of negation and disjunction.)

~q ~ ~p; t h a t is, an implication is logically equivalent q Show t h a t -P+ to its contrapositive. SOLUTION:

Once again, construct a t r u t h table, with columns headed by p , q , p --+ q, ~q, ~p, and ~q ~ ~p. Use the t r u t h tables for implication and negation to complete the last four columns. The resulting table (see Table 1.12) shows t h a t the columns headed by p --~ q and ~q -~ ~p are identical; therefore, p~

Table

1.12

q-~q

~

~p.

P

q

P --* q

~q

~p

T T F F

T F T F

T F T T

F T F T

F F T T

~--

~q

identical columns

~ ~p

T F T T _3

m

This is an extremely useful, powerful result t h a t plays an i m p o r t a n t role in proving theorems, as we will see in Section 1.5. It follows by this example t h a t q -+ p - u p __~ ~q (Why?); t h a t is, the converse of an implication and its inverse have identical t r u t h values. Using t r u t h tables, the laws of logic in Table 1.13 can be proved. We shall prove one of the De Morgan laws and leave the others as routine exercises. Table

1.13

Laws of Logic

Let p, q, and r be any three propositions. Let t denote a tautology and f a contradiction. Then: Idempotent 1. p A p =_p

laws 2. p V p - - p

Identity laws 3. p A t = p

4. p v f - - p

Inverse laws 5. p A ( ~ p ) - - f

6. p V ( ~ p ) - - t

Domination 7. p v t = t

laws 8. p A f - - f Continued

Chapter I The Language of Logic

22

T a b l e 1.13 (continued)

C o m m u t a t i v e laws 10. p v q = q v p

9. p A q - - q A p

Double negation 11. ~ ( ~ p ) - p A s s o c i a t i v e laws 12. p A ( q A r ) - - = ( p A q ) A r

13. p V ( q V r ) = ( p V q ) V r

D i s t r i b u t i v e laws 14. p A (q v r) -- (p A q) v (p A r)

15. p v ( q A r ) = ( p v q ) A ( p v r )

De Morgan's laws 16. ~ ( p A q ) _ = ~ p v ~ q

17. ~ ( p v q ) - ~ p A ~ q

I m p l i c a t i o n c o n v e r s i o n law 18. p - - , q - = ~ p v q C o n t r a p o s i t i v e law 19. p ~ q = _ ~ q ~ p R e d u c t i o ad absurdum law 20. p - - - , q - ( p A ~ q ) ~ f

We can make a few observations about some of the laws. The comm u t a t i v e laws imply t h a t the order in which we take the conjunction (or disjunction) of two propositions does not affect their t r u t h values. The associative laws say t h a t the way we group the c o m p o n e n t s in a conjunction (or disjunction) of t h r e e or more propositions does not alter the t r u t h value of the resulting proposition; accordingly, p a r e n t h e s e s are not needed to indicate the grouping. In o t h e r words, the expressions p A q A r and p v q v r are no longer ambiguous, b u t do make sense. Nonetheless, p a r e n t h e s e s are essential to indicate the g r o u p i n g s in the distributive laws. For instance, if we delete the p a r e n t h e s e s in law 14, t h e n its left-hand side (LHS) becomes p A q v r = (p A q) v r, since A has higher priority t h a n v. But (p A q) v r ~ p A (q v r). (You m a y verify this.) We now verify De M o r g a n ' s law 16 in the following example. ~

Verify t h a t ~ ( p A q) - ~ p v '~q, SOLUTION: C o n s t r u c t a t r u t h table with columns headed b y p , q , p Aq, ~ ( p Aq), ~p, ~q, and ~ p v ~q. Since c o l u m n s 4 and 7 in Table 1.14 are identical, it follows t h a t ~(p A q) - ~ p v ~q. m

1.2

23

Logical Equivalences

A u g u s t u s De M o r g a n (1806-1871) was born in Madurai, Tamil Nadu, India, where his father was a colonel in the Indian army. When the young De Morgan was 7 months old, the family moved to England. He attended private schools, where he mastered Latin, Greek, and Hebrew and developed a strong interest in mathematics. After graduating in 1827 from Trinity College, Cambridge, he pondered a career either in medicine or law, but pursued mathematics. His professional career began in 1828 at University College, London. Three years later, however, when the college dismissed a colleague in anatomy without explanation, De Morgan resigned on principle. He returned to Trinity in 1836 when his successor died and remained there until a second resignation in 1866. A fellow of the Astronomical Society and a founder of the London Mathematical Society, De Morgan greatly influenced the development of mathematics in the 19th century. He exuded his passion for the subject in his teaching, stressing principles over techniques. An incredibly prolific writer, De Morgan authored more than 1000 articles in more than 15journals, as well as a number of textbooks, all characterized by clarity, logical presentation, and meticulous detail. De Morgan's original contributions to mathematics were mainly in analysis and logic. In 1838, he coined the term mathematical induction and gave a clear justification to this proof method, although it had been in use. His The Differential and Integral Calculus (1842) gives the first precise definition of a limit and some tests for convergence of infinite series. De Morgan was also interested in the history of mathematics. He wrote biographies of Sir Isaac Newton and Edmund Halley. His wife wrote De Morgan's biography in 1882. His researches into all branches of knowledge and his prolific writing left him little time for social or family life, but he was well-known for his sense of humor.

Table

1.14

p

q

p ^ q

~ ( p ^ q)

,~p

~q

T T F F

T F T F

T F F F

F T T T

F F T T

F T F T

L

~p

identical columns

v ,-~q

T F T T __2

De Morgan's laws, although important, can be confusing, so be careful when you negate a conjunction or a disjunction. The negation of a conjunction (or disjunction) of two s t a t e m e n t s is the disjunction (or conjunction) of their negations. The next two examples illustrate De Morgan's laws. ~

Let

p" Peter likes plain yogurt

and q: Peter likes flavored yogurt.

24

Chapter I

The Language of Logic

Then p A q: P e t e r likes plain y o g u r t and flavored y o g u r t . p v q: P e t e r likes plain y o g u r t or flavored yogurt. By De M o r g a n ' s laws, ~ ( p A q) -

~p

v

~q" P e t e r does not like plain y o g u r t or does not like flavored y o g u r t

and ~ ( p v q) -

~ p A ~q" P e t e r likes n e i t h e r plain y o g u r t n o r flavored yogurt.

m

De M o r g a n ' s laws can be used in reverse order also; t h a t is, ~ p v ~ q --(p A q) and ~ p A ~ q - ~ ( p v q). For instance, the sentence, Claire does not like a sandwich or does not like pizza for lunch can be r e w r i t t e n as It is false that Claire likes a sandwich and pizza for lunch. Likewise, the sentence, The earth is not fiat a n d not round can be r e s t a t e d as It is false

that the earth is fiat or round. D e t e r m i n e w h e t h e r or not the s t a t e m e n t x ~-- x + 1 will be executed in the following sequence of s t a t e m e n t s : O ~-- 7; b ~-- 4 i f ~ [ ( a < b) v x~--x+l

(b >__ 5)]

then

SOLUTION: The s t a t e m e n t x ~-- x + 1 will be executed if the value of the boolean expression ~ l ( a < b) v (b >__5)1 is true. By De M o r g a n ' s law, ~ l ( a < b) v (b >__5)1 - ~ ( a < b) A ~(b >__5) - (a > b) A (b < 5) Sincea-7andb-4, botha>_bandb < 5 are true; so, (a >__b ) A(b < 5) is true. Therefore, the a s s i g n m e n t s t a t e m e n t will be executed, m One of the elegant applications of the laws of logic is employing t h e m to simplify complex boolean expressions, as the next example illustrates. Using the laws of logic simplify the boolean expression (p A ~q) v q v (~p Aq). SOLUTION" [The justification for every step is given on its r i g h t - h a n d - s i d e (RHS). ] (p A ~q) v q v (~p A q) - [(p A ~q) v q] v (--p A q) --[qV(pA~q)]V(~pAq)

assoc, law comm. law

25

1.2 Logical Equivalences - [(q v p) A (q v ~q)] v (~p A q)

dist. law

-- [(q V p) A t] V (~p A q)

qv~q-t

= (q V p) V (~p A q)

rAt--r

= (~p

comm. law

A

q)

v

(p

v

q)

-- [~p v (p v q)]

A

[q v (p v q)]

dist. law

= [(~p v p) v q]

A

[q v (p v q)]

assoc, law

-_- (t

v

q)

A

[q

v

(p

v

q)]

~p vp-t

=- t A [q V (p Vq)]

tvq-t

-- q V (p V q)

tAr----r

=_qV(qVp)

comm. law

=(qVq)

assoc, law

Vp

=_qvp

idem. law

--pVq

comm. law

For any propositions p, q, and r, it can be shown t h a t p --~ (q v r) (p A~q) --~ r (see Exercise 12). We shall employ this result in Section 1.5. Here are two e l e m e n t a r y but elegant applications of this equivalence. Suppose a and b are any two real n u m b e r s , and we would like to prove the following theorem: I f a . b = O, then either a = 0 or b = 0. By virtue of the above logical equivalence, we need only prove the following proposition: I f a . b = 0 a n d a V: O, t h e n b = 0 (see Exercise 43 in Section 1.5). Second, suppose a and b are two a r b i t r a r y positive integers, and p a prime number. Suppose we would like to prove the following fact: Ifp[ab,* then either p la or p lb. Using the above equivalence, it suffices to prove the following equivalent s t a t e m e n t : I f plab a n d p Xa, t h e n p]b (see Exercise 37 in Section 4.2). We shall now show how useful symbolic logic is in the design of switching networks.

Equivalent Switching Networks (optional) Two switching n e t w o r k s A and B are equivalent if they have the same electrical behavior, either b o t h open or both closed, symbolically described by A - B. One of the i m p o r t a n t applications of symbolic logic is to replace an electrical network, w h e n e v e r possible, by an equivalent simpler network to minimize cost, as illustrated in the following example. To this end,

*xly m e a n s

"x is a f a c t o r o f y . "

Chapter 1 The Language of Logic

26

let A be any circuit, T a closed circuit, and F an open circuit. T h e n A A T -A, A A A' - F, A v T -= T, and A v A' -- T (see laws 3 t h r o u g h 8). Likewise, laws 1 t h r o u g h 11 can also be extended to circuits in an obvious way. ~

Replace the switching n e t w o r k in Figure 1.5 by an equivalent simpler network.

|

F i g u r e 1.5

@ 9

SOLUTION: The given network is r e p r e s e n t e d by (A A B') v [(A A B) v C]. Let us simplify this expression using the laws of logic. (The reason for each step is given on its RHS.) (A A B') v I(A A B) v CI -= I(A A B') v (A A B)I v C

assoc, law

- [A A (B' v B)I v C

dist. law

-- (A A T) v C

B'vB-T

_=AvC

AAT=A

Consequently, the given circuit can be replaced by the simpler circuit in Figure 1.6.

|

F i g u r e 1.6

9 We close this section with a brief introduction to fuzzy logic.

Fuzzy Logic (optional) "The binary logic of m o d e r n computers," wrote Bart Kosko and Satoru Isaka, two pioneers in the development of fuzzy logic systems, "often falls short when describing the vagueness of the real world. Fuzzy logic offers more graceful alternatives." Fuzzy logic, a b r a n c h of artificial intelligence, incorporates the vagueness or value j u d g e m e n t s t h a t exist in everyday life, such as "young," "smart," "hot," and "cold." The first company to use a fuzzy system was F. L. Smidth and Co., a cont r a c t i n g company in Copenhagen, Denmark, which in 1980 used it to r u n a

27

1.2 LogicalEquivalences

B a r t Kosko holds degrees in philosophy and economics from the Universityof Southern California, an

M.S. in applied mathematics, and a Ph.D. in electrical engineering from the University of California, Irvine. Currently, he is on the faculty in electrical engineering at the University of Southern California. S a t o r u I s a k a received his M.S. and Ph.D. in systems science from the University of California, San Diego. He specializes in fuzzy information processing at Omron Advanced Systems at Santa Clara, and in the application of machine learning and adaptive control systems to biomedical systems and factory automation.

cement kiln. Eight years later, Hitachi used a fuzzy system to r u n the subway system in Sendai, Japan. Since then Japanese and American companies have employed fuzzy logic to control hundreds of household appliances, such as microwave ovens and washing machines, and electronic devices, such as cameras and camcorders. (See Figure 1.7.) It is generally believed t h a t fuzzy, common-sense models are far more useful and accurate t h a n standard mathematical ones.

F i g u r e 1.7

JuST TeEw~y (;INA LIKESIT

JUST THE WHY TEl) LIKES ir dUSTTHEWAY THE FEDEX 6UT LIKES IT

In fuzzy logic, the t r u t h value t(p) of a proposition p varies from 0 to 1, depending on the degree of its truth; so 0 _< t(p) _< 1. For example, the s t a t e m e n t "The room is cool" may be assigned a t r u t h value of 0.4; and the s t a t e m e n t "Sarah is smart" may be assigned a t r u t h value of 0.7.

(~hapter 1 The Language of Logic

28

Let 0 < x , y _< 1. T h e n t h e o p e r a t i o n s A, V, a n d ' a r e defined as follows" x A y -- min{x,y} x v y -- max{x,y} !

x -1-x w h e r e min{x,y} d e n o t e s t h e m i n i m u m o f x a n d y, a n d max{x,y} d e n o t e s t h e m a x i m u m of x a n d y. N o t all p r o p e r t i e s in p r o p o s i t i o n a l logic are valid in fuzzy logic. F o r i n s t a n c e , t h e l a w o f e x c l u d e d m i d d l e , p v ~ p is true, does n o t hold in fuzzy logic. To see this, let p be a simple p r o p o s i t i o n w i t h t(p) = 0.3. T h e n t(p') = 1 - 0 . 3 = 0. 7; so t ( p v p ' ) = t ( p ) v t ( p ' ) = 0 . 3 v 0 . 7 = m a x { 0 . 3 , 0.7} = 0 . 7 # 1. T h u s p v p ' is not a t a u t o l o g y in fuzzy logic, l In p r o p o s i t i o n a l logic, t(p v p') = 1; so p v p' is a t a u t o l o g y . T h i n k of 1 r e p r e s e n t i n g a T a n d 0 r e p r e s e n t i n g an F. I Likewise, t(p A p') = t(p) A t(p') = 0.3 A 0. 7 = min{0.3, 0.7} = 0 . 3 # 0; so p A p ' is not a c o n t r a d i c t i o n , u n l i k e in p r o p o s i t i o n a l logic. N e x t we p r e s e n t briefly an i n t e r e s t i n g application* of fuzzy logic to decision m a k i n g . It is based on t h e Y a g e r m e t h o d , developed in 1981 by R o n a l d R. Yager of Iona College, a n d e m p l o y s fuzzy i n t e r s e c t i o n a n d i m p l i c a t i o n -~, defined by p --~ q = u p v q.

Fuzzy Decisions S u p p o s e t h a t from a m o n g five U.S. c i t i e s - - B o s t o n , Cleveland, M i a m i , N e w York, a n d San D i e g o - - w e would like to select t h e b e s t city to live in. We will use seven c a t e g o r i e s C I t h r o u g h C7 to m a k e t h e decision; t h e y are climate, cost o f h o u s i n g , cost o f living, o u t d o o r activities, e m p l o y m e n t , crime, a n d culture, respectively, a n d are j u d g e d on a scale 0 - 6 : 0 = terrible, 1 = bad, 2 = pool', 3 = average, 4 = fairly good, 5 = very good, a n d 6 = excellent. T a b l e 1.15 shows t h e relative i m p o r t a n c e of each c r i t e r i o n on a scale 0 - 6 a n d t h e r a t i n g for e a c h city in each category. T a b l e 1.15

Category

Importance

Boston

Cleveland

Miami

New York

San Diego

C1 C2 C3 C4 C5 C6 C7

6 3 2 4 4 5 4

3 1 3 5 4 2 6

2 5 4 3 3 4 3

5 4 3 6 3 0 3

1 0 1 2 4 1 6

6 1 5 6 3 3 5

*Based on M. Caudill, "Using Neural Nets: Fuzzy Decisions," A/Expert, Vol. 5 (April 1990), pp. 59-64.

1.2 Logical Equivalences

29

The ideal city to live in will score high in the categories considered m o s t i m p o r t a n t . In order to choose the finest city, we need to evaluate each city by each criterion, weighing the relative i m p o r t a n c e of each category. Thus, given a p a r t i c u l a r category's i m p o r t a n c e , we m u s t check the city's score in t h a t category; in o t h e r words, we m u s t c o m p u t e the t r u t h value of i --+ s - ~ i v s for each city, w h e r e i denotes t h e i m p o r t a n c e r a n k i n g for a p a r t i c u l a r category and s the city's score for t h a t category. Table 1.16 shows the r e s u l t i n g data. Now we take the conjunction of all scores for each city, using the m i n function (see Table 1.16). The lowest combined score d e t e r m i n e s t h e city's overall ranking. It follows from the table t h a t San Diego is clearly t h e winner.

T a b l e 1.16

Category

~i

C1 C2 C3 C4 C5 C6 C7

0 3 4 2 2 1 2

Intersection

N e w York

San D i e g o

s

Boston ~ivs

s

Cleveland ,~ivs

s

,.~ivs

s

s

~ivs

3 1 3 5 4 2 6

3 3 4 5 4 2 6

2 5 4 3 3 4 3

2 5 4 3 3 4 3

5 4 3 6 3 0 3

5 4 4 6 3 1 3

1 0 1 2 4 1 6

6 1 5 6 3 3 5

6 3 5 6 3 3 5

2

Miami

2

1

,~ivs

1 3 4 2 4 1 6 1

next best choices

3 winner

Finally, suppose we add a sixth city, say, Atlanta, for consideration. T h e n the Yager m e t h o d e n s u r e s t h a t the revised choice will be the existing choice (San Diego) or Atlanta; it c a n ' t be any of the others. T h u s the p r o c e d u r e allows i n c r e m e n t a l decision making, so m a n a g e a b l e subdecisions can be combined into an overall final choice.

Exercises 1.2 Give the t r u t h value o f p in each case. 1. p -- q, and q is not true.

2. p - q, q - r, and r is true.

Verify each, w h e r e f denotes a contradiction. (See Table 1.14.)

3. ~ ( ~ p ) =-p

4. p Ap ------p

5. p Vp = p

6. p A q = - - q A p

7. p V q = q V p

8. " ~ ( p V q ) = - - ~ p A ' - q

9. ~ ( p --> q) - p

A ~q

30

Chapter I

The Language of Logic

10. p ~ q = ( p A ~ q ) - - ~ f

11. p A ( q A r ) = - - ( p A q ) A r

12. p ~ ( q V r ) - - ( p A ' ~ q ) ~ r

13.

(p v q) ~

r -- (p --~ r) A (q ~

r)

U s e De M o r g a n ' s laws to e v a l u a t e e a c h b o o l e a n e x p r e s s i o n , w h e r e x = 2, y = 5, a n d z = 3. 14. ~ [ ( x < z ) 16.

A(y
"~[(X _> y) V (3' < Z)]

15.

~[(y<X)

A(Z<X)]

17.

~[(X < Z) V (z < y)]

D e t e r m i n e w h e t h e r t h e a s s i g n m e n t s t a t e m e n t c ~- c + 1 will be e x e c u t e d by t h e if-statement or while-loop, w h e r e x ~- 5, y ~-- 3, a n d z ~- 7. 18. I f ~ l ( x < y) A (y _< z)] t h e n c+--c+l 20.

If~l(x >__ y) v (x c~-c+l

<

z)! t h e n

z) A (x _< Y)! t h e n

19. If ~ [ ( x = c~c+l 21.

If ~ [ ( x _< z) v (y = c~c+l

z)i t h e n

2 2 . W h i l e "~[(x _< z) v (x < 3)! do c~-c+l

23. While ~l(x c<--c+l

> 6) A (y = 4)! do

2 4 . W h i l e ~ l ( x = y) v (y = z)l do c~--c+l

25. While ~l(x c~-c+l

= 6) v (y = z)l do

T h e logical o p e r a t o r s N A N D (not and) a n d N O R (not or) a r e defined as follows: p N A N D q _= ~ ( p A q) p N O R q = ~ ( p v q) C o n s t r u c t a t r u t h table for each proposition. 26. p N A N D q

27. p N O R q

M a r k each s e n t e n c e as t r u e or false, w h e r e p , q , statements, t a tautology, and f a contradiction. 28. p A q = _ q A p

29. p v q - q w p

30. p A t = p

32. pv--p=t

33. pA~p--=f

34.

35.

38.

39. I f p = q a n d q = r , If p v q - _ - p v r ,

31. p v f = _ p

~(pAq)=upA~q

36. p ~ q = - - q ~ p

~(pvq)~pv~q

37. p = p

41.

and r are arbitrary

thenp=r.

Ifp=_q, thenq=p.

40. I f p A q = p A r ,

thenq=r.

thenq-r.

Use De M o r g a n ' s laws to verify each. (Hint: p ~ q = ~ p v q). 42.

~ ( ~ p A ~ q ) -- p V q

43.

~ ( ~ p V q) =_ p A ~ q

45.

~ ( p A ~ q ) -- ~ p V q

46. ~ ( p ~ q) - - p A ~ q

44.

~(~p v ~q) - p A q

47. p --~ ~ q -- ~ ( p A q )

1.2 Logical Equivalences

31

48. Show t h a t the connectives A, -+, and ~ can be expressed in t e r m s of v and ~. (Hint: Use Exercise 44, law 18, and Tables 1.6 and 1.7.) Simplify each boolean expression. 49. p A ( p A q )

50. p v ( p v q )

51. p V ( ~ p A q )

52. ( p A ~ q ) v ( p A q ) V r *54. (p A ~q) V (~p A q) V (~p A ~ q )

*53. p A (p V ~q) A (~p V ~q)

C o n s t r u c t an equivalent, simpler n e t w o r k for each switching network.

55.

@

@

9 56.

|

@

|

@

The S h e f f e r s t r o k e I is a binary operator** defined by the following t r u t h table. P

q

Plq

T T F F

T F T F

F T T T

(Note: On page 25 we used the vertical bar ] to m e a n is a f a c t o r of. The actual m e a n i n g should be clear from the context. So be careful.) Verify each. (Note: Exercise 58 shows t h a t the logical operators I and NAND are the same.)

57. ~ p = pip

58. plq -- ~ ( P A q)

59. p A q - (Plq)l(Plq)

60. p v q =_ (PLP)l(q]q)

61. p ~ q =_ ((P[P)I(P[P))[(q[q)

62. ~ ( p v q) - ((P[P)l(qlq))l((P[P)l(qlq))

*63. Express p XOR q in t e r m s of the Sheffer stroke. (Hint: p XOR q - [(p v q) A ~ ( p A q)]. ) **The Sheffer stroke, named after the American logician Henry M. Sheffer (1883-1964), was devised by the American logician Charles S. Peirce (1839-1914).

32

Chapter I The Language of Logic

*64. Express p o q in terms of the Sheffer stroke. (Hint: p ~ q (p ~ q) A (q --~ p).) [Note: Exercises 57-64 indicate t h a t all boolean operators can be expressed in t e r m s of the Sheffer stroke!] Exercises 6 5 - 7 8 deal w i t h p r o p o s i t i o n s in f u z z y logic.

Let p, q, and r be simple propositions with t(p) = 1, t(q) - 0.3, and t(r) = 0.5. Compute the t r u t h value of each, where s' denotes the negation of the s t a t e m e n t s. 65. (p')'

66. p A q

67. p v r

68. q v q'

69. q A q'

70. p' v q

71. (p A q)'

72. p' v q'

73. (p v q)'

74. p A q'

75. q V r'

76. ( p V q ) A ( p ' V q )

Let p be a simple proposition with t(p) = x and p' its negation. Find each. 77. t(p v p')

78. t(p A p')

We now investigate a class of propositions different from those presented in the preceding sections. Take a good look at the following propositions: 9 All people are mortal. 9 Every computer is a 16-bit machine. 9 No birds are black. 9 Some people have blue eyes. 9 There exists an even prime number. Each contains a word indicating quantity such as all, every, none, some, and one. Such words, called q u a n t i f i e r s , give us an idea about how m a n y objects have a certain property. There are two different quantifiers. The first is all, the u n i v e r s a l q u a n tifier, denoted by V, an inverted A. You may read V as for all, for each, or for every. The second quantifier is some, the e x i s t e n t i a l q u a n t i f i e r , denoted by 3, a backward E. You may read 3 for some, there exists a, or for at least one. Note t h a t the word some means at least one. The next two examples d e m o n s t r a t e how to write quantified propositions symbolically. Let x be any apple. T h e n the sentence A l l apples are green can be written as For every x, x is green. Using the universal quantifier V, this sentence can be represented symbolically as (Vx)(x is green) or (Vx)P(x) where P(x) :x is green. (Note: x is j u s t a d u m m y variable.) m

1.3

Quantifiers

33

Predicate

Here P(x), called a p r e d i c a t e , states the property the object x has. Since P(x) involves just one variable, it is a u n a r y predicate. The set of all values x can have is called the u n i v e r s e o f d i s c o u r s e (UD). In the above example, the UD is the set of all apples. Note t h a t P(x) is not a proposition, but just an expression. However, it can be transformed into a proposition by assigning values to x. The t r u t h value of P(x) is predicated on the values assigned to x from the UD. The variable x in the predicate P(x) is a f r e e variable. As x varies over the UD, the t r u t h value of P(x) can vary. On the other hand, the variable x in (Vx)P(x) is a b o u n d variable, bound by the quantifier V. The proposition (u has a fixed t r u t h value. Rewrite the sentence Some chalkboards are black, symbolically. SOLUTION: Choose the set of all chalkboards as the UD. Let x be an a r b i t r a r y chalkboard. Then the given sentence can be written as: There exists an x such t h a t x is black. Using the existential quantifier, this can be symbolized as (3x)b(x), where b(x): x is black, m The next example illustrates how to find the t r u t h values of quantified propositions. The a b s o l u t e v a l u e of a real n u m b e r x, denoted by Ixl, is defined by

I x l - {x --X

if x > 0 ifx < 0

Determine the t r u t h value of each proposition, where the UD = set of all real numbers: (1) ( V x ) ( x 2 > 0)

(2) (Vx) (ixl > 0)

SOLUTION: (1) Since the square of every real n u m b e r is nonnegative, the t r u t h value of (Vx) (X2 > 0) is T. (2) It is not true that the absolute value of every n u m b e r is positive, since 10i - 0, not greater t h a n zero. So the t r u t h value of (Vx) (Ixl > 0) is F. m A predicate may contain two or more variables. A predicate t h a t contains two variables is a b i n a r y predicate. For instance, P(x,y) is a binary predicate. If a predicate contains n variables, it is an n - a r y predicate. The next two examples involve binary predicates.

34

Chapter I The Language of Logic Rewrite each proposition symbolically, w h e r e UD - set of real n u m b e r s . (1) (2) (3) (4)

For each i n t e g e r x, t h e r e exists an i n t e g e r y such t h a t x + y - 0. T h e r e exists an i n t e g e r x such t h a t x + y = y for every i n t e g e r y. For all integers x a n d y, x .y = y . x. T h e r e are i n t e g e r s x a n d y such t h a t x + y = 5.

SOLUTION: (1) (2) (3) (4)

(u (3x)(Vy)(x (Vx)(Vy)(x (3x)(3y)(x

+ y - 0)), which is usually w r i t t e n as (Vx)(3y)(x + y - 0). + y = y) . y = y . x) + y = 5) m

The order of the variables x a n d y in (Vx)(Vy) a n d (3x)(3y) can be c h a n g e d w i t h o u t affecting the t r u t h values of t h e propositions. For instance, (Vx)(Vy)(xy = yx) ==_ (Vy)(Vx)(xy = yx). Nonetheless, t h e order is import a n t in (Vx)(3y) and (3y)(Vx). For example, let P(x,y): x < y w h e r e x and y are integers. T h e n (Vx)(3y)P(x,y) m e a n s For every integer x, there is a suitable i n t e g e r y such t h a t x < y; y = x + 1 is such an integer. Therefore, (Vx)(3y)P(x,y) is true. But (3y)(Vx)P(x,y) m e a n s There exists an integer y, say, b, such that (Vx)P(x,b); t h a t is, every integer x is less than b. Clearly, it is false. Moral? T h e proposition (Qlx)(Q2y)P(x,y) is e v a l u a t e d as (Qlx)[ (Q2y)P(x,y)], w h e r e Q1 and Q2 are quantifiers. A graphical approach can be helpful in finding the t r u t h values of propositions in the form (Qlx)(Q2y)P(x,y), w h e r e x and y are real n u m b e r s , as the next example illustrates. (optional*) D e t e r m i n e t h e t r u t h value of each proposition, w h e r e P(x,y)" y < x 2, and x and y are real n u m b e r s . (1) (Vx)(Vy)P(x,y) (4) (Vy)(3x)P(x,y)

(2) (Sx)(3y)P(x,y) (5) (3x)(Vy)P(x,y)

(3) (Vx)(3y)P(x,y) (6) (3y)(Vx)P(x,y)

SOLUTION: The g r a p h of the e q u a t i o n y = x 2 is a parabola, as s h o w n in F i g u r e 1.8. The p a r a b o l a is shown as a b r o k e n g r a p h since no points on it satisfy the i n e q u a l i t y y < x 2. The shaded region r e p r e s e n t s t h e solutions of the inequality. (1) Is y < x 2 for all x and y? In o t h e r words, is the e n t i r e plane shaded? Since this is not t h e case, proposition (1) is false. (2) (3x)(3y)P(x,y) is t r u e i f y < x 2 for some real n u m b e r s x a n d y; t h a t is, if and only if some portion of the c a r t e s i a n plane is shaded. Since this is true, proposition (2) is true.

*Based on E. A. Kuehls, "The Truth-Value of {V,3, P(x,y)}: A Graphical Approach," Mathematics Magazine, Vol. 43 (Nov. 1970), pp. 260-261.

35

1.3 Quantifiers F i g u r e 1.8

y y =X 2 I I I I / / /

\

/

/

y

<X 2

/

\ \

/

(3) (Vx)(ay)P(x,y) is t r u e if t h e r e is a point (x,y) in the shaded a r e a corresponding to every x; t h a t is, it is t r u e if every vertical line i n t e r s e c t s the shaded area. Since this is the case, the proposition is true. (4) (Vy)(ax)P(x,y) is t r u e since every horizontal line intersects the s h a d e d area. (5) (ax)(Vy)P(x,y) m e a n s t h e r e is a n x such t h a t y < x 2 for ally. Therefore, the proposition is t r u e if t h e r e is a vertical line which lies wholly w i t h i n the shaded region. Since no such line exists, the proposition is false. (6) (ay)(Vx)P(x,y) is t r u e if t h e r e is a horizontal line which lies wholly within the shaded area. Since t h e r e are such lines, the proposition is true.

I Note: This graphical approach elucidates the difference b e t w e e n (Vx)(3y) and (3x)(u and also b e t w e e n (u and (::lx)(u (3x)(u d e m a n d s a fixed x, w h e r e a s (u does not d e m a n d such a fixed x. Next we discuss how to negate quantified propositions. Recall from E x a m p l e 1.24 t h a t the proposition All apples are green can be symbolized as (u w h e r e P(x): x is green. Its negation is: It is false t h a t all apples are green. T h a t is, t h e r e exists an apple t h a t is not green. In symbols, this can be w r i t t e n as (3x)(~P(x)). Thus, ~[(Vx)P(x)] _= (3x)[~P(x)]. Similarly, ~[ (3x)P(x)] - (Vx)[~P(x)]. These two properties are De M o r g a n ' s laws for n e g a t i n g quantifiers.

De M o r g a n ' s l a w s 9 ~[(Vx)P(x)l ~ (3x)[~P(x)l 9 ~[('v'x)P(x)l ~ (Vx)[~P(x)l

m

Chapter I The Language of Logic

36

By virtue of these laws, be careful w h e n n e g a t i n g quantified propositions. W h e n you negate the universal quantifier u it becomes the existential quantifier 3; w h e n you negate the existential quantifier, it becomes the universal quantifier. In Section 1.5, we discuss a nice application of the first law to disproving propositions.

Negate each proposition, where the UD - set of integers. (1) (Vx) (X 2

--

X)

(2) (3X) (IXl -- X)

SOLUTION: 9 ~l(Vx)(x 2 - x)l - (3x)l~(x 2 - x)! -

(3x)(x 2 r x).

9 "-l(3x)(Ixl - x)l = (Vx)l-~( xl - x)! - (Vx)(Ixl # x). Negate each quantified proposition. (1) (2) (3) (4)

Every c o m p u t e r is a 16-bit machine. Some girls are blondes. All chalkboards are black. No person has green eyes.

SOLUTION: Their negations are: (1) (2) (3) (4)

Some c o m p u t e r s are not 16-bit machines. No girls are blondes. Some chalkboards are not black. Some people have green eyes.

m

In closing, we should point out t h a t w h a t we discussed in Sections 1.1 and 1.2 is p r o p o s i t i o n a l logic; it deals with u n q u a n t i f i e d propositions. However, as we saw t h r o u g h o u t this section, not all propositions can be symbolized in propositional logic, so quantifiers are needed. The area of logic t h a t deals with quantified propositions is p r e d i c a t e logic.

Exercises 1.3 D e t e r m i n e the t r u t h value of each proposition, where the UD consists of the n u m b e r s +1, i 2 , and 0. 1.

(VX)(X 2 - -

4)

4. (Vy)(y4 + 3y 2 -- 2)

2.

( 3 X ) ( X 3 + 2X 2 - - X ~-

5. ~(u

3 -x)

2) 3. (Vx)(x 5 + 4x

-

5 x 3)

6. (Vx)l~(x 5 - 4x)l

1.3

Quantifiers

37

L e t P(x): x 2 > x, Q(x): x 2 = x, a n d t h e U D = set of i n t e g e r s . D e t e r m i n e t h e t r u t h v a l u e of e a c h p r o p o s i t i o n . 7. (Vx)l~P(x)]

8. ( 3 x ) l ~ P ( x ) ]

10. (Vx)[P(x) A Q(x)]

9. ( ~ x ) l P ( x ) A Q(x)!

11. ( 3 x ) [ P ( x ) v Q(x)]

12. (Vx)[P(x) v Q(x)]

R e w r i t e e a c h s e n t e n c e symbolically, w h e r e P(x): x is a 16-bit m a c h i n e , Q(x): x u s e s t h e ASCII** c h a r a c t e r set, a n d t h e U D = set of all c o m p u t e r s . 13. T h e r e is a c o m p u t e r t h a t is a 16-bit m a c h i n e a n d u s e s t h e A S C I I c h a r a c t e r set as well. 14. We can find a 16-bit c o m p u t e r t h a t does n o t u s e t h e A S C I I c h a r a c t e r set. 15. We can find a c o m p u t e r t h a t is e i t h e r a 16-bit m a c h i n e or d o e s n o t u s e t h e ASCII c h a r a c t e r set. 16. T h e r e exists a c o m p u t e r t h a t is n e i t h e r a 16-bit m a c h i n e n o r u s e s t h e ASCII c h a r a c t e r set. N e g a t e each p r o p o s i t i o n , w h e r e x is a n a r b i t r a r y i n t e g e r . 17. (Vx)(x 2 > 0) 18. ( 3 x ) ( x 2 r 5x - 6) 19. E v e r y s u p e r c o m p u t e r is m a n u f a c t u r e d in J a p a n . 20. T h e r e a r e no w h i t e e l e p h a n t s . R e w r i t e each s e n t e n c e numbers.

symbolically,

where

the

UD

consists

of real

21. T h e p r o d u c t of a n y two real n u m b e r s x a n d y is positive. 22. T h e r e a r e real n u m b e r s x a n d y s u c h t h a t x = 2y. 23. F o r each real n u m b e r x, t h e r e is s o m e real n u m b e r y s u c h t h a t x .y - x. 24. T h e r e is a real n u m b e r x s u c h t h a t x + y = y for e v e r y real n u m b e r y. 25-28.

F i n d t h e t r u t h v a l u e of e a c h p r o p o s i t i o n in E x e r c i s e s 2 1 - 2 4 .

R e w r i t e e a c h in words, w h e r e U D = set of i n t e g e r s . 29. (VX)(X2 >_ 0)

30. ~(3X)(X 2 = 2)

31. ( 3 X ) ( 3 y ) ( x + y = 7)

32. (VX)(3y)(xy = 3)

33. (3X)(Vy)(y -- X = y)

34. (VX)(Vy)(x + y = y + X)

35-40.

F i n d t h e t r u t h v a l u e of e a c h p r o p o s i t i o n in E x e r c i s e s 2 9 - 3 4 .

L e t U D = set of i n t e g e r s , P(x,y): x is a m u l t i p l e of y, a n d Q(x,y) : x > y. D e t e r m i n e t h e t r u t h v a l u e of e a c h p r o p o s i t i o n .

**ASCII is the acronym for American Standard Code for Information Interchange.

38

Chapter I The Language of Logic

42. (Vx)P(x, 2)

41. (3x)P(15,x)

43. ~(3x)P(x, 5)

44. (3x)[P(x, 3) A Q(x, 3)1

45. (3x)[P(x, 2) v Q(x, 6)1

46. (Vx)(3y)P(x,y)

47. (Vx)(3y)Q(x, y)

48. (Vx)[P(x, 3) ~ Q(x, 3)]

49. (3x)[Q(x, 3) --~ P(x, 3)]

Let UD = set of real n u m b e r s and P(x,y): y2 < x. D e t e r m i n e the t r u t h value of each proposition. 50. (Vx)(Vy)P(x,y)

51. (3x)(3y)P(x,y)

52. (Vx)(3y)P(x,y)

53. (Vy)(3x)P(x,y)

54. (3x)(u

55. (3y)(Vx)P(x,y)

A third useful quantifier is the u n i q u e n e s s q u a n t i f i e r 3!. T h e proposition (3!x)P(x) means There exists a unique (meaning exactly one) x such that P(x). Determine the t r u t h value of each proposition, where UD - set of integers. 56. (3!x)(x + 3 = 3)

57. (3!X)(X 2 : 1)

58. (3!x)(3!y)(xy = 1)

59. (3!x)(Vy)(x + y = y)

60. (3!x)(3!y)(2x = 3y)

61. (Vx)(3!y)(x + y = 4)

Determine the t r u t h value of each, where P(s) denotes an a r b i t r a r y predicate. 62. (3x)P(x) --* (3!x)(P(x)

63. (3!x)P(x) --* (3x)P(x)

64. (Vx)P(x) --* (3!x)P(x)

65. (3x)P(x) ~ (Vx)P(x)

66. (Vx)P(x) ~ (3x)P(x)

67. (3!x)P(x) ~ (3!y)P(y)

*68. Define the quantifier 3! in t e r m s of the quantifiers 3 and v.

Suppose we are given a finite set of propositions (called h y p o t h e s e s ) H1, H 2 , . . . , Hn, all assumed true. Also assume t h a t from these p r e m i s e s , we can arrive at a conclusion C t h r o u g h reasoning (or argument). Such a discussion can be written in i n f e r e n t i a l f o r m as follows, where the symbol .'. means therefore" H1

H2

hypotheses

Hn .'. C

~ conclusion

1.4 Arguments(optional)

39

What does it mean to say that our reasoning in such a discussion is logical-that is, that the argument is valid?

Valid and Invalid Arguments An argument is v a l i d if the conjunction of the hypotheses H1, H 2 , . . . , Hn logically implies the conclusion C: that is, the implication H1 A H2 A ... A Hn -~ C is a tautology. Otherwise, the argument is i n v a l i d , a fallacy. Thus, an argument is valid if and only if the conclusion is a logical consequence of the hypotheses. In other words, if the hypotheses are assumed true, then the conclusion must follow logically from them. True hypotheses always lead to a true conclusion by a valid argument. We begin checking the validity of arguments by using a well-known logic puzzle, due to R. M. Smullyan. Test the validity of the following argument. H1 : There are more residents in New York City than there are hairs in the head of any resident. H2 : No resident is totally bald. .'. At least two residents must have the same number of hairs on their heads. SOLUTION: (The argument contains two hypotheses. We always assume they are true and need to check whether the given conclusion follows logically from them.) Suppose there are n residents in New York City. By H1, the number of hairs on the head of every resident is less n; by H2 every resident has at least one hair on his head. If each person has a different number of hairs, there must be n positive integers less than n, which is impossible. Therefore, at least two residents must have the same number of hairs on their heads. Since the logical conclusion agrees with the given conclusion, the argument is valid. (This example is an application of the pigeonhole principle presented in Section 3.4.) m The next example presents another well-known logic puzzle, again due to Smullyan. There are two kinds of inhabitants, knights and knaves, on an island. Knights always tell the truth, whereas knaves always lie. Every inhabitant is either a knight or a knave. One day three i n h a b i t a n t s - - A , B, and C - were standing together in a garden. A nomad came by and asked A, "Are you a knight or a knave?" Since A answered rather indistinctly, the stranger could make nothing out of his reply. So he asked B, "What did A say?" B replied, "A said, he is

40

Chapter 1

The Language of Logic

Henry Ernest Dudeney (1857-1930), England's greatest puzzlist and perhaps the greatest puzzlist who ever lived, was born in Mayfield, Sussex, England. Dudeney and Sam Loyd, the American puzzle genius, used to exchange puzzles and collaborate on puzzle articles in magazines and newspapers. Dudeney authored six books on puzzles, beginning with The Canterbury Puzzles (1907). Three of his collections were published posthumously.

a knave." At this point C j u m p e d into the conversation and said, "Don't believe B; he is lying." What are B and C? SOLUTION: A knight would never say, "I'm a knave," since he never lies. A knave would not say that either since he never tells the truth. Therefore, A did not say he was a knave. So B lied to the nomad and hence is a knave. Consequently, C was telling the truth, so C is a knight. Thus B is a knave and C is a knight. (This example is pursued f u r t h e r in the exercises.) m The next puzzle* is a variation of a brainteaser developed by the English puzzlist, Henry Dudeney. Its solution does not employ any logic variables, but illustrates a clever problem-solving technique. ~

Smith, Jones, and Robinson are the brakeman, engineer, and fireman o n a train, not necessarily in that order. Riding on the train are three passengers with the same last names who are identified by a "Mr." before their names. Assuming the following premises are true, determine who the engineer is. No two passengers live in the same city. Mr. Robinson lives in New York. The b r a k e m a n lives in Dallas. Mr. Jones has forgotten all the algebra he learned in high school. H5: The passenger whose last name is the same as the b r a k e m a n ' s lives in Los Angeles. H6 : The b r a k e m a n and one of the passengers, a m a t h e m a t i c a l genius, attend the same local church. H7: Smith beats the fireman in golf.

HI: H2: H3: H4 :

*Based on M. Gardner, Mathematical Puzzles and Diversions, The University of Chicago Press, Chicago, 1987.

1.4 Arguments(optional)

41

SOLUTION: We begin with two three-by-three arrays of empty cells with labels as in Figure 1.9. Use the premises to fill in the cells with O's and l's; enter a i in a cell with headings x and y if x has property y and enter a 0 otherwise. Premise H7 implies Smith is not a fireman, so enter a 0 in the u p p e r right cell in Figure 1.9b. Since Mr. Robinson lives in New York (H2), place a i in the lower left cell in Figure 1.9a and O's in the r e m a i n i n g cells of the same row and column (why?). It now follows t h a t either Mr. Smith or Mr. Jones lives in Dallas. Does Mr. Jones live there?

F i g u r e 1.9

J Mr. Smith

Smith

Mr. Jones

Jones

Mr. Robinson

Robinson (a)

(b)

Since Mr. Jones cannot be a mathematical genius by H4, the passenger genius m u s t be Mr. Smith. By H6, Mr. Smith and the b r a k e m a n live in the same city. Since it m u s t be Dallas by H3, enter a 1 in the upper middle cell in Figure 1.9a and O's in the r e m a i n i n g cells of the same row and column. It now follows t h a t Mr. Jones lives in Los Angeles, so place a 1 in the middle cell of the third column in Figure 1.9a and O's everywhere else. Figure 1.10a displays the resulting array. By premise H5, the b r a k e m a n and the passenger who lives in Los Angeles have the same last name, so the b r a k e m a n m u s t be Jones; therefore, put a 1 in the first cell of the middle row in Figure 1.9b and O's in the r e m a i n i n g cells of the same row and column. By now, the top row in Figure 1.9b contains two O's, so the middle cell must occupy a i (why?); so the middle cell in the bottom row a 0; hence the lower right cell m u s t occupy a 1. Figure 1.10b shows the resulting array.

F i g u r e 1.10 Mr. Smith

0

1

0

Smith

0

1

0

Mr. Jones

0

0

1

Jones

1

0

0

Mr. Robinson

1

0

0

Robinson

0

0

1

(a)

(b)

It follows from Figure 1.10b t h a t Smith is the engineer,

m

42

Chapter I The Language of Logic

B e r t r a n d A r t h u r William Russell (1872-1970), a British philosopher and mathematician, was born into a prominent, aristocratic, and progressive-minded family near Trelleck, Wales. His mother died in 1874 and his father two years later; so the young Russell was brought up by his father's parents. Russell was home-educated by tutors. In 1890 he entered Trinity College, Cambridge, where he excelled in both mathematics and the moral sciences. In 1895, he was awarded a fellowship for his original dissertation on the foundations of geometry, published in 1897. After graduation, he worked briefly in the British embassy in Paris and then he went to Germany, where he wrote his first book, German Social Democracy (1896). In 1910, Trinity appointed him a lecturer in logic and the philosophy of mathematics. Russell's outspokenness and liberal views often landed him in controversies. Around 1907, Russell fought hard for women's suffrage in the United Kingdom. During World War I, he was dismissed by Trinity for his protests and pacifist views. In 1918, he was imprisoned for 6 months for an article that was branded seditious. While in prison, he wrote Introduction to Mathematical Philosophy. When he was about 90 years old, he was imprisoned again for campaigning for nuclear disarmament. In 1925, Trinity, realizing that the 1916 dismissal was excessively harsh, invited Russell back. He served there as a fellow from 1944 until his death. Russell wrote more than 40 books on diverse subjects, including philosophy and physics; his greatest work is the three-volume Principia Mathematica (1910-1913), which he coauthored with the Cambridge philosopher Alfred North Whitehead (1861-1947). It describes the logical construction of the foundations of mathematics from a set of primitive axioms. Russell won the 1950 Nobel prize for literature "as a defender of humanity and freedom of thought."

The following example is yet another well-known puzzle, the b a r b e r p a r a d o x , presented by the British mathematician and philosopher Bertrand Russell in 1918. There is a male barber in a certain town. He shaves all those men and only those men who do not shave themselves. Does the barber shave himself?. SOLUTION: Suppose the barber shaves himself. Then he belongs to the class of men who shave themselves. But no one in this class is shaved by the barber, so the barber does not shave himself, which is a contradiction. On the other hand, suppose the barber does not shave himself. Since the barber shaves all those men who do not shave themselves, he shaves himself, again a contradiction. Thus either case leads to a paradox: If the barber shaves himself, he does not shave himself; and conversely if he does not shave himself, then he shaves himself. So, logically, no such barber exists, m The symbols and the laws of logic can often be applied to check the validity of an argument, as the next two examples illustrate. To this end, follow the steps below: 9 Rewrite the hypotheses symbolically.

1.4 Arguments(optional)

43

9 Assume the hypotheses are true. 9 If the i n f e r e n c e rules in Table 1.17 and/or the laws of logic can be used to reach the given conclusion, t h e n the given a r g u m e n t is valid; otherwise, it is invalid; t h a t is, the a r g u m e n t contains a flaw.

Table 1.17

Inference Rules 1. p A q ~

conjunction simplification addition law of detachment law of the contrapositive disjunctive syllogism hypothetical syllogism

(pAq)

2. p A q - - + p 3. p - - + p V q 4. [p A (p ~ q)l ~ q

5. [(p ~ q) A (~q)] ~ ~p 6. [(p Vq) A (~P)l -+ q 7. [ ( p o q ) A ( q ~ r ) l ~ ( p ~ r )

A few words of explanation about each rule: The c o n j u n c t i o n rule says t h a t if both p and q are true, then p A q is t r u e - - a fact you already knew. According to the s i m p l i f i c a t i o n rule, i f p A q is true, t h e n p is true. The a d d i t i o n r u l e says t h a t ifp is true, then p v q is true regardless of the t r u t h value of q. By the l a w of d e t a c h m e n t , if an implication p -~ q is true and the premise p is true, then you can always conclude t h a t q is also true; in other words, a true premise leads to a true conclusion logically. The l a w of the c o n t r a p o s i t i v e says t h a t if an implication p -~ q is true, but the conclusion q is false, then the premise p m u s t be false. The two syllogisms can be interpreted similarly. It is obvious t h a t the inference rules play a central role in d e t e r m i n i n g the validity of an argument. These rules, which are tautologies, can be established using t r u t h tables. Try a few. Each of the inference rules can be written in inferential form. For instance, the law of d e t a c h m e n t can be r e w r i t t e n as follows: P p~q .'.

q

Check the validity of the following argument. If the computer was down Saturday afternoon, t h e n Mary went to a matinee. Either Mary went to a matinee or took a nap S a t u r d a y afternoon. Mary did not take a nap t h a t afternoon. .'. The computer was down S a t u r d a y afternoon.

44

Chapter I The Language of Logic

SOLUTION: To avoid our emotions' playing any role in the way we reason, first t r a n s l a t e the discussion into symbols. Let p: The c o m p u t e r was down S a t u r d a y afternoon. q: M a r y went to a m a t i n e e S a t u r d a y afternoon. r: M a r y took a nap S a t u r d a y afternoon. T h e n the given a r g u m e n t can be symbolized as follows:

Hl "p --~ q H2 q v r

hypotheses

H3 " ~ r .'. p

~ conclusion

Every step in our logical r e a s o n i n g and the c o r r e s p o n d i n g justification are given below: 1. 2. 3. 4. 5.

~ r is true. q v r is true. q is true. p -~ q is true. T h e n p may be true or false.

hypothesis H3 hypothesis H~ step 1, step 2, and disjunctive syllogism hypothesis H 1 step 4, step 5, and definition of implication.

Since our logical conclusion does not agree with the given conclusion, the given a r g u m e n t is invalid. (Using a t r u t h table you may verify t h a t I(p --~ q) A (q V r) A (~r)i --~ p is not a tautology. This provides an a l t e r n a t e d e m o n s t r a t i o n t h a t this a r g u m e n t is invalid.) Note: Trivial steps may be omitted from such a r e a s o n i n g w i t h o u t jeopardizing the logical progression, m We conclude this section with an example from Lewis Carroll's famous book Symbolic Logic. Two additional examples appear in the exercises. Check the validity of the following a r g u m e n t . Babies are illogical. Nobody is despised who can m a n a g e a crocodile. Illogical persons are despised. .'. Babies cannot m a n a g e crocodiles.

SOLUTION: First t r a n s l a t e the sentences into if-then form using symbols, so we let p: q: r: s:

Harry Harry Harry Harry

is a baby. is illogical. can m a n a g e a crocodile. is despised.

1.4 Arguments(optional)

9.: ;[email protected]":. .....'i'. ,' :,!~u'~':"., , . , '..~." "',;~4.~'.~i, ..,,. .~'. ~ .... ....., :-.,. :...,

~" :., ... ~:..

~.,.:,.~ "~ ,~. . ,~

.:~.,,. v,.,.....~,~';.... . .:%, . ,:~.,:.... 9 ,,-,,7% .i{?}~" ;;:,

" . :!.;..'.'.~.~: ~.,v~-,.~.

45

L e w i s C a r r o l l (1832-1898) (a pseudonym of Charles Lutwidge Dodgson) was the son of a clergyman and was born in Daresbury, England. He graduated from Christ Church College, Oxford University, in 1854 and began teaching mathematics at his alma mater in 1855, where he spent most of his life. He became a deacon in the Church of England in 1861. Carroll's famous Alice in Wonderland and its sequel, Through the LookingGlass and What Alice Found There, have provided a lot of pleasure to both children and adults all over the world. Alice in Wonderland is available in more than 30 languages, including Arabic and Chinese, and also in braille. The character is named for Alice Liddell, a daughter of the dean of Christ Church College.

, ":{.~.t~::"

T h e n t h e a r g u m e n t can be w r i t t e n as: Hl:p~q H2: r ~ ~ s H3: q --+ s .'. p --~ ~ r E v e r y step of o u r logical r e a s o n i n g is given below: 1. 2. 3. 4. 5.

(p --~ q) A (q --~ s) is true. p --+ s is t r u e . s --+ ~ r is t r u e . (p --~ s) A (S --+ ~ r ) is t r u e . p --~ ~ r is t r u e .

c o n j u n c t i o n rule h y p o t h e t i c a l syllogism law of t h e c o n t r a p o s i t i v e c o n j u n c t i o n rule h y p o t h e t i c a l syllogism

Since t h e given conclusion a g r e e s w i t h t h e logical conclusion, t h e a r g u m e n t is valid, m

Exercises 1.4 R e w r i t e each i m p l i c a t i o n in i n f e r e n t i a l form. 1. [(p ~ q) A ( ~ q ) ] - + ~P

2. [(p -+ q) A (q ~ r ) ] - ~ (p -~ r)

Verify t h a t each i n f e r e n c e rule is a tautology. 3. p--+ ( p v q )

4. [ ( p ~ q )

T e s t t h e validity of each a r g u m e n t . 5. p v q qvr ~r .'. p

6. p ~ q ~.pvr ~r

.'. ~ q

A(qor)l~

(p~r)

46

Chapter I The Language of Logic

7. If Bill likes cats, he dislikes dogs. Bill likes dogs. .. Bill dislikes cats. 8. If P a t passes this course, she will g r a d u a t e this year. P a t does not pass this course. .. P a t will not g r a d u a t e this year. 9. F r a n k b o u g h t a personal c o m p u t e r or a video cassette recorder (VCR). If he b o u g h t a VCR, t h e n he likes to watch movies at home. He does not like to watch movies at home. .. F r a n k b o u g h t a personal computer. 10. If P e t e r is married, he is happy. If he is happy, t h e n he does not read the c o m p u t e r magazine. He does read the c o m p u t e r magazine. .'. Peter is u n m a r r i e d .

(Exercises 11 and 12 come from Lewis Carroll's Symbolic Logic.) "11. All philosophers are logical. An illogical person is always obstinate. .'. Some obstinate persons are not philosophers. "12. No ducks waltz. No officers ever decline to waltz. All my poultry are ducks. .'. My poultry are not officers. Give the simplest possible conclusion in each a r g u m e n t . A s s u m e each premise is true. 13. p o q up v r ~r

14. p ~ q p v ~r r

15. p - ~ ~ q ~ r --~ q p

16. p - ~ q ~ r --* ~q ~r

17. The p r o g r a m is r u n n i n g if and only if the c o m p u t e r is working. The c o m p u t e r is w o r k i n g or the power is off. The power is on. 18. Linda has a video cassette recorder (VCR). If she has a personal computer, t h e n she does not have a VCR. If she does not have a personal computer, t h e n she has a calculator.

1.4 Arguments (optional)

47

19. Carol is a baby if and only if she is illogical. Either she is illogical or unhappy. But she is happy. 20. Three persons took a room for $30 at a hotel. Soon after they checked out, the room clerk realized she had overcharged them since the room rents for $25. She sent a bellhop to them with a $5 reimbursement, but he returned to them only $3, keeping $2 for himself. Thus the room cost $30 - $3 -- $27 and $27 + $2 - $29, so what happened to the extra dollar? 21. Aaron, Benjamin, Cindy, and Daphne 28, and 27 years old, not necessarily to the oldest person. Aaron is older Daphne. Who is married to whom and Teacher, 1990)

are all friends. They are 34, 29, in that order. Cindy is married than Cindy, but younger t h a n how old are they? (Mathematics

22. A family party consisted of one grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-inlaw, and one daughter-in-law. A total of 23 people, apparently. But no; there were only seven people at the party. How could this be possible? (B. Hamilton, 1992) 23. Three g e n t l e m e n ~ Mr. Blue, Mr. Gray, and Mr. W h i t e ~ have shirts and ties that are blue, gray, and white, but not necessarily in that order. No person's clothing has the same color as his last name. Mr. Blue's tie has the same color as Mr. Gray's shirt. What color is Mr. White's shirt? (Mathematics Teacher, 1986) 24. Three men and their wives were given $5400. The wives together received $2400. Sue had $200 more than Jan, and Lynn had $200 more than Sue. Lou got half as much as his wife, Bob the same as his wife, and Matt twice as much as his wife. Who is married to whom? (Mathematics Teacher, 1986) There are seven lots, 1 through 7, to be developed in a certain city. A builder would like to build one bank, two hotels, and two restaurants on these lots, subject to the following restrictions by the city planning board (The Official LSAT PrepBook, 1991): used. If lot 2 is used, lot 4 cannot be used. If lot 5 is used, lot 6 cannot be built on |1 The bank can be built only on lot 5, 6, or 7. A hotel cannot be lot 5. A restaurant_ can _be built._ only on lot~ 1, 2, 3, or 5. ]

J

25. Which of the following is a possible list of locations for building them? A. The bank on lot 7, hotels on lots 1 and 4, and restaurants on lots 2 and 5.

48

Chapter I The Language of Logic

B. T h e b a n k on lot 7, hotels on lots 3 a n d 4, a n d r e s t a u r a n t s on lots 1 a n d 5. C. T h e b a n k on lot 7, hotels on lots 4 a n d 5, a n d r e s t a u r a n t s on lots 1 and 3. 26. If a r e s t a u r a n t is built on lot 5, which of t h e following is not a possible list of locations? A. A hotel on lot 2 a n d lot 4 is left undeveloped. B. A r e s t a u r a n t on lot 2 a n d lot 4 is left undeveloped. C. A hotel on lot 2 a n d lot 3 is left undeveloped. Exercises 27-31 refer to E x a m p l e 1.32 a n d are based on S m u l l y a n ' s What is the name of this book ? A and B are i n h a b i t a n t s of the island. W h a t are t h e y if A says each of t h e following? 27. "At least one of us is a k n a v e . " 28. " E i t h e r I ' m a knave or B is a k n i g h t . " 29. A, B, and C are i n h a b i t a n t s of the island. Two r e s i d e n t s are of the same type if they are both k n i g h t s or both knaves. A says, "B a n d C are of the s a m e type." Someone t h e n asks C, "Are A and B of t h e s a m e type?" W h a t does C answer? 30. A says, "All of us are knaves," and B says, " E x a c t l y one of us is a k n i g h t . " W h a t are A, B, and C? 31. A says, "All of us are knaves," and B says, "Exactly one of us is a knave." W h a t is C? Every i n h a b i t a n t on a m y s t e r i o u s p l a n e t is e i t h e r red or green. In addition, each i n h a b i t a n t is e i t h e r male or female. E v e r y red m a n always tells the t r u t h , w h e r e a s every green m a n always lies. T h e women, on t h e o t h e r hand, are opposite: every green w o m a n tells the t r u t h and every red w o m a n lies. Since the natives always disguise t h e i r voices, and w e a r m a s k s a n d gloves, it is impossible to identify t h e i r sex or color. B u t a clever a n t h r o p o l o g i s t from M a t h l a n d m e t a native who m a d e a s t a t e m e n t from which he was able to deduce t h a t the native was a green w o m a n . (R. S m u l l y a n , Discover, 1993) 32. W h a t could the native have said? J u s t i f y y o u r answer. 33. The second native t h e a n t h r o p o l o g i s t interviewed also m a d e a statem e n t from which he was able to conclude t h a t the native was a m a n (but not his color). Give a s t a t e m e n t t h a t would work. Again, justify y o u r answer.

1.5 Proof Methods

49

Four women, one of whom was known to have committed a serious crime, made the following statements when questioned by the police: (B. Bissinger, Parade Magazine, 1993) Fawn: Kitty: Bunny: Robin:

"Kitty did it." "Robin did it." "I didn't do it." "Kitty lied."

34. If exactly one of these statements is true, identify the guilty woman. 35. If exactly one of these statements is false, identify the guilty woman. *36. "How is it, Professor Whipple," asked a curious student, "that someone as notoriously absentminded as you are manages to remember his telephone number?" "Quite simple, young man" replied the professor. "I simply keep in mind that it is the only seven-digit number such that the number obtained by reversing its digits is a factor of the number." What is Professor Whipple's telephone number? (A. J. Friedland, 1970) *37. Five angry cowgirls, standing in a field, accuse each other of rustling. No two distances between every two women are the same. Each has one bullet in her gun. At the count of ten, each shoots the nearest person in the toe. Will any cowgirl be shot or will at least one escape injury? (M. Gardner, Parade Magazine, 1993)

Proofs, no matter how simple or complicated they are, are the heart and soul of mathematics. They play a central role in the development of mathematics and guarantee the correctness of mathematical results and algorithms (see Chapters 4 and 5). No mathematical results or computer algorithms are accepted as correct unless they are proved using logical reasoning. A t h e o r e m in mathematics is a true proposition. Many theorems are implications H1 A H2 A . . . A Hn ---> C. P r o v i n g such a theorem means verifying that the proposition HI A H2 A.-. A Hn --> C is a tautology. This section presents six standard methods for proving theorems: v a c u o u s p r o o f , t r i v i a l p r o o f , d i r e c t p r o o f , i n d i r e c t p r o o f , p r o o f b y cases, and e x i s t e n c e proof. Vacuous and trivial proofs are, in general, parts of larger and complicated proofs, as will be seen in Chapters 4 and 5. Vacuous Proof

Suppose the hypothesis H of the implication H ~ C is false. Then the implication is true regardless of whether C is true or false. Thus if the

50

Chapter I The Language of Logic hypothesis H can be shown to be false, the t h e o r e m H ~ C is t r u e by default; such a proof is a v a c u o u s p r o o f . Vacuous proofs, a l t h o u g h rare, are necessary to h a n d l e special cases, as will be seen in C h a p t e r 5. Since the hypothesis of the s t a t e m e n t I f I = 2, t h e n 3 = 4 is false, the proposition is vacuously true. m

Trivial Proof Suppose the conclusion c of the implication H ~ C is true. Again, the implication is true irrespective of the t r u t h value of H. Consequently, if C can be shown to be true, such a proof is a t r i v i a l p r o o f . Let P(n)" Ifx is a positive real n u m b e r and n any nonnegative integer, then (1 + x ) n > 1 + n x . Since (1 + x) ~ > 1 + 0. x always, the proposition P(0) is true. Thus the theorem is trivially true when n = 0. In this trivial proof we did not use the premise t h a t x > 0. m

Next we pursue a n o t h e r proof method.

Direct Proof In the d i r e c t p r o o f of the t h e o r e m H1 A H2 A . . . A Hn ~ C, assume the given hypotheses Hi are true. Using the laws of logic or previously known facts, establish the desired conclusion C as the final step of a chain of implications: H ~ C1, C1 ~ C2,... ,Cn -* C. Then, by the repeated application of the hypothetical syllogism, it follows t h a t H ~ C. The next example illustrates this method. Often, theorems are stated in terms of sentences, so we need to first rewrite t h e m symbolically and t h e n work with the symbols, as the next example demonstrates. Prove directly t h a t the product of any two odd integers is an odd integer.

PROOF: Let x and y be any two odd integers. Then there exist integers m and n such t h a t x = 2m + 1 and y = 2n + 1. Thus, x . y - (2m + 1). (2n + 1) = 4mn

+ 2m + 2n + 1

= 2(2mn + m + n) + 1 =2k+1 where k - 2 m n + m + n is an integer. Therefore, x y is an odd integer. This concludes the proof. (Can you rewrite this proof as a chain of implications?) m

1.5 ProofMethods

51

Indirect Proof T h e r e are two kinds of i n d i r e c t p r o o f s for the t h e o r e m H1 A H2 A . . - A Hn ---> C: p r o o f o f t h e c o n t r a p o s i t i v e and p r o o f b y c o n t r a d i c t i o n . The first m e t h o d is based on the law of the contrapositive, H1 A H2 A - . . A Hn ---> C - ~ C -+ ~(H1 A H2 A-.. A Hn). [You m a y recall, by De M o r g a n ' s law, t h a t ~(H1 A H2 A . . . A Hn) -= ~H1 v ~ H 2 v . . . v ~Hn.] In this method, a s s u m e the desired conclusion C is false; t h e n using the laws of logic, establish t h a t some hypothesis Hi is also false. Once you have done this, the t h e o r e m is proved. The next example enlightens this method. Prove indirectly: If the square of an integer is odd, t h e n the integer is odd.

PROOF OF THE CONTRAPOSITIVE Let x be any integer such t h a t X 2 is odd. We would like to prove t h a t x m u s t be an odd integer. In the indirect method, we a s s u m e the conclusion is false; t h a t is, x is not odd; in o t h e r words, a s s u m e x is an even integer. Let x - 2k for some integer k. T h e n x 2 - (2k) 2 - 4k 2 - 2(2k2), which is an even integer. This makes our hypothesis t h a t x 2 is an odd integer false. Therefore, by the law of the contrapositive, our a s s u m p t i o n m u s t be wrong; in other words, x m u s t be an odd integer. Thus, ifx 2 is an odd integer, t h e n x is also an odd integer, m P r o o f b y c o n t r a d i c t i o n , the other variation of indirect proof, is based on the law of reductio ad absurdum: H1 A H2 A . . . A H,~ --~ C ~ IH1 A H2 A . . . A H,~ A (~C)| ~ F. In this method, a s s u m e the given hypotheses Hi are true, but the conclusion C is false. T h e n argue logically and reach a contradiction F. The next example illustrates this method, where a p r i m e n u m b e r p is a positive integer with exactly two positive factors, 1 and p. Prove by contradiction: T h e r e is no largest prime n u m b e r ; t h a t is, there are infinitely m a n y prime n u m b e r s . P R O O F BY C O N T R A D I C T I O N (Notice that the theorem has no explicit hypothesis.) Suppose the given conclusion is false; t h a t is, t h e r e is a largest prime n u m b e r p. So the prime n u m b e r s we have are 2, 3, 5 , . . . ,p; a s s u m e t h e r e are k such primes, p l , p 2 , . . . , and Pk. Let x denote the product of all of these prime n u m b e r s plus one: x = ( 2 . 3 - 5 . . . p ) + 1. Clearly, x > p. W h e n x is divided by each of the primes 2, 3, 5 , . . . ,p, we get 1 as the remainder. So x is not divisible by any of the primes. Hence either x m u s t be a prime, or if x is composite t h e n x is divisible by a prime q # Pi. In either case, there are more t h a n k primes. But this contradicts the a s s u m p t i o n t h a t there are k primes, so our a s s u m p t i o n is false. In other words, t h e r e is no largest prime number, m Now we t u r n to yet a n o t h e r proof technique.

Chapter I

52

The Language of Logic

Proof by Cases Suppose we would like to prove a t h e o r e m of the form H1 v H2 v . . . v Hn -+ C. Since H I v H2 v . . . v H a -+ C --- ( H I ---> C) A (H2 ---> C) A . . - A (Hn ~ C), the s t a t e m e n t H1 v H2 v . . . v Hn ~ C is t r u e if a n d only if each i m p l i c a t i o n Hi --> C is true. Consequently, we need only prove t h a t each

implication is true. Such a proof is a p r o o f b y c a s e s , as i l l u s t r a t e d in the following example, due to R. M. Smullyan. Let A, B, and C be three i n h a b i t a n t s of the island described in Example 1.32. Two i n h a b i t a n t s are of the s a m e type if they are b o t h k n i g h t s or both knaves. Suppose A says, "B is a knave," and B says, "A a n d C are of the same type." Prove t h a t C is a knave.

PROOF BY CASES Although this t h e o r e m is not explicitly of the form I-I1 v H2 v . . . v I-In ~ C, we artificially create two cases, namely, A is a k n i g h t and A is a knave.

Case 1

Suppose A is a knight. Since k n i g h t s always tell the t r u t h , his s t a t e m e n t t h a t B is a knave is true. So B is a knave and hence B's s t a t e m e n t is false. Therefore, A and C are of different types; t h u s C is a knave.

Case 2

Suppose A is a knave. T h e n his s t a t e m e n t is false, so B is a knight. Since k n i g h t s always tell the t r u t h , B's s t a t e m e n t is true. So A and C are of the same type; t h u s C is a knave. T h u s in both cases, C is a knave,

m

Existence Proof Finally, t h e o r e m s of the form (~x)P(x) also occur in m a t h e m a t i c s . To prove such a theorem, we m u s t establish the existence of an object a for which P(a) is true. Accordingly, such a proof is an e x i s t e n c e p r o o f . T h e r e are two kinds of existence proofs: the c o n s t r u c t i v e e x i s t e n c e proof and the n o n c o n s t r u c t i v e e x i s t e n c e p r o o f . If we are able to find a m a t h e m a t i c a l object b such t h a t P(b) is true, such an existence proof is a c o n s t r u c t i v e p r o o f . The following example elucidates this method. Prove t h a t there is a positive integer t h a t can be expressed in two different ways as the sum of two cubes.

CONSTRUCTIVE PROOF By the discussion above, all we need is to produce a positive integer b t h a t has the required properties. Choose b - 1729. Since 1729 - 13 + 123 = 93 + 103, 1729 is such an integer.* m

*A fascinating anecdote is told about the number 1729. In 1919, when the Indian mathematical genius Srinivasa Ramanujan (1887-1920) was sick in a nursing home in England, the eminent

1.5 ProofMethods

53

A n o n c o n s t r u c t i v e existence proof of the t h e o r e m (3x)P(x) does not provide us with an e l e m e n t a for which P(a) is true, b u t r a t h e r establishes its existence by an indirect m e t h o d , usually contradiction, as i l l u s t r a t e d by the next example. Prove t h a t t h e r e is a p r i m e n u m b e r > 3.

NONCONSTRUCTIVE PROOF Suppose t h e r e are no primes > 3. T h e n 2 and 3 are the only primes. Since every integer >_ 2 can be expressed as a p r o d u c t of powers of primes, 25 m u s t be expressible as a p r o d u c t of powers of 2 and 3, t h a t is, 25 - 2 i 3 j for some integers i and j. But n e i t h e r 2 nor 3 is a factor of 25, so 25 c a n n o t be w r i t t e n in the form 2 i 3 j , a contradiction. Consequently, t h e r e m u s t be a p r i m e > 3. II We invite you to give a constructive proof of the s t a t e m e n t in the example. We conclude this section with a brief discussion of counterexamples.

Counterexample Is the s t a t e m e n t E v e r y g i r l is a b r u n e t t e t r u e or false? Since we can find at least one girl who is not a b r u n e t t e , it is false! More generally, suppose you would like to show t h a t the s t a t e m e n t (Vx)P(x) is false. Since ~[(Vx)P(x)] _= (3x)[~P(x)] by De M o r g a n ' s law, the s t a t e m e n t (Vx)P(x) is false if t h e r e exists an item x in the UD for which the predicate P(x) is false. Such an object x is a c o u n t e r e x a m p l e . Thus, to disprove the proposition (Vx)P(x), all we need is to produce a c o u n t e r e x a m p l e c for which P(c) is false, as the next two examples d e m o n s t r a t e . N u m b e r theorists d r e a m of finding formulas t h a t g e n e r a t e p r i m e n u m b e r s . One such f o r m u l a was found by the Swiss m a t h e m a t i c i a n L e o n h a r d E u l e r (see C h a p t e r 8), namely, E ( n ) - n 2 - n + 41. It yields a p r i m e for n = 1, 2, . . . , 40. Suppose we claim t h a t the f o r m u l a g e n e r a t e s a p r i m e for every positive integer n. Since E(41) = 412 - 41 + 41 = 412 is not a prime, 41 is a counterexample, t h u s disproving the claim. II A r o u n d 1640, F e r m a t conjectured t h a t n u m b e r s of the form f ( n ) - 22'' + 1 are prime n u m b e r s for all n o n n e g a t i v e integers n. For instance, f(0) = 3, f(1) = 5, f(2) = 17, f(3) = 257, and f(4) = 65,537 are all primes. In 1732, however, E u l e r established the falsity of F e r m a t ' s conjecture by producing a counterexample. He showed t h a t f(5) - 22'~ + 1 - 641 • 6700417, a composite n u m b e r . (Prime n u m b e r s of the form 22'' + I are called F e r m a t primes.) I1 English mathematician Godfrey Harold Hardy (1877-1947) visited him. He told Ramanujan that the number of the cab he came in, 1729, was "a rather dull number" and hoped that it wasn't a bad omen. "No, Hardy," Ramanujan responded, "It is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways."

Chapter I The Language of Logic

54

Exercises 1.5 D e t e r m i n e if each implication is vacuously t r u e for the indicated value of n. 1. If n > 1, t h e n 2 n > n; n - 0 2. If n > 4, t h e n 2 n >_ n2"n, - O, 1, 2, 3 D e t e r m i n e if each implication is trivially true. 3. If n is a p r i m e n u m b e r , t h e n n 2 -+- n is an even integer. 4. If n > 41, t h e n n 3 - n is divisible by 3. Prove each directly. 5. The s u m of a n y two even i n t e g e r s is even. 6. The s u m of a n y two odd i n t e g e r s is even. 7. The s q u a r e of an even i n t e g e r is even. 8. The product of any two even i n t e g e r s is even. 9. The s q u a r e of an odd integer is odd. 10. The product of any two odd i n t e g e r s is odd. 11. The product of any even i n t e g e r a n d any odd integer is even. 12. The s q u a r e of every integer of t h e form 3k + 1 is also of t h e s a m e form, w h e r e k is an a r b i t r a r y integer. 13. The s q u a r e of every integer of t h e form 4k + 1 is also of t h e s a m e form, w h e r e k is an a r b i t r a r y integer. 14. T h e a r i t h m e t i c

mean

~-~ of a n y two n o n n e g a t i v e real n u m b e r s a

and b is g r e a t e r t h a n or equal to t h e i r g e o m e t r i c m e a n j~ab. IHint" consider (v/-a- v/-b)2 > 0.] Prove each u s i n g the law of the contrapositive. 15. If the s q u a r e of an integer is even, t h e n the integer is even. 16. If the s q u a r e of an integer is odd, t h e n the i n t e g e r is odd. 17. If the product of two integers is even, t h e n at least one of t h e m m u s t be an even integer. 18. If the product of two integers is odd, t h e n both m u s t be odd integers. Prove by contradiction, w h e r e p is a p r i m e n u m b e r . 19. j ~ is an irrational n u m b e r . 21. v/~ is an irrational n u m b e r .

20. v ~ is an i r r a t i o n a l n u m b e r . *22. log102 is an i r r a t i o n a l n u m b e r .

Prove by cases, w h e r e n is an a r b i t r a r y integer and Ixl denotes t h e absolute value of x.

1.5 ProofMethods

55

23. n 2 + n is a n e v e n i n t e g e r .

25. n 3 - n is divisible by 3.

24. 2n 3 + 3n 2 + n is a n e v e n i n t e g e r .

( H i n t : A s s u m e t h a t e v e r y i n t e g e r is of t h e f o r m

3k, 3k + 1, or 3k + 2.)

26. ] - x [ = [ x [

27. [ x . y [ = [ x [ . [ y ]

28. [x +y] _< [x[ + [y[

P r o v e by t h e e x i s t e n c e m e t h o d . 29. T h e r e a r e i n t e g e r s x s u c h t h a t x 2 = x. 30. T h e r e a r e i n t e g e r s x s u c h t h a t [x] = x. 31. T h e r e a r e i n f i n i t e l y m a n y i n t e g e r s t h a t c a n be e x p r e s s e d as t h e s u m o f t w o c u b e s in t w o d i f f e r e n t ways. 32. T h e e q u a t i o n x 2 + y2 = z 2 h a s infinitely m a n y i n t e g e r s o l u t i o n s . Give a c o u n t e r e x a m p l e to d i s p r o v e e a c h s t a t e m e n t , w h e r e P(x) d e n o t e s a n arbitrary predicate. 33. T h e a b s o l u t e v a l u e of e v e r y real n u m b e r is positive. 34. T h e s q u a r e of e v e r y real n u m b e r is positive. 35. E v e r y p r i m e n u m b e r is odd. 36. E v e r y m o n t h h a s exactly 30 days. 37. (3x)P(x) ~

(3!x)P(x)

38. (~x)P(x) -~ (Vx)P(x) 39. F i n d t h e flaw in t h e following "proof": L e t a a n d b be real n u m b e r s s u c h t h a t a = b. T h e n a b = b 2. Therefore, a 2 - ab = a 2 - b 2 F a c t o r i n g , a ( a - b) - ( a + b ) ( a - b) C a n c e l a - b f r o m b o t h sides: a=a+b

Since a - b, t h i s yields a - 2a. C a n c e l a f r o m b o t h sides. T h e n we get 1 = 2. L e t a, b, a n d c b e a n y real n u m b e r s . T h e n a < b if a n d only if t h e r e is a positive real n u m b e r x s u c h t h a t a + x - b. U s e t h i s fact to p r o v e each. 40. If a < b a n d b < c, t h e n a < c. ( t r a n s i t i v e 41. I f a < b ,

thena+c
property)

56

Chapter I The Language of Logic

42. I f a + c < b + c ,

thena
43. Let a and b be any two real n u m b e r s such t h a t a . b - 0. T h e n e i t h e r a - 0 or b - O. [ H i n t : p ~ (q v r ) - ( p A ~ q ) ~ r.] *44. The f o r m u l a f ( n ) - n 2 - 79n + 1601 yields a p r i m e for 0 _< n < 10. Give a c o u n t e r e x a m p l e to disprove t h e claim t h a t t h e f o r m u l a yields a p r i m e for every n o n n e g a t i v e i n t e g e r n. *45. P r i m e n u m b e r s of the form f ( n ) = 2 n - 1, w h e r e n is a positive integer, are called M e r s e n n e p r i m e s , after the F r a n c i s c a n m o n k M a r i n M e r s e n n e (1588-1648). For example, f(2) = 3,f(3) = 7, a n d f ( 5 ) = 31 are M e r s e n n e primes. Give a c o u n t e r e x a m p l e to disprove t h e claim t h a t if n is a prime, t h e n 2 n - 1 is a prime.

This c h a p t e r p r e s e n t e d the f u n d a m e n t a l s of symbolic logic and the s t a n d a r d techniques of proving t h e o r e m s . Proposition 9 A p r o p o s i t i o n is a declarative sentence t h a t is either t r u e or false, b u t not both (page 2). 9 A c o m p o u n d p r o p o s i t i o n can be formed by c o m b i n i n g two or more simple propositions, using logical operators" A, v, 4 , 4 , and ( page 5). 9 The c o n j u n c t i o n of two propositions is t r u e if and only if b o t h compon e n t s are true; their d i s j u n c t i o n is t r u e if at least one c o m p o n e n t is true. An i m p l i c a t i o n is false only if the premise is t r u e and the conclusion is false. A b i c o n d i t i o n a l is t r u e if and only if both c o m p o n e n t s have the same t r u t h value (pages 5-14). 9 The t r u t h tables for the various logical operations can be combined into a single table, as in Table 1.18.

T a b l e 1.18

p

q

pAq

T T F F

T F T F

T F F F

(pVq)

,.~p

T T T F

F F T T

p--.q

T F T T

p~q

T F F T

9 T h r e e new implications can be c o n s t r u c t e d from a given implication: c o n v e r s e , i n v e r s e , and c o n t r a p o s i t i v e (page 11).

Chapter Summary

57

9 Various types of sentences and propositions can be s u m m a r i z e d in a tree diagram, as in Figure 1.11.

Figure 1.11

sentence ~

interrogative

declarative

~

~

imperative matory

proposition

simple

non-proposition

compound

conjunction

biconditional

disjunction

inclusive

exclusive

conditional

implication

converse

inverse

contrapositive

9 A tautology is a compound s t a t e m e n t t h a t is always true. A contradiction is a compound s t a t e m e n t t h a t is always false. A c o n t i n g e n c y is a proposition t h a t is neither a tautology nor a contradiction (page 16). 9 Two compound propositions, p and q, are l o g i c a l l y e q u i v a l e n t if t h e y have identical t r u t h values, symbolized by p - q (page 20). 9 The i m p o r t a n t laws of logic are listed in Table 1.13 on page 21.

Argument 9 An a r g u m e n t H1 A H2 A . . - A Hn ---> C is valid if the implication is a tautology; otherwise, it is invalid (page 39). 9 The i m p o r t a n t inference rules are listed in Table 1.15 on page 43.

Quantifiers 9 There are two quantifiers: u n i v e r s a l quantifier (V) and e x i s t e n t i a l quantifier (3) (page 32). 9 A predicate P(x) is a sentence about the properties of the object x. The set of all values of x is the u n i v e r s e o f d i s c o u r s e (UD) (page 33).

9 De Morgan's laws:

--~[(Vx)P(x)] = (3x)[~P(x)l ~[(3x)P(x)] ~ (Vx)[~P(x)l

(page 35)

Chapter I The Language of Logic

58 Proof Methods

9 T h e r e are six c o m m o n l y used proof techniques: v a c u o u s proof, trivial proof, direct method, indirect method, proof by cases, a n d existence proof (pages 49-53).

Figure 1.12

proof vacuous

trivial

direct

contrapositive

indirect

by cases

contradiction

exi~sence

constructive

nonconstructive

9 In a direct proof, a s s u m e the given h y p o t h e s e s are true. T h e n t r y to reach the given conclusion logically (page 50). 9 For indirect proof by contrapositive, a s s u m e the given conclusion is false. T h e n establish directly t h a t the given hypothesis is also false (page 51). 9 For indirect proof by contradiction, a s s u m e the given hypothesis is true, but the given conclusion is false. T h e n try to reach a contradiction ( page 51). 9 For a constructive existence proof of a t h e o r e m (3x)P(x), produce an element b such t h a t P(b) is t r u e (page 52). In a n o n c o n s t r u c t i v e existence proof, establish the existence of such an element b by an indirect m e t h o d (page 53).

Counterexample 9 To disprove the proposition (Vx)P(x), it suffices to produce an object c for which P(c) is false (page 53).

Review Exercises C o n s t r u c t a t r u t h table for each proposition. 1. (p vq) A (~q)

2. (p ~ q) ~ r

3. p --. (q --* r)

4. (p ~ q) V r

Evaluate each boolean expression, where a - 3, b -- 7, c - 2, and d - 11.

5. (a_ c ) v ( b > d ) ]

6. l ( a > b ) A ( b < c ) l v ( c < d )

7. ( c > d ) v l ( b < c ) A ( d < b ) l

8. ( b < c ) v ~ [ ( a < c ) A ( c < d ) l

Represent each sentence symbolically, numbers. 9. I f w < x a n d y < z ,

thenw+y<x+z.

where w , x , y ,

and z are real

Chapter Summary

59

10. If w - x a n d y - z, t h e n w .y = x . z. D e t e r m i n e if t h e a s s i g n m e n t s t a t e m e n t x ~ y + z will be e x e c u t e d in e a c h s e q u e n c e of s t a t e m e n t s , w h e r e i ~ 5, j ~-- 3, a n d k ~ 7. 11. If (i < j ) v (j _< k) t h e n

12. If (i > j ) A (j > k) t h e n

x+--y+z 13. I f ~ [ ( i

x+--y+z

> j ) v (j

<

k)] t h e n

x+--y-z

14. O d d +--0 w h i l e (odd < 2) A (i < 4) do

x+--y+z

else

x+--y+z R e p r e s e n t e a c h n e t w o r k symbolically. 15.

| |

| | 9

16.

-@

|

Let t be a t r u e s t a t e m e n t a n d p a n a r b i t r a r y s t a t e m e n t . F i n d t h e t r u t h v a l u e of each. 17. p v

~ t --+p

18. p

v

t ~ ~t

19. t A ( p V t )

20. p A t -->pVt

U s e t h e g i v e n i n f o r m a t i o n to d e t e r m i n e t h e t r u t h v a l u e of e a c h s t a t e m e n t . 21. p ~ q, i f p v q is false.

22. p --+ q, if ~ p v q is false.

23. p --* ~ q , if q ~

24. p A q, i f p - . q is false.

~ p is false.

25. p v q, i f p ~ q is false.

26. p ~ q, i f p A q is t r u e .

27. p ~ q, i f p A ~ q is t r u e .

28. p v ~ q , if q A ~ p is t r u e .

29. p A (q A r), if r --= s a n d s is n o t t r u e . Give t h e c o n v e r s e , i n v e r s e , a n d c o n t r a p o s i t i v e of e a c h i m p l i c a t i o n . 30. If P a t is a girl, t h e n she h a s g r e e n eyes. 31. I f x < y ,

thenx+z
60

Chapter 1 The Language of Logic

Write t h e contrapositive of each implication. 32. If Ixl < 3, t h e n x < 3 a n d x > - 3 . 33. If Ixl > 3, t h e n e i t h e r x > 3 o r x < - 3 .

D e t e r m i n e if each is a logical equivalence. 34. p A q = ~ ( p o

35.

uq)

36. p A (p v q ) = p v (p A q )

~ ( p A ~ q ) = u p v q.

37. p --~ (q --~ r) -- (p --~ q) --~ r

D e t e r m i n e if each is a tautology. 38. p v (p A q ) ~ p

39. p A (p v q ) o p

40.

41. (p --~ ~q) ~ (q ~ u p )

(p A q) ~ u ( p ~ ~ q )

M a r k t r u e or false, w h e r e p, q, r, and s are a r b i t r a r y s t a t e m e n t s . 42. I f q _ = r , t h e n p A q - - = p A r .

43. I f q - r ,

thenpvq-pvr.

44. I f p _ = q , t h e n p ~

45. I f q - r ,

thenp~q-p~r.

r=q~r.

46. If p ,-. q is a tautology, t h e n

47. I f p --~ q - p --. r, t h e n q - r.

p-q.

C o n s t r u c t an equivalent simpler switching n e t w o r k for each circuit. 48.

~

49.

Test the validity of each a r g u m e n t . 51. p __~ u q (q A r) ~ u s

50. p v q q-.r ur

rAS

... u p

... u p

52. E i t h e r J a n e is not J o h n ' s sister or M a r y is not H a r r y ' s wife. M a r y is H a r r y ' s wife or J a n e is not m a r r i e d . J o h n goes to school if and only if J a n e is not m a r r i e d . J o h n does not go to school. .'. J a n e is not J o h n ' s sister. D e t e r m i n e the t r u t h value of each, w h e r e the UD consists of the integers 0 and 1. 53. ( 3 x ) ( x 3 ~= X) 55. ( 3 y ) l ( y - 1) 2 # y 2 _

54. (Vx)I(X + 1) 2 : X2 + 11 11

56. (Vx)(Vy)l(x + y ) 2 = x 2 +y21

Chapter Summary

61

Let UD - set of integers, P(x)" x < 3, and Q(x): x >_ 3. D e t e r m i n e the t r u t h value of each. 57. (Vx)lP(x) A Q(x)l

58. (Vx)[P(x) v Q(x)]

59. (3y)[P(y) A Q(y)]

60. (Sz)lP(z)vQ(z)l

61. (Vx)[~P(x)]

62. (3z)[~Q(z)]

Prove each, where a, b, c, d, and n are any integers. 63. The product of two consecutive integers is even. 64. n 3 + n is divisible by 2. 65. n 4 -- n 2 is divisible by 3. 66. I f a 12, t h e n e i t h e r a > 6 o r b

>6.

68. I f a b - ac, t h e n either a - 0 or b - c. [ H i n t : p ---> (qvr) - (p A~q) ~ r.] 69. If a 2 -- b 2, t h e n either a - b or a - - b . [ H i n t : p --> (q v r) -

( p A ~ q ) --+ r.]

70. Give a c o u n t e r e x a m p l e to disprove the following s t a t e m e n t : If n is a positive integer, t h e n n 2 + n + 41 is a prime n u m b e r . [Note: In 1798 the e m i n e n t F r e n c h m a t h e m a t i c i a n Adrien-Marie Legendre (1752-1833) discovered t h a t the formula L ( n ) - n 2 Zr- n + 41 yields distinct primes for 40 consecutive values of n. Notice t h a t L ( n ) - E ( - n ) ; see Example 1.45. ] The propositions

Let p , q , t(r)-0.5.

in Exercises

71-81 are fuzzy logic.

and r be simple propositions with t ( p )

-

1, t ( q ) -

0.3,

and

C o m p u t e the t r u t h value of each, where s' denotes the negation of the s t a t e m e n t s. 71. p A ( q v r )

72. p V ( q A r )

73. ( p A q ) V ( p A r )

74. ( p V q ) A ( p V r )

75. p'

76. (p

A

q'

77. (p v q')' v q

v

q')

v

(p

A

q)

78. (p A q)' A (p V q)

79. Let p be a simple proposition with t ( p ) = x and p' its negation. Show t h a t t ( p v p') = 1 if and only if t ( p ) = 0 or 1. Let p and q be simple propositions with t ( p ) 0 <_ x , y _< 1. Verify each.

-

x and t ( q )

-

80. ( p A q)' -- p ' V q' [ H i n t : Show t h a t t ( ( p A q ) ' ) -- t ( p ' ) v t(q').] 81. ( p v q)' -- p ' A q' [ H i n t : Show t h a t t ( ( p v q ) ' ) = t ( p ' ) A t(q').]

y, where

62

Chapter I

The Language of Logic

Supplementary Exercises W r i t e t h e converse, inverse, a n d c o n t r a p o s i t i v e of e a c h i m p l i c a t i o n . 2. Iflxl > a , t h e n x < - a o r x > a .

1. Iflxl < a , t h e n - a < x < a . Simplify each b o o l e a n expression.

3. (p

v

~q)

A

*5. (p

A

~q)

v

~(p (~p

A A

*4. [p v q v (~p A -~q)] v (p A ~q)

q)

q)

v

(~p

A

*6. (p V q) A "~(p A q) A (~p V q)

~q)

7. Let p - q a n d r - s. D e t e r m i n e i f p --+ (p A r) -- q --+ (q A S). N e g a t e each proposition, w h e r e U D = set of real n u m b e r s .

9. (Vx)(Vy)(xy = yx)

8. (VX)(3y)(xy >__1)

11. (Vx)(3y)(3z)(x + y = z)

10. (Vx)(Vy)(3z)(x + y = z) P r o v e each.

12. T h e e q u a t i o n x 3 + y3 = z 3 h a s infinitely m a n y i n t e g e r solutions. "13. Let n be a positive integer. T h e n n(3n 4 + 7n 2 + 2) is divisible by 12. "14. Let n be a positive integer. T h e n n(3n 4 + 13n 2 + 8) is divisible by 24. "15. In 1981 O. H i g g i n s discovered t h a t t h e f o r m u l a h(x) = 9x 2 471x + 6203 g e n e r a t e s a p r i m e for 40 c o n s e c u t i v e v a l u e s of x. Give a c o u n t e r e x a m p l e to show t h a t not every value of h(x) is a p r i m e . "16. T h e f o r m u l a g ( x ) = x 2 - 2999x + 2248541 yields a p r i m e for 80 consecutive v a l u e s ofx. Give a c o u n t e r e x a m p l e to disprove t h a t every value o f g ( x ) is a prime. In a t h r e e - v a l u e d l o g i c , developed by t h e Polish logician J a n L u k a s i e w i c z (1878-1956), t h e possible t r u t h values of a p r o p o s i t i o n a r e 0, u, a n d 1, w h e r e 0 r e p r e s e n t s F, u r e p r e s e n t s u n d e c i d e d , a n d 1 r e p r e s e n t s T. T h e logical c o n n e c t i v e s A, V, ', --~, a n d o are defined as follows: A

0

v

U

0

!

u

0

0

0

0

0

u

0

1

u

0

u

u

u

u

u

u

1

0

u

1

1

1

1

0

--+

0

u

0 u 1

1 u 0

1 1 u

0 u 1

0

u

1 u 0

u 1 u

Let p a n d q be a r b i t r a r y p r o p o s i t i o n s in a t h r e e - v a l u e d logic, w h e r e r' d e n o t e s t h e n e g a t i o n of s t a t e m e n t r a n d t(r) d e n o t e s t h e t r u t h value of r.

Chapter Summary

63

17. If t(p v p') = 1, show t h a t t(p) = 0 or 1. 18. Show t h a t p A q ~ p v q is a three-valued tautology. 19. Show t h a t (p --+ q) ~ (p' v q) is not a three-valued tautology. 20. Show t h a t (t9 ~ q) ~ (~q ~ ~p) is a three-valued tautology. 21. D e t e r m i n e if [p A (p ~ q)] ~ q is a three-valued tautology. Verify each. 22. (p A q)' =_p' v q'

[Hint: Show t h a t t((p A q)') = t(p') V t(q').]

23. (p V q)' -- p' A q'

[Hint: Show t h a t t((p v q)') = t(p') A t(q').]

Computer Exercises Write a p r o g r a m to perform each task. C o n s t r u c t a t r u t h table for each proposition. 1. ( p v q ) A ~ q

2. p N A N D q

3. p N O R q

4. (p ~ q) ~ ( ~ p v q)

5. (p --+ q) ~ r

6. (p --+ q) <-+ ( ~ q ~

~p)

D e t e r m i n e if each proposition is a tautology, by c o n s t r u c t i n g a t r u t h table. 7. p A ( p o q ) - - + q

9. p A ( p v q )

op

11. p A q --~ p V q

8. ( p V q ) A ( ~ q ) - - + p

10. ( p ~ q )

A ( ~ q ) - ~ ~p

12. (p --~ q) A (q --~ r) --~ (p -~ r)

D e t e r m i n e if the given propositions are logically equivalent, by c o n s t r u c t i n g t r u t h tables. 13. ~ ( p A q), ~ p A ~ q

14. p --~ q, ~ q --~ --p

15. p A ( q A r ) , (pAq) Ar

16. p A ( q V r ) , (pAq) V ( p A r )

17. (p --~ q) --~ r, p -~ (q --~ r)

18. p --~ (q v r), p A (~q) -+ r

Exploratory Writing Projects Using library and I n t e r n e t resources, write a t e a m r e p o r t on each of the following in y o u r own words. Provide a well-documented bibliography. 1. Write an essay on the c o n t r i b u t i o n s of G. Boole and W. Leibniz to m a t h e m a t i c a l logic. 2. Explain how (symbolic) logic helps you in everyday life. Give concrete examples. 3. Explain why proofs are i m p o r t a n t in m a t h e m a t i c s and c o m p u t e r science. Do they help you in everyday life? In problem-solving? In a work e n v i r o n m e n t ? Give examples of proofs in c o m p u t e r science.

Chapter I The Language of Logic

64

4. Give a detailed history of Fermat's last theorem. Include biographies of mathematicians who have worked on the problem. 5. Collect a number of well-known conjectures from number theory and explain recent advances toward establishing them or disproving them. Study a number of puzzles from R. M. Smullyan's Alice in Puzzle-land, What is the name of this book ?, and The Lady or the Tiger. 6. Write each as an argument and test the validity of it. 7. Write each as a theorem and establish it. 8. List a number of applications of fuzzy logic to everyday life. How do they enrich our lives? 9. Write a biography of H. M. Sheffer, C. S. Peirce, and A. M. Legendre. 10. Collect several examples on arguments from Lewis Carroll's Symbolic Logic and test the validity of each. Explain with examples the use of Euler diagrams in the analysis of arguments. 11. Collect several logic puzzles from recent issues of Discover magazine and Parade magazine. Solve each. 12. Study logic problems in the recent edition of The official L S A T PrepBook and solve them. 13. List several attempts to develop formulas for generating prime numbers. 14. Write an account of Fermat primes, Mersenne primes, the infinitude of each family, and their applications. 15. Investigate the pentomino puzzle, developed in 1954 by S. W. Golomb of the University of Southern California. Enrichment Readings

1. L. Carroll, Symbolic Logic and the Game of Logic, Dover, New York, 1958. 2. L. Carroll, Alice's Adventures in Wonderland and Through the Looking-Glass and What Alice Found There, Oxford, New York, 1982. 3. N. Falletta, The Paradoxicon, Wiley, New York, 1990. 4. M. Gardner, Mathematical Puzzles and Diversions, The University of Chicago Press, Chicago, 1987. 5. J. T. Johnson, "Fuzzy Logic," Popular Science, Vol. 237 (July 1990), pp. 87-89. 6. B. Kosko and S. Isaka, "Fuzzy Logic," Scientific American, Vol. 269 (July 1993), pp. 76-81.

Chapter Summary

65

7. H. T. Nguyen and E. A. Walker, A First Course in Fuzzy Logic, 2nd ed., Chapman and Hall/CRC, Boca Raton, FL, 2000. 8. R. M. Smullyan, Alice in Puzzle-land, Penguin, New York, 1984. 9. R. M. Smullyan, What is the name of this book?, Prentice Hall, Englewood Cliffs, NJ, 1978. 10. R. M. Smullyan, The Lady or the Tiger?, Random House, New York, 1992. 11. D. Solow, How to Read and Do Proofs, Wiley, New York, 1982.

Chapter 2

The L a n g u a g e of S e t s The essence of mathematics lies in its freedom. -- GEORG

CANTOR

T

he concept of a set is so f u n d a m e n t a l t h a t it unifies m a t h e m a t i c s and its cognates. It has revolutionized m a t h e m a t i c a l thinking, enabling us to express ourselves in clear and concise terms. The foundation of set theory was laid by the e m i n e n t G e r m a n mathematician Georg Cantor during the latter part of the 19th century. "Today, Cantor's set theory has p e n e t r a t e d into almost every b r a n c h of m a t h e m a t ics," as the m a t h e m a t i c a l historian Howard Eves writes in An Introduction to the History of Mathematics. In this chapter we present the language of sets. We introduce the concept of a set, the various ways of describing a set and of constructing new sets from known sets, a variety of applications, and a brief introduction to fuzzy sets. The following are some of the problems we shall p u r s u e in this chapter: 9 Find the n u m b e r of positive integers < N and divisible by a, b, or c. 9 How m a n y subsets does a finite set with n elements have? 9 How would you define the set of legally paired parentheses? 9 How m a n y sequences of legally paired p a r e n t h e s e s can be formed using n pairs of left and right parentheses?

This section introduces the concept of a set, various methods of defining sets, and relationships between sets.

67

Chapter2 The Language of Sets

68

Georg Cantor (1845-1918) was born in St. Petersburg, Russia, where his father was a successful merchant and broker. Cantor showed great interest in mathematics from early childhood. In 1856, the family moved to Germany. Six years later, he entered the University of Zurich, but in the following year he moved to the University of Halle to study mathematics, physics, and philosophy. There he was greatly influenced by the eminent mathematician Karl Weierstrass (1815-1897). Although his father wanted him to become an engineer, Cantor relentlessly pursued his interest in mathematics and received his doctorate of philosophy at 22 from the University of Berlin for his work in number theory. In 1869, Cantor began his professional career as an unsalaried lecturer at the University of Halle. Five years later, he published his revolutionary work on set theory. Cantor developed an arithmetic of transfinite numbers analogous to that of finite numbers, thus creating another area of mathematical study. He proved that the set of real numbers is uncountable and he also established the existence of infinitely many different transfinite cardinal numbers by ingenious methods. He also made significant contributions to indeterminate equations and trigonometric series. Deeply religious, Cantor was also interested in art, music, and philosophy. Being unhappy with his low salary at the University, Cantor tried to secure a better-paid position at the University of Berlin, but was sabotaged by Leopold Kronecker (1823-1891), an eminent mathematician at the University, who severely criticized Cantor's views on sets. Relentless attacks by contemporary mathematicians intensified the manic depression he suffered from. Cantor died in a mental hospital in Halle in 1918. Cantor was "one of the greatest intellects of the nineteenth century," according to Bertrarzd Russell. He "was an imaginative genius whose work has inspired/evel;y aspect of/mathematical thought," Hazel Perfect of the UnJiversity of Sheffield wrote in 1994.

Set A s e t is a collection of well-defined objects,* called e l e m e n t s (or m e m b e r s ) of the set. T h e r e should be no a m b i g u i t y in d e t e r m i n i n g w h e t h e r or not a given object belongs to the set. For example, the vowels of the E n g l i s h a l p h a b e t form a (well-defined) set, w h e r e a s beautiful cities in the U n i t e d S t a t e s do not form a set since its m e m b e r s h i p would be debatable. Sets are denoted by capital letters and t h e i r e l e m e n t s by lowercase letters. If an object x is an e l e m e n t of a set A, we write x E A; otherwise x CA. For example, let A be the set of New E n g l a n d states. T h e n Connecticut E A, w h e r e a s M i c h i g a n r A. T h e r e are two m e t h o d s of defining sets.

Listing Method A set can s o m e t i m e s be described by listing its m e m b e r s w i t h i n braces. For instance, the set B of New E n g l a n d states can be described as B = {ME, VT, NH, MA, CT, RI}. *To be precise, this is a circular definition; set is an undefined term, like point and line in geometry.

2.1 The Concept of a Set

69

The order in which the elements are e n u m e r a t e d is immaterial. Thus B can also be written as {VT, RI, MA, CT, NH, ME}. If an element is repeated, it is not counted more t h a n once. For example, { x , x , y , x , y , z } {x,y,z}. A set with a large n u m b e r of elements t h a t follow a definite p a t t e r n is often described using ellipses (...) by listing a few elements at the beginning. For example, the set of letters of the alphabet can be written as {a, b, c , . . . , z} and the set of odd positive integers as {1, 3, 5 , . . . }. Set-Builder Notation

Another way of describing a set is by using the s e t - b u i l d e r n o t a t i o n . Its general form is {x IP(x)}, where P(x) is a predicate indicating the p r o p e r t y (or properties) the object x has. You may read {x I P(x) } as the set c o n s i s t i n g o f all objects x such that x has the property P(x). Here the vertical bar "1" means such that. (Again, the m e a n i n g of the vertical bar should be clear from the context.) Let B be the set of all m o n t h s of the year with exactly 30 days. Then B - {xlx is a m o n t h of the year with exactly 30 days }

= {September, April, June, November}

I1

Next we present another of Russell's paradoxes introduced in 1901, which is quite similar to the barber paradox. Russell's Paradox

Let S = {X IX r X}; that is, S consists of all sets that do not belong to themselves as elements. Does S c S? I f S ~ S, then, by definition, S ~ S; on the other hand, if S r S, then, again by definition, S E S. Thus, in either case, we have a contradiction. This paradox shows, not every predicate defines a set; t h a t is, there is no set of all sets. Next we present several relationships between sets. Subset

If every element of A is also an element of B, A is a s u b s e t of B, denoted by A c_ B. In symbols, (A c_ B) o (Vx)(x ~ A --* x e B). If A c_ B, we also say that B c o n t a i n s A and write B D A. If A is not a subset of B, we write A ~ B; thus (A ~ B) ~ (3x)(x c A A x r B). Let A = set of states in the United States, B - set of New England states, and C - set of Canadian provinces. Then B g A, but B ~ C and A ~ C. I1 To show t h a t a set X is a subset of Y, select an arbitrary element x in X; then using the laws of logic and known facts, show that x is in Y also. We shall apply this technique in later sections. To show t h a t X ~ Y, all you need is to find an element x e X which does not belong to Y.

70

Chapter 2 The Language of Sets

Equal Sets Two sets A and B are e q u a l , denoted by A = B, if they contain the same I elements. In other words, A - B if (A _ B)/x (B c__A). (We shall use this ] property to prove the equality of sets.) If A c B and A 7~ B, t h e n A is a p r o p e r s u b s e t of B, denoted by A c B. Consider the sets A - {x ]x is a vowel of the alphabet}, B - {a, e, i, o, u}, C - {2, 3, 4}, and D - { x l x is a digit in the n u m e r a l 23432.} Then A - B, and C - D . m Does a set have to contain any element? Can there be a set with no elements? Suppose Fred went h u n t i n g in a nearby jungle and r e t u r n e d home with great tales, but no animals. The set of animals he caught is null. This leads us to the following definition.

Empty Set The set containing no elements is the empty (or n u l l ) set; it is denoted by 0 o r {}. The set of pink elephants is empty. So are the set of m o u n t a i n s on the earth t h a t are 50,000 feet tall and the set of prime n u m b e r s between 23 and 28. m Many people mistakenly believe t h a t {~} = IZI; this is n o t true, since {0} contains an element ~, whereas ~ = {} contains no elements. T h u s

Logically, it can be proved t h a t ~ is a subset of every set; t h a t is, 121 _ A for every set A. Besides, although m a n y people t h i n k t h a t there are m a n y empty sets, it can be proved t h a t it is unique, m e a n i n g there is exactly one empty set. (See Exercises 53 and 54.)

Universal Set It is always possible to choose a special set U(r ~) such that every set u n d e r discussion is a subset of U. Such a set is called a u n i v e r s a l set, denoted by U. Thus A c_ U for every set A. Suppose we wish to discuss something about the sets {a}, {b, c, d}, and {b, d, e, f}. Then U - {a, b, c, d, e, f} may be chosen as a valid universal set. (There are other valid choices also.) m

2.1

The Concept of a Set

71

(optional) Programming languages such as Pascal support the data type SET, although the implementations have a limit on the number of elements on the base-type of the set, that is, on the size of the universal set. For example, consider the Pascal declarations: TYPE

MONTHS = (JAN, FEB, MAR, APR, MAY, JUN, JUL, AUG, SEP, OCT, NOV, DEC); SETOFMONTHS = SET OF MONTHS; VAR

SPRING,SUMMER,FALL,WINTER: SETOFMONTHS;

Here the universal set is SETOFMONTHS = { JAN, FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC}.

The above variable declarations define four set variables, namely, SPRING, SUMMER, FALL, and WINTER. The set values assigned to them must be subsets of SETOFMONTHS. For instance, SPRING := [JAN,FEB,MAR];

is a legal Pascal assignment, although it is preposterous. The set membership operator in Pascal is IN and can be used to determine if an element belongs to a set. For example, FEB IN SPRINGis a legal boolean expression. Likewise, the set inclusion and containment operators are < = and > =, respectively, m

Disjoint Sets Sets need not have common elements. Two such sets are d i s j o i n t sets. For example, the sets {Ada, BASIC, FORTRAN} and {C++, Java} are disjoint; so are the sets { + , - , . , / } and {A, v , - ~ , ~ }.

Venn Diagrams Relationships between sets can be displayed using Venn diagrams, named after the English logician John Venn. In a Venn diagram, the universal set U is represented by the points inside a rectangle and sets by the points enclosed by simple closed curves inside the rectangle, as in Figure 2.1. Figure 2.2 shows A c B, whereas Figure 2.3 shows they are not disjoint.

F i g u r e 2.1

U

Chapter 2

72 Figure

2.2

The Language of Sets

U

AcB.

Figure

2.3

A and B m a y have c o m m o n elements.

9 . ~:~,~" ,,,~,

J o h n V e n n (1834-1923) was born into a philanthropic family in Hull, England. After attending the high schools at Highgate and Islington, in 1853 he entered Gonville and Caius College, Cambridge, and graduated in mathematics three years later. He was elected a fellow of the College, a position he held until his death. 9 'x~ In 1859 Venn was ordained in the Church of England, but after a brief period of church work, he returned to Cambridge as a lecturer on moral sciences. In 1883 he gave up his priesthood. The same year, he received a D.Sc. . . . . . !,. ~ ,::.,~, from Cambridge and was elected a fellow of the Royal Society of London. Venn was greatly influenced by Boole's work in ,symbolic logic. Venn's masterpiece, Symbolic Logic (1881), clarifies the inconsistencies and ambiguities in Boole's ideas and notations. He employed geometric diagrams to represent logical arguments, a technique originated by Leibniz and developed further by Euler. Venn added a rectangle to represent the universe of discourse. Venn published two additional books, The Logic of Chance (1866) and The Principles of Empirical Logic (1889).

~

.,. ;~:.,=~.,"

I

Can the elements of a set be sets? Certainly. { {a}, {b, c} }, and {0, {0}, {a, b } } are two such sets. In fact, the subsets of a set can be used to build a new set.

Power Set

The family of subsets of a set A is the p o w e r Find the power set P(A) of the set A - {a, b}.

s e t of A, denoted by P(A).

2.1 The Concept of a Set

73

SOLUTION: Since 0 is a subset of every set, 0 e P(A). Also, {a} a n d {b} are s u b s e t s of A. F u r t h e r , every set is a s u b s e t of itself, so A E P(A). T h u s , t h e v a r i o u s e l e m e n t s of P(A) are O, {a}, {b}, a n d A; t h a t is, P(A) - {0, {a}, {b},A}. m Sets can be classified as finite a n d infinite sets, as defined below.

Finite and Infinite Sets A set w i t h a definite n u m b e r of e l e m e n t s is a f i n i t e set. A set t h a t is n o t finite is i n f i n i t e . T h e sets {a,b,c} a n d t h e set of c o m p u t e r s in t h e w o r l d are finite, b u t t h e set of i n t e g e r s a n d t h e set of points on a line are infinite, m It m a y s o m e t i m e s be difficult to k n o w t h e exact n u m b e r of e l e m e n t s in a finite set. B u t t h a t does not affect its finiteness. F o r example, t h e set of r e s i d e n t s in California at a given t i m e is finite, a l t h o u g h it is difficult to d e t e r m i n e the actual count. It is impossible to list all t h e e l e m e n t s of an infinite set. C o n s e q u e n t l y , the e n u m e r a t i o n m e t h o d w i t h ellipsis or t h e set-builder n o t a t i o n is u s e d to define infinite sets. In t h e f o r m e r case, t h e ellipsis would come at t h e end of t h e list, for example, 1~ - {1, 2, 3 , . . . }. T h e following are some special infinite sets we will be u s i n g f r e q u e n t l y : Z = set of i n t e g e r s = {..., - 2 , - 1, 0, 1, 2 , . . . } 1~ = Z + = set of positive i n t e g e r s = {1, 2, 3 , . . . } Z - = set of negative i n t e g e r s - {..., - 3 , - 2 , - 1 } W = set of whole n u m b e r s - {0, 1, 2, 3 , . . . } Q - set of r a t i o n a l n u m b e r s - { p / q l p , q ~ Z A q # 0} R = set of real n u m b e r s R + = set of positive real n u m b e r s = {x ~ RIx > 0} IR- = set of negative real n u m b e r s - {x E IRIx < 0} A few additional subsets of R, called i n t e r v a l s , will prove useful in o u r discussions. T h e y are given below, w h e r e a < b"

closed interval

[a,b] - {x ~ IRla _< x _< b}

closed-open

interval

[a,b) - {x c Rla _< x < b}

open-closed

interval

(a,b] - {x ~ IRla < x _< b}

open interval

(a,b) -

{x ~ I~la < x < b}

Chapter2 The Language of Sets

74

-,q

~.,~,~.

.

9

,,

D a v i d H i l b e r t (1862-1943) was born and educated in KSnigsberg, Germany (now in Russia). He made significant contributions to algebra, analysis, geometry, and mathematical physics. He described the importance of set theory in the development of mathematics: "No one shall expel us from the paradise which Cantor has created for us."

..

.$. ~.,~'

A bracket at an endpoint indicates it is included in the set, whereas a parenthesis indicates it is not included. The set {x c R ix > a} is denoted by [a, oc) using the i n f i n i t y s y m b o l o~. Likewise, the set {x ~ R Ix < a} is denoted by ( - ~ , a]. Next we present two interesting paradoxes related to infinite sets and proposed in the 1920s by the G e r m a n m a t h e m a t i c i a n David Hilbert.

The Hilbert Hotel Paradoxes Imagine a grand hotel in a major city with an infinite n u m b e r of rooms, all occupied. One m o r n i n g a visitor arrives at the registration desk looking for a room. "I'm sorry, we are full," replies the manager, "but we can certainly accommodate you." How is this possible? Is she contradicting herself?. To give a room to the new guest, Hilbert suggested moving the guest in Room 1 to Room 2, the guest in Room 2 to Room 3, the one in Room 3 to Room 4, and so on; Room 1 is now vacant and can be given to the new guest. The clerk is happy t h a t she can accommodate him by moving each guest one room down the hall. The second paradox involves an infinite n u m b e r of conventioneers arriving at the hotel, each looking for a room. The clerk realizes t h a t the hotel can make a fortune if she can somehow accommodate them. She knows she can give each a room one at a time as above, but t h a t will involve moving each guest constantly from one room to the next, resulting in total chaos and frustration. So Hilbert proposed the following solution: move the guest in Room 1 to Room 2, the guest in Room 2 to Room 4, the one in Room 3 to Room 6, and so on. This puts the old guests in even-numbered rooms, so the new guests can be checked into the odd-numbered rooms. Notice t h a t in both cases the hotel could accommodate the guests only because it has infinitely m a n y rooms.

2.1 The Concept of a Set

75

A third paradox: Infinitely m a n y hotels with infinitely m a n y rooms are leveled by an earthquake. All guests survive and come to Hilbert Hotel. How can they be accommodated? See Example 3.23 for a solution. We close this section by introducing a special set used in the study of formal languages. Every word in the English language is an a r r a n g e m e n t of the letters of the alphabet {A, B , . . . ,Z, a, b , . . . , z}. The alphabet is finite and not every a r r a n g e m e n t of the letters need make any sense. These ideas can be generalized as follows.

Alphabet A finite set Z of symbols is an alphabet. (E is the uppercase Greek letter sigma.) A w o r d (or s t r i n g ) o v e r E is a finite a r r a n g e m e n t of symbols from E. For instance, the only alphabet understood by a computer is the b i n a r y a l p h a b e t {0,1 }; every word is a finite and unique a r r a n g e m e n t of O's and l's. Every zip code is a word over the alphabet {0,... ,9}. Sets such as {a, b, c, ab, bc} are not considered alphabets since the string ab, for instance, can be obtained by juxtaposing, that is, placing next to each other, the symbols a and b.

Length of a Word The l e n g t h of a word w, denoted by llwli, is the n u m b e r of symbols in it. A word of length zero is the e m p t y w o r d (or the null w o r d ) , denoted by the lowercase Greek letter ~ ( l a m b d a ) ; It contains no symbols. For example, llabll = 2, llaabba]l - 5, and IIs = 0. The set of words over an alphabet E is denoted by Z*. The empty word belongs to Z* for every alphabet Z. In particular, if Z denotes the English alphabet, then Z* consists of all words, both meaningful and meaningless. Consequently, the English language is a subset of Z*. More generally, we make the following definition.

Language A l a n g u a g e over an alphabet Z is a subset of E*. The following two examples illustrate this definition. The set of zip codes is a finite language over the alphabet E -- { 0 , . . . , 9}. m Let E - {a, b}. Then E* - {~, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, b a a , . . . }, an infinite set. Notice that {aa, ab, ba, bb} is a finite language over E, whereas {a, aa, aba, bab, aaaa, a b b a , . . . } is an infinite language, m Words can be combined to create new words, as defined below.

Chapter2 The Language of Sets

76 Concatenation

The c o n c a t e n a t i o n of two words x a n d y over an a l p h a b e t , d e n o t e d by x y , is obtained by a p p e n d i n g t h e word y at t h e end of x. T h u s if x = x l . . . X m a n d y = Yl...Yn, xy - Xl . . .XmYl . . .Yn. For example, let E be t h e E n g l i s h alphabet, x = CAN, a n d y = ADA; t h e n x y = CANADA. Notice t h a t c o n c a t e n a t i o n is n o t a c o m m u t a t i v e operation; t h a t is, x y 7/= y x . It is, however, associative; t h a t is, x ( y z ) = ( x y ) z = x y z . Two i n t e r e s t i n g p r o p e r t i e s are satisfied by t h e c o n c a t e n a t i o n operation: 9 The c o n c a t e n a t i o n of a n y word x with )~ is itself; t h a t is, ~x - x - x~ for every x e E*. 9 Let

x,y

~

E*. T h e n

[]xY]l -

][xll +

ilYi]. (See Section 5.1 for a proof.)

For example, let E = {a,b }, x = aba, and y - bbaab. T h e n x y - a b a b b a a b and [[xyl[ - 8 - 3 + 5 - IIxl[ + IlyJl. A useful notation: As in algebra, t h e e x p o n e n t i a l n o t a t i o n can be employed to e l i m i n a t e the r e p e a t i n g of symbols in a word. Let x be a symbol and n an i n t e g e r >_ 2; t h e n x n denotes the c o n c a t e n a t i o n x x . . . x to n - 1 times. U s i n g this compact notation, the words a a a b b a n d a b a b a b can be a b b r e v i a t e d as a3b 2 a n d (ab) 3, respectively. Notice, however, t h a t (ab) 3 = ababab ~: a3b 3 - aaabbb.

Exercises 2.1 Rewrite each set using the listing method. 1. The set of m o n t h s t h a t begin with the l e t t e r A. 2. The set of letters of the word GOOGOL. 3. The set of m o n t h s with exactly 31 days. 4. The set of solutions of t h e e q u a t i o n x 2 - 5x + 6 - 0. Rewrite each set u s i n g the set-builder notation. 5. The set of integers b e t w e e n 0 and 5. 6. The set of J a n u a r y , F e b r u a r y , May, a n d July. 7. The set of all m e m b e r s of t h e U n i t e d Nations. 8. {Asia, Australia, Antarctica} D e t e r m i n e if t h e given sets are equal. 9.

{x,y,z},

{x,z,y}

11.

{xlx 2 -

x},

10.

{xlx 2

= 1}, {xlx 2 - x}

12. {x, {y}}, {{x},y}

{0, 1}

M a r k each as t r u e or false.

13. a E {alfa}

14. b _ {a,b,c}

15. {x} _ {x,y, z}

2.1 The Concept of a Set

77

16. { 0 } - 0

17. 0 ~ 0

18. { 0 } - 0

19. {0} = 0

20. 0 __ 0

21. 0 ~ {0}

22. {xlx ~ x } - 0

23. {x,y} - {y,x}

24. {x} ~ {{x},y}

25. 0 is a subset of every set.

26. E v e r y set is a subset of itself.

27. Every n o n e m p t y set has at least two subsets. 28. The set of people in the world is infinite. 29. The set of words in a dictionary is infinite. Find the power set of each set. 30. 0

31. {a}

32. {a,b,c}

33. Using Exercises 30-32, predict the n u m b e r of subsets of a set with n elements. In Exercises 34-37, n denotes a positive integer less t h a n 10. Rewrite each set using the listing method. 34. {nln is divisible by 2}

35. {nln is divisible by 3}

36. {nln is divisible by 2 and 3}

37. {nln is divisible by 2 or 3}

Find the family of subsets of each set t h a t do not contain consecutive integers. 38. {1,2}

39. {1,2,3}

40. Let an denote the n u m b e r of subsets of the set S - {1, 2 , . . . , n} t h a t do not contain consecutive integers, where n > 1. F i n d a 3 and a4. In Exercises 41-46, a language L over E -- {a, b} is given. Find five words in each language. 41. L - {x e E ' I x begins with and ends in b.} 42. L - {x ~ E*lx contains exactly one b. } 43. L - {x E E*fx contains an even n u m b e r of a's. } 44. L - {x e E ' I x contains an even n u m b e r of a ' s followed by an odd n u m b e r of b's. } C o m p u t e the length of each word over {a, b }. 45. aab

46. aabbb

47. ab 4

48. a3b 2

A r r a n g e the b i n a r y words of the given length in increasing order of magnitude. 49. L e n g t h two.

50. L e n g t h three.

Chapter 2 The Language of Sets

78

A t e r n a r y w o r d is a word over the alphabet {0, 1, 2}. A r r a n g e the t e r n a r y words of the given length in increasing order of magnitude. 51. Length one.

52. Length two.

Prove each. *53. The empty set is a subset of every set. (Hint: Consider the implication x e ~ ~ x e A.) *54. The empty set is unique. (Hint: Assume there are two empty sets, 01 and ~2. T h e n use Exercise 53.) *55. Let A, B, and C be a r b i t r a r y sets such t h a t A c B and B c C. T h e n

AcC. (transitive property) *56. If E is a n o n e m p t y alphabet, t h e n E* is infinite. (Hint: Assume Z* is finite. Since Z # 0, it contains an e l e m e n t a. Let x e E* with largest length. Now consider xa.)

J u s t as propositions can be combined in several ways to construct new propositions, sets can be combined in different ways to build new sets. You will find a close relationship between logic operations and set operations.

Union The u n i o n of two sets A and B, denoted by A u B, is obtained by merging them; t h a t is, A u B - {xl(x e A) v (x e B)}. Notice the similarity between union and disjunction.

~

Let A - {a, b, c}, B - {b, C, d , e } , a n d C - { x , y } . T h e n A u B - { a , b ,

BUAandBUC-

{b,c,d,e,x,y}-CUB.

C, d , e } -

m

The shaded areas in Figure 2.4 r e p r e s e n t the set A u B in t h r e e different cases.

Intersection The i n t e r s e c t i o n of two sets A and B, denoted by A N B, is the set of elements common to both A and B; t h a t is, A 5 B - {xl(x e A) v (x e B)}.

79

2.2 Operationswith Sets F i g u r e 2.4

U

A UB

U

AUB=B

A U B, where A and B are disjoint

Notice the relationship between intersection and conjunction.

Let A {Nov, Dec, Jan, Feb}, B - {Feb, Mar, Apr, May}, and C - {Sept, Oct, Nov, Dec}. T h e n A n B - {Feb} - B N A a n d B n C - 0 - C A B . (Notice t h a t B and C are disjoint sets. More generally, two sets are disjoint if and only if their intersection is null.) m

F i g u r e 2.5

O9 O9

Berkeley Street intersection

Figure 2.5 shows the intersection of two lines and t h a t of two streets, and Figure 2.6 displays the set A n B in three different cases.

F i g u r e 2.6

U

U

G ANB

Let A - {a, b, c, d, g}, B and (A u B) N (A U C).

ANB=O

{b, c, d, e, f}, and C -

ANB=A

{b,c,e,g,h}.FindAu(BnC)

Chapter 2 The Language of Sets

80

SOLUTION: (1)

BNC

= {b,c,e} {a,b, c, d, e, g}

AU(BNC)-

A UB = { a , b , c , d , e , f,g}

(2)

AuC-{a, (AuB) N(AuC)-

b, c, d, e, g, h} {a,b,c,d,e,g}

= A u (B n C) See the Venn diagram in Figure 2.7. F i g u r e 2.7

m A third way of combining two sets is by finding their difference, as defined below. Difference The d i f f e r e n c e of two sets A and B (or the r e l a t i v e c o m p l e m e n t of B in A), denoted by A - B (notice the order), is the set of e l e m e n t s in A t h a t are not in B. T h u s A - B = {x ~ AIx r B}. L e t A - { a , . . . , z , 0 , . . . , 9 } , and B - {0,...,9}. T h e n A - B - { a , . . . , z } and B-A=O. The shaded areas in Figure 2.8 r e p r e s e n t the set A - B in t h r e e different cases.

F i g u r e 2.8

U

U

A-B

A - B =A

A-B

m F o r any set A r U, a l t h o u g h A - U - ~, the difference U - A ~: ~. This shows yet a n o t h e r way of obtaining a new set.

2.2 Operationswith Sets

81

Complement T h e difference U - A is the (absolute) c o m p l e m e n t of A, d e n o t e d by A' (A prime). T h u s A ' = U - A = {x ~ U Ix r A}. I F i g u r e 2.9 r e p r e s e n t s t h e c o m p l e m e n t of a set A. ( C o m p l e m e n t a t i o n c o r r e s p o n d s to negation.)

F i g u r e 2.9

U

A'

~

{ a , . . . , Z }. F i n d the c o m p l e m e n t s of the sets A - {a, e, i, O~U} a n d Let U B = {a, c, d, e, . . . , w}. T h e n A' = U - A = set of all c o n s o n a n t s in t h e alphabet, and B' = U - B = {b, x, y, z}. m Let A = { a , b , x , y , z } , B F i n d (A u B)' and A' N B'. SOLUTION: (1)

{c, d, e, x, y, z}, and U -

{a,b,c,d,e,w,x,y,z}.

A U B = {a, b, c, d, e, x, y, z} (A u B)' = {w}

(2)

A' = {c, d, e, w}

B' = {a, b, w} A' N B' = {w} = (A U B)' See F i g u r e 2.10.

F i g u r e 2.10

U

m Since as a rule, A - B r B - A , new set.

by t a k i n g t h e i r u n i o n we can form a

82

Chapter2 The Language of Sets Symmetric Difference The s y m m e t r i c d i f f e r e n c e of A and B, denoted by A @ B, is defined by A @ B - (A - B) U (B - A ) . LetA - {a,...,z,0,...,9} andB - {0,...,9,+,-,.,/}. ThenA-B { a , . . . , z } and B - A = {+,-,.,/}. SoA@B - (A-B) u (B-A) { a , . . . , z, + , - , . , / } .

a

The s y m m e t r i c difference of A and B is pictorially displayed in Figure 2.11 in three different cases.

Figure 2.11

A|

AOB=AUB

A|

=A-B

Set and Logic Operations Set operations and logic operations are closely related, as Table 2.1 shows.

Table 2.1

Set operation AuB ANB A' A@B

Logic operation pvq pvq ~p

pXORq

The i m p o r t a n t properties satisfied by the set operations are listed in Table 2.2. (Notice the similarity between these properties and the laws of logic in Section 1.2.) We shall prove one of them. Use its proof as a model to prove the others as routine exercises. We shall prove law 16. It uses De M o r g a n ' s law in symbolic logic, a n d the fact t h a t X = Y if and only if X ___ Y and Y __C_X.

PROOF: In order to prove t h a t (AUB)' - A' NB', we m u s t prove two parts" (A UB)' c A' N B' and A' n B' c (A u B)'. 9 To prove t h a t (A u B)' c_ (A' n B')Let x be an a r b i t r a r y element of (A u B)'. T h e n x r (A u B). Therefore, by De M o r g a n ' s law, x r A and x r B; t h a t is, x e A' and x e B'. So x e A' NB'. T h u s every element of (A UB)' is also an element ofA' NB'; t h a t is, (A u B)' ___A' n B'.

83

2.2 Operations with Sets

T a b l e 2.2

Laws of Sets Let A, B, and C be any three sets and U the universal set. Then:

I d e m p o t e n t laws 2. A n A = A

1. A u A = A

I d e n t i t y laws 4. A n U = A

3. A u O = A

Inverse laws

6. A n A I = O

5. A u W = U

D o m i n a t i o n laws 8. A n O = O

7. A u U = U

C o m m u t a t i v e laws 10. A n B = B n A

9. A u B = B u A

Double c o m p l e m e n t a t i o n law 11. (At)t = A

A s s o c i a t i v e laws 12. A u ( B u C ) - ( A u B ) u C

13. A n ( B n C ) - ( A n B ) n C

D i s t r i b u t i v e laws 14. A u ( B n C ) = ( A u B ) n ( A u C )

15. A n ( B u C ) - ( A n B ) u ( A n C )

De Morgan's laws 16. (A u B)' - A' n B'

17. (A n B)' - A' u B'

A b s o r p t i o n laws 18. A u ( A n B ) - A

19. A n ( A u B ) = A (Note: The following laws have no names.)

20. I f A c _ B , t h e n A n B = A .

21. I f A c _ B , t h e n A u B = B .

22. I f A c _ B , t h e n B tC_A t.

23. A - B - A n B

t

24. A |

9 To p r o v e t h a t A' c~ B' c_ (A u B ) " Let x be a n y e l e m e n t of A' n B'. T h e n x E A' a n d x ~ B'. T h e r e f o r e , x r A a n d x r B. So, by De M o r g a n ' s law, x r (A u B). C o n s e q u e n t l y , x ~ (A w B)'. T h u s , since x is a r b i t r a r y , A' • B' c (A u B)'. T h u s , (A u B)' - A' n B'. See t h e V e n n d i a g r a m s in F i g u r e 2.12 also.

Note" L a w 23 is a v e r y u s e f u l r e s u l t a n d will be u s e d in t h e n e x t section.

A few words of explanation" T h e c o m m u t a t i v e laws i m p l y t h a t t h e o r d e r in w h i c h t h e u n i o n (or i n t e r s e c t i o n ) of t w o sets is t a k e n is i r r e l e v a n t . T h e associative laws i m p l y t h a t w h e n t h e u n i o n (or i n t e r s e c t i o n ) of t h r e e or m o r e

84

Chapter 2 The Language of Sets

F i g u r e 2.12

(A U B)' = shaded area

A' n B' = cross-shaded area

I

sets is taken, the way the sets are grouped is immaterial; in other words, such expressions without p a r e n t h e s e s are perfectly legal. For instance, A U B u C - A u (B U C) = (A u B) U C is certainly valid. The two De M o r g a n ' s laws in propositional logic play a central role in deriving the corresponding laws in sets. Again, as in propositional logic, p a r e n t h e s e s are essential to indicate the groupings in the distributive laws. For example, if you do not parenthesize the expression A N (B U C) in law 15, t h e n the LHS becomes A N B U C = (A N B ) U C = (A U C) N (B U C) r (A N B ) U (A N C).

Notice the similarity between the set laws and the laws of logic. For example, properties 1 t h r o u g h 19 and 22 have their c o u n t e r p a r t s in logic. Every corresponding law of logic can be obtained by replacing sets A, B, and C with propositions p, q, and r, respectively, the set operators N, U, a n d ' with the logic operators A, v, and ~ respectively, and equality (=) with logical equivalence (=_). Using this procedure, the absorption law A w (A N B) = A, for instance, can be t r a n s l a t e d as p v (p A q) - p, which is the corresponding absorption law in logic. J u s t as t r u t h tables were used in Chapter I to establish the logical equivalence of compound statements, they can be applied to verify set laws as well. The next example illustrates this method. Using a t r u t h table, prove t h a t (A u B)' - A' N B'. SOLUTION. Let x be an a r b i t r a r y element. T h e n x may or may not be in A. Likewise, x may or may not belong to B. E n t e r this information, as in logic, in the first two columns of the table, which are headed by x E A and x e B. The table needs five more columns, headed by x ~ (A u B), x ~ (A u B)', x ~ A', x ~ B', and x ~ (A' n B') (see Table 2.3). Again, as in logic, use the entries in the first two columns to fill in the r e m a i n i n g columns, as in the table.

85

2.2 Operations with Sets

T a b l e 2.3

x e A

x e B

x e (A U B )

x e (A u B)'

x e A'

x 9 B'

x 9 (A' n B')

T T F F

T F T F

T T T F

F F F T

F F T T

F T F T

F F F T

Note: The shaded columns are identical Since the columns headed by x e (A u B)' and x e (A' n B') are identical, it follows t h a t (A u B)' = A' n B'. I Using t r u t h tables to prove set laws is purely mechanical a n d elem e n t a r y . It does not provide any insight into the d e v e l o p m e n t of a m a t h e m a t i c a l proof. Such a proof does not build on previously k n o w n set laws, so we shall not r e s o r t to such proofs in s u b s e q u e n t discussions. .

.

.

.

.

.

J u s t as the laws of logic can be used to simplify logic expressions a n d derive new laws, set laws can be applied to simplify set expressions a n d derive new laws. In order to be successful in this art, you m u s t k n o w the laws well and be able to apply t h e m as needed. So, practice, practice, practice.

Using set laws, verify t h a t (X - Y) - Z = X - (Y u Z). PROOF(X - Y) - Z - (X - Y) n Z'

A-B

=AnB'

= (X n Y') n Z'

A-B

=AnB'

= X n (Y' n Z')

associative law 13

= X n (Y u Z)'

De M o r g a n ' s law 16

=X-(YuZ)

A-B

=AnB'

Simplify the set expression (A n B') U (A' n B) U (A' n B'). SOLUTION: (You m a y supply the justification for each step.) (A n B') u (A' n B) u (A' n B') - (A n B') u [(A' n B) u (A' n B')] = ( A n B') u [A' n (B u B')] = ( A n B') u (A' n U) = (A riB') u A '

I

(Jhapter2 The Languageof Sets

86

= A' u (A N B') = (A' u A) n (A' u B') = U n (A' u B')

=A'uB'

m

Often subscripts are used to n a m e sets, so we now t u r n o u r a t t e n t i o n to such sets.

Indexed Sets Let I, called the i n d e x s e t , be the set of subscripts i used to n a m e the sets Ai. T h e n the u n i o n of the sets Ai as i varies over I is d e n o t e d b y . u Ai. Similarly, tEI

n Ai denotes the i n t e r s e c t i o n of the sets Ai as i r u n s over I. In p a r t i c u l a r ,

iEI

let I - {1, 2 , . . . , n }. T h e n u Ai - A1 uA2 u . . . UAn, which is often w r i t t e n as n U

i=1

iEI

n

Ai or simply U1 Ai. Likewise, N

iEI

Ai -

n n Ai i=I cx.)

I f I - N, the expression

n ? Ai - A1 n A2 N...

NAn.

('x.)

u Ai is w r i t t e n as u Ai - ~ Ai, u s i n g the infinity iEb~

/el

symbol oc; similarly, i ?b~ A i - i=ln A i - N1 A i . Before we proceed to define a new b i n a r y o p e r a t i o n on sets n , we define an o r d e r e d set. i=1

Ordered Set Recall t h a t the set {al, a 2 , . . . , a , } is an u n o r d e r e d collection of elements. Suppose we assign a position to each element. T h e r e s u l t i n g set is an o r d e r e d s e t with n e l e m e n t s or an n - t u p l e , d e n o t e d by ( a l , a 2 , . . . , an). (Notice the use of p a r e n t h e s e s versus braces.) The set (al, a2) is an o r d e r e d pair. Two n-tuples are e q u a l if and only if t h e i r c o r r e s p o n d i n g e l e m e n t s are equal. T h a t is, (al, a 2 , . . . , a,,) = (bl, b 2 , . . . , b,) if a n d only if ai = bi for every i.

•

Every n u m e r a l and word can be considered an n-tuple. F o r instance, 345 = (3, 4, 5) l

t

$

ones tens hundreds

c o m p u t e r = (c, o, m, p, u, t, e, r) 1001011 = (1, 0, 0, 1, 0, 1, 1) 11010010 = (1, 1, 0, 1, 0, 0, 1, 0)

ASCII* code for letter K EBCDIC** code for letter K

*American Standard Code for Intbrmation Interchange. **Extended Binary Coded Decimal Interchange Code.

m

87

2.2 Operationswith Sets

Rend D e s c a r t e s (1596-1650) was born near Tours, France. At eight, he entered the Jesuit school at La Fleche, where because of poor health he developed the habit of lying in bed thinking until late in the morning; he considered those times the most productive. He left the school in 1612 and moved to Paris, where he studied mathematics for a brief period. After a short military career and travel through Europe for about 5years, he returned to Paris and studied mathematics and philosophy. He then moved to Holland, where he lived for 20years writing several books. In 1637 he wrote Discours, which contains his contributions to analytic geometry. In 1649 Descartes moved to Sweden at the invitation of Queen Christina. There he contracted pneumonia and died.

~Z ( .::

We are now ready to define the next and final operation on sets.

Cartesian Product The c a r t e s i a n p r o d u c t of two sets A and B, denoted by A x B, is the set of all ordered pairs (a,b) with a 9 A and b 9 B. T h u s A x B = {(a,b)l a 9 A A b 9 B}.A x A is denoted b y A 2. It is n a m e d after the F r e n c h philosopher and m a t h e m a t i c i a n Ren~ Descartes. Let A -

{a, b } and B -

{x, y, z}. T h e n

A x B = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)} B z A = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)} A 2 - A x A = {(a, a), (a, b), (b, a), (b, b)}

m

(Notice t h a t A x B r B x A.)

The various elements of A x B in Example 2.22 can be displayed in a r e c t a n g u l a r fashion, as in F i g u r e 2.13, and pictorially, using dots as in Figure 2.14. The circled dot in row a and column y, for instance, r e p r e s e n t s the element (a, y). The pictorial r e p r e s e n t a t i o n in F i g u r e 2.14 is the g r a p h ofA x B .

F i g u r e 2.13

a Elements of A b

(a, x)

(a, y)

(a, z)

(b, x)

(b, y)

(b, z)

x

y Elements of B

z

88

Figure

2.14

Pictorial representation ofA •

Chapter 2

The Language of Sets

a

9

|

9

x

y

z

Figure 2.15 shows the graph of the infinite set 1~ 2 - - 1~ x N. The circled dot in column 4 and row 3, for instance, represents the element (4,3). The horizontal and vertical dots indicate that the pattern is to be continued indefinitely in both directions. Figure

2.15

9

9

9

1

2

3

},

4

.

.

.

More generally, R 2 - - R • R consists of all possible ordered pairs (x,y) of real numbers. It is represented by the familiar x y - p l a n e or the c a r t e s i a n p l a n e used for graphing (see Figure 2.16).

Figure

2.16

The cartesian plane R 2"

(-3,4)

(0,3)

(5,0) The following example presents an application of cartesian product. ~

Linda would like to make a trip from Boston to New York and then to London. She can travel by car, plane, or ship from Boston to New York, and by plane or ship from New York to London. Find the set of various modes of transportation for the entire trip. SOLUTION:

Let A be the set of means of transportation from Boston to New York and B the set from New York to London. Clearly A - {car, plane, ship} and B - {plane, ship}. So the set of possible modes of transportation is given by

89

2.2 Operationswith Sets F i g u r e 2.17

London N Boston

e

plane

w

~

-.~---"~'~ship

A x B - {(car, plane), (car, ship), (plane, plane), (plane, ship), (ship, plane), (ship, ship)}. See Figure 2.17. I The definition of the product of two sets can be extended to n sets. The c a r t e s i a n p r o d u c t o f n s e t s A 1 , A 2 , . . . ,An consists of all possible n -tuples (al, a 2 , . . . , an), where ai ~ Ai for every i; it is denoted by A1 x A2 x ... x An. If all Ai's are equal to A, the product set is denoted by An. LetA-

{x},B - {y,z}, and C -

A x B x C-

{1,2,3}. Then

{(a,b,c)la ~ A , b ~ B, a n d c ~ C}

= {(x, y, 1), (x, y, 2), (x, y, 3), (x, z, 1), (x, z, 2), (x, z, 3)} Finally, take a look at the map of the continental United States in Figure 2.18. It provides a geographical illustration of partitioning, a concept that can be extended to sets in an obvious way.

F i g u r e 2.18

I

Partition

Consider the set S = {a, b, c, d, e, f, g, h, i} and the subsets $1 = {a, b }, $ 2 - {c}, $3 = {d, e, f}, $4 - {g, h}, and $5 - {i}. Notice t h a t these subsets have three interesting properties: (1) They are nonempty; (2) they are pairwise disjoint; that is, no two subsets have any common elements; (3) their union

90

Chapter 2 The Language of Sets is S. (See Figure 2.19.) T h e set P - { S 1 , $ 2 , $ 3 , $ 4 , $ 5 } of S.

is called a p a r t i t i o n

F i g u r e 2.19

S1

More generally, let I be an index set and P a family of s u b s e t s Si of a n o n e m p t y set S, w h e r e i ~ I. T h e n P is a p a r t i t i o n of S if: 9 Each set Si is n o n e m p t y . 9 The subsets are pairwise disjoint; t h a t is, Si n S j - ~,~ if i ~=j. 9 The union of the subsets Si is S; t h a t is, u Si = S. i~I

(Each subset Si is a b l o c k of the partition.) T h u s a p a r t i t i o n of S is a collection of n o n e m p t y , pairwise disjoint subsets of S whose u n i o n is S. ~

Let Z,. denote the set of integers which, w h e n divided by 5, leave r as the r e m a i n d e r . T h e n 0 < r < 5 (see Section 4.1): w

Z0 = { . . . , - 5 , 0 , 5 , . . . } Zl = { . . . , - 4 , 1 , 6 , . . . } Z 2 -- { . . . , - 3 , 2 , 7,...} Z3 = { . . . , - 2 , 3 , 8 , . . . } Z4 = { . . . , - 1 , 4 , 9 , . . . } P = {Z0, Zl, Z2, Z3, Z4} is a partition of the set of integers. See F i g u r e 2.20. (This example is discussed in more detail in Section 7.4.) m

F i g u r e 2.20 Set of integers Z.

2.2 Operationswith Sets

91

The sports pages of newspapers provide fine examples of partitions, as the next example illustrates. ~

In 2003, the set of teams S in the National Football League was divided into two conferences, American and National, and each conference into four divisions m East, South, North, and West. Let El, $1, N1, and W1 denote the set of teams in East, South, North, and West Divisions in the American Conference, respectively, and E2, $2, N2, and W2 the corresponding sets in the National Conference. Then: E1 - {Buffalo, Miami, New England, NY Jets} $1 -- {Indianapolis, Tennessee, Houston, Jacksonville} N1 - {Baltimore, Cincinnati, Cleveland, Pittsburgh} W1 - {Denver, Kansas City, Oakland, San Diego} E2 = {Washington, Philadelphia, Dallas, NY Giants} $2 - {Atlanta, Tampa Bay, Carolina, New Orleans} N2 = {Chicago, Detroit, Minnesota, Green Bay } W2 = {Arizona, Seattle, St. Louis, San Francisco } Clearly, P - {E1,S1,N1, W 1 , E 2 , S 2 , N 2 , W2} is a partition of S. We close this section with a brief introduction to fuzzy sets.

Fuzzy Sets (optional) Fuzzy sets, a generalization of ordinary sets, were introduced in 1965 by Lotfi A. Zadeh of the University of California at Berkeley. They have applications to h u m a n cognition, communications, decision analysis, psychology, medicine, law, information retrieval, and, of course, artificial intelligence. Like fuzzy logic, they model the fuzziness in the natural language for example, in terms like young, healthy, wealthy, and beautiful. In fuzzy set theory, every element x in the universal set U has a certain degree of membership du(x), where 0 < du(x) < 1;du(x) indicates the degree of fuzziness. Accordingly, a fuzzy set S is denoted by listing its elements along with their degrees of membership; an element with zero degree of membership is not listed. For example, let Ube the fuzzy set of wealthy people and S - {Tom 0.4, Dick 0.7, Harry 0.6}. Then Harry belongs to S with degree of membership 0.6; ds(Harry) - 0.6 measures Harry's degree of wealthiness. The concept of an ordinary subset can be extended to fuzzy sets also.

Fuzzy Subset Let A and B be fuzzy sets. Then A is a f u z z y s u b s e t of B if A ___B and dA(x) < dB(x) for every element x in A.

Chapter2 The Language of Sets

92

~

":t .... ."'

" ....

Lotfi A. Z a d e h (1921-) was born in Baku, Azerbaijan. An alumnus of the University of Tehran (1942) and the Massachusetts Institute of Technology (1946), he received his Ph.D. from Columbia University in 1949 for his dissertation on frequency analysis of time-varying networks. He began his professional career in the Department of Electrical Engineering at Columbia. In 1959, he joined the Department of Electrical Engineering and Computer Science at the University of California, Berkeley, serving as its chair during the years 19631968. Currently, he is a professor at Berkeley and Director of Berkeley Initiative in Soft Computing. Zadeh's earlier "work was centered on systems analysis, decision analysis, and information systems. Since then his current research has shifted to the theory of fuzzy sets and its applications to artificial intelligence (AI). His research interest now is focused on fuzzy logic, soft computing, computing with words, and the newly developed computational theory of perceptions and precisiated natural language, "according to the University of California Web site. A truly gifted mind and an expert on AI, Zadeh has authored about 200journal articles on a wide variety of subjects relating to the conception, design, and analysis of information~intelligent systems. He serves on the editorial boards of more than 50 journals and on the advisory boards of a number of institutions related to AI. Zadeh is a recipient of numerous awards and medals, including the IEEE Education Medal, IEEE Richard W. Hamming Medal, IEEE Medal of Honor, the A S M E Rufus Oldenburger Medal, B. Bolzano Medal of the Czech Academy of Sciences, Kampe de Feriet Medal, AACC Richard E. Bellman Central Heritage Award, the Grigore Moisil Prize, Honda Prize, Okawa Prize, AIM Information Science Award, IEEE-SMC J. P. Wohl Career Achievement Award, SOFT Scientific Contribution Memorial Award of the Japan Society for Fuzzy Theory, IEEE Millennium Medal, and the ACM 2000 Allen Newell Award. He has received honorary doctorates from many universities from around the world. 9

2

"~

For example, letS = {Betsey 0.6, Mat 0.5} and T - {Betsey 0.8, J o n a t h a n 0.3, Mary 0.5, Mat 0.7} by fuzzy sets of smart people. Then S is a fuzzy subset of T. Operations on ordinary sets can be extended to fuzzy sets as well. Operations on Fuzzy Sets Let A and B be any fuzzy set. The u n i o n of A and B is AUB, where dA uB(x) = max{dA(x), ds(x)}; their i n t e r s e c t i o n is A N B, where dANB(X) = m i n { d A ( x ) , d s ( x ) } ; and the c o m p l e m e n t of A is A', where dA,(X) = 1 - dA(x); in A' only the degrees of membership change. Using the sets S and T above, S u T = {Betsey 0.8, J o n a t h a n 0.3, Mary 0.5, Mat 0.7 } S n T = {Betsey 0.6, Mat 0.5} S' = {Betsey 0.4, Mat 0.5} Additional opportunities to practice the various operations are given in the exercises.

2.2

Operations with Sets

93

Exercises 2.2

L e t A = {a, e, f, g,i}, B = {b, d, e, g, h}, C = {d, e, f, h, i}, and U = { a , b , . . . ,k}. Find each set. 1. C l

2. B Q C '

3. C Q A '

4. ( A U B ) '

5. (B n C)'

6. (A u C')'

7. (B N C')'

8. A ~ B

9. ( A - B ) - C

10. A - ( B - C )

11. (A u B) - C

12. (A N B) - C

Using the Venn d i a g r a m in Figure 2.21 find each set. 13. ( A U B ) N C

14. A n (B U C)

15. A -

16. ( A @ B ) U C

17. A Q ( B O C )

18. A - ( B @ C )

F i g u r e 2.21

(B-C)

U a

A w

f

Let A = {b, c}, B = {x}, and C = {x,z}. Find each set. 19. A x B

20. B x A

21. A x 0

22. A x B x O

23. A x ( B u C )

24. A x ( B N C )

25. A x B x C

26. A x C x B

M a r k each as true or false, where A, B, and C are a r b i t r a r y sets and U the universal set. 27. A - ~ ) = A

28. 0 - A = - A

29. O - O = 0

30. A - A = O

31. A - B = B - A

32. A - A ' = O

33. (A')' = A

34. (A N B ) ' = A' NB'

35. (A U B)' = A' U B'

36. A c A u B

37. A c A N B

38. B D (A - B ) = O

Give a counterexample to disprove each proposition. 39. ( A - B ) - C = A - ( B - C )

40. A U ( B - C ) = ( A U B ) - ( A U C )

41. A u ( B @ C ) = ( A u B ) G ( A u C )

42. A O ( B N C ) = ( A O B ) N ( A G C )

D e t e r m i n e if each is a partition of the set { a , . . . , z, 0 , . . . , 9}. 43. { { a , . . . , z } , { 0 , . . . , 9 } , 0 } 44. { { a , . . . , j } , { i , . . . , t } , { u , . . . , z } , { 0 , . . . , 9 } }

94

Chapter 2 The Language of Sets 45. {{a,... ,1}, { n , . . . , t } , {u,...,z}, {0,...,9}} 46. {{a,...,u}, {v,...,z}, {0,3}, { 1 , 2 , 4 , . . . , 9 } Prove each, where A, B, and C are any sets. 47. (A')' = A

48. A u (A n B ) = A

49. A N ( A u B ) = A

50. (A N B)' = A' u B'

51. A @ A = O

52. A @ U = A '

53. A @ B = B @ A

54. A - B = A A B '

55. (A u B u C)' = A' n B' n C'

56.

(A A B n C)' = A' u B' u C'

Simplify each set expression. 57. A A (A - B)

58. (A - A') u (B - A)

59.

(A - B') - (B - A')

60.

(A u B) u (A A B')'

61.

(A u B) - (A n B)'

62.

(A U B)' N (A n B')

63.

(A n B)' u (A u B')

64.

(A u B')' A (A' c~ B)

65.

(A' u B')' u (A' ~ B)

*66. State De Morgan's laws for sets Ai, i ~ I. (I is an index set.) *67. State the distributive laws using the sets A and Bi, i ~ I. The s u m of two fuzzy sets A and B is the fuzzy set A | B, where dA ~ B(X) = 1A IdA (x) + d B ( x ) l; their d i f f e r e n c e is the fuzzy set A - B , where dA-B(X) = 0 v IdA(x) -- dB(x)i; and their c a r t e s i a n p r o d u c t is the fuzzy set A x B, where dA xB(X,Y) = dA(x) AdB(x). Use the fuzzy s e t s A = {Angelo 0.4, Bart 0.7, Cathy 0.6} and B = {Dan 0.3, Elsie 0.8, F r a n k 0.4} to find each fuzzy set. 68. A U B

69. A A B

70. A'

71. A u B '

72. A N B '

73. A ~ A '

74. A |

75. A - B

76. B - A

77. A x B

78. B x A

79. A x A

Let A and B be any fuzzy sets. Prove each. *80.

(A u B)' = A' • B'

"81.

(A • B)' = A' u B'

Sets and the various set operations can be implemented in a c o m p u t e r in an elegant manner. Computer Representation

Although the elements of a set have no i n h e r e n t order, when the set is represented in a computer, an order is imposed upon t h e m to p e r m i t

2.3 ComputerOperations with Sets (optional)

95

implementation. The universal set U with n elements is represented as an array with n cells, each containing a 1: n-1

2

U ] 1[ 1]

...

1

0

]1111111]

The elements are represented by the binary digits (or bits) 0 and 1 in the right-to-left fashion. Subsets of U are represented by assigning appropriate bits to the various cells. A bit 1 in a cell indicates the corresponding element belongs to the set, whereas a 0 would indicate the element does n o t belong to the set. ~

UsingU={a,b,...,h},representthesetsA={a,b,g}andB-{c, 8-bit strings.

e, h } a s

SOLUTION: Remember, the elements are represented in the right-to-left order. Thus:

h

g

f

e

d

c

b

a

lllllrll ll llll AlOlllOlOlOlOlllll lllOlOlllOlllOlOl

m

Next we discuss how the various subsets of a finite set can be found methodically. T a b l e 2.4

Subset

Bit String

0 {x} {y} {x,y} {z} {x,z} {y,z} {x,y,z}

000 001 010 011 100 101 110 111

Interestingly enough, there is a close relationship between sets and bit strings. Table 2.4, for instance, lists the various subsets of the set {x, y, z }. Notice that the table contains all possible three-bit strings and their decimal

96

Chapter 2

The Language of Sets

values increase from 0 to 7. (See Section 4.3 for a discussion of nondecimal bases.) Next we present a systematic procedure to find the bit string of the subset that "follows" a given subset with bit string b 2 b l b o . Such a recipe for solving a problem in a finite number of steps is called an algorithm.*

Next-Subset Algorithm Take a good look at each string in Table 2.4. Can you find a rule to obtain each, except 000, from the preceding string? It is fairly simple: From right to left, locate the first 0. Change it to 1 and the l's to its right to 0's. For example, suppose you would like to find the subset following {x, y} with bit string b 2 b l b o -- 011. From right to left, the first 0 is b2. Change it to 1, and bl and b0 to 0's. The resulting string is 100 and the corresponding subset is {z }. This rule can be generalized and translated into an algorithm. See Algorithm 2.1. Use it to find the subsets following {z} and { y, z}. Algorithm next-subset (bn-lbn-2 ... bo) (* This algorithm finds the b i t string of the subset that follows a given subset of an n-element set S. *) Begin (* next-subset *) find the f i r s t 0 from the r i g h t change i t to I replace the bits to i t s r i g h t with O's End (* next-subset *) Algorithm 2.1

The next-subset algorithm can be employed to find all subsets of a finite set S. Algorithm 2.2 shows the steps involved. Use it to find the subsets of {x,y,z}. Algorithm subsets (S) (* Using the next-subset algorithm, this algorithm finds the b i t representations of all subsets of an n-element set S. *) Begin (* subsets *) bn-lbn-2 . . . b o <- 0 0 . . . 0 (* i n i t i a l i z e string *) done ,-- false (* boolean flag *) while not done do begin (* while *) find the subset following bn_lbn_ 2 . . . b 0 . *The word algorithm is derived from the last name of the ninth-century Arabian astronomer and mathematician Abu-Abdullah M u h a m m e d ibn-Musa al-Khowarizmi (Muhammed, the father of Abdullah and the son of Moses of Khwarizm). He was a teacher in the m a t h e m a t ical school in Baghdad, Iraq. His last name indicates he or his family originally came from Khwarizm (now called Khiva) in Uzbekistan. His books on algebra and Indian numerals had a significant influence in Europe in the 12th century through their Latin translations. The term algebra is derived from the title of his algebra book Kitab al-jabr w'al-muqabalah.

2.3 ComputerOperations with Sets (optional) if

97

every b i t bi = 1 then (* t e r m i n a t e the loop *) done <-- true

endwh i I e End (* subsets *)

Algorithm

2.2

Next we show how the set o p e r a t i o n s can be i m p l e m e n t e d in a c o m p u t e r .

Computer Operations The r e p r e s e n t a t i o n of sets as n-bit strings allows us to use logic operations to p e r f o r m set operations. They are i m p l e m e n t e d t h r o u g h t h e bit operations m AND, OR, XOR, C O M P - - defined by Table 2.5, w h e r e C O M P indicates one's complement: comp(1) = 0 and comp(0) = 1.

T a b l e 2.5

I bit

I AND

OR

XOR

0

1

0

1

0

1

0 0

0 1

0 1

1 1

0 1

1 0

COMP

~--

logic operators

1 0

The various set operations are accomplished by p e r f o r m i n g t h e corresponding logic operations, as s h o w n in Table 2.6. Notice t h a t t h e logic operation c o r r e s p o n d i n g to A - B m a k e s sense since A - B = A N B', by law 23 in Table 2.2.

T a b l e 2.6

Set operations

Logic operations

ANB

AAND B A ORB COMP(A) AXORB A AND (COMP(B))

A uB Af

A@B A- B

Let U = { a , b , . . . , h } , A = { a , b , c , e , g } , and B = {b,e,g,h}. U s i n g bit r e p r e s e n t a t i o n s , find the setsANB, AUB, A@B, B', a n d A - B as 8-bit words.

SOLUTION: A-01010111 B-11010010

Chapter 2 The Language of Sets

98

U s i n g Tables 2.5 a n d 2.6, we have: (1) A N B - O 1010010 (3) A @ B - 1 0 0 0 0 1 0 1 (5) A-0 i 010111 B'-00101101 So A - B - A AND (COMP(B)) =00000101

(2) A U B - 1 1 0 1 0 1 1 1 (4) B'-00101101

U s i n g the bit r e p r e s e n t a t i o n s , you m a y verify t h a t A N B = {b, e, g}, A u B - {a, b, c, e, g, h}, A @ B - {a,c,h}, B' - {a,c,d,f}, a n d A - B {a, c}. I

Exercises 2.3 U s i n g the universal set U = { a , . . . , h}, r e p r e s e n t each set as an 8-bit word. 1. {a,c,e,g}

2. {b,d,f}

3. { a , e , f , g , h }

4. 0

Use A l g o r i t h m 2.1 to find the subset of the set {so,sl,s2,s3} t h a t follows t h e given subset.

5. {s:~}

6. {so,s3}

7. {s2,s3}

8. {so,s2,s3}

Using Algorithm 2.2, find the subsets of each set. 9.

10. {so, sl,s2,s3}

{8(),81}

Using the sets A -- {a,b,e,h}, B = { b , c , e , f , h } , C U = { a , . . . , h}, find the b i n a r y r e p r e s e n t a t i o n of each set.

{c,d,f,g}, a n d

11. A N B

12. A u B

13. B'

14. A - B

15. C - B

16. A |

17. B @ C

18. C @ A

19. A N C '

20. A u B '

21. A N ( B N C )

22. A u ( B N C )

23. A - ( B |

24. ( A @ B ) - C

25. A O ( B O C )

26. ( A o B ) O C

This section p r e s e n t s four formulas involving finite sets, which we shall use frequently. Recall t h a t every finite set has a fixed n u m b e r of e l e m e n t s , so we m a k e the following definition.

Cardinality The c a r d i n a l i t y

of A, denoted by IAI, is the n u m b e r of e l e m e n t s in it.*

*It should be clear from the context whether the symbol "1 I" refers to absolute value or cardinality.

2.4 The Cardinality of a Set

99

For example, I~1 = 0, I{0}1 = 1, a n d I{a, b, c}l = 3. Let A and B be any two finite sets. How is IA u BI related to IAI a n d IBI? First, let's study an example. Let A - {a, b, c} and B - {b, c, d, e, f}. Clearly, IAI = 3, IBI - 5, tA u BI - 6, and IA n BI = 2, so IA u BI = IAI + I B I - IA n BI.

m

More generally, we have the following result" ( I n c l u s i o n - E x c l u s i o n P r i n c i p l e ) Let A and B be two finite sets. T h e n IA u B I - IAI + I B I - IA NBI.

PROOF: Suppose IA N BI - k. Since A N B c A and A N B _c B, we can a s s u m e t h a t IAI - k + m and BI - k + n for some n o n n e g a t i v e integers m and n (see Figure 2.22). Then: IAUBI-m+k+n = ( m + k) + (n + k) - k

= IAI + I B I - I A n B I

F i g u r e 2.22

This completes the proof.

m

In addition, if A and B are disjoint sets, t h e n IA n BI IA u BI - IAI + IBI. T h u s we have the following result. C O R O L L A R Y 2.1 (Addition Principle) IAI + IBI.

I01 -

0, so

L e t A and B be finite disjoint sets. Then IAUBI =

The next example d e m o n s t r a t e s the inclusion-exclusion principle. Find the n u m b e r of positive integers < 300 and divisible by 2

or

3.

SOLUTION: Let A {x c l~lx _< 300 and is divisible by 2} and B {x ~ l~lx < 300 and is divisible by 3 }. T h e n A 5 B consists of positive integers _< 300 t h a t are divisible by 2 and 3, t h a t is, divisible by 6. T h u s A - {2, 4 , . . . , 300},

I00

Chapter 2

The Language of Sets

B = {3, 6 , . . . , 300}, a n d A n B = {6, 1 2 , . . . , 300}. Clearly, Inl - 150, IBI 100, and IA n BI - 50, so by T h e o r e m 2.1, IA u B I - I A I + I B I - I A n B I - 150 + 100 - 50 - 200 T h u s t h e r e are 200 positive integers < 3 0 0 and divisible by 2 or 3. (See E x a m p l e s 3.11, and 3.12 in Section 3.2.) T h e o r e m 2.1 can be extended to any finite n u m b e r of finite sets. F o r instance, the next example derives the formula for t h r e e finite sets. [ ] ~ ~ ~ ~ ~

Let A, B, and C be t h r e e finite sets. Prove t h a t I A u B u C I - IAI + IBI + ICI - IA n B I - I B n CI - t C n A I + IA n B n CI. PROOF:

IA UB U CI - IA W (B U C)I : IAI + IB u CI - IA n (B u C)I

by T h e o r e m 2.1

= IAI + IB u CI - I(A n B) u ( A n C)I

by the distributive law

- IAI + (IBI + ICI - IB n C I ) - I IA n B I + IA n CI - I(A n B) n (A n C)Ii

= IAI + IBI + ICI - IA n B I - IB n CI - IC h A l + IA n B n CI, s i n c e A n C = C n A and (A riB) n ( A n C) = A n B n C.

I

The next example shows how useful sets are in d a t a analysis. ~

A survey a m o n g 100 s t u d e n t s shows t h a t of the t h r e e ice cream flavors vanilla, chocolate, and s t r a w b e r r y , 50 s t u d e n t s like vanilla, 43 like chocolate, 28 like s t r a w b e r r y , 13 like vanilla and chocolate, 11 like chocolate and s t r a w b e r r y , 12 like s t r a w b e r r y and vanilla, and 5 like all of them. Find the n u m b e r of s t u d e n t s surveyed who like each of the following flavors. (1) Chocolate but not s t r a w b e r r y . (2) Chocolate and s t r a w b e r r y , but not vanilla. (3) Vanilla or chocolate, b u t not s t r a w b e r r y . SOLUTION: Let V, C, and S symbolize the set of s t u d e n t s who like vanilla, chocolate, and s t r a w b e r r y flavors, respectively. Draw t h r e e i n t e r s e c t i n g circles to r e p r e s e n t t h e m in the most general case, as in Figure 2.23. O u r first goal is to distribute the 100 s t u d e n t s surveyed into the various regions. Since five s t u d e n t s like all flavors, IV n C •SI = 5. Twelve s t u d e n t s like both s t r a w b e r r y and vanilla, so IS n VI = 12. But five of t h e m like chocolate also. Therefore, I(S n V) - CI = 7. Similarly, I(V n C) - SI = 8 and I(C n S) - VI = 6.

2.4

The Cardinality of a Set

Figure 2.23

101

U

10

Of the 28 s t u d e n t s who like s t r a w b e r r y , we have a l r e a d y a c c o u n t e d for 7 + 5 + 6 = 18. So t h e r e m a i n i n g 10 s t u d e n t s belong to t h e set S - (V u C). Similarly, IV - (C u S)[ = 30 a n d [C - (S u V)[ - 24. T h u s far, we have accounted for 90 of the 100 s t u d e n t s . T h e r e m a i n i n g 10 s t u d e n t s lie outside t h e region V u S u C, as in F i g u r e 2.23. T h e r e q u i r e d a n s w e r s can now be directly read from this V e n n diagram: (1) ] C - S ] - 24 + 8 strawberry.

32. So 32 s t u d e n t s like chocolate b u t not

(2) I(C n S ) - VI - 6. Therefore, 6 s t u d e n t s like both chocolate a n d s t r a w b e r r y , b u t not vanilla. (3) 30 + 8 + 24 - 62 s t u d e n t s like vanilla or chocolate, b u t not s t r a w b e r r y . T h e y are r e p r e s e n t e d by the region (V u C) - S. m Finally, suppose a set contains n elements. How m a n y subsets does it have? Before we a n s w e r this partially, let us study t h e next example, which uses the addition principle. ] ~ ~ ~ ~ ~

Let s3 denote the n u m b e r of subsets of t h e set S - {a, b, c}. Let S* - S - { b } . We shall use the subsets of S* in a clever way to find s3 a n d all s u b s e t s of S. Let A denote the subsets of S*. T h e n A = {~, {a}, {c}, {a, c} }. Clearly every e l e m e n t of A is also a subset of S. Now add b to every e l e m e n t in A. Let B denote the r e s u l t i n g set: B - {{b}, {b, a}, {b, c}, {b, a, c} }. E v e r y subset of S e i t h e r contains b or does not contain b; so, by the addition principle, 8 3 - - IAI + [BI = 4 + 4 = 8. m More generally, we have the following result.

~

Let

8n

denote the n u m b e r of subsets of a set S with n elements. T h e n where n > 1.

Sn -- 2Sn_l,

PROOF: Let x e S. Let S* = S - {x}. T h e n S* contains n - 1 e l e m e n t s a n d hence has Sn-1 subsets by definition. Each of t h e m is also a subset of S. Now i n s e r t x in each of t h e m . The r e s u l t i n g Sn-1 sets are also subsets of S. Since every subset of S e i t h e r contains x or does not contain x, t h e addition principle indicates a total of 8 n - 1 q" 8 n - 1 --- 2Sn-1 subsets of S. (Notice t h a t so = 1. Why?) m

Chapter 2 The Language of Sets

102

Consequently, if you know the n u m b e r of subsets of a set with n - 1 elements, this theorem can be employed to compute the n u m b e r of subsets of a set with n elements. For instance, by Example 2.33, a set with three elements has eight subsets; therefore, a set with four elements has 2.8 = 16 subsets. The technique used in the proof of Theorem 2.2 can be applied to write an algorithm for finding the power set of a set S. See Algorithm 2.3. It uses the fact t h a t if A is a subset of S and s E S, t h e n A u {s} is also a subset of S. Algorithm subsets(S) (* This algorithm finds the power set of a set S with n elements Sl, s2 . . . . . Sn. Sj denotes the j t h element in the power set. *) Begin (* subsets *) power set <- {0} (* i n i t i a l i z e power set *) numsubsets <- I (* i n i t i a l i z e the number of subsets *) for i = I to n do (* si denotes the ith element in S *) begin (* for *) j <-- 1 (* j - t h element in P(S) *) temp ~- numsubsets (* temp is a temporary variable*) while j _< temp do (* construct a new subset *) begin (* while *) add Sj U {si} to the power set j ~-j+1 numsubsets <-- numsubsets + 1

endwhi I e endfor End (* subsets *) Algorithm 2.3

I

Although Theorem 2.2 does not give us an explicit formula for the n u m b e r of subsets, it can be used to find the formula. The next theorem gives us the explicit formula, which we shall prove in Section 4.4 (see Example 4.18). A set with n elements has 2 ~ subsets, where n >_ 0. For example, a set with four elements has 24 = 16 subsets! Exercises 2.4

Find the cardinality of each set. 1. The set of letters of the English alphabet. 2. The set of letters of the word T W E E D L E D E E . 3. The set of m o n t h s of the year with 31 days. 4. The set of identifiers in J a v a t h a t begin with 3.

m

2.4

The Cardinality of a Set

103

Let A and B be two sets such t h a t ]AI - 2a - b, IBI = 2a, ]AN B] = a - b, and ]U] = 3a + 2b. F i n d the cardinality of each set.

5. A U B

6. A - B

7. B'

8. A - A '

9. Find ]AI if IA] - IB], ]A u BI - 2a + 3b, and ]A ~ BI - b. 10. F i n d l A n B l i f l A i - a + b -

BlandlAuBI-2a+2b.

11. Find ]AN BI if ]AI - 2a, IB] - a, and ]A u B] - 2a + b. Let A and B be finite sets such t h a t A ___B, IAI - b, IBI - a + b. Find the cardinality of each set. 12. A U B

13. A - B

14. B - A

15. A ~ B

Let A and B be finite disjoint sets, where IA] - a, and IBI - b. Find the cardinality of each set. 16. A U B

17. A - B

18. B - A

19-21. Find the cardinality of each set in Exercises 16-18, where A _c B, B is finite, IAI - a, and IBJ - b. 22. A survey conducted recently a m o n g 300 adults in O m e g a City shows 160 like to have their houses painted green, and 140 like t h e m blue. Seventy-five adults like both colors. How m a n y do not like either color? 23. A survey was t a k e n to d e t e r m i n e the preference between two l a u n d r y detergents, Lex and Rex. It was found t h a t 15 people liked Lex only, 10 liked both, 20 liked Rex only, and 5 liked n e i t h e r of them. How m a n y people were surveyed? Find the n u m b e r of positive integers < 500 and divisible by: 24. Two or three.

25. Two, three, or five.

26. Two or three, but not six.

27. N e i t h e r two, three, nor five.

Find the n u m b e r of positive integers _< 1776 and divisible by: 28. Two, three, or five.

29. Two, three, or five, but not six.

30. Two, three, or five, but not 15.

31. Two, three, or five, but not 30.

According to a survey a m o n g 160 college students, 95 s t u d e n t s take a course in English, 72 take a course in French, 67 take a course in G e r m a n , 35 take a course in English and in French, 37 take a course in F r e n c h and in G e r m a n , 40 take a course in G e r m a n and in English, and 25 take a course in all three languages. Find the n u m b e r of s t u d e n t s in the survey who take a course in: 32. English, but not G e r m a n .

33. English, French, or German.

104

Chapter2 The Language of Sets 34. English or French, but not German.

35. English and French, German.

b u t not

36. English, but neither F r e n c h nor German. 37. Neither English, French, nor German. A recent survey by the MAD corporation indicates t h a t of the 700 families interviewed, 220 own a television set but no stereo, 200 own a stereo but no camera, 170 own a camera but no television set, 80 own a television set and a stereo but no camera, 80 own a stereo and a camera but no television set, 70 own a camera and a television set but no stereo, and 50 do not have any of these. Find the n u m b e r of families with: 38. Exactly one of the items.

39. Exactly two of the items.

40. At least one of the items.

41. All of the items.

Using Algorithm 2.3, find the power set of each set. List the elements in the order obtained. 42. {a,b}

43. {a,b, c}

A finite set with a elements has b subsets. Find the n u m b e r of subsets of a finite set with the given cardinality. 44. a + l

45. a + 2

46. a + 5

47. 2a

Let A, B, and C be subsets of a finite set U. Derive a formula for each.

48. IA'NB'I

49. IA'NB'NC'I

*50. State the inclusion-exclusion principle for four finite sets A i , 1 <_ i < 4. (The formula contains 15 terms.) "51. Prove the formula in Exercise 50. **52. State the inclusion-exclusion principle for n finite sets A i , 1 < i < n.

A new way of defining sets is using recursion. (It is a powerful problemsolving technique discussed in detail in Chapter 5.) Notice t h a t the set of n u m b e r s S interesting characteristics:

-

{2,22,222, 22~,...} has three

(1) 2 E S . (2) I f x ~ S , t h e n 2 x e S . (3) Every element of S is obtained by a finite n u m b e r of applications of properties 1 and 2 only.

2.5 RecursivelyDefined Sets

105

P r o p e r t y 1 identifies explicitly the p r i m i t i v e e l e m e n t in S a n d hence e n s u r e s t h a t it is n o n e m p t y . P r o p e r t y 2 establishes a s y s t e m a t i c p r o c e d u r e to c o n s t r u c t new e l e m e n t s from k n o w n e l e m e n t s . How do we know, for instance, t h a t 222E S? By p r o p e r t y 1, 2 E S; then, by p r o p e r t y (2), 22 E S; now choose x - 22 a n d apply p r o p e r t y 2 again; so 222E S. P r o p e r t y 3 guarantees t h a t in no o t h e r way can the e l e m e n t s of S be constructed. T h u s t h e various e l e m e n t s of S can be obtained s y s t e m a t i c a l l y by a p p l y i n g t h e above properties. These t h r e e c h a r a c t e r i s t i c s can be generalized a n d m a y be employed to define a set S implicitly. Such a definition is a recursive definition.

Recursively Defined Set A r e c u r s i v e d e f i n i t i o n of a set S consists of t h r e e clauses: 9 The b a s i s c l a u s e explicitly lists at least one p r i m i t i v e e l e m e n t in S, e n s u r i n g t h a t S is n o n e m p t y . 9 The r e c u r s i v e c l a u s e establishes a s y s t e m a t i c recipe to g e n e r a t e new e l e m e n t s from k n o w n elements. 9 The t e r m i n a l c l a u s e g u a r a n t e e s t h a t t h e first two clauses are the only ways the e l e m e n t s of S can be obtained. The t e r m i n a l clause is generally o m i t t e d for convenience. Let S be the set defined recursively as follows. (1) 2 E S .

(2) I f x E S ,

thenx 2ES.

Describe the set by the listing method. SOLUTION: 9 2 E S, by the basis clause. 9 Choose x = 2. T h e n by t h e recursive clause, 4 E S. 9 Now choose x = 4 a n d apply the recursive clause again, so 16 E S. C o n t i n u i n g like this, we get S = {2, 4, 16,256, 6 5 5 3 6 , . . . }. I The next t h r e e e x a m p l e s f u r t h e r elucidate the recursive definition. Notice t h a t the l a n g u a g e L -

{a, aa, ba, aaa, aba, baa, b b a , . . . } consists of

words over the a l p h a b e t E = {a, b} t h a t end in t h e letter a. It can be defined recursively as follows. 9 aEL. 9 If x E L, t h e n ax, bx E L. For instance, the word aba can be c o n s t r u c t e d as follows: 9 a E L. Choosing x = a, bx = ba E L.

106

Chapter 2 The Language of Sets 9 Now choose x = ba. Then ax = aba E L.

The tree diagram in Figure 2.24 illustrates systematically how to derive the words in L. F i g u r e 2.24

a

aa aaa

ba baa

aba

bba I

(Legally Paired Parentheses)An important problem in c o m p u t e r science is to determine w h e t h e r or not a given expression is legally parenthesized. For example, (()), ( ) (), and (() ()) are validly paired sequences of parentheses, but ) (), ( ) (, and ) ( ) ( are not. The set S of sequences of legally paired parentheses can be defined recursively as follows: 9 ()ES. 9 Ifx,y

E S , then x y and (x) belong to S.

The tree diagram in Figure 2.25 shows the various ways of constructing the elements in S. F i g u r e 2.25

()

()() ()

(()) ())

()()()()

(()

((()))

(())(())

I

A simplified recipe to determine if a sequence of parentheses is legally paired is given in Algorithm 2.4. Algorithm Legally Paired Sequence (* This algorithm determines i f a nonempty sequence of parentheses is l e g a l l y paired. Count keeps track of the number of parentheses. I t is incremented by I i f the current parenthesis is a l e f t parenthesis. and decremented by I i f i t is a r i g h t parenthesis. *) Begin (* algorithm *) count +-- 0 (* i n i t i a l i z e *) read a symbol i f symbol = l e f t paren then

2.5

Recursively Defined Sets

107

while not the end of the sequence do begin (* while *) i f symbol = l e f t paren then count +-- count + I else (* symbol = right parenthesis *) count ~-- c o u n t - 1 read the next symbol endwhi ] e i f count = 0 then legal sequence else invalid sequence else invalid sequence (* begins with a right paren *) End (* algorithm *)

Algorithm 2.4

This e x a m p l e is s t u d i e d f u r t h e r in C h a p t e r s 6 a n d 9. ~

m

A legal e x p r e s s i o n in p r o p o s i t i o n a l logic is called a w e l l - f o r m e d f o r m u l a (wff). F o r convenience, we r e s t r i c t o u r d i s c u s s i o n to t h e logical v a r i a b l e s p, q, a n d r, a n d t h e o p e r a t o r s A, v, a n d ~. T h e n t h e set of w e l l - f o r m e d f o r m u l a s can be defined recursively: 9 T h e logic v a r i a b l e s are wffs. 9 I f x a n d y a r e wffs, t h e n so a r e (x), ~(x), (x A y), a n d (x v y). F o r instance, t h e e x p r e s s i o n ((p) A ((-~(q)) V (r))) is a wff, b u t (q A ( ~ r ) ) is not (why?). ( P a r e n t h e s e s are often o m i t t e d w h e n a m b i g u i t y is impossible.) m

Exercises 2.5 In Exercises 1-6, a set S is defined recursively. F i n d four e l e m e n t s in e a c h case. 1.

3.

i) I E S ii) x E S ~ 2 x E S

i) e E S

2.

i) I E S ii) x E S ~ 2 x E S

4.

i) 3 E S ii) x E S ~ l g x c S t

6.

i))~L ii) x c L -~ axb E L

ii) x E S ~ e x c S 5.

i))~EL ii) x E L ~ xbb E L

tlg x means log2 x.

108

Chapter 2 The Language of Sets In Exercises 7-10, identify the set S t h a t is defined recursively. 7. 9.

i) l e S ii) x , y ~ S --+ x + y ~ S

8.

i) 2 e S i i ) x , y ~ S --+ x + y ~ S

10.

i) l e S ii) x , y ~ S --> x •

~S

i) 0 e S ii) x e X , A e S

--> { x } u A e S

Define each language L over the given alphabet recursively. 11. {0, 00, 10, 100, 110, 0000, 1010,...}, E -- {0, 1}. 12. L = {1, 11,111, 1111, 1 1 1 1 1 , . . . } , E -- {0, 1}.

13. L -

{x ~ E * l x - b n a b n , n > 0}, E -- {a,b}.

14. The language L of all palindromes over E - {a, b}. (A p a l i n d r o m e is a word that reads the same both forwards and backwards. For instance, abba is a palindrome.) "15. {b, bb, bbb, bbbb,... }, E - {a, b}. * 16. {b, aba, aabaa, a a a b a a a , . . . }, E -- {a, b}. * 17. {a, aaa, aaaaa, a a a a a a a , . . . }, E - {a, b}. "18. {1, 10, 11,100, 1 0 1 , . . . } , E - {0, 1}.

Determine if each sequence of parentheses is legal. 19. (()())

20. (())(

21. (()( )

22. (()( ))( )

The nth C a t a l a n n u m b e r Cn, named after the Belgian mathematician, Eugene Charles Catalan (1814-1894), is defined by Cn=

(2n)! n!(n +

1)!'

n>0 -

where n! (n f a c t o r i a l ) is defined by n! - n ( n - 1)... 3 . 2 . 1 and 0! - 1. Catalan n u m b e r s have many interesting applications in computer science. For example, the n u m b e r of well-formed sequences of n pairs of left and right parentheses is given by the n t h Catalan number. Compute the n u m b e r of legally paired sequences with the given pairs of left and right parentheses. 23. Three

24. Four

25. Five

26. Six

27. List the well-formed sequences of parentheses with three pairs of left and right parentheses. 28. Redo Exercise 27 with four pairs of left and right parentheses. Using Example 2.37, determine if each is a wff in propositional logic. 29. (p A ((~(q)) v r))

30. ((~(p)) v ((q) A (~r))

Chapter Summary

109

31. (((~p) v q) A (~q) V (~p)))

32. ((p V q) A ((~(q)) V (~(r))))

33. Determine if the following recursive definition yields the set S of legally paired parentheses. If not, find a validly paired sequence t h a t cannot be generated by this definition. i) () e S.

ii) I f x e S, t h e n ( )x, (x), x( ) ~ S.

34. Define the set of words S over an alphabet Z recursively. A s s u m e s (Hint: use concatenation.) 35. Let E be an alphabet. Define E* recursively. (Hint: use concatenation.) *36. Define the language L of all binary representations of n o n n e g a t i v e integers recursively.

This chapter presented the concept of a set, different ways of describing a set, relations between sets, operations with sets and their properties, a n d formal languages. How sets and set operations work in a typical c o m p u t e r were also discussed.

Set 9 A s e t is a well-defined collection of objects

(page 68).

9 A set can be described using words, listing the elements, or by t h e set-builder notation (page 69). 9 A _c B if and only if every element of A is also an e l e m e n t of B (page 69). 9 ( A - B) ~ (A c B) A (B c_ A)

(page 70).

9 The n u l l s e t 0 contains no elements

(page 70).

9 The

universal

set

U contains

all elements

9 A and B are d i s j o i n t s e t s if A N B = O

under

discussion (page 70). (page 71).

9 The p o w e r s e t P(A) of a set A is the family of all subsets of A (page 72). 9 A set with a definite n u m b e r of elements is finite; if a set is not finite, it is i n f i n i t e (page 73).

Chapter 2 The Language of Sets

II0

Formal Language 9 An alphabet Z is a finite set of symbols; {0,1} is t h e b i n a r y a l p h a b e t (page 75). 9 A w o r d over E is a finite a r r a n g e m e n t of s y m b o l s f r o m E. A w o r d of l e n g t h zero is t h e e m p t y w o r d ~ (page 75). 9 E* consists of all possible w o r d s over Z

(page 75).

9 A formal l a n g u a g e over Z is a s u b s e t of E*

(page 75).

9 T h e c o n c a t e n a t i o n of two w o r d s x a n d y is t h e w o r d xy.

(page 76).

9 Union

A w B -xl(x

(page 78).

9 Intersection

A N B = {xl(x ~ A) A (x ~ B)}

(page 78).

9 Difference

A - B = {x ~ A l x ~ B}

(page 80).

A' = U - A = {x e UIx r A}

(page 81).

Set Operations

9 Complement

~ A ) v (x ~ B )

9 S y m m e t r i c difference A G B = (A - B) u (B - A)

(page 82).

9 Cartesian product

(page 87).

A x B = {(a, b)l(a ~ A) A (b ~ B)}

9 T h e f u n d a m e n t a l p r o p e r t i e s of set o p e r a t i o n s a r e listed in T a b l e 2.2 (page 83).

Partition 9 A partition of a set S is a finite collection of n o n e m p t y , p a i r w i s e disjoint s u b s e t s of S w h o s e u n i o n is S (page 90).

Computer Implementation 9 Set o p e r a t i o n s a r e i m p l e m e n t e d in a c o m p u t e r u s i n g t h e bit o p e r a t i o n s in T a b l e 2.5 a n d t h e logic o p e r a t i o n s in T a b l e 2.6. (page 97).

Cardinality 9 Inclusion-exclusion

principle IA u BI -

IAI + I B I - [A N BI (page 99).

9 Addition principle [A u B[ = IA[ + IBI, w h e r e A N B = 0 9 A set w i t h n e l e m e n t s h a s 2 n s u b s e t s

(page 99). (page 102).

Recursion 9 T h e recursive definition of a set consists of a basis clause, r e c u r s i v e clause, a n d a t e r m i n a l clause (page 105).

Chapter Summary

111

Review Exercises U s i n g the V e n n d i a g r a m in F i g u r e 2.26, find each.

1. A - ( B N C )

2. ( A U B ) - C

3. A - ( B - C )

4. ( A - B ) - C

5. A @ B

6. ( A - B )

7. A - (B G C)

8. A u (B O C')

Figure 2.26

x (B-C)

U

9. Find t h e s e t s A a n d B i f A N B ' = {d, f}, a n d A' n B' - {i}.

{a,c},BNA'=

{b,e,g},ANB-

Let A, B, and C be sets such t h a t A - (B u C) - {b,e}, B - (C u A) - {k}, C - (A u B) - {h},A N B = { f , g } , B N C - {j}, C N A - {i}, a n d A N B N C = 0 . Find each set. 10. A - ( B N C )

11. ( A G B ) - C

12. A ( 9 ( B G C )

F i n d the power set of each set. 13. { 0 , { 0 } }

"14. {2,{3},{2, 3}}

15. Let A - {n ~ Ntn < 20 a n d n is d i v i s i b l e b y 2}, B - {n ~ NIn < 20 and n is divisible by 3}, a n d C - {n e NIn < 20 and n is divisible by 5}. D e t e r m i n e if t h e y form a p a r t i t i o n of the set {n e NIn < 20}. Let U - {1 , . . . , 8}, A - {1, 3, 5, 7, 8}, a n d B - {2, 3, 6, 7}. F i n d t h e b i n a r y r e p r e s e n t a t i o n of each set. 16. A - ( A N B )

17. A - B '

18. A - ( A • B )

19. A G ( A G B )

A survey found t h a t 45% of w o m e n like plain yogurt, 55% like flavored yogurt, and 23% like both. C o m p u t e the p e r c e n t a g e of w o m e n w h o like each. 20. Plain yogurt, b u t not flavored. 21. Plain or flavored yogurt, b u t not both. A survey was t a k e n a m o n g t h e s t u d e n t s on c a m p u s to find out w h e t h e r t h e y prefer vanilla or s t r a w b e r r y ice c r e a m and w h e t h e r t h e y prefer c h o c o l a t e or

112

Chapter2 The Language of Sets Pudding

T a b l e 2.7 Chocolate

Ice Cream

Tapioca

Neither

Total

Vanilla

68

53

12

133

Strawberry

59

48

9

116

Neither

23

21

7

51

Total

150

122

28

300

tapioca pudding. The results are s u m m a r i z e d in Table 2.7. Find the n u m b e r of s t u d e n t s who: 22. Like s t r a w b e r r y ice cream and tapioca pudding. 23. Do not like pudding. 24. Like at least one of the ice cream flavors. 25. Like neither ice cream nor pudding. Find the n u m b e r of positive integers _< 4567 and divisible by: 26. Two, three, or five.

27. Two, five, or seven, but not 35.

Find four elements in each set S defined recursively. 28.

ii) x E S ~ 30.

29.

i) I E S

ii)x E S ~ l g x E S

I+xES

i)v~ES ii) x E S o , / 2 §

i) 3 E S

31.

i) 1 E S ii) x E S o

~/I+2xES

Define each set S recursively. 32. {2, 4, 16,256,...}

33. { 1 , 3 , 7 , 1 5 , 3 1 , . . . }

34. {b, ba 2, ba 4, b a 6 , . . . }

35. {~, ba, b2a 2, b3a3,... }

Find five words in each language L over the alphabet E = {a, b}. 36. {x E E*lx contains exactly one a} 37. {x E E*lx contains an odd n u m b e r o f a ' s } Define each language L over the given alphabet recursively. 38. {x E Z* Ix contains exactly one a }, E -- {a, b}. 39. {x E E*[x ends in ab}, E = {a,b}. 40. { 2 , 3 , 4 , 5 , 6 , . . . } , E --{2,3}. 41. { 1, 010, 00100, 0001000, 000010000, . . . }, E = {0, 1 }.

Chapter Summary

113

Determine if each is a well-formed formula. 42. (p A ((~(q)) v (r)))

43. (((p) A (q)) v (~(q) A (r)))

Let A, B, and C be any sets. Prove each. *44. A ~ ( B u C ) - - ( A n B ) u ( A A C ) *45. A u ( B n C ) = ( A u B ) n ( A u C )

*46. A u ( B - C ) = ( A u B ) - ( C - A ) *47. A n ( B - C ) = ( A n B ) - ( A n C ) Simplify each set expression. 48. (A' u B')' u (A' ~ B)

*49. [A - (B u C)] N [(B N C) - A]

Consider the fuzzy sets, where A - {Mike 0.6, Andy 0.3, Jeff 0.7} and B -- {Jean 0.8, J u n e 0.5}. Find each fuzzy set. 50. A u B'

51. A ' A B

52. A @ B'

53. A x B

Let A and B be any fuzzy sets. Prove each. *54. (A u B)' = A' - B

*55. ( A - B ) ' = A ' u B

Supplementary Exercises Prove each, where A, B, and C are a r b i t r a r y sets. 1. A - (B u C) - (A - B) n (A - C) 2. [ A n ( A - B ) ] U ( A ' U B ) ' = A - B *3. A n ( B @ C ) = ( A n B ) @ ( A n C ) *4. A @ ( B @ C ) - ( A @ B ) @ C Simplify each set expression. 5. ( A ~ B ) n ( B n C ) n ( C ~ A ) *7. (A u B') n (A' u B) N (A' u B')

*6. [(A u B) ~ C] u [A N (B u C)] *8. [(A u B') u (A' u B)]' A (A' n B')

Find the n u m b e r of positive integers < 1000 and n o t divisible by" 9. 2,3, o r 5 .

*10. 2 , 3 , 5 , o r 7 .

11. Define recursively the language { o n l n ] n >_ 0} over E - {0, 1}. 12. Define recursively a word w over a finite alphabet E. Let x = x l x 2 . . . Xn ~ ~ * . Then the string X n . . . X2Xl is called the r e v e r s e of x, denoted by x R. For example, the reverse of the binary word 01101 is 10110. Let x, y ~ ~ n . Prove each. 13. (xy) R - y R x R

114

Chapter2 The Language of Sets 14. The string x is palindromic if and only if X R

--

X.

15. The word x x R is palindromic.

Computer Exercises Write a program to do each task, where n denotes a positive integer < 20. 1. Read in k subsets of the set S - {1, 2 , . . . , n } and determine if the subsets form a partition of S. 2. Read in two sets A and B, where U - { 1 , 2 , 3 , . . . ,n}. Print the bitrepresentations of A and B. Use them to find the elements in A u B, A ~ B, A', A - B, A @ B, and A x B, and their cardinalities. 3. Find all subsets of the set {1, 2 , . . . , n}. 4. Read in sequences of left and right parentheses, each containing at most 25 symbols. Determine if each word consists of legally paired parentheses. 5. Print the Catalan numbers Co through Cn.

Exploratory Writing Projects Using library and Internet resources, write a team report on each of the following in your own words. Provide a well-documented bibliography. 1. Write an essay on the life and contributions of G. Cantor. 2. Explain the various occurrences of the ordered pair notation in everyday life. 3. Explain how the addition principle is used to define the addition of positive integers. Give concrete examples. 4. Explain how the concept of partitioning is used in everyday life. In sports. In computer science. Give concrete examples. 5. Study a number of mathematical paradoxes and explain them. 6. Discuss the various string operations and list the programming languages that support them. 7. Describe fuzzy sets and their applications, and L. A. Zadeh's contributions to them. 8. Write a biography of Abu-Abdullah Muhammed Khowarizmi and the origin of the word a l g o r i t h m .

ibn-Musa

al-

9. Extend the concept of the cardinality of a finite set to infinite sets. Describe the arithmetic of transfinite cardinal numbers. 10. Discuss the halting problem.

Chapter Summary

115

Enrichment Readings

1. R.R. Christian, Introduction to Logic and Sets, Blaisdell, Waltham, MA, 1965. 2. M. Guillen, Bridges to Infinity: The Human Side of Mathematics, J. P. Tarcher, Inc., Los Angeles, 1983, pp. 41-60. 3. P. R. Halmos, Naive Set Theory, Van Nostrand, New York, 1960. 4. S. Sahni, Concepts in Discrete Mathematics, 2nd ed., Camelot, Fridley, MN, 1985, pp. 93-111. 5. P. C. Suppes, Axiomatic Set Theory, Van Nostrand, New York, 1960. 6. R.L. Wilder, Evolution of Mathematical Concepts: An Elementary Study, Wiley, New York, 1968.

Chapter 3

Functions and Matrices To know him [Sylvester] was to know one of the historic figures of all time, one of the immortals; and when he was really moved to speak, his eloquence equaled his genius. G. B. HALSTED

T

his chapter presents two mathematical entities in some detail: functions and matrices. The concept of a function is central to every branch of mathematics and to m a n y other areas of learning as well. We will look at the notion of a function and study a few exotic functions. In addition, we will discuss a few i m p o r t a n t properties of special functions and a few techniques for constructing new functions from known ones. Matrices find their applications in diverse fields such as computer science, engineering, the natural sciences, and the social sciences. A few of the interesting problems we shall study in this chapter are: 9 Find the n u m b e r of leap years beyond 1600 and not exceeding a given year N. 9 Find the first day and the n u m b e r of F r i d a y - t h e - t h i r t e e n t h s in a given year, and the date for Easter Sunday of the year. 9 If we select 367 students from a campus, will at least two of t h e m have the same birthday? 9 Suppose every pair of nonadjacent vertices of a hexagon is joined by a line segment, and each line segment is colored red or blue. Will the line segments form at least one monochromatic triangle?

The concept of a function is so f u n d a m e n t a l t h a t it plays the role of a unifying t h r e a d t h a t intertwines every branch of mathematics. It is used in your everyday life as well. For example, when you compute your electric or water bill, you are using the concept of a function, perhaps unknowingly. 117

118

Chapter 3 Functions and Matrices

Here is an example of a function from the academic world. Consider five m a t h e m a t i c s m a j o r s - - B e n s o n , Goldberg, Hall, Rawlings, and Wilcox. Their quality-point averages (QPA) on a 0-4 scale are 3.56, 3.80, 2.65, 3.56, and 2.23, respectively. Each element in the set A = {Benson, Goldberg, Hall, Rawlings, Wilcox} is assigned a unique element from the set B - {2.23, 2.65, 3.56, 3.80}, as shown in Figure 3.1. F i g u r e 3.1

I

/

~

1__~3.56

9

A

B

This assignment has two interesting properties: 9 Every major is assigned a QPA. 9 Every major has a unique (meaning exactly one) QPA. Such an assignment is a function. More generally, we make the following definition. Function

Let X and Y be any two n o n e m p t y sets. A f u n c t i o n from X to Y is a rule that assigns to each element x ~ X a unique element y ~ Y. Functions are usually denoted by the letters f, g, h, i, etc. If f is a function from X to Y, we write f : X --* Y. The set X is the d o m a i n of the function f and Y the c o d o m a i n of f, denoted by dom(f) and codom(f), respectively. If X = Y, then f is said to be a function o n X. The next example elucidates these definitions. Determine whether or not the assignments in Figures 3.2-3.4 are functions. F i g u r e 3.2 a

b

X

Y

119

3.1 The Concept of a Function

F i g u r e 3.3

a

1

b

X

Y

F i g u r e 3.4

vl ~2 v3 '4 X

Y

SOLUTION: 9 The assignment in Figure 3.2 describes a function f from X to Y, since every element in X is assigned to exactly one element in Y. Dom(f) = X and codom(f) = Y. (Notice t h a t the definition does not prohibit two or more distinct elements in X being paired with the same element in Y. Also, it does not require t h a t every element of Y be used.) 9 On the other hand, the assignment in Figure 3.3 is not a function since not every element in X is assigned an element in Y. 9 The "pairing" in Figure 3.4 is also not a function since b e X is not assigned a unique element in Y. m Let f : X --* Y, so every element x E X is paired with a unique element y e Y, as in Figure 3.5. T h e n y is the v a l u e (or i m a g e ) of the function f at x, denoted by f(x), and x is a p r e - i m a g e of y u n d e r f (see Figure 3.5); y is also known as the o u t p u t corresponding to the i n p u t (or a r g u m e n t ) x. Thus y = f(x).* Read f(x) as f o f x .

F i g u r e 3.5 f

~

-

~

input (or argument)

output X

Y

*This functional notation is due to Euler. See Chapter 8 for a biography of Euler.

120

Chapter3 Functions and Matrices

Warning: (1) f(x) does not m e a n f t i m e s x. It simply denotes t h e i t e m y e Y t h a t x e X is paired with. (2) Let f 9 X --> Y and x a n y e l e m e n t in X. Then, for convenience, we m a y call f(x) the function, a l t h o u g h it is incorrect. R e m e m b e r , f is the function and f(x) is j u s t a value!

There is an a l t e r n a t e way of defining a function f : X ~ Y. Since every x e X d e t e r m i n e s a u n i q u e e l e m e n t y = f(x) in Y, we can form t h e o r d e r e d pair (x~) which belongs to X • Y. T h e set of all such pairs (x,y) can be used to define f.

Next we define a useful subset of the codomain of a function. Range of a Function

Let f 9X ~ Y a n d A __ X. T h e n f(A) denotes the set {f(a)ia e A}. In particular, f ( X ) is the r a n g e of f, denoted by r a n g e ( f ) . T h u s r a n g e ( f ) - f(X) {f(x) ix e X}. Notice t h a t range(f) c_ Y. Consider the function f in F i g u r e 3.2. T h e n f(a) - f(c) - f(d) - 2 and f ( b ) - 3. L e t A - {b,c}. T h e n f ( A ) - {f(b),f(c)} - {3,2}. Also, r a n g e ( f ) {2, 3 } r codom(f), i P r o g r a m m i n g languages provide built-in functions. For example, R O U N D and T R U N C are two such functions. Both are functions from to Z; R O U N D r o u n d s off a real n u m b e r to the n e a r e s t integer, w h e r e a s T R U N C chops off the fractional part. The F O R T R A N , C + +, and J a v a functions M A X and M I N select the largest and the smallest of n integers (or real numbers); t h e y are functions from Z n to Z (or R" to R). F u n c t i o n s are often defined using formulas; t h a t is, by s t a t i n g t h e i r general behavior. M a n y of the formulas you are familiar with are, in fact, examples of functions. For instance, the formula C(r) - 2 n r (circumference of a circle of radius r) defines a function. We now p r e s e n t a few examples of a b s t r a c t functions defined by formulas. Let Z - {a, b, c}. Let f 9 Z* --* W, defined by f(x) = ilx li. T h e n f(~) - 0, f(abc) - 3, and f(aibJc h) - i + j + k. i ~

Let S be the set of binary words defined recursively as follows: i) l e S .

ii) I f x e S t h e n x 0 ,

xleS.

S consists of b i n a r y r e p r e s e n t a t i o n s of positive integers with no leading zeros. (See Section 4.3 for a discussion of b i n a r y numbers.) Let g" S --~ lq defined by g(x) - decimal value of x. T h e n g(100) - 4, g(110) - 6, and g(101001) - 41. i

3.1

~

The Concept of a Function

121

The c h a r a c t e r sets ASCII, m u l t i n a t i o n a l 1, box drawing, t y p o g r a p h i c a l s y m bols, math/scientific symbols, and G r e e k symbols, used by WordPerfect* a r e denoted by t h e c h a r a c t e r set n u m b e r s 0, 1, 2, 3, 4, 6, and 8, respectively (see Appendix A.1). Let A - {0, 1, 2, 3, 4, 6, 8}. E a c h c h a r a c t e r in a c h a r a c t e r set is associated w i t h a u n i q u e decimal n u m b e r , called its o r d i n a l n u m b e r (or its relative position). For example, t h e ordinal n u m b e r of t h e c h a r a c t e r '&' in ASCII is 38.** Let B - {32, 33, 3 4 , . . . , 60}, t h e set of ordinal n u m b e r s . T h e n we can define a function f : A x B ~ C defined by f(i,j) - c, w h e r e c is the c h a r a c t e r w i t h ordinal n u m b e r j in t h e c h a r a c t e r set i. F o r e x a m p l e , f(0,36) = '$' a n d f(8,38) = ' E ' . m

Piecewise Definition

The above definitions of functions consist of j u s t one formula. In fact, t h e definitions of m a n y of the real-world functions consist of m o r e t h a n one formula. Such a definition is a p i e c e w i s e d e f i n i t i o n .

~

A town in M a s s a c h u s e t t s charges each h o u s e h o l d a m i n i m u m of $75 for up to 4000 cubic feet (ft 3) of w a t e r every 6 m o n t h s . In addition, each household has to pay 60r for every 100 ft 3 of w a t e r in excess of 4000 ft 3. E x p r e s s the w a t e r bill f(x) as a function of the n u m b e r of cubic feet of w a t e r x u s e d for 6 m o n t h s . SOLUTION:

The m i n i m u m charge is $75 for up to 4000 ft 3 of water, so f(x) 0 < x < 4000. Suppose you used more t h a n 4000 ft 3 of water. T h e n

Cost for the excess -

x - 4000 100

75 if

(0.60) - 0.006(x - 4000)

Then Total cost - m i n i m u m charge + cost for t h e excess - 75 + 0.006(x - 4000) Thus, the w a t e r bill f(x) in dollars can be c o m p u t e d u s i n g t h e piecewise definition" i f 0 < x < 4000

f (x) -- 175

/ 75 +

0.006(x - 4000)

if x > 4000

*WordPerfect is a wordprocessing program marketed by Corel Corporation. **A character within single quotes indicates a literal character.

m

Chapter 3 Functions and Matrices

122

~

Let A - {0, 1 , . . . , 127}, the set of ordinal n u m b e r s in ASCII. Let f 9 A --* ASCII be defined by n o n p r i n t a b l e control c h a r a c t e r uppercase letter lowercase letter other printable c h a r a c t e r

f(n)-

if 0 < n < 31 or n - 127 if 65 < n < 90 i f 9 7 < n < 122 otherwise

Clearly, f is defined piecewise.

m

F u n c t i o n s defined piecewise are w r i t t e n as i f - t h e n - e l s e s t a t e m e n t s in most p r o g r a m m i n g languages. For example, the function in E x a m p l e 3.6 can be w r i t t e n as follows: if

(x _> O) and (x _< 4000) then f ( x ) ~- 50 else f ( x ) ~-- 50 + O.O06(x - 4000)

The geometrical r e p r e s e n t a t i o n of a function, called a g r a p h , is often used to study functions. Remember, a picture is w o r t h a t h o u s a n d words. Since every function f : X --. Y is a set of ordered pairs (x,y), the g r a p h of f consists of points corresponding to the ordered pairs in f, as the next example illustrates. ~

G r a p h each function. (1) Let f 9 Z ~ Z defined by f(x) (2) Let g 9 R --* R defined by

-

x 2.

X --1 3X + 4

g(x)--

ifx > 0 if--2<X<0 otherwise

SOLUTION: The g r a p h s of the functions are displayed in Figures 3.6 and 3.7, respectively. Notice t h a t the graph of f is a discrete collection of points. F i g u r e 3.6

-,~

~

~I

x

3.1 The Concept of a Function F i g u r e 3.7

123

y

q{

~x

I~ m N e x t we define a n d i l l u s t r a t e t w o w a y s to c o n s t r u c t n e w f u n c t i o n s f r o m k n o w n ones.

Sum and Product L e t f 9 X -~ R a n d g 9 Y ~ I~. T h e y c a n be c o m b i n e d to c o n s t r u c t n e w f u n c t i o n s . T h e s u m and p r o d u c t of f a n d g, d e n o t e d by f + g a n d f g , respectively, a r e d e f i n e d as follows: ( f + g ) ( x ) = f (x) + g ( x ) ( f g ) ( x ) - f (x) . g ( x )

T h e f u n c t i o n s f + g a n d f g a r e d e f i n e d w h e r e v e r both f a n d g a r e d e f i n e d . T h u s d o m ( f + g) = d o m ( f g ) = d o m ( f ) n dom(g). Let f(x)= x 2 and g(x) = ~x [ 1, c~). T h e n

1, w h e r e d o m ( f ) - ( - ~ , c ~ )

and dom(g) =

( f + g ) ( x ) - f (x) + g ( x ) -- x 2 + ~/x - 1

and

( f g ) ( x ) - f ( x ) . g ( x ) = x 2 ~/x

"1

Since dom(f) n d o m ( g ) = [ 1 , ~ ) , b o t h f(x) a n d g ( x ) a r e d e f i n e d o n l y w h e n x > 1, so d o m ( f + g) - dom(fg) = [1,c~). m Finally, t w o f u n c t i o n s f 9 A ~ B a n d g 9 C ~ D a r e e q u a l if A = C, B - D, a n d f(x) - g ( x ) for e v e r y x ~ A. W e shall u s e t h i s d e f i n i t i o n i n t h e n e x t section.

Exercises 3.1 T h e Celsius a n d F a h r e n h e i t scales a r e r e l a t e d by t h e f o r m u l a F = 9 C + 32. 1. E x p r e s s - 4 0 ~

on t h e F a h r e n h e i t scale.

124

Chapter 3 Functions and Matrices

2. Express 131~ on the Celsius scale.

Letg(x)-

3. g ( - 3 . 4 )

21x]+3 5 -x 2

4. g(0)

if x < 0 if0<x<3. otherwise

C o m p u t e each.

5. g(0.27)

6. g(4.5)

Using Example 3.6 c o m p u t e the w a t e r bill for each a m o u n t of water. 7. 1000 ft 3

8. 4000 ft 3

9. 5600 ft 3

10. 7280 ft 3

Let E = {0, 1}. Let f : E* ~ W defined by f(x) = Ilxl]. E v a l u a t e f(x) for each value of x. 11. 000101

12. 1010100

13. 0001000

14. 00110011

Let Z denote the English alphabet. L e t f : E* • Z* --* E* defined b y f ( x , y ) = xy, the concatenation of x and y. Find f ( x # ) for each pair of words x and y. 15. combi, natorics

16. net, w o r k

Let A = {32, 3 3 , . . . , 126}. Let f : A --~ ASCII defined by f(n) = c h a r a c t e r with ordinal n u m b e r n. Find f(n) for each value of n. 17. 38

18. 64

19. 90

20.

123

Let g : A S C I I --~ A defined by g(c) = n, where A = { 3 2 , 3 3 , . . . , 126} and n denotes the ordinal n u m b e r of the c h a r a c t e r c. Find g(c) for each c h a r a c t e r c. 21. '+'

22. '<'

23. 'z'

24. '{'

Let f : Z • Z --~ Z defined by f ( x # ) = 2x + 3y - 6xy. C o m p u t e the following. 25. f(2,3)

26. f ( - 3 , 0 )

27. f ( - 2 , 3 )

28. f ( - 3 , - 5 )

Let Z denote the English alphabet. Let g : Z* ~ Z* defined by f(w) = awa. The function prefixes and suffixes each word with a. Find f(w) for each word w. 29. zale

30. mbrosi

31. rom

32. nesthesi

Using the function in Example 3.4 evaluate each, if defined. 33. f(101)

34. f(1010)

35. f(o01)

36. f(11011)

Let n e N. A positive integer d is a p r o p e r f a c t o r of n if d is a factor of n and d < n. For example, the proper factors of 12 are 1, 2, 3, 4, and 6. Let a: N ~ N defined by a(n) = s u m of the proper factors of n. (a is the lowercase Greek letter, sigma.) C o m p u t e a(n) for each value of n, where p and q are distinct primes. [A positive integer n such t h a t a(n) - n is a perfect number.] 37. 6

38. 12

39. pq

40. p2

3.2 SpecialFunctions

125

U s i n g t h e functions f ( x ) = 2x + 3 a n d g ( x ) 41. ( f + g ) ( - 3 )

42. (fg)(2)

= x 2 -

l, find t h e following.

43. ( f + g ) ( x )

44. ( f g ) ( x )

Let f 9X ~ Y and A, B c X t. Prove each. 945. f(A u B) - f(A) u f(B)

*46. f(A ~ B) ___f(A) n f(B)

9 47. If B _c A c X, t h e n f(A) - f ( B ) ~ f ( A -

B).

H e r e we t u r n our a t t e n t i o n to some i m p o r t a n t functions used in d i s c r e t e mathematics.

Polynomial Function A function f 9 I~ ~ IR defined by f ( x ) = a n x n + a n - 1 x n - 1 + ' " - F a l + a0, w h e r e a0, a l , . . . , an E ]~, n e W, and an ~= O, is a p o l y n o m i a l f u n c t i o n . T h e expression an xn + a n _ l x n - 1 + . . . + a l x + ao is a p o l y n o m i a l o f d e g r e e n in x. W h e n n - 1, f is a l i n e a r f u n c t i o n ; w h e n n - 2, f is a q u a d r a t i c function.

Exponential and Logarithmic Functions Let a e •+, a ~= 1, and x a n y real n u m b e r . T h e function f: I~ ~ R + defined by f ( x ) - a x is an e x p o n e n t i a l f u n c t i o n w i t h base a. T h e m o s t f r e q u e n t l y used base in c o m p u t e r science is two. F i g u r e 3.8 shows t h e g r a p h of t h e exponential function f ( x ) -- 2 x. F i g u r e 3.8

y 2x

J 0

~X

Let a e IR+, a r 1, a n d x and y a n y real n u m b e r s such t h a t y - a x. T h e n x is called the l o g a r i t h m of y to the base a, d e n o t e d by log ay. T h u s (logay - x) <-->(y -- aX). Accordingly, the function f: R + -~ R defined t S, T c_X means S c_X and T ~ X.

Chapter3 Functionsand Matrices

126

by f(x) = log ax is the l o g a r i t h m i c f u n c t i o n with base a. (See Appendix A.3 for a brief discussion of exponential and logarithmic functions.) R e m e m b e r t h a t the most commonly used base in c o m p u t e r science is two. The corresponding logarithm is denoted by lg. T h u s lg x - log2 x, and Figure 3.9 shows the graph of the logarithmic function f(x) = lg x.

F i g u r e 3.9

y

~X

C

Absolute Value Function The a b s o l u t e v a l u e f u n c t i o n is a function f: R -~ R defined by f(x) - Ix l. Its graph is displayed in Figure 3.10. (Languages such as F O R T R A N and Java provide a built-in function, ABS, for finding absolute values.)

F i g u r e 3.10

y y= Ixl

0

WX

Floor and Ceiling Functions The floor and ceiling functions are often used in the study of algorithms, as will be seen in the next two chapters. The floor of a real n u m b e r x, denoted by LxJ, is the greatest integer _< x. The c e i l i n g of x, denoted by Ix], is the least integer >__x. The floor ofx r o u n d s d o w n x while the ceiling ofx r o u n d s up. Accordingly, if x r Z, the floor of x is the nearest integer to the left o f x on the n u m b e r line and the ceiling of x is the nearest integer to the right o f x, as shown in Figure 3.11. The floor f u n c t i o n f(x) - lxJ and the

F i g u r e 3.11

lxJ [xl-1

[xJ+ 1 x

Ix]

32 SpecialFunctions

127

.-..., ,~," ,.2

,,

'~,~ .

,.,..-:.

,.

<...,.-

?'

K e n n e t h E. Iverson (1920-) was born at Camrose, Alberta, Canada. After graduating from Queen's University, Kingston, Ontario, in 1950, he received his M.A. from Harvard University in the following year. Three years later, he received his Ph.D. in applied mathematics from Harvard, where he taught until 1960. The programming language APL (A Programming Language) was his brainchild. He conceived the idea while a student at Harvard. After joining IBM in 1960, Iverson and Adin D. Falkoff developed APL into a full-fledged programming language. Iverson's many honors include the Harry Goode Award from American Federation of Information Processing Society (1975), the Turing Award from the Association of Computing Machinery (1979), and an honorary degree from York University, Toronto (1997).

c e i l i n g f u n c t i o n g(x) = [xl are also known as the g r e a t e s t i n t e g e r f u n c t i o n and the l e a s t i n t e g e r f u n c t i o n , respectively. For example, LnJ = 3, Llg 3J = 1, L-3.5J = - 4 , L-2.7J = - 3 , [n l = 4, [lg 31 = 2, [ - 3 . 5 1 = - 3 , [-2.71 = - 2 , and L3J = 3 = F31. These two notations and the names floor and ceiling were introduced by Kenneth E. Iverson in the early 1960s. Both notations are variations of the notation Ix], which was used in n u m b e r theory. Figures 3.12 and 3.13 show the graphs of the floor and ceiling functions. The p r o g r a m m i n g languages PL/1, C, and J a v a provide the floor and ceiling functions as built-in functions, namely, F L O O R and C E I L . BASIC supports an intrinsic function called INT, which is in fact the floor function. The floor function comes in handy when real n u m b e r s are to be t r u n c a t e d or rounded off to a desired n u m b e r of decimal places. For example, the real n u m b e r n = 3.1415926535... t r u n c a t e d to three decimal places is given by L1000nJ/1000 -- 3141/1000 -- 3.141; on the other hand, n rounded to t h r e e decimal places is L1000n + 0.5J/1000 = 3.142. Figure 3.12 Graph of the floor function. '

0

o

'0

~X

128

Chapter3 Functionsand Matrices

F i g u r e 3.13

y

G r a p h of the ceiling function.

O

~X

0

O

The next example p r e s e n t s an application of the ceiling function to everyday life. In 2003, the postage rate for a first-class letter of weight x not more t h a n 1 ounce was 37r the rate for each additional ounce or a fraction t h e r e o f up to 11 ounces was an additional 23r Thus, the postagep(x) for a first-class letter is given b y p ( x ) = 0.37 + 0.23 Ix - 17, where0<x< 11. For instance, the postage for a letter weighing 7.8 ounces is p(7.8) = 0.37 + 0 . 2 3 [ 7 . 8 - 17 = $1.98. m

~

(The post-office f u n c t i o n )

Some properties of the floor and ceiling functions are listed below. We shall prove part 3, leaving the o t h e r parts as r o u t i n e exercises. ~

L

e

x be any real n u m b e r and n any integer. Then:

t

(1) LnJ = n = In[ (3) L x + n J = [ x J + n (5)

(2) Ix7 = [xJ + l ( x C Z ) (4) I x + n 7 = Ix7 + n

~n] n- 1 ifn isodd ~ = 2

(6) [ 2 7

-

n+ 1 ifn isodd 2

PROOF" 3) Every real n u m b e r x can be w r i t t e n as x = k + x' where k - LxJ and 0 < x' < 1. T h e n x + n = k + n + x ' - (n + k ) + x ' So

Ix+n] =n+k,

since0<x'<

1

- LxJ + n as desired,

m

The floor function can be used to d e t e r m i n e the n u m b e r of positive integers less t h a n or equal to a positive integer a and divisible by a positive integer b, as the next t h e o r e m shows.

3.2

Special Functions

129

Let a and b be any positive integers. T h e n the n u m b e r of positive i n t e g e r s < a and divisible by b is La/bJ. PROOF:

Suppose t h e r e are k positive integers < a and divisible by b. We n e e d to show t h a t k - La/bJ. The positive multiples of b less t h a n or equal to a are b, 2 b , . . . , kb. Clearly, k b < a or k < a/b. F u r t h e r , (k + 1)b > a. So k + 1 > a/b or a/b - 1 < k. So a a l
II

For example, the n u m b e r of positive integers _< 1776 and divisible by 13 is [1776/13J - L136.615...J - 136. The next two examples employ T h e o r e m 3.2 and the i n c l u s i o n - e x c l u s i o n principle. Find the n u m b e r of positive integers < 3000 and n o t divisible by 7 or 8. SOLUTION: Let A - {x ~ Nix < 3000 and divisible by 7} and B - {x e Nix < 3000 a n d divisible by 8}. We need to find IA' N B'I" IA' N B ' I -

I(n u B)'I

= IUf-

IA u B I

= I U I - I A I - IBI + IA n BI = 3000 - L3000/7A - L3000/8J + L3000/56] = 3000 - 428 - 375 + 53 - 2250

II

Find the n u m b e r of positive integers _< 2076 and divisible by 3, 5, or 7. SOLUTION: Let A, B, and C denote the sets of positive integers < 2076 and divisible by 3, 5, and 7, respectively. By the inclusion-exclusion principle, I A u B u CI - IAI + IBI + ICI - IA NBI - IB n CI - ICNAI + IA N B n CI

= 692 + 415 + 296 -

138 - 59 - 98 + 19 -

1127

II

130

Chapter 3

Functions and Matrices

In October 1582, Fr. C h r i s t o p h e r Clavius a n d Aloysius Giglio i n t r o d u c e d the G r e g o r i a n c a l e n d a r at the r e q u e s t of Pope Gregory XIII to rectify t h e errors of t h e J u l i a n calendar. In t h e G r e g o r i a n calendar, w h i c h is universally accepted, a nonleap y e a r c o n t a i n s 365 days and a leap y e a r c o n t a i n s 366 days. (A y e a r is a l e a p y e a r if it is a c e n t u r y divisible by 400, or if it is a n o n c e n t u r y and divisible by 4. For example, 1600 a n d 1976 w e r e leap years, w h e r e a s 1778 a n d 1900 were not.) T h e next example shows how to derive a f o r m u l a to c o m p u t e t h e n u m b e r of leap y e a r s beyond 1600 a n d not exceeding a given y e a r y. Prove t h a t the n u m b e r of leap y e a r s ~ after 1600 and not e x c e e d i n g a given y e a r y is given by

=L J_L PROOF: Let n be a y e a r such t h a t 1600 < n < y. To derive the f o r m u l a for t~ we proceed step-by-step:

9 To f i n d the n u m b e r o f years n in the range divisible by 4" Let 4nl be such a year. T h e n 1600 < 4n1 < Y" t h a t is, 400 < n l

Therefore, t h e r e are nl =

<

y

- 400 such years.

9 To f i n d the n u m b e r o f centuries in the range 1600 < n < y: Let 100n2 be a c e n t u r y such t h a t 1600 < 100n2 < y.

Then 16
Y-~. 100 Therefore, t h e r e are n2 -

~ LSj

16 c e n t u r i e s beyond 1600 a n d _
9 To f i n d the n u m b e r o f centuries in the range divisible by 400: Since t h e y are of the form 400n3, we have 1600 < 400n3 _< y. T h e n

4
4--0-0' s o n 3 -

~

-4.

9 Therefore, t~ - n l - n2 A- n3

--

§

-L J-LI- J- L -

~J

L

-4

- 388

I

The t e c h n i q u e employed for c o m p u t e r r e p r e s e n t a t i o n of sets (Section 2.3) is a consequence of the next function.

3.2 Special Functions

131

Characteristic Function Let U be a universal set and S an a r b i t r a r y subset of U. T h e n we can define a function f s : U ~ {0, 1} as follows: fs(x)-

/1 /0

ifxES otherwise

The function f s is called the c h a r a c t e r i s t i c f u n c t i o n of S. The following example illustrates this definition. Let U -

{ a , b , c , d , e , f}, A -

{a,c, d, e}, a n d B - {a,b, d}. T h e n

fA (x) -- / 1

/0

w h e n x - a, c, d, e otherwise

In other words, fA(a) = fA(c) = fA(d) = fA(e) = 1 and fA(b) = fA(f) = 0 (see Figure 3.14). Similarly, fB(a) = fB(b) = fB(d) = 1 and fB(c) = fB(e) = fB (f) ----0. m

F i g u r e 3.14

fA a

d

-]

A

B

The characteristic function f s assigns the value 1 or 0 to each e l e m e n t of the universe. So f s and hence the set S can be uniquely identified by an n-bit word and vice versa, where IUI = n. This fact enabled us to r e p r e s e n t sets as n-bit words in Section 2.3. For example, the characteristic function fA and hence the set A in the above example uniquely d e t e r m i n e the 6-bit word 011101, w h e r e we have listed the bits from right to left for consistency. Similarly, fB d e t e r m i n e s the word 001011. The characteristic function satisfies the following properties. Let A and B be any two sets, and U the universe. Let f s denote t h e characteristic function of a subset S of U and x an a r b i t r a r y e l e m e n t in U. Then: (1) (2) (3) (4)

fAnB(x) fAuB(x) fA,(X) = fAeB(X)

= 1 =

fA(x) "fB(x) fA (x) + fB(x) -- fAnB(x) -- fA (x) fA(x) + fB(x) -- 2fAnS(x)

132

Chapter3 Functionsand Matrices PROOF: We shall prove p a r t 1 and leave the other parts as exercises. Casel LetxeANB. T h e n f A ~ ( X ) = 1, by the definition of the characteristic function. Since x e A n B, x e A and x e B. Therefore, fA(x) = 1 -- f s ( x ) ; SO, f A ( x ) " f B ( x ) = 1. 1. Thus f A n S ( x ) = 1 -- fA (x) 9f s ( x ) . Case2 Letx~ANB. T h e n fAnB(X) - O. Since x r A n B , x ~ A or x r B. Therefore, e i t h e r fA (x) = 0 or f s ( x ) = 0. So, in any case, fA ( x ) . f s ( x ) -- 0. T h u s f A n B ( x ) = 0 = fA (x) "fB(x). Thus f A n B ( x ) = f A ( x ) " fB(X) for every x ~ U.

I

Mod and Div Functions The m o d f u n c t i o n f(x, y) = x m o d y denotes the r e m a i n d e r when an integer x is divided by a positive integery. The d i v f u n c t i o n g ( x ~ y ) - x divy denotes the quotient when x is divided by y. P r o g r a m m i n g languages often provide two such built-in operators, r o o d and div; in C+ +, the mod o p e r a t o r is denoted by the percent symbol %, and the div operator by the forward slash/. For example, 23 mod 5 - 3, 18 mod 6 - 0, 23 div 5 - 4, and 5 div 6 - 0. A scientific calculator, such as the TI-86, can be used to compute x mod y using the keys [MATH], [Num], [MORE1, and ~--~. Consult the m a n u a l for your calculator to check if it supports the operator. The mod function can determine the day of the week in n days from a given day. In 7 days, 14 days, and so on from a given day, it will again be the same day. Consequently, all we need do is remove the m a x i m u m n u m b e r of 7's from n. Let r be the r e m a i n d e r when n is divided by 7. T h e n the r t h day from the given day is the day we are looking for, as the next example demonstrates. Today is Thursday. W h a t day of the week will it be in 100 days from today? SOLUTION: 100 mod 7 - 2. Two days from T h u r s d a y is Saturday, so it will be S a t u r d a y in 100 days from Thursday. I The following example is a simple application of both div and mod operators.

C a r d D e a li n g Consider a s t a n d a r d deck of 52 playing cards. They are originally assigned the n u m b e r s 0 t h r o u g h 51 in order. Use the suit labels 0 = clubs, 1 diamonds, 2 = hearts, and 3 = spades to identify each suit, and the card labels 0 - ace, 1 -- deuce, 2 = t h r e e , . . . , and 12 -- king to identify the cards

3.2 Special Functions

133

in each suit. S u p p o s e c a r d x is d r a w n at r a n d o m f r o m a well-shuffled deck, w h e r e 0 < x < 51. H o w do we i d e n t i f y t h e card? First, we n e e d to d e t e r m i n e t h e suit to w h i c h t h e c a r d belongs. It is given by x div 13. Next, we n e e d to d e t e r m i n e t h e c a r d w i t h i n t h e suit; t h i s is given by x m o d 13. T h u s c a r d x is c a r d (x m o d 13) in s u i t (x div 13). F o r e x a m p l e , let x - 50. Since 50 div 13 - 3, t h e c a r d is a spade. N o w 50 m o d 13 = 11, so it is a q u e e n . T h u s c a r d 50 is t h e q u e e n of spades, m N e x t we p u r s u e a n i n t r i g u i n g a p p l i c a t i o n of t h e floor f u n c t i o n a n d t h e m o d o p e r a t o r to t h e g a m e of chess.

The Two Queens P u z z l e T h e r e are two q u e e n s on a n 8 x 8 c h e s s b o a r d . O n e c a n c a p t u r e t h e o t h e r if t h e y are on t h e s a m e row, c o l u m n , or diagonal. T h e 64 s q u a r e s on t h e b o a r d a r e n u m b e r e d 0 t h r o u g h 63. S u p p o s e one q u e e n is in s q u a r e x a n d t h e o t h e r in s q u a r e y, w h e r e 0 < x, y < 63. C a n one q u e e n c a p t u r e t h e o t h e r ? Since t h e s q u a r e s a r e labeled 0 t h r o u g h 63, we can label e a c h r o w w i t h t h e n u m b e r s 0 t h r o u g h 7, a n d e a c h c o l u m n w i t h t h e s a m e n u m b e r s 0 t h r o u g h 7. In fact, each row label - [r/8J a n d e a c h c o l u m n label = c m o d 8, w h e r e 0 < r, c < 63. See F i g u r e 3.15. T h u s , t h e q u e e n in s q u a r e x lies in r o w [x/8J a n d c o l u m n x m o d 8, a n d t h a t in s q u a r e y lies in r o w Ly/8J a n d c o l u m n y m o d 8. C o n s e q u e n t l y , t h e t w o q u e e n s will be in t h e s a m e r o w if a n d only if Lx/8J - [y/8J a n d in t h e s a m e c o l u m n if a n d only if x m o d 8 - y m o d 8. F o r example, if x - 41 a n d y - 47, t h e two q u e e n s lie on t h e s a m e row.

F i g u r e 3.15

< 0

0

1

2

3

4

5

6

7

1

8

9

10

11

12

13

14

15

2

16

17

18

19

20

21

22

23

3

24

25

26

27

28

29

30

31

4

32

33

34

35

36

37

38

39

5

4O

41

42

43

44

45

46

47

6

48

49

50

51

52

53

54

55

7

56

57

58

59

60

61

62

63

column label

T

row label H o w do we d e t e r m i n e if t h e y lie on t h e s a m e diagonal? T h e r e a r e 15 n o r t h e a s t diagonals a n d 15 s o u t h e a s t diagonals. W i t h a bit of p a t i e n c e , we can show t h a t t h e q u e e n s lie on t h e s a m e d i a g o n a l if a n d only if t h e a b s o l u t e value of t h e difference of t h e i r r o w labels e q u a l s t h a t of t h e d i f f e r e n c e of t h e i r c o l u m n labels; t h a t is, if a n d only if ILx/8J - Ly/8Jl - Ix m o d 8 y m o d 81.

Chapter 3

134

Functions and Matrices

F o r example, let x = 51 and y - 23; see F i g u r e 3.15. T h e n JL51/8J L23/8J i = ]6-2] = 4 = ]3-70 = I51 mod 8 - 2 3 mod 8i, so one q u e e n c a p t u r e s the other. On the other hand, ifx = 49 a n d y = 13, t h e n I [49/8J - L13/8J ] r 149 mod 8 - 13 mod 8i; so one queen c a n n o t c a p t u r e the other, m Exercises 3.2

E v a l u a t e each, w h e r e n is an integer. 1. Ln + 1/2J

2. In/2]

3. In + 1/2]

4. [n/2]

Let x - 3.456 and y - 2.789. C o m p u t e each. 5. Lx + y J

9. L-xJ

6. LxJ + [yJ

7.

10.-LxJ

Lxyj

8. Lxj[yJ

11. F x + y ]

12. Fx] + Fy]

Find the r a n g e of each function on IR. 13.

f ( x ) - LxJ + [-xJ

14.

f ( x ) - Fx] + F-x]

Find the n u m b e r of positive integers < 3076 and divisible by: 15. 3 or 4

16. 3, 5, or 7

17. 3, 5, or 6

18. N e i t h e r 3 n o r 5

C o m p u t e the n u m b e r of leap years after 1600 and not beyond each year. 19. 2000 Let U -

20. 2020

21. 3076

22. 4050

{ a , . . . , g}. Define the characteristic function h of each set.

23. {a,c,d,f}

24. {a,e,g}

25. {b,c,g}

26. { a , c , d , f , g }

fs is given

Let U - { a , . . . , h}. In Exercises 27-30, a characteristic function as an 8-bit word. Find the c o r r e s p o n d i n g set S. 27. 11010100

28. 00101101

29. 10101010

30. 01010101

Find the day of the week in each case. 31. 234 days from Monday

32. 365 days from F r i d a y

33. 1776 days from W e d n e s d a y

34. 2076 days from S a t u r d a y

Let S = { t r u e , false }. Define a b o o l e a n f u n c t i o n if y e a r n is a leap y e a r and false otherwise. F i n d 35. 1996

36.

2020

37. 2076

f 9 N --* S by

f(n)

-

true

f(n) for each y e a r n. 38.

3000

39. J a n u a r y 1, 2000, falls on a S a t u r d a y . W h a t day of the week will J a n u a r y 1, 2020, be? (Hint: Look for leap years.) 40. J a n u a r y 1, 1990, was a Monday. W h a t day of the week was J a n u a r y 1, 1976? (Hint: Again, look for leap years.)

3.2 Special Functions

135

Each day of the week, beginning with Sunday, can be identified by a code x, where 0 < x < 6. J a n u a r y 1 of any year y can be determined using the following formula**. x-(y+[Y41]

-[10:]+[400])

mod7

(3.1)

Using this formula determine the first day in each year. 41. 2000

42. 2020

43. 2076

44. 3000

The n u m b e r of Friday-the-thirteenths in a given year y can be computed using formula (1) above and Table 3.1. For example, suppose t h a t J a n u a r y 1 of a year y falls on a Sunday(0). If it is not a leap year, there will be two Friday-the-thirteenths: J a n u a r y 13 and October 13; if it is a leap year, there will be three: J a n u a r y 13, April 13, and July 13. Compute the n u m b e r of Friday-the-thirteenths in each year. 45. 2000 T a b l e 3.1

46. 2020

47. 2076

Code x

January 1

N o n l e a p year y

0 1 2 3 4 5 6

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

January, October April, July September, December June February, March, November August May

48. 3076

Leap year y

January, April, July September, December June March, November February, August May October

( E a s t e r S u n d a y ) The date for Easter Sunday in any year y can be computed as follows. Let a - y mod 19, b - y mod 4, c - y mod 7, d - (19a + 24) mod 30, e - (2b + 4c + 6d + 5) mod 7, and r - (22 + d + e). If r < 31, then Easter Sunday is March r; otherwise, it is April [r (mod 31)]. Compute the date for Easter Sunday in each year. 49. 1996

50. 2000

51. 2076

52. 3000

Prove each, where x c R and n ~ Z. 53. [ 2 ] - n - l i f n 2

is odd.

54. [ 2 ] - n + 1 2

if n is odd.

**Based on G. L. Ritter et al., "An Aid to the Superstitious," Mathematics Teacher, Vol. 70, May 1977, pp. 456-457.

136

Chapter 3 Functions and Matrices

l

n21

55.

~ 59.

T

n

n2 - 1 -

4

if n is odd.

56.

n

I n21

n2 4+ 3 if n is odd.

-

58. F x T = L x ] + l

§ F l--n

60.

[xl = - [ - x J

[x+nl

=

(xCZ)

[xl + n

Let A and B be any two sets, and U the universe. Let f s denote the characteristic function of a subset S of U and x an arbitrary element in U. Prove each. "61. f A u B ( x ) : f A ( x ) + f B ( x ) -- f A n B ( X ) *62. fA,(X) = 1 -- fA (x) *63. f A e B ( X ) =

f A ( x ) + f B ( x ) -- 2 f A n B ( X )

Let x, y E R. Let max{x,y} denote the m a x i m u m o f x and y, and min{x,y} denote the m i n i m u m of x and y. Prove each. *64. max{x,),} + min{x~} = x + y

*65. max{x,y} - min{x,y} = Ix - Y l

Functions satisfy a n u m b e r of properties and we begin with the identity function. Identity F u n c t i o n

A function f on X is the i d e n t i t y f u n c t i o n if f ( x ) = x for every x in X. It is denoted by lx and leaves every input unchanged. The graph of the identity function on R is the 45~ y = x. Let S be an ordered set. ORD(x) denotes the ordinal n u m b e r of each element x in S, the first ordinal n u m b e r being 0. For example, using ASCII, ORD(' <,)t __ 60 and ORD('$') = 36. (Pascal, for instance, provides such a built-in function.) If the a r g u m e n t x, however, is an integer n, ORD(n) - n. Thus ORD is the identity function on W. m

Injection

A function f 9X ~ Y is an i n j e c t i o n (or o n e - t o - o n e f u n c t i o n ) if different input values yield different output values. Thus f is injective, if x l r x2 implies f ( x l ) r f(x2); equivalently, f is injective if f ( x l ) - f(x2) implies Xl = x2 (why?). The next two examples illustrate this definition. t x within single quotes indicates the character x.

3.3 Propertiesof Functions

137

L e t A - {0, 1 , 2 , . . . , 127}. Let C H R : A ~ ASCII defined by CHR(n) = ASCII c h a r a c t e r with ordinal n u m b e r n. For example, CHR(59) - ';' and CHR(43) = ' + ' . Since distinct ordinal n u m b e r s correspond to different characters, CHR is injective. (C, C + + , and Java, for instance, provide such a built-in function.) m ~

Let f " E* ~

W defined by f ( x ) - Ilxil, w h e r e E -- {a, b , c } . The words a a a

and b a b are distinct words with the same length, so f is not injective,

m

How do you c h a r a c t e r i z e the g r a p h of an injective function f : IR --~ IR? The function is not injective if t h e r e are two distinct i n p u t elements a, b ~ IR such t h a t f ( a ) = f ( b ) , t h a t is, if a horizontal line intersects its g r a p h in two distinct points ( a , f ( a ) ) and ( b , f ( b ) ) . T h u s f is injective if and only if no horizontal line intersects the g r a p h in more t h a n one point. For example, the function f ( x ) = x 2 is not injective (see Figure 3.16).

F i g u r e 3.16

(b, f(b))

(a, f ( a) )

O~

}. x

Surjection A function f" X ~ Y is a s u r j e c t i o n (or an o n t o f u n c t i o n ) if for every y in Y t h e r e exists an x in X such t h a t f ( x ) = y , t h a t is, if every element in Y has at least one pre-image in X. In other words, f is surjective if range(f) - Y. The following two examples clarify this definition. ~

Let E be a n o n e m p t y alphabet. Let f" E* ~ W defined by f ( x ) - Ilxll. Let n E W. T h e n x n ~ E* and IIx ~ II - n. Thus, given any n ~ W, there exists an e l e m e n t u = x n ~ E* such t h a t f ( u ) = n . Consequently, f is surjective, i

~

D e t e r m i n e if the function f ( x ) - x 2 on R is surjective. SOLUTION: For every y in R, does t h e r e exist a real n u m b e r x such t h a t x 2 = y? No, for instance, t h e r e is no real n u m b e r x such t h a t x 2 = - 1 , so f is n o t surjective, m

Bijection A function f" X -~ Y is a b i j e c t i o n (or o n e - t o - o n e c o r r e s p o n d e n c e ) it is both injective and surjective.

if

138

Chapter3 Functionsand Matrices Let A be t h e set of p r i n t a b l e ASCII c h a r a c t e r s a n d B = {32, 3 3 , . . . , 126}. Let f 9 A ~ B defined by f(c) - ordinal n u m b e r of c h a r a c t e r c. Since f is b o t h injective a n d surjective, f is bijective. (Notice t h e d e l i b e r a t e choice of B to m a k e t h e function surjective.) m Notice t h a t 23 m o d 5 - 3 = 48 mod 5, b u t 23 # 48; therefore, t h e m o d function is not injective. However, w h e n an i n t e g e r a is divided by m, t h e r e are m possible r e m a i n d e r s , namely, 0, 1, 2 , . . . , m - 1 (see Section 4.1); so, given a n y n o n n e g a t i v e i n t e g e r r less t h a n m, we can always find an i n t e g e r a such t h a t r = a mod m; t h u s t h e mod function is surjective.

Hashing We are now r e a d y to e x a m i n e an i n t e r e s t i n g application of t h e m o d function in everyday life. B a n k s use nine-digit account n u m b e r s to create a n d m a i n t a i n c u s t o m e r accounts. C u s t o m e r records are stored in an a r r a y in a c o m p u t e r a n d can be accessed fairly easily and quickly u s i n g t h e i r u n i q u e k e y s , which in this case are the account n u m b e r s . Access is often accomplished u s i n g t h e h a s h i n g f u n c t i o n h ( x ) = x mod m, w h e r e x denotes t h e key (account n u m b e r ) and m the n u m b e r of cells in the array; h ( x ) d e n o t e s t h e h a s h a d d r e s s of the c u s t o m e r record with key x. See F i g u r e 3.17.

F i g u r e 3.17

f f - customer record with key x

0

1

2

m - 1 ~-- hash address x mod m

In particular, let m = 1009 and x record is stored in location

= 207630764. T h e c o r r e s p o n d i n g

h(207630764) - 207630764 mod 1009 - 762 Likewise, h(307620765) = 307620765 mod 1009 - 881 Since the h a s h i n g function is not injective (why?), theoretically different c u s t o m e r records can be assigned to t h e s a m e location. F o r example, h(207630764) -- 762 - h(208801204) This results in a c o l l i s i o n .

3.3 Properties of Functions

139

One simple way to resolve a collision is to do a s e q u e n t i a l search for the next available cell, b e g i n n i n g w i t h t h e cell w h e r e t h e collision has occurred. T h e n we store t h e i t e m in t h e available cell. If we come to t h e end of the a r r a y w i t h o u t any success, t h e n we would continue the search back at the b e g i n n i n g of the array, as if t h e a r r a y were circular. This way of resolving a collision is called l i n e a r p r o b i n g . So we would store t h e d a t a with t h e account n u m b e r 208801204 in location 763 ( a s s u m i n g t h a t is available), m Obviously, the t e c h n i q u e i l l u s t r a t e d in this example can be a d a p t e d to a variety of situations. For example, t h e various identifiers in a c o m p u t e r p r o g r a m can be stored in a symbol table u s i n g t h e i r first letters as keys; s t u d e n t records can be stored in a h a s h table u s i n g t h e i r social s e c u r i t y n u m b e r s ; a n d p a t i e n t s ' medical records can be m a i n t a i n e d in a table u s i n g t h e i r social security n u m b e r s as keys. Next we p r e s e n t a few simple and useful properties of functions associated with finite sets. Let X and Ybe a n y two finite sets with IX] - I Y] - n. A function f 9X --~ Y is injective if and only if f is surjective. PROOF: Let X = {x 1, x 2 , . . . , x,~ } and Y = {y 1, Y 2 , . . . , yn }. Suppose f i s injective. T h e n f ( x l ) , . . . ,f(xn) are n distinct elements. So t h e y m u s t be the s a m e e l e m e n t s y 1,... ,y~ in some order. Therefore, f is surjective. Conversely, suppose f i s surjective. T h e n f ( x l ) , . . . , f(x,~) = Y. Since I Y] = n, the e l e m e n t s f ( x l ) , . . . , f(xn ) m u s t be different, so f m u s t be injective, m

Let A and B be two finite sets with t h e s a m e cardinality. Suppose we would like to show t h a t a function f 9 A --~ B is bijective. Then, by T h e o r e m 3.4, it suffices to show t h a t f is e i t h e r injective or surjective. Two finite sets have the same cardinality if and only if t h e r e exists a bijection b e t w e e n t h e m . PROOF: LetXand Y b e two finite sets with ]X[ - m and ]Y] - n. L e t X = {Xl,... ,Xm}. Let f 9X --~ Y be bijective. Since f i s injective, f ( x l ) , . . . ,f(Xm) are m distinct e l e m e n t s in Y. Consequently, m < n. Since f is surjective, every e l e m e n t y in Y has at least one i n p u t in X, so n _< m. T h u s ]X I - I Y[. Conversely, suppose m - n and Y - {Yl,... ,Ym}. Define a function f 9 X --~ Y by f(xi) - Y i for every i. We will now show t h a t f is injective. Let xj and xk be two e l e m e n t s in X such t h a t f(xj) - f(xk). Then, by definition, Yj - Yk; so, j - k and hence xj - xk. Therefore, f is injective and hence, by T h e o r e m 3.4, f is bijective, m

Let f 9X --~ Y, w h e r e X and Y are finite sets and IX] > ]YI. T h e n w h a t can we say a b o u t the function f? (Obviously, f c a n ' t be bijective.) This is a n s w e r e d in the next section.

Chapter3 Functionsand Matrices

140

Cardinality of an Infinite Set (optional) Before closing this section, we e x t e n d t h e concept o f c a r d i n a l i t y of finite sets to infinite sets, a topic of g r e a t i m p o r t a n c e to t h e o r e t i c a l c o m p u t e r science a n d c e r t a i n l y to m a t h e m a t i c s . Recall t h a t two finite sets h a v e t h e s a m e c a r d i n a l i t y if t h e r e is a bijection b e t w e e n t h e m . This leads to t h e following definition. Two sets X a n d Y have t h e s a m e c a r d i n a l i t y if t h e r e exists a bijection from X to Y, d e n o t e d by IXI = I YI. This definition can be u s e d to p a r t i t i o n t h e family of infinite sets into two disjoint classes. To this end, we m a k e t h e following definition.

Countable and Uncountable Sets A set S is c o u n t a b l y infinite if t h e r e exists a bijection b e t w e e n S a n d N. A set t h a t is finite or c o u n t a b l y infinite is c o u n t a b l e . A set t h a t is n o t c o u n t a b l e is u n c o u n t a b l e . T h e c a r d i n a l i t y of N is d e n o t e d by ~0 (read " a l e p h - n a u g h t , " " a l e p h " b e i n g t h e first l e t t e r of t h e H e b r e w a l p h a b e t . This s y m b o l w a s i n t r o d u c e d by Cantor). T h u s a set S is c o u n t a b l y infinite if ISI - ~0 = INI. I f A a n d B are finite sets such t h a t A c B, t h e n IAI < IBI. This, h o w e v e r , need not be t r u e in t h e case of infinite sets. F o r e x a m p l e , E c N, w h e r e E d e n o t e s t h e set of even positive integers; n o n e t h e l e s s , IEI - I NI = ~0, as s h o w n by t h e p a i r i n g s in F i g u r e 3.18.

F i g u r e 3.18

1

2

3

4

5

.

.

.

n

.

.

.

2

4

6

8

10

.

.

.

2n

.

.

.

Show t h a t N • 1~ is c o u n t a b l y infinite.

PROOF: A l t h o u g h we shall not give a f o r m a l proof, t h e a r r o w s in F i g u r e 3.19 show how t h e v a r i o u s e l e m e n t s of N • N can be listed as t h e first, second, third, a n d so on in a s y s t e m a t i c way, s h o w i n g t h a t N x N is c o u n t a b l y infinite.

F i g u r e 3.19

~(5,

(1,1)~

(1,2)

(2, 1)

(2, 2)

~, (1,3) (2, 3) ~

(1,4) (2, 4) I ~ ' .

. . . . .

(3, 1) j ' ~ " (3,2) ~ I ~ (3,3) ~

(3,4)

9 9 9

(4, 1) ~ ' I

(4, 4)

" 9 .

(4, 2) ~

(4, 3)

1)/~"

m

3.3

Properties of Functions

141

It follows by Example 3.22 t h a t the set of positive rational n u m b e r s is countable. Consequently, the set of negative rational n u m b e r s is also countable. Since the union of two countable sets is countable (see Exercise 49), it follows t h a t Q is countable. It may seem improbable t h a t t h e r e exist infinite sets t h a t are u n c o u n t able. For instance, the open interval (0,1) is such a set, as the next example shows. Show t h a t the open interval (0,1) is uncountable. P R O O F (by contradiction): Assume t h a t the interval (0,1) is countable. T h e n every real n u m b e r between 0 and i can be listed as al, a2, a3,. 99 Each a i has a unique decimal expansion (for n u m b e r s with two different decimal expansions, choose the expansion with trailing 9's. For example, a l t h o u g h 0.5 = 0.5000 . . . . 0.4999..., select 0.4999... for our discussion.): al

= O.alla12a13a14...

a 2 = O. a 2 1 a 2 2 a 2 3 a 2 4

999

a 3 = O. a 3 1 a 3 2 a 3 3 a 3 4

999

a 4 - - O. a 4 1 a 4 2 a 4 3 a 4 4

999

where each a i j is a digit. Now construct a real n u m b e r b = bi-

O.blb2b3b4...

l1

if all

#1

!2

if all

-- 1

as follows:

Clearly, 0 < b < 1; therefore b m u s t be one of the n u m b e r s in the above list al, a2, a3, a4, .... However, since bi # a i i for every i, b cannot be in the list. This leads to a contradiction. Therefore, the real n u m b e r s between 0 and 1 cannot be listed and hence the interval (0,1) is uncountable. (The technique employed is called C a n t o r ' s d i a g o n a l i z a t i o n p r o c e d u r e . ) m Since the interval (0,1) is uncountable, it follows t h a t R is also uncountable. So, although both 1~ and IR are infinite, IIRI > ~0. Exercises 3.3

Determine if each function is the identity function. x

a

b

c

d

1.

x

a

b

c

d

2. f(x)

a

b

c

d

a

b

c

d

a

b

c

c

0

f(x)

b

c

d

a

f(x)

142

Chapter3

Functions and Matrices

D e t e r m i n e if each function is injective, w h e r e t r u n c ( x ) d e n o t e s t h e i n t e g r a l p a r t of the real n u m b e r of x. 4. f(x) = Ix l , x ~ R

5. g ( x ) = 2

x,xeR

6. h(x) = lg x, x ~ IR+

7. f ( x ) =

LxJ, x c

8. g ( x ) = [ x ] , x ~ R

9. h ( x ) = trunc(x), x ~ R

10. f : S --~ W defined by f ( A ) = IAf, w h e r e S is the family of all finite sets. D e t e r m i n e if each function from R to Z is surjective. 11. f ( x ) =

Ix l

12. g ( x ) =

[xJ

13. h ( x ) =

Ix]

14. h ( x ) = lg Ix], x ~: 0

15. ORD: ASCII --~ W defined by ORD(c) = ordinal n u m b e r of t h e c h a r a c t e r c. 16. Let f : R ~ • defined by f(x) = ax + b, w h e r e a,b ~ R a n d a r 0. Show t h a t f is surjective; t h a t is, find a real n u m b e r x such t h a t f(x) = c. D e t e r m i n e if each function f :A --~ B is bijective. 17. f ( x ) = x2, A = B = R

18. f ( x ) = v/-x,A = R +, B = R

19. f(x) = ] x I , A = B = R

20. f(x) = LxJ,A = B = R

2 1 . f(x) = [ x ] , A = B = R

2 2 . f(x) = 2 IxI,A = B = R

D e t e r m i n e if the functions in Exercises 23-30 are bijective. If t h e y are not bijective, explain why. 23. f : E* ~ W defined by fix) = decimal value of x, w h e r e E = {0,1 }. 24. f : E* • Z* --~ E* defined by f(x,v) = xy, w h e r e E d e n o t e s t h e English alphabet. 25. g : E* --~ E* defined by g(w) = a w a , where E = {a,b,c }. 26. f : R • IR -* R • R defined by f ( x , y ) = ( x , - y ) . 27. The ORD function on ASCII. 28. The p r e d e c e s s o r f u n c t i o n ( P R E D ) and s u c c e s s o r f u n c t i o n ( S U C C ) are two i m p o r t a n t functions used in c o m p u t e r science. T h e y are defined on ordered sets. Ifc is a p r i n t a b l e ASCII c h a r a c t e r , PRED(c) denotes the predecessor of c a n d SUCC(c) denotes the successor of c; for example, P R E D ( ' ? ' ) = '@' a n d SUCC(':') = ';'. D e t e r m i n e if P R E D and SUCC are bijective. U s i n g the hash function in E x a m p l e 3.2, c o m p u t e the location corresponding to the given key. 29. 012398745

30. 430358856

S t u d e n t records are m a i n t a i n e d in a table u s i n g the h a s h i n g function h(x) = x mod 9767, w h e r e x denotes the s t u d e n t ' s social s e c u r i t y n u m b e r .

3.3 Properties of Functions

143

Compute the location in the table corresponding to the given key, where the record is stored. 31. 012-34-5678

32. 876-54-3210

33-34. Redo Exercises 31 and 32 if h ( x ) - first part in x mod 13. 35. Store the following two-letter abbreviations of states in the United States in a hash table with 26 cells, using the hashing function h ( x ) = first letter in x: NY, OH, FL, AL, MA, CA, MI, AZ 3{}. Redo Exercise 35 with the following state abbreviations: MD, CT, ID, MA, NB, NJ, MI, WI, CA, IA, WA, MN, NH, IN, NC, WY, NM, MS, MO, CO, NY, IL, NV, WV, ND, MT Two sets A and B are e q u i v a l e n t , denoted by A ~ B, if there exists a bijection between them. Prove each. 37. A ~ A (reflexive p r o p e r t y ) 38. A ~ A x

{1}

39. IfA ~ B, t h e n A x {1} ~ B x {2} 40. Z ~ O, the set of odd integers Prove each. 41. A bijection exists between any two closed intervals [a, b] and [c, d], where a < b and c < d. ( H i n t : Find a suitable function that works.) 42. The set of odd positive integers is countably infinite. 43. The set of integers is countably infinite. 44. Any subset of a countable set is countable. 45. A set A is infinite if and only if there exists a bijection between A and a proper subset of itself. 46. The open interval (a,b) is uncountable. [Hint: Find a suitable bijection from (0,1) to (a,b).] 47. The set Q+ of positive rational numbers is countable. 48. The set of irrational numbers is uncountable. ( H i n t : Prove by contradiction.) *49. A countable union of countable sets is countable. "50. The cartesian product of two countable sets is countable. "51. If E is a finite alphabet, then E* is countable.

144

Chapter3 Functions and Matrices

Suppose m pigeons fly into n pigeonholes to roost, where m > n. T h e n obviously at least two pigeons m u s t roost in the same pigeonhole (see F i g u r e s 3.20 and 3.21). This property, called the pigeonhole principle, can be s t a t e d in t e r m s of functions, as the next t h e o r e m shows.

F i g u r e 3.20

F i g u r e 3.21

J J

J

dS f J

( T h e P i g e o n h o l e P r i n c i p l e ) L e t f" X ~ Y, where X and Y are finite sets, IX I - m, I Y I - n, and m > n. T h e n t h e r e exist at least two distinct e l e m e n t s Xl and x2 in X such t h a t f ( x l ) - f(x2).

PROOF" Let X - {Xl,... ,Xm}. Suppose f is injective. T h e n f ( x l ) , . . . , f ( x m ) are distinct elements in Y. So m _< n. But this contradicts the a s s u m p t i o n t h a t m > n. Therefore, f is not injective and there m u s t be at least two distinct elements x l and x2 such t h a t f ( x l ) - f(x2). Hence the theorem, m The pigeonhole principle is a simple but i m p o r t a n t c o u n t i n g principle t h a t we shall use in C h a p t e r s 4, 7, and 8. The pigeonhole principle, which can be applied in a variety of situations, can be restated as follows: If m objects are placed into n boxes, t h e n at least one box m u s t contain two or more objects, where m > n. Accordingly, the pigeonhole principle is also called the D i r i c h l e t B o x P r i n c i p l e after the G e r m a n m a t h e m a t i c i a n P e t e r G u s t a v Lejeune Dirichlet, who used it extensively in his work on n u m b e r theory. Although the principle looks simple and straightforward, to apply it successfully you m u s t choose the pigeons and pigeonholes appropriately, as the next few examples illustrate. Suppose we select 367 s t u d e n t s from campus. Show t h a t at least two of t h e m m u s t have the same birthday.

SOLUTION: The m a x i m u m n u m b e r of days in a year is 366, and this occurs in a leap year. T h i n k of s t u d e n t s as pigeons and days of the year as pigeonholes. Let A be the set of s t u d e n t s and B the set of days, where IA[ = m = 367 and ]B] = n = 366. Let f : A -~ B defined by f ( a ) = b i r t h d a y of s t u d e n t a.

3.4 The Pigeonhole Principle

145

Gustav Peter Lejeune Dirichlet (1805-1859) was born in Duren, Germany. The son of a postmaster, he first attended a public school and then a private school that emphasized Latin. After attending the Gymnasium in Bonn for 2 years, Dirichlet entered a Jesuit college in Cologne where he received a strong background in theoretical physics under the physicist Georg Simon Ohm. In May 1822, he moved to the University of Paris. In 1826, Dirichlet returned to Germany and taught at the University of Breslau. Three years later, he moved to the University of Berlin where he spent the next 27 years. Dirichlet's primary interest in mathematics was number theory, inspired by Gauss' masterpiece, Disquisitiones Arithmeticae (1801). He established Fermat's Last Theorem for n = 14. Among the many results he discovered include the proof of a theorem presented to the Paris Academy of Sciences on algebraic number theory in 1837: The sequence {an + b} contains infinitely many primes, where a and b are relatively prime. In 1855, when Gauss died, Dirichlet moved to the University of GSttingen. Three years later, he went to Montreaux, Switzerland, to deliver a speech in honor of Gauss. While there, he suffered a heart attack and was barely able to return home. During his illness his wife succumbed to a stroke, and Dirichlet died.

Since m > n, by the pigeonhole principle, there should be at least two students al and a2 such t h a t f(al) - f(a2); t h a t is, at least two students have the same birthday, m The next example* is geometric, d e m o n s t r a t i n g t h a t the pigeonhole principle can pop up in seemingly unusual situations. Suppose five l a t t i c e p o i n t s , t h a t is, points with integer coordinates, are selected on the cartesian plane and each pair of points is joined by a line segment. Show t h a t at least one of the line segments m u s t contain a lattice point between its endpoints. SOLUTION: The set of lattice points can be partitioned into four n o n e m p t y disjoint classes according to the p a r i t y (evenness or oddness) of their coordinates: (odd,odd), (odd,even), (even,odd), and (even,even). Since there are five points (pigeons) and four classes (pigeonholes), by the pigeonhole principle, at least two of t h e m - - s a y , A(a,b) and B(c,d)--must belong to the same class. By the m i d p o i n t f o r m u l a in analytic geometry, the midpoint M of the line segment AB is (a~c, b+d)2 . Since the sum of any two odd or even integers /

is an even integer, it follows t h a t M is also a lattice point. Thus AB contains a lattice point M different from its endpoints, m *Based on C. T. Long, "On Pigeons and Problems," Mathematics Teacher, Vol. 81 (January 1988), pp. 28-30, 64.

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Functions and Matrices

It is well known that the decimal expansions of rational n u m b e r s are periodic. Using the pigeonhole principle, we shall establish this, but first a few words of explanation may be helpful. Using the familiar long division method, you may verify that 4111 = 0. 12345345345345... 33300 Although the decimal expansion is nonterminating, it is p e r i o d i c ; t h a t is, a certain block of digits, namely, 345, gets repeated. Accordingly, the expansion is usually written as 0.12345, using a bar over the first repeating block. The number of digits in the smallest repeating block is the p e r i o d of the expansion; here it is 3. We are now ready to prove the above proposition. ~

Prove that the decimal expansion of a rational number is periodic. PROOF"

a

Consider, for convenience, a positive rational number ~, where 0 < a < b. a

Let ~ - O . d l d 2 d 3 . . . where, by the division algorithm (see Section 4.1), we have: 10a - b d l + rl 10rl - bd2 + r2 10r2 - bd3 + r3 (3.2) lOrj -- bdj+ 1 + rj+ 1

and 0 _< ri < b for every i. (Note: The digits a l l , d 2 , . . , in the decimal expansion are the quotients when 10a, 10rl,... are divided by b. Since a remainder has only b choices, by the pigeonhole principle, two of the remainders r l , r 2 , . . . ,rb+l must be equal; that is, rj - rk for s o m e j and k, where 1 < j < k _< b + 1. Consequently, dk+l -- dj+l, dk+2 - d]+2,..., d2k-j -- d k , d 2 k - j + l -- dj+l, and so on. Thus d j + l . . , dk is the smallest block getting repeated and the period of the decimal expansion is k - j . m The next example, a rather sophisticated application of the pigeonhole principle, is due to the Hungarian mathematician Paul ErdSs. ~

( E r d S s T h e o r e m ) Ifn + 1 integers are selected from the set {1, 2 , . . . , 2n}, one of them divides another integer that has been selected. PROOF"

Let al, a2,... ,an+l denote the integers selected. Write each of them as a product of a power of 2 and an odd integer; that is, ai - 2e/bi, where 1 0. The integers bl, b2,... ,bn+l are odd positive integers < 2n.

147

3.4 The Pigeonhole Principle

.. (..~,.v,.o:

.r

....-~"i~(,: . ~.~.~ ""'"'~.~"-~.:~ ~".:~'~ " " t ..~ . ..~. ~..'~. . Jo,~, , f~t..~. ~. .;~, ." ...... "~z;i!~ ~:: ~ ....~,_,~-:.~,_:~: ......... ~ " ~ ~is~~ ! ' .'~'- "" ~ " ' - - ~ " ~,: " ~j N ~

P a u l E r d 6 s (1913-1996) was born in Budapest, Hungary. Except for about three years in schools, Erd6s (pronounced air-dosh) was taught at home, mostly by his father, who had returned from a Siberian prison after 6 years. A child prodigy, Erd6s, at age 3, discovered negative numbers for himself. In 1930 Erd6s entered the Peter Pazmany University in Budapest. Three years later, he discovered a beautiful proof of the celebrated Chebyshev theorem that there is a prime between any positive integer n and 2n. In 1934 he received his Ph.D. from the university. ~i , ~ . , ~ ~ An author of about 1500 articles and coauthor of about 500, Erd6s was 9~ one of the mostprolific writers in mathematics. A tribute in 1983 described him as "the prince of problem-solvers and the absolute monarch of problem-posers." As "the Euler of our time," he contributed extensively to number theory, combinatorics, function theory, complex analysis, set theory, group theory, and probability, the first two areas being closest to his heart. "Always searching for mathematical truths," he deemed worldly possessions a nuisance, so he never had a home, a car, checks, or even an address. Always traveling from meeting to meeting, carrying a half-empty suitcase, he would stay with mathematicians wherever he went and donate the honoraria he earned as prizes to students. A recipient of many honors, Erd6s died of a heart attack while attending a mathematics meeting in Warsaw.

Since t h e r e are exactly n odd positive i n t e g e r s < 2n, by t h e pigeonhole principle, two of t h e e l e m e n t s bl, b 2 , . . . , bn+l m u s t be equal, say, bi - bj. T h a t is, aj - 2~1b/ - 2(:Jbi. Thus, if ei < e] t h e n ai aj, a n d if ei < ei t h e n aj [ai.* m T h e pigeonhole principle tells us t h a t if m pigeons are d i s t r i b u t e d into n pigeonholes, w h e r e m > n, at least two pigeons m u s t s h a r e t h e s a m e pigeonhole. In fact, if m o r e t h a n 2m pigeons are assigned to m pigeonholes, t h e n at least t h r e e pigeons m u s t s h a r e t h e s a m e pigeonhole. T h u s t h e pigeonhole principle can be generalized as follows. ( T h e G e n e r a l i z e d P i g e o n h o l e P r i n c i p l e ) If m pigeons are assigned to n pigeonholes, t h e r e m u s t be a pigeonhole c o n t a i n i n g at least [ ( m - 1)/nJ + 1 pigeons.

P R O O F (by c o n t r a d i c t i o n ) : Suppose no pigeonhole c o n t a i n s m o r e t h a n [(m - 1)/nJ pigeons. Then: m a x i m u m n u m b e r of pigeons - n . [(m - 1)/nJ

m--1

=m--1 *a/b means a is a factor of b. See Section 4.2.

148

Chapter3 Functions and Matrices

This contradicts our assumption that there are m pigeons. Thus, one pigeonhole must contain at least [(m - 1)/nJ + i pigeons, m This generalized version of the pigeonhole principle is illustrated in the following examples. If we select any group of 1000 students of them must have the same birthday.

on

campus, show that at least three

SOLUTION: The maximum number of days in a year is 366. Think of students as pigeons and days of the year as pigeonholes. Then, by the generalized pigeonhole principle, the minimum number of students having the same birthday is [ ( 1 0 0 0 - 1)/366J + 1 = 2 + 1 = 3. m The next example provides an interesting application of the generalized version to geometry. We shall revisit it in Chapter 8. Suppose every pair of vertices of a hexagon is joined by a line segment, which is colored red or blue. Prove that the line segments form at least one monochromatic triangle, that is, a triangle with all its sides having the same color. PROOF: Let the letters A through F denote the vertices of a hexagon. Five line segments (pigeons) emanate from each vertex (see Figure 3.22). Without loss of generality, consider the line segments at A. Since there are exactly two colors (pigeonholes), by the generalized pigeonhole principle, at least three of the line segments at A must be monochromatic, say, red. Suppose they are AB, AD, and AF, indicated by the solid line segments in Figure 3.23. F i g u r e 3.22

E

D

F

C

E.

F i g u r e 3.23

F

~ A"

Case 1

i /~D -C v.B

Suppose DF is colored red. Then AADF is monochromatic.

3.4 The Pigeonhole Principle

149

C a s e 2 Suppose DF is not red. Then it is blue, indicated by the broken line segment in Figure 3.23. If BD is red, the AABD is monochromatic. If BD is not red, consider BF. If BF is red, the AABF is a red triangle. If BF is blue, then ABDF is a blue triangle. Thus the line segments form at least one monochromatic triangle,

m

Additional examples of the pigeonhole principle are presented in Section 3.7, as well as Chapters 4, 6, 7, and 8. Look for them. Exercises 3.4

1. Show t h a t in any 11-digit integer, at least two digits are the same. 2. Show that in any 27-letter word, at least two letters are the same. 3. Six positive integers are selected. Show t h a t at least two of t h e m will have the same r e m a i n d e r when divided by five. 4. A C§ + identifier contains 37 alphanumeric characters. Show t h a t at least two characters are the same. 5. Show t h a t in any group of eight people, at least two m u s t have been born on the same day of the week. 6. Show that in any group of 13 people, at least two m u s t have been born in the same month. 7. There are six m a t c h i n g pairs of gloves. Show t h a t any set of seven gloves will contain a matching pair. 8. The sum of nine integers in the range 1-25 is 83. Show t h a t one of t h e m m u s t be at least 10. 9. The total cost of 13 refrigerators at a d e p a r t m e n t store is $12,305. Show that one refrigerator m u s t cost at least $947. 10. Mrs. Zee has 19 skirts and would like to arrange t h e m in a chest t h a t has four drawers. Show t h a t one drawer m u s t contain at least five skirts. 11. Show t h a t the repeating decimal 0 . a l a 2 . . . a i b l b 2 . . . bj is a rational number. 12. Let n 9 H. Suppose n elements are selected from the set {1, 2 , . . . , 2n}. Find a pair of integers in which one is not a factor of another integer.

Use the pigeonhole principle to prove the following. 13. If five points are chosen inside a unit square, t h e n the distance between at least two of t h e m is no more t h a n v/2/2. 14. Five points are chosen inside an equilateral triangle of unit side. The distance between at least two of t h e m is no more t h a n 1/2.

150

Chapter3 Functionsand Matrices 15. If 10 points are selected inside an e q u i l a t e r a l t r i a n g l e of u n i t side, t h e n at least two of t h e m are no m o r e t h a n 1/3 of a u n i t apart. 16. Let f 9X ~ Y a n d y ~ Y. Define f - l ( y ) _ {x ~ X i f ( x ) - y}. In o t h e r words, f - l ( y ) consists of all p r e - i m a g e s ofy. Use E x a m p l e 3.1 to find f - 1(y) for every y ~ Y. "17. Prove the following a l t e r n a t e version of t h e generalized pigeonhole principle: Let f 9X --~ Y, w h e r e X a n d Y are finite sets, IX] > k 9 ]Y], and k c IN. T h e n t h e r e is an e l e m e n t t ~ Y such t h a t f - l ( t ) c o n t a i n s m o r e t h a n k elements. 18. Prove t h a t any set S of t h r e e integers contains at least two i n t e g e r s whose s u m is even. (Hint: Define a suitable function f 9 S ~ {0, 1} and use Exercise 17.) "19. U s i n g the pigeonhole principle, prove t h a t the c a r d i n a l i t y of a finite set is unique.

Besides adding and m u l t i p l y i n g functions, t h e r e is a very f u n d a m e n t a l way of c o n s t r u c t i n g new functions. Consider the f u n c t i o n s f , g : R ~ I~ defined b y f ( x ) = 2 x + 3 a n d g ( x ) = x 2. Let x be an input into f. T h e n f(x) = 2x + 3 is a real n u m b e r a n d hence can be considered an i n p u t into g. The r e s u l t i n g o u t p u t is g ( f ( x ) ) (see F i g u r e 3.24). T h u s the functions f and g can be employed to define a new function, called the c o m p o s i t e of f and g, as shown in F i g u r e 3.25.

F i g u r e 3.24

f(x)

g(f(x) )

F i g u r e 3.25

go/"

This leads us to the following definition.

3.5

151

Compositionof F u n c t i o n s

Composition Let f :X ~ Y and g : Y ~ Z. The c o m p o s i t i o n o f f and g, denoted by g o f (notice the order of the functions), is a function from X to Z, defined by (g o f)(x) = g ( f ( x ) ) . Read g o f as g circle f or the composition o f f a n d g. [In general, dom(g) need not be the same as codom(f); all t h a t is needed is t h a t range (f) c_ dom(g).] L e t f , g " R ~ IRt defined b y f ( x ) ( f o g)(x).

2x + 3 a n d g ( x ) - x 2. Find ( g o f ) ( x ) a n d

SOLUTION: f ( x ) - 2x + 3

Then

(g o f)(x) - g ( f (x))

= g(2x + 3) - (2x + 3) 2 g(x)

D

So

--

x 2

( f o g)(x) - f (g(x))

= f(x 2) - 2(x 2) + 3 =

m

2x2+3

It follows from Example 3.31 that, in general, f o g r g o f; in other words, composition is not a commutative operation. For instance, putting clothes in a washing machine and t h e n in a dryer does not yield the same result as p u t t i n g t h e m in a dryer and then in a washing machine!

(optional) Composition is easily accomplished in computer science. To illustrate this, study the following algorithm fragment, where x ~ ]R: i.

if

2. 3.

x < 4 then x<--x+2

else

4.

x <-- x -

5. 6.

if

7.

else

8.

3

x < 5 then x +--x 2 x ~-- 2 x -

1

Find the value of x resulting from the execution of this fragment with the initial values of x - 2 and x - 5. t f, g . X --+ Y is a n a b b r e v i a t i o n for f 9X ~

Y a n d g 9X ~

Y.

152

Chapter 3 Functions and Matrices

SOLUTION:

9 Suppose x - 2. Since 2 < 4, line 2 is executed and hence x <-- 4. T h e n the condition in line 5 is tested. Since 4 < 5, line 6 is executed. So x gets the value 16 from line 6. 9 Suppose x - 5 initially; t h e n line 2 is skipped. So x ~-- 2 by line 4. Since 2 < 5, line 6 is executed. Therefore, x ~ 4. m To see t h a t this example employs composition, let f ( x ) a n d g ( x ) denote the functions defined by the i f - t h e n - e l s e s t a t e m e n t s in the above a l g o r i t h m fragment. Then: f(x) _ { ~ + 2

ifx<4

-3

ifx>4

and

g(x) -

! x2

ifx <5

! 2x-1

if x > 5

The o u t p u t r e s u l t i n g from the f r a g m e n t is given by the composition of f and g. You m a y verify t h a t (g o f) (2) - 16 and ( g o f)(5) - 4; in fact g o f is defined by (x + 2) 2

ifx < 3

2x+3

if3<x<4

(x-3) 2

if4<x<8

2x- 7

otherwise

(g o f)(x) -

(See Exercise 54.) A few simple properties satisfied by the composition operation follow. Their proofs are fairly s t r a i g h t f o r w a r d ; we shall prove p a r t 3 and leave the others as exercises. Let f " X --+ Y and g . Y -+ Z. Then: (1) (3) (4) (5)

folz=f (2) 1 y o f = f If f and g are injective, t h e n g o f is injective. If f and g are surjective, t h e n g o f is surjective. I f f and g are bijective, t h e n g o f is bijective.

PROOF:

(3) Let Xl, x2 c X such t h a t (g o f)(xl) - (g o f)(x2). T h e n g(f(xl))

Then Consequently,

- g ( f ( x 2 ) ) , by definition.

f ( x l ) - f(x2), since g is injective.

x 1 - x2, since f is injective.

Thus, if (g o f ) ( x l ) - ( g o f)(x2), t h e n X l - - X 2 , SO g o f is injective. (Exercises 44-46 provide partial converses to the results 3 t h r o u g h 5.) m

153

3.5 Compositionof Functions

Before we define the inverse of a function and discuss its properties, let us study the next example. Let f (x) - a x + b and g ( x ) -

Xa--~b on R, where a ~ 0. Find ( g o f ) ( x ) and

( f o g)(x).

SOLUTION: Let x e I~. Then: (1) (g o f ) ( x ) - g ( f ( x ) )

(2) ( f o g ) ( x ) - f ( g ( x ) )

=g(ax+b)

=f(x-b)a

= (ax+b)-ba

=a(X-b)a

--X

:X

+b

In this example, ( g o f ) ( x ) - x - ( f o g ) ( x ) for allx. T h a t is, g o f - f o g = 1R, the identity function. In other words, one function u n d o e s w h a t the other has done. This leads to the following definition, m

Inverse Function Let f" X -~ Y. Suppose there is a function g" Y -~ X such t h a t ( g o f ) ( x ) - x for every x ~ X and ( f o g ) ( y ) - y for every y c Y; it is called the i n v e r s e of f, denoted by f - l ; t h a t is, g - f - 1 . [Note" dom(f) - codom(f -1) and codom(f) - d o m ( f - 1 ) ; also f - l ( x ) r 1/f(x).] It can be shown t h a t the inverse of f is unique (see Exercise 52). The function f - 1 does just the opposite of what f has done, as illustrated in Figure 3.26. A function t h a t has an inverse is said to be i n v e r t i b l e .

F i g u r e 3.26

f

X

Y

The next two examples illustrate this definition. ~

With the functions f and g in Figure 3.27, notice that: ( g o f ) ( a ) - g ( f ( a ) ) g(3) = a, ( g o f ) ( b ) - g ( f ( b ) ) = g(2) - b, ( g o f ) ( c ) - g ( f ( c ) ) - g ( 1 ) - c, and ( g o f ) ( d ) = g ( f ( d ) ) = g(O) - d. Thus, ( g o f ) ( x ) - x for every x in X and, similarly, ( f o g ) ( y ) = y for every y in Y. So g = f - 1 . m Consider the functions ORD 9 ASCII ~ {0, 1 , . . . , 127} and CHR" {0, 1 , . . . , 127} --~ ASCII. ORD(c) gives the ordinal n u m b e r of the character c in ASCII, whereas CHR(n) r e t u r n s the character with ordinal n u m b e r n.

Chapter 3 Functions and Matrices

154 Figure 3.27

f

g )

a

X

Y

Y

X

Let CH denote a c h a r a c t e r variable and n a valid ordinal n u m b e r . T h e n CHR(ORD(CH)) = CH and ORD(CHR(n)) = n. T h u s C H R a n d ORD are inverse functions, m U n f o r t u n a t e l y , n o t every function is invertible. T h e n e x t t h e o r e m gives a necessary and sufficient condition for invertibility.

~

A function f 9X --, Y is invertible if and only if it is bijective. PROOF: Suppose f is invertible. We would like to show f is bijective. 9 T o p r o v e t h a t f is injective:

Let Xl and x2 be any two e l e m e n t s in X such t h a t f ( x l ) = f(x2). Since f is invertible, f - 1 exists. T h e n f-l(f(xl))

= f-l(f(x2))

( f - 1 o f)(xl) = ( f - 1 o f)(x2) Xl : X 2 Therefore, f is injective. 9 To p r o v e t h a t f is s u r j e c t i v e :

Let y be any element in Y. We have to produce a suitable e l e m e n t x in X such t h a t f ( x ) = y. Choose x = f - l ( y ) (see F i g u r e 3.26). T h e n f(x) - f(f-l(y))

_ ( f o f - 1 ) ( y ) _ y.

T h u s f is both injective and surjective; therefore, it is bijective. Conversely, suppose f is bijective. T h e n every e l e m e n t x in X is paired with a u n i q u e e l e m e n t y in Y and vice versa. Define a function g : Y --, X as follows: g ( y ) = the u n i q u e e l e m e n t x in X such t h a t f ( x ) = y. T h e n ( g o f ) ( x ) = g ( f ( x ) ) = g ( y ) = x and ( f o g ) ( y ) = f ( g ( y ) ) - f ( x ) = y. Therefore, g = f - 1 and hence f is invertible, m The next two examples use T h e o r e m 3.9 to d e t e r m i n e the invertibility of a function.

3.5 Composition of Functions

~

155

The exponential function f 9 I~ ~ R + defined by f ( x ) = 2x is bijective, so it is invertible. Its inverse is the logarithmic function g : I~+ ~ R defined by g ( x ) = lg x. m L e t f " E* --~ W defined byf(x) - Ilxll, where E denotes the English alphabet. Since f is not bijective (why?), f is not invertible, m We close this section with a list of additional properties satisfied by the inverse of a function and leave their proofs as exercises for you to pursue.

~

Let f " X ~ Y and g " Y ~ Z be invertible functions. Then: 9 f - 1 o f = 1x

9 f o f - 1 = 1y

9 (f-l)-1

9 (g o f ) - i

= f

.

f - 1 is bijective.

= f-1 o g-1

m

Exercises 3.5

Let f , g 9 R ~ R be defined by f ( x ) = 2x - 1 and g ( x ) 1. (g of)(2)

2. (f o g ) ( - 1 )

3.

=

1. Find:

x 2 +

(g o f)(x)

4.

(f o g)(x)

Let f ( x ) - Lx] and g ( x ) - [xl, where x ~ I~. Compute each. 5. ( g o f ) ( - 2 . 3 )

Let f , g each.

6. ( f o g ) ( - 2 . 3 )

7. ( g o f ) ( - 4 . 1 )

8. ( f o g ) ( - 3 . 9 )

9 W ~ W defined by f ( x ) - x mod 5 and g ( x ) - x div 7. Evaluate

9. (g of)(17)

10. (f og)(23)

11. (g of)(97)

12. (f og)(78)

Determine if the function g is the inverse of the corresponding function f. 13. f (x) - x 2, x _> 0; g ( x ) - v/-x, x >_ 0

14. f ( x ) - x 2, x _< 0; g ( x ) - - v ~ ,

x >_ 0

Define the inverse g of each function f. x

a

b

c

d

15.

x

a

b

c

d

f(x)

b

c

d

a

16. f(x)

4

1

3

2

Determine if the given function is invertible. If it is not invertible, explain why. 17. ORD on Z. 18. f" ASCII ~ W defined by f ( c ) - ordinal n u m b e r of the character c. 19. f" W -~ W defined by f ( n ) = n (mod 5). 20. f" E* --~ E* defined by f ( w ) = awa, where E - {a, b, c}. 21. f" S ~ N defined by f ( x ) - decimal value of x, where S is the set of binary r e p r e s e n t a t i o n s of positive integers with no leading zeros.

156

Chapter3 Functions and Matrices

22. f 9 E* ~ W defined by f ( x ) = decimal value of x, where E = {0,1}. n

23. Let f 9 ~n

__.>

W defined by f ( x ) -

~ xi, where

]~n

denotes the set

i=l

of words of length n over E -- {0, 1, 2} and x w e i g h t of x; for example, f(10211) = 5.]

-

XlX2...Xn

. If(x)

is the

Mark each sentence as true or false. Assume the composites and inverses are defined: 24. The composition of two injections is injective. 25. The composition of two surjections is surjective. 26. The composition of two bijections is a bijection. 27. Every function is invertible. 28. Every injective function is invertible. 29. Every invertible function is injective. 30. Every invertible function is surjective. 31. Every invertible function is bijective. 32. Every bijection is invertible. 33. The composition of two invertible functions is invertible. Using the algorithm fragment in Example 3.32, compute the o u t p u t resulting from each initial value of x. 34. - 5

35. 0

36. 3

37. 7

Let f 9X ~ Y and g 9 Y ~ Z. Prove each. 38. f o l z - f

39. 1 y o f = f

40. If f and g are injective, then g o f is injective. 41. If f and g are surjective, then g o f is surjective. 42. If f and g are bijective, then g o f is bijective. 43. The identity function 1x is bijective. 44. I f g o f is injective, then f is injective. 45. I f g o f is surjective, then g is surjective. 46. If g o f is bijective, then f is injective and g is surjective. Let f 9X --~ Y and g 9 Y --, Z be invertible functions. Prove each. 47. f - l o f _ l x

48. f o f - l - l y

49. f - 1 is bijective.

50. ( f - 1 ) - I _ f

51. (g o f ) - i _ f - 1 o g - 1

*52. The inverse o f f is unique.

3.6 Sequencesand the Summation Notation

157

53. L e t f ' A - - + B , g ' B - - + C, a n d h ' C - - + D . Provethatho(gof)(h o g) o f ( a s s o c i a t i v e property). [Hint: Verify t h a t (h o (g o f ) ) ( x ) = ((h o g) o f ) ( x ) for every x in A.] *54. Let f a n d g denote t h e functions defined by t h e i f - t h e n - e l s e statem e n t s in E x a m p l e 3.31. Show t h a t g o f is defined as given in t h e example. (Hint" Consider the cases x _< 4 a n d x > 4, a n d t h e n two subcases in each case.) Prove each, w h e r e X ~ Y implies set X is e q u i v a l e n t to set Y. *55. I f A ~ B, t h e n B ~ A ( s y m m e t r i c property). *56. I f A ~ B and B ~ C, t h e n A ~ C ( t r a n s i t i v e property). Let f 9X -+ Y be bijective. Let S a n d T be subsets of Y. Prove each. *57. f - l ( s u T) - f - l ( S ) u f - l ( T )

*58. f - l ( s n T) - f - l ( s ) n f - l ( T )

Sequences and the s u m m a t i o n n o t a t i o n play a key role in the next t h r e e chapters, so we p r e s e n t t h e m here. Let a be a whole n u m b e r and X - {a,a + 1,a + 2,...}. A function s with d o m a i n X or a subset of X is called a s e q u e n c e . Let n e X. T h e n s(n) is called a t e r m of the sequence, denoted by Sn. T h e various t e r m s of the sequence can be listed as Sa, Sa+l, 8a+2,... in i n c r e a s i n g order of subscripts. In particular, let X = l~. T h e n the t e r m s of the sequence are: 81,82,83,.

9 9 ,8n,.

9 9

t

general term

The n t h t e r m 8n is t h e general term of the sequence; the sequence is often denoted by {Sn}~ or simply {Sn}. (It should be clear from the context w h e t h e r the braces indicate a set or a sequence.) T h e g e n e r a l t e r m is often used to define a sequence. Consider the sequence {Sn}, w h e r e Sn - 2n - 1. T h e various t e r m s of t h e sequence are 1, 3, 5, 7, . . . . Formally, the sequence is t h e function s : l~ -+ 1~ defined by s(n) = 2n - 1. I Let an be t h e b i n a r y r e p r e s e n t a t i o n of t h e positive i n t e g e r n with no leading zeros. The various t e r m s of the sequence {an} are 1, 10, 11, 100, 101, 110, 111, . . . . I Sequences can be classified as finite or infinite, as t h e next definition shows.

158

Chapter3 Functionsand Matrices Finite and Infinite Sequences A sequence is f i n i t e if its domain is finite; otherwise, it is i n f i n i t e . Thus, a finite sequence is made up of a finite n u m b e r of terms, a n d an infinite sequence contains infinitely m a n y terms. Both types are useful in mathematics and computer science as well. Every word over an alphabet can be considered a finite sequence. F o r instance, the binary word 010110111 is a finite sequence c o n t a i n i n g nine terms. The elements of a finite language form a finite sequence; for example, the words of length _< 2 over the alphabet {a, b, c} form a finite sequence, namely, k, a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc; on the other hand, k, a, a2 a3 is an infinite sequence We now t u r n to the s u m m a t i o n notation you will find very useful t h r o u g h o u t the remainder of the book.

The Summation Notation Often we need to work with sums of terms of n u m b e r sequences {a,z }. S u m s such as ak + ah+l + " + am can be written in a compact form u s i n g the s u m m a t i o n s y m b o l ~ , which denotes the word s u m . The s u m m a t i o n notation was introduced in 1772 by the brilliant French m a t h e m a t i c i a n Joseph Louis Lagrange. (Recall t h a t E denoted an alphabet in C h a p t e r 2; its actual meaning should be clear from the context.) A typical term in the above sum can be denoted by ai, so it is the s u m of i=m

the terms ai as i runs from k to m. It is denoted by ~ ai. T h u s i=k

i----m

E ai = ak + ak+l + i=k

"

'

"

+ am

The variable i is the s u m m a t i o n i n d e x . The values k and m are the l o w e r and u p p e r l i m i t s of the index i. The "i - " above the ~ is usually omitted; in fact, the indices above and below the ~ are also omitted w h e n there is no confusion. Thus i - - - DI

Dl

Eai--Eai-i =k i =k

171

~-~ai k

For example,

6 ~-~i-- 1 + 2 + 3 + 4 + 5 + 6 - - 2 1 i=1

3.6

Sequences and the Summation Notation

159

J o s e p h L o u i s L a g r a n g e (1736-1813) ranks with Leonhard Euler (see Chapter 8) as one of the greatest mathematicians of the 18th century. The eldest of 11 children in a wealthy family in Turin, Italy, Lagrange was forced to pursue a profession after his father, an influential cabinet official, lost all his wealth by engaging in unsuccessful financial speculations. While studying the classics at the College of Turin, the 17-year-old Lagrange found his interest in mathematics kindled by an essay by the astronomer Edmund Halley on the superiority of the analytical methods of calculus over geometry in the solution of optical problems. In 1754, he began corresponding with several outstanding mathematicians in Europe. The following year, he was appointed professor of mathematics at the Royal Artillery School in Turin. Three years later, he helped to found a society that later became the Turin Academy of Sciences. While at Turin, Lagrange developed revolutionary results in the calculus of variations, mechanics, sound, and probability, winning the prestigious Grand Prix of the Paris Academy of Sciences in 1764 and 1766. In 1766, when Euler left the Berlin Academy of Sciences, Frederick the Great wrote to Lagrange that "the greatest king in Europe" would like to have "the greatest mathematician of Europe" at his court. Accepting the invitation, Lagrange moved to Berlin to head the Academy and remained there for 20 years. When Frederick died in 1786, Lagrange moved to Paris at the invitation of Louis XVI. He was appointed professor at the Ecole Normale and then at the E,cole Polytechnique, where he taught until 1799. He died in Paris. Lagrange made significant contributions to analysis, analytic mechanics, calculus, probability, and number theory, as well as helping to establish the French metric system.

2 i(i- 1)= (-1)(-1-

1)+ 0(0- 1)+ 1(1- 1)+ 2(2- 1)- 4

i=-1 T h e index i is a d u m m y variable; you can use a n y variable as t h e index w i t h o u t affecting t h e value of t h e sum, so m

m

m

Eai=Eaj=Eak i=~

j=e

k=e

3 E v a l u a t e ~ i 2. i=-2 SOLUTION: 3 ~ i 2 - ( - 2 ) 2 + ( - 1 ) 2 + 0 2 + 12 + 2 2 + 3 2 - 19

m

i=-2 The following r e s u l t s are e x t r e m e l y useful in e v a l u a t i n g finite sums. T h e y can be proved u s i n g m a t h e m a t i c a l i n d u c t i o n (Section 4.4).

160

Chapter 3

Functions and Matrices

Let n e N a n d c e R. L e t sequences. Then:

al,

a n d b l , b 2 , . . . , be a n y two n u m b e r

a2, . . . ,

n

9

E

c -

(3.3)

nc

i=1 n

9

(cai)

c

(3.4)

ai

i=1

"

E (ai + bi) i=1

ai

+

bi

(3.5)

(These r e s u l t s can be e x t e n d e d for a n y lower limit k ~ Z.)

m

T h e n e x t e x a m p l e i l l u s t r a t e s this t h e o r e m . 2

E v a l u a t e ~ [(5j) 3 - 2j]. j=l

SOLUTION: E j=-

1(5j)3 - 2j] -1

(5j) 3 -

j

=-1

= 125

j3 =-1

_ 2 ~j j=-i

= 125[(-1) a + 0a + i a + 23]- 2(-1+

0 + 1 + 2)

= 996

m

Indexed Summation

T h e s u m m a t i o n n o t a t i o n can be e x t e n d e d to s e q u e n c e s w i t h index sets I as t h e i r d o m a i n s . F o r instance, ~ ai d e n o t e s t h e s u m of t h e values ai as i ieI

r u n s over t h e v a r i o u s values in I. As a n example, let I = {0, 1, 3, 5}. T h e n ~ (2i + 1) r e p r e s e n t s t h e s u m of iEI

t h e values of 2i + 1, so (2i + 1) = ( 2 . 0 + 1) + ( 2 . 1 +

1) + ( 2 . 3 + 1) + ( 2 . 5 + 1) - 22

i~I

O f t e n we need to e v a l u a t e s u m s of t h e form ~ aij, w h e r e t h e s u b s c r i p t s i p a n d j satisfy c e r t a i n p r o p e r t i e s P. (Such s u m m a t i o n s are used in C h a p t e r s 4 a n d 6.)

3.6 Sequences and the Summation Notation

161

For example, let I = {1, 2, 3, 4}. Then

~

(2i + 3j) denotes the sum

l
of the values of 2i + 3j, where 1 < i < j < 4. This can be abbreviated as (2i + 3j) provided the index set is obvious from the context. To find this

i<j sum, we must consider every possible pair (i,j), where i,j ~ I and i < j. Thus: ~--:(2i + 3j) = ( 2 . 1 + 3 . 2 ) + ( 2 . 1 + 3 - 3 ) + ( 2 . 1 + 3 . 4 )

i<j

~-

(2.2 + 3.3) + (2.2 + 3.4) + (2.3 + 3.4)

= 80 Evaluate

~

d, where d16 indicates that d is a factor of 6.

d>l d16

SOLUTION: d - sum of positive integers d, where d is a factor of 6. d>l dJ6

= sum of positive factors of 6 = 1+2+3+6=

12

m

Multiple summations arise often in mathematics. They are evaluated in the right-to-left fashion. For example, the double summation ~ ~ aij i

j

is evaluated as E (~-~aij) and the triple summation E E ~-~aijk as i j i j k E [ E ( E aijk)] 9 i j k We close this section with an example of a double summation. 1

2

Evaluate E

E (2i +3j).

i=-lj=0

SOLUTION: 1

2

~--~(2i + 3 j ) i=-lj=0

(2i + 3j) i = - 1 [_j=0 1

[(2i + 3 . 0 ) + (2i + 3 . 1 ) + (2i + 3 . 2 ) ] i=-1 1

= ~

(6i + 9)

i=-1

= [6. ( - 1 ) + 91 + ( 6 . 0 + 9) + ( 6 . 1 + 9) =27

m

Chapter3 Functionsand Matrices

162

Exercises 3.6 Evaluate each sum. 4

6

2.

1. E i i=1

2

Y~ (3n - 2)

j=-2

3

E3k k=-2 5

3

3(k 2)

0

(3k) 2

0

k=-2

k=-I

4

5

0;-2) 2

11.

~ (3-2k)k

12. ~ (0.1)i

j=-i

k=l

2)

E J(J-

0

n=0

4

10.

j=o

4

5.

E3 i=-1

7.

3. ~ ( j - 1)

k=O

4

4.

4

~ (3+k)

(0.9)5-i

i=0

Rewrite each s u m u s i n g the s u m m a t i o n notation.

14. 3 1 + 3 2 + . . . + 3 1 ~

13. 1 + 3 + 5 + . . . + 2 3 15. 1 . 2 + 2 . 3 + . . . +

16. 1 ( 1 + 2 ) + 2 ( 2 + 2 ) + . . - + 5 ( 5 + 2 )

11.12

D e t e r m i n e if each is t r u e or false. n

n

n

i= m

n

18. E x i - E

17. ~--~ i - ~-~ (n + m - i )

i=m

i= m

xn+m-i i=m

n

19. S u m s of the form S -

E

t h a t S - an - am.

i=m+ l

(ai- ai-1)

20. Using Exercise 19 and the identity

are

telescoping

1 + i(i 1)

1

i

sums.

1 + i 1

Show

derive a

n 1 formula for ~ i(i + 1)" i=1

21. Using Exercise 19 and the identity (i + 1) 2 - i 2 - 2i + 1, find a f o r m u l a r/

forE/. i=1

Evaluate each sum, where ~ij is defined as follows.

~/j =

1

ifi-j

0

otherwise

[~ij is called K r o n e c k e r ' s delta, after the G e r m a n m a t h e m a t i c i a n Leopold Kronecker (1823-1891).]

3.6 Sequences and the Summation Notation 5

163

6

3

22. E E (2i + 3j)

i=lj=l

i=lj=l 6

5

6

24. E E (2i + 3j)

i=1j=1

6

6

26. E E (i2 _ j + 1)

j=1i=1

5

3

28. E E ~ij

5

29. E E (2 + 3~ij)

i=1j=1 6

5

27. E ~ ( i 2 - j + 1)

i=lj=l 5

5

25. E E ( i 2 - i )

j = l i=1 5

i

23. E ~ ( ] + 3 )

i=lj=l

7

30. Y~. y~ (i2 _ 3i + ~ij) i=lj=l

Just as ~ is used to denote sums, the product akak+l.., am is denoted by m

ai. The p r o d u c t s y m b o l Yl is the Greek capital letter pi. For example, i=k

n

n! - [-I i. Evaluate each product. i=1 3

31.

5

~ (i+l)

32.

i=1

50

~ (j2+l)

33.

j=3

50

1-I 1

34.

I-I ( - 1 ) k

j=-5

k=0

Evaluate each sum and product, where p is a prime and I = { 1, 2, 3, 5 }. 3

35.

~--:k!

36.

}--~p

37.

p_<10

k=0

39. 38. 1--[ ( 3 i - 1) ieI

~

d

40.

~

d>_l

d>_l

d112

dJl2

41.

43. 42.

d>l

El ij

[-I

i,jeI

i<j

ilj

4

46. ~ ( 3 j - 3 j - l ) j=l

Expand each. 3

2

48. ~ aij j--1

i-1 3

2

49. E E aij i=lj=l

51.

~ (ai + aN) 1
(i + 2j)

i<j

(2 i + 3/)

45. i,jd

47. ~a~/

(~)

i,j6I p<25

d118

44.

1-[ P p_<10

2

3

50. E E aij j = l i=1

52.

~ l
(ai + aj)

164

Chapter 3 Functions and Matrices

A r t h u r Cayley (1821-1895) was born in Richmond, England. At 14 he entered King's College, London. His teachers, recognizing his superb mathematical talents, encouraged him to be a mathematician. At 17, Cayley entered Trinity College, Cambridge, where he was rated to be in a class by himself, "above the first." By age 25, he had published 25 papers, the first one at age 20. In 1846, he left his position at Cambridge to study law and became a successful lawyer. Feeling unfulfilled, he left the law after 14 years, although during this period he had published more than 200 papers. In 1863 Cayley rejoined the faculty at Cambridge University. He pursued his mathematical interests, until his death.

James Joseph Sylvester (1814-1897) attended Cambridge University, :,';.,~:"

~..;

.-~

' , '.

.,.:.<~,,..:L~,.~q.

which for several years denied him the degrees he earned, because he was Jewish. At 24, he became professor of natural philosophy at the University of London. Three years later, he taught at the University of Virginia for a year and then returned to England to become an actuary while continuing his mathematical investigations. Sylvester was professor of mathematics at Johns Hopkins University from 1876 to 1883. In 1878 he founded The American Journal of Mathematics.

;lli;~ ~

,

,

53.

~ IAi NA/I l_
~

54.

IAi N A/ n Ak l

l
Matrices were discovered jointly by two English mathematicians, Arthur Cayley and James Joseph Sylvester. Matrix notation allows data to be summarized in a very compact form and manipulated in a convenient way. The sports pages of every newspaper provide fine examples of matrices. For example, during the National Hockey League 2001-2002 regular season, the Boston Bruins won 43 games, lost 24 games, tied 6 games, and had 9 overtime losses; the New York Rangers won 36 games, lost 38 games, tied 4 games, and had 4 overtime losses; the Detroit Red Wings won 51 games, lost 17 games, tied 10 games, and had 4 overtime losses; and the Los Angeles Kings won 40 games, lost 27 games, tied 11 games, and had 4 overtime losses. These data can be arranged in a compact form:

Boston New York Detroit Los Angeles

won

lost

tied

o v e r t i m e loss

43 36 51 40

24 38 17 27

6 4 10 11

9 4 4 4

3.7

Matrices

165

S u p p o s e y o u k n o w t h a t t h e first r o w r e f e r s to Boston, t h e s e c o n d r o w to N e w York, a n d so on, a n d t h e first c o l u m n r e f e r s to t h e n u m b e r of wins, t h e second c o l u m n to t h e n u m b e r of ties, a n d so on. T h e n t h e r o w a n d c o l u m n h e a d i n g s c a n be deleted. Call t h e r e s u l t i n g a r r a n g e m e n t A: 43 37 39 51

A

24 11 10 7

6 36 35 26

9 4 4 4

S u c h a r e c t a n g u l a r a r r a n g e m e n t of n u m b e r s is called a m a t r i x . M o r e generally, we h a v e t h e following definition.

Matrix

A m a t r i x is a r e c t a n g u l a r a r r a n g e m e n t of n u m b e r s enclosed by b r a c k e t s . A m a t r i x w i t h m rows a n d n c o l u m n s is a n m x n (read m by n) m a t r i x , its s i z e b e i n g m x n. If m = 1, it is a r o w v e c t o r ; a n d if n - 1, it is a c o l u m n v e c t o r . If m = n, it is a s q u a r e m a t r i x o f o r d e r n. E a c h n u m b e r in t h e a r r a n g e m e n t is a n e l e m e n t of t h e m a t r i x . M a t r i c e s a r e d e n o t e d by u p p e r c a s e letters. F o r e x a m p l e , let 1 A=

and 1

0

B-

-6

0

-3

4

-2

7

-1

4 2

A is a 2 x 3 m a t r i x , w h e r e a s B is a s q u a r e m a t r i x of o r d e r 3. T h e e l e m e n t s of t h e r o w v e c t o r [0 3 - 7 ] a r e 0, 3, a n d - 7 . T h e double s u b s c r i p t n o t a t i o n is e x t r e m e l y u s e f u l in n a m i n g t h e elem e n t s of a n m x n m a t r i x A. L e t a i j d e n o t e t h e e l e m e n t in r o w i a n d c o l u m n j of A. T h e n t h e m a t r i x h a s t h e f o r m all a21

a12 a22

... . 99

alj

...

aln

a2j

99

a2n

ai l

ai2

. . .

aij

. . .

ain

A m

_ aml

am2

...

amj

t

...

<

row i

amn

columnj

F o r convenience, it is a b b r e v i a t e d as A - ( a i j ) m x n , or s i m p l y ( a i j ) if t h e size is clear f r o m t h e context. H o w do we d e t e r m i n e if two m a t r i c e s a r e equal? T h i s is a n s w e r e d by t h e n e x t definition.

Chapter3 Functionsand Matrices

166

Equality of Matrices Two matrices A = (aij) and B = (bij) are e q u a l if they have the same size and aij = bij for every i and j. For example, if 1

x

-3

1

0

-3

2

0

y

z

0

-1

then x = 0,y = 1, and z = 2. The following definition presents two special matrices.

Zero and Identity Matrices If every element of a matrix is zero, then it is a zero m a t r i x , denoted by O. Let A - (aij)nxn. Then the elements a l l , a 2 2 , . . . ,ann form the m a i n d i a g o n a l of the matrix A. Suppose

aij -

1

ifi=j

0

otherwise

Then A is the i d e n t i t y m a t r i x of order n; it is denoted by 1,2, or I when there is no ambiguity. For example, [0 0 0] and I0

0] are zero matrices.

A = ( a i j ) n x n is the identity matrix In if every element on its main diagonal

is 1, and every element above and below it is 0. For example, [0

01] is the

identity matrix of order 2, namely, I2. Just as propositions and sets can be combined to construct new propositions and new sets, matrices also can be combined to produce new matrices. The various matrix operations are presented and illustrated below.

Matrix Addition The s u m of the matricesA = (aij)m• a n d B = (bij)m• is defined b y A + B = (aij + bij)m• (We can add only matrices of the same size.) Let

A-

[2_371 0

1

and

B-

1

2

0

0]

-1

Then A+B-

=

0+2

1+0

1+(-1)

2

1

0

1

3.7 Matrices

167

Negative of a Matrix The n e g a t i v e or ( a d d i t i v e i n v e r s e ) of a m a t r i x A - (aij), denoted by - A , is defined by - A = ( - a i j ) . For instance, the negative of A -

[23_4] 0

-5

is

-A -

[_2_34]

6

0

5

-6

You may verify that A + (-A) = O.

Matrix Subtraction The d i f f e r e n c e A - B of the matrices A = ( a i j ) m x n and B - ( b i j ) m x n is defined by A - B = (aij - b i j ) m x n . (We can subtract only matrices of the same size.) For example, using the matrices A and B in Example 3.44,

A-B-

[

2-(-1)

(-3)-5

0-2

1-0

7-0 1-(-1)

] I

3

-8

-2

1

7 2

J

The next example introduces us to the fourth matrix operation. Suppose you bought 12 coconut donuts, 15 b u t t e r n u t donuts, and 6 cinnamon donuts from shop I, and you bought 9 coconut donuts, 12 b u t t e r n u t donuts, and 16 cinnamon donuts from shop II. Then the n u m b e r of donuts of each kind you bought from each shop is given by the matrix coconut A =

butternut

cinnamon

shop I shop II

9

12

16

Suppose each donut costs 75r Then the cost of each type of donut at each shop is obtained by multiplying each entry of A by 75. The resulting matrix is denoted by 75A. Thus 75A -

E7512 75.15 70.01[000 1125450] 75.9

75.12

75.16

675

900

1200

I

This example leads to the next definition.

Scalar Multiplication Let A = (aij) be any matrix and k any real num ber (called a s c a l a r ) . Then kA = (kaij).

Chapter 3

168

Functions and Matrices

The f u n d a m e n t a l properties of the various matrix operations are stated in the following theorem. We shall prove two of them, and leave the others as routine exercises. ~ ~ ~ ~ ~ ~

Let A, B, and C be any rn x n matrices, O the m • n zero matrix, and c and d any real numbers. Then: 9 A+B=B+A

9 A+(B+C)=(A+B)+C

9 A+O=A=O+A

9 A+(-A)=O=(-A)+A

9 (-1)A =-A

9 c(A +B)

9 (c + d ) A

=cA

=cA

+cB

9 (cd)A=c(dA)

+dA

PROOF:

Let A - (aij)m

•

9 To prove that A + (-A)

A + (-A)

= 0 = (-A)

- (aij)m•

+ (--aij)m•

+ A:

negative of A

= (aij + (--a~i))m •

matrix addition

= (0),,, •

a~] + ( - a i j ) - 0

=0

zero matrix

Similarly, ( - A ) + A - O. T h u s A + ( - A ) - O - ( - A ) + A. 9 T o p r o v e t h a t (c + d ) A - c A + d A :

(c + d ) A - (c + d ) ( a i j ) m •

definition of A

= ((c + d ) a i j ) m •

scalar multiplication

= (caij + d a i j ) m •

dist. prop. of n u m b e r s

= (caij)m xn + ( d a i j ) m xn

matrix addition

= c(aij)m•

scalar multiplication

= cA + dA

+ d(aij)m•

definition of A

This concludes the proofs.

m

Before we define matrix multiplication, let us study the next example.

~

D i s c o u n t G a s sells regular, unleaded, and p r e m i u m gasoline at two gasoline

stations X and Y. Matrix A shows the price (in dollars) of a gallon of each kind of gasoline; matrix B, the average n u m b e r of gallons sold at each

3.7

Matrices

169

location:

A

S

__

m

regular

unleaded

premium

[2.50

2.75

3.00]

X

Y

regular

3000

3500

unleaded

4000

3750

premium

1500

2000

SOLUTION:

Notice that: Revenue from location X - 2.50(3000) + 2.75(4000) + 3.00(1500) = $23,000.00 Revenue from location Y - 2.50(3500) + 2.75(3750) + 3.00(2000) = $25,062.50 These two values can be used to form the matrix X

Y

[23,000.00

25,062.50]

Each of its elements can be obtained by multiplying each element of A by the corresponding element in each column of B and adding them up, as shown below:

[2.50

2.75

3.00]

o00 i 1 4000 1500

3750 2000

= 2.50(3000) + 2.75(4000) + 3.00(1500) = 23,000.00 [2.50

2.75

3.00]

I

3000 4000 1500

3500 1 3750 2000

= 2.50(3500) + 2.75(3750) + 3.00(2000) = 25,062.50

170

Chapter3 Functionsand Matrices

The matrix [23,000.00 25,062.50] is the p r o d u c t of the matrices A and B, denoted by AB. Thus J ~ 3 0 0 0 AB = [2.50 2.75 3~0] ]-4000 1500

3500 3750 2000

= [23,000.000

25,062.50]

More generally, we define the product of two matrices as follows,

ll

Matrix Multiplication

The p r o d u c t A B of the matricesA = ( a i j ) m • a n d B - (bij)n• is the matrix C - (cij)m • where cij is the sum of the products of the corresponding elements in row i of A and c o l u m n j of B, as shown below: all

a12

999 a l n

bll

b12

9

by

9

blp

a21

a22

999 a2n

b21

b22

9

b2j

9

b2p

ail

ai2

...

bil

bi2

9

bij

9

bip

_ a,,, 1

a,,, 2

a m n _ _b,~l

bn2

9

bnj

9

bnp

C/j

9

ain

99

_

II

where cij - a i l b l j + a i 2 b 2 j + . . . + ainb,~j = ~ a i k b k j . k=l The product C = AB is defined only if the n u m b e r of columns in A equals the n u m b e r of rows in B. The size of the product is m x p. The next example illustrates this definition. Let

A=[10

-24

Find AB and B A , if defined.

-31]

and

B-

I - i 3] _ 0 1

3.7

Matrices

171

SOLUTION:

Since the number of columns of A equals the n u m b e r of rows of B, the product AB is defined. Furthermore, the size of AB is 2 x 2:

4

[3 -i]

-

-1

0 3+4.0+(-1)

(-1) 0 ( - 2 ) +

.1+(-1)

L0 4] The product

BA

-

BA

is also defined (why?) and its size is 3 x 3:

0

-2

-1

I -

-

4

3.1+(-2). 0.1-+-1.0 (-1). 1 +0.0 0

-1

0

14

-

3. ( - 2 ) + ( - 2 ) . 0.(-2)+1.4 (-1). (-2) + 0 . 4

4

3 . 3 + ( - 2 ) . ( - 1 ) -1 0.3+1.(-1) (-1). 3 + 0 . (-1)

4

2

You may notice that AB r

BA.

m

We can use the definition of matrix multiplication to develop an algorithm to find the product of two matrices in an obvious way, as Algorithm 3.1 shows. Algorithm product (A,B,C) (* Let A - (aij)m• and B : ( b i j ) n • p . This algorithm shows how to find t h e i r product C = ( c i j ) m • *) Begin (* product *) for i = i to m do for j = I to p do begin (* for j *) Cij ~-- 0 (* i n i t i a l i z e

*)

for k = 1 to n do Cij <--- Cij + aikbkj

(* update c U *)

endfor End (* product *)

Algorithm 3.1

Chapter 3

172

Functions and Matrices

The fundamental properties of matrix multiplication are stated in the next theorem. They can be proved without much difficulty using the summation notation. ~

Let A, B, and C be three matrices. Then: (1) A ( B C ) = ( A B ) C (3) A ( B + C) = A B + A C

(2) A I - A =IA (4) (A + B ) C = A C + B C

provided the indicated sums and products are defined,

m

We close this section with an example to illustrate Theorem 3.13. ~

part

3 of

Let A-

[25

-~]

,

B-

[1 2

0-1] -3 5 '

and

C -

0

-2 0

[- 3

1 4

J

Show that A ( B § C) - A B § A C . SOLUTION: First notice that both B and C are the same size, so B + C is defined and is of size 2 x 3. Furthermore, since A is 2 x 2 and B + C is 2 x 3, A ( B + C) is defined. Similarly, AB + A C is also defined. B+C-

[: o_1] -3

15 +

o_~1 1_~ [-:3 0 4 1[- - 1 - 3 x]

-~~]

~+~-[~~ -~o][_11 _~~ ~] [~ 1o

.

[~ ~][~ 0, -

5

0

-

~=[~~ ~][~ ~01] [~ 10 ~o] +

-

4

[~ ~0 ~:][~ lO

~o]

= A ( B + C)

Solve the following equations. 0 -3

y+3 1

4 z+ 2

~ ~~]

~o

m

E x e r c i s e s 3.7

1.

[~

-

0 -3

-1 1

4 -2

3.7

2.

Matrices

[x y 1 0][3 -3 4

y-z -5

2 z -x

-

-3 4

-1 -4 -5

173

0] 2 1

Find the additive inverse of each matrix.

. [~0 4~] LetA=

4 [~ ~~

[10 ~] [0~ 0

2

3 ,B-

6. A - B

0

7. B + C

o [i

-2

-5

i]

~1] an~ [ ~in~eac~0 ~ ~ ~ 8. A §

11. 2A § 3B

10. 2 B - C

-

12. 3B - 2C

9. - 2 B

13. 3A § ( - 2 ) B

Let A be an m x n matrix, B a p x q matrix, and C an r x s matrix. U n d e r what conditions is each defined? Find the size of each when defined. (Note: A 2 means AA.) 14. A + B

15. B -

18. A ( B § C)

19. A B - A C

22.

C

16. B C

17. A 2

20. AB § C

21. A ( B C )

A team in the N H L earns 2 points, I point, or 0 points for a win, tie or overtime loss, or a loss, respectively. Using matrices, find the n u m b e r of points earned by each team listed at the beginning of this section.

A s u m m e r vacation lodge in s u n n y California expects four guests: A, B, C, and D. They plan to stay at the lodge for 7, 14, 21, and 28 days, respectively. Each of t h e m has diabetes. Since the nearest drugstore is several miles away, the m a n a g e r of the lodge decides to store three different types of i n s u l i n - semi-lente, lente, and u l t r a - - needed by these guests. Their daily insulin r e q u i r e m e n t s are s u m m a r i z e d in Table 3.2. Table 3.2

Guests Insulin

A

B

C

D

Semi-lente Lente Ultra

25 20 20

40 0 0

35 15 30

0 15 40

Each gram of insulin of the three types costs 10, 11, and 12 cents, respectively t. Using matrices, compute each: t Based on R. F. Baum, "Insulin Requirements as a Linear Process in Time," in Some Mathematical Models in Biology, R. M. Thrall, ed., The University of Michigan Press, Ann Arbor, MI, 1967, pp. 0L2.1-0L2.4.

174

Chapter 3 Functions and Matrices

23. The n u m b e r of g r a m s of each type of insulin needed. 24. The total cost of the insulin. 25. The insulin r e q u i r e m e n t s if the guests decide to stay an additional 3, 5, 8, and 13 days, respectively. 26. The insulin r e q u i r e m e n t s if the guests decide to stay t h r e e times t h e i r original time. Let A, B, and C be any m x n matrices, O the rn • n zero matrix, a n d c a n d d any real n u m b e r s . Prove each (see T h e o r e m 3.12). 27. A + B = B + A

28. A + (B + C) = (A + B ) + C

29. A + O = A = O + A

30. ( - 1)A = - A

31. c(A + B) = c A + cB

32. ( c d ) A = c ( d A )

Let A, B, and C be any square matrices of order 2. Prove each. 33. A ( B C ) = ( A B ) C 34. A(B + C) = A B + A C 35. (A + B ) C -- A C + B C

The t r a n s p o s e

of a m a t r i x A -

(aii)m•

denoted by A T, is defined as

A T - (aji)nxm. Find the t r a n s p o s e of each.

36.

112 31 2

-2

0

1

-1

37.

0

Iab c1 d

e

f

f

g

h

38. A square m a t r i x A is s y m m e t r i c i f A T - A. W h a t can you say a b o u t the elements of a s y m m e t r i c matrix A? 39. Let A be a square matrix. Prove t h a t (AT) T -- A. Let A, B, and C be square matrices of order 2. Prove each. 40. ( A + B ) T = A T + B T

41.

42. (fiAT) T = A n T

43. (ABC) T -- C T B T A T

(AB)

T --

BTA T

A square m a t r i x A of order n is i n v e r t i b l e if there is a m a t r i x B such t h a t A B - In = B A . T h e n B is the i n v e r s e of A, denoted by A - i . In Exercises 44 and 45, verify t h a t B = A-1. Assume t h a t k = a d - bc r O.

44A iac db] , B = ~ 1[ - cd b]a

Chapter Summary

45. A =

3 1

175

1 2

-1 -3

,B=

-8 -5

3 4

-1 -7

Find each product. 46.

E

2

3

0

0lE l 1

1 2

-2 -3

3 4

Y

Rewrite each linear system as a m a t r i x e q u a t i o n AX = B. 49. 2x + 3y = 4 4x+5y = 6

50.

x - 2y = 4 3x+y-z=-5 x+ 2y-3z=6

5 1 - 5 2 . Using Exercises 44 and 45, solve t h e linear s y s t e m s in Exercises 49 and 50, respectively. Let M denote the set of 2 x 2 m a t r i c e s over W. Let f 9 1~ ~ M defined by f(n) = 53. 2

. C o m p u t e f(n) for each value o f n . 54. 3

55. 4

56. 5

Prove each. *57. The inverse of a s q u a r e m a t r i x A is unique. (Hint: Assume A has two inverses B and C. Show t h a t B - C.) *58. If A is an invertible matrix, t h e n (A- 1)- 1 _ A. *59. If A and B are two invertible m a t r i c e s of order n, t h e n (AB) -1 = B-1A-1. 60. Write an a l g o r i t h m to c o m p u t e the s u m of the matrices A = (aij) m x n and B = (bij)m •

This c h a p t e r p r e s e n t e d the concept of a function, the s u m m a t i o n notation, and matrices. F u n c t i o n s can be defined by the ordered pair n o t a t i o n , tables, or graphs. Several properties of functions and some exotic functions were examined, including how to c o n s t r u c t new functions from k n o w n ones. Function 9 A f u n c t i o n f" X ~ Y is a p a i r i n g of every e l e m e n t x in X w i t h a u n i q u e e l e m e n t y in Y. D o m ( f ) - X, codom(f) = Y, and r a n g e ( f ) {f(x) c YIx ~ X} (page 118).

Chapter3 Functionsand Matrices

176

Special Functions n

(an ~ O)

(page 125).

a x (a > O , a r 1)

(page 125).

9 P o l y n o m i a l function f ( x ) = ~ a i x i l=O

9 E x p o n e n t i a l function f ( x ) =

9 L o g a r i t h m i c function f ( x ) - log a x (a > 0, a ~: 1)

(page 126).

9 Absolute value function f ( x ) - Ixl

(page 126).

9 Floor function f ( x ) = Ix]

(page 126).

9 Ceiling function f ( x ) = Ix]

(page 126). 1

9 Characteristic

functionfA(x)-

0

ifx ~A otherwise

(page 131 ).

9 Mod function f ( x , y ) = x mod y

(page 132).

9 Div function g ( x , y ) = x divy

(page 132).

Properties of Functions 9 A function f : X -~ X is the i d e n t i t y f u n c t i o n every x ~ X 9 f " X ~ Y is i n j e c t i v e , i f x l r

X2 ~

f ( x l ) r f(x2)

9 f 9X --. Y is s u r j e c t i v e , if r a n g e ( f ) - Y

on X if f ( x ) = x for (page 136). (page 136). (page 137).

9 f " X --. Y is b i j e c t i v e , if it is both injective a n d surjective (page 137). 9 I f X and Y are finite sets with IX[ = [Y I, f : X --~ Y is injective if a n d only if it is surjective (page 139). 9 Two sets have the s a m e cardinality if and only if a bijection exists between them (page 140).

The Pigeonhole Principle 9 S i m p l e v e r s i o n If m pigeons fly into n pigeonholes to roost, w h e r e m > n, t h e n at least two pigeons m u s t roost in t h e s a m e pigeonhole (page 147 ). 9 G e n e r a l i z e d v e r s i o n If m pigeons fly into n pigeonholes to roost, w h e r e m > n, one pigeonhole m u s t contain at least L(m - 1)/nJ + 1 pigeons (page 147).

Composition 9 T h e c o m p o s i t i o n of the functions f : X ~ by ( g o f ) ( x ) = g ( f ( x ) ) for every x in X

Y a n d g : Y ~ Z is given (page 151).

Chapter Summary

177

9 The composition of two bijections is bijective

(page 152).

9 The function g : Y ~ X is the inverse of the function f : X -~ Y if g o f = 1z and f o g - 1y (page 153). 9 A function is invertible if and only if it is bijective

(page 154).

S e q u e n c e s and the S u m m a t i o n Notation

9 A sequence {Sn} is a function with domain X = {a,a + 1,a + 2,...} or a finite subset of X, where a 9 W (page 157). 9 Using the s u m m a t i o n symbol ~ , the sum ak + a k + l + " " + a m is w r i t t e n i=m

as ~ ai

(page 158).

i=k

Matrix

9 An m x n m a t r i x (aij)m • is a r e c t a n g u l a r a r r a n g e m e n t of elements, where aij denotes the element in row i and c o l u m n j (page 165).

Review Exercises

Find the n u m b e r of positive integers < 1776 and divisible by each: 1. 5 o r 7

2. 5 b u t n o t 7

3. 5, 6, or 7

4. 3 or 5, but not 7

Find the day of the week in each case. 5. 1024 days from Sunday

6. 1948 days from T h u r s d a y

Find the m o n t h of the year in each case. 7. 256 m o n t h s from March

8. 1976 m o n t h s from August

Using formula (3.1) in Exercises 3.2, determine the first day in each year. 9. 2048

10. 4076

11. 7776

12. 7997

Using the formula in Exercises 3.2, compute the date for E a s t e r S u n d a y in each year. 13. 2550

14. 3443

15. 4076

16. 6666

Determine if each function is injective. 17. f ( x ) -- - ] x l ,

x e R

18. g ( x ) = v/-x, x 9 IR +

Determine if each function f :A -~ B is surjective. 19. f(x) = - ~ / ~ , A = IR+, B - JR-

20. f(x) = 2 x, A = R , B = R +

178

Chapter 3

Functions and Matrices

D e t e r m i n e if each function f :A ~ B is bijective. 21. f ( x ) = 2

IxI,A=B=I~

22. f = O R D ,

A=B-Z

Let A and B be finite sets with IAI - 3 and IBI - 2. Find the n u m b e r of: 23. Functions t h a t can be defined from A to B. 24. Constant functions t h a t can be defined from A to B. 25. Injections t h a t can be defined from A to B. 26. Surjections t h a t can be defined from A to B. 27. Bijections t h a t can be defined from A to B. S t u d e n t records are m a i n t a i n e d in a table using the h a s h i n g function h ( x ) = x mod 9767, where x denotes the s t u d e n t ' s social security n u m b e r . Compute the location in the table corresponding to the given key, w h e r e the record is stored. 28. 011-53-1212

29. 212-44-7557

3 0 - 3 1 . Redo Exercises 28 and 29 if h ( x ) = first p a r t in x mod 23. 32. The confirmation n u m b e r for flight reservations made with an airline over the I n t e r n e t consists of three letters followed by a digit and then two letters. Store the following confirmation n u m b e r s in a hash table of 26 cells using the hashing function h ( x ) = first letter in x: VPS3SL, NBC4GK, CBS1AA, AQX5CD, CBA3BA, NCR4SK, CNN 1TK, ABC5ZZ 33. Redo Exercise 32 if h ( x ) = last letter in x. 34. Redo Exercise 32 using a hash table of 10 cells and h ( x ) - digit in x. 35. Redo Exercise 32 using a hash table of 10 cells and h ( x ) = digit in x m o d 5. 36. Show t h a t in any group of seven positive integers, at least two of t h e m leave the same r e m a i n d e r when divided by six. 37. The total cost of mailing six letters is $19. Show t h a t the mailing charge for at least one letter is $3 or more. Let f, g : R ~ R defined by f ( x ) = 2[x7+ 1 and g ( x ) = 3Lx] - 2. Compute each. 38. (g o f)(2.56)

39. (f o g ) ( - 3 . 4 5 )

40. (g o f ) ( - 4 . 6 7 )

41. (f o g ) ( - 5 . 7 3 )

Chapter Summary

179

Let f, g : W ~ W defined by f ( x ) = x m o d 6 a n d g ( x ) - x div 6. E v a l u a t e each. 42. (g o f)(31)

43. ( f o f)(49)

44. ( f o g ) ( 1 7 6 )

45. (g o g ) ( 1 3 3 1 )

M a r k each s e n t e n c e as t r u e or false, w h e r e x a n d y are a r b i t r a r y real numbers. 46.

Ix + yj = [xj + [yj

47.

Ix + Y l = [xl + [Yl

48.

LxyJ = LxJ [yJ

49.

[xyl = Ix] [y]

Give a c o u n t e r e x a m p l e to disprove each proposition, w h e r e x, y e R a n d neZ. 50.

[x+yJ = [xJ+[yJ

51.

[x+yl

= [xl+[yl

53.

[xyl = [x][yl

54.

LnxJ = n LxJ

52.

[xyJ = [xJ[yJ

55.

[nxl = n f x ]

F i n d t h e first four t e r m s of t h e s e q u e n c e w i t h t h e given g e n e r a l t e r m , w h e r e = (1 + v/5)/2 a n d fl = (1 - j ~ ) / 2 . (The n u m b e r a is t h e g o l d e n r a t i o . ) 56.

an-

L'" J

57.

bn -

['" l

58.

L n - oen + r '~ , n >__ 1

59.

fn -

- ~ (or n - r n )

--~ + 89 , n >__ l

_~ _1

, n >__ l

60. A r r a n g e t h e t e r m s of t h e s e q u e n c e of t e r n a r y w o r d s of l e n g t h < 2 over the a l p h a b e t {0,1,2} in i n c r e a s i n g o r d e r of t h e i r n u m e r i c values. E v a l u a t e each. 10

10

i

i

z [.]

.1. FIL J

i=1

17

i

17

F1L'J

i

E I.]

"

i=1

65. Let f 9 A ~ injective.

B a n d g 9 B ~ A such t h a t f o g -

lB. P r o v e t h a t g is

66. Let f 9 A o surjective.

B and g 9 B o

1A. P r o v e t h a t g is

A such t h a t g o f -

Supplementary Exercises Let A - ( a i j ) n x n a n d B - ( b i j ) n x n . P r o v e each. 1. (A + B ) T - A T + B T

2. (AB) T - B T A T

3. A A T is s y m m e t r i c .

4. Let G - {0, 1} a n d d" G n x G n --+ W defined by d ( x , y ) - n u m b e r of c o m p o n e n t s in w h i c h t h e w o r d s x a n d y differ, d ( x , y ) is called t h e H a m m i n g d i s t a n c e b e t w e e n t h e n-bit w o r d s x a n d y. Is d bijective?

Chapter 3

180

-,,,-

.o

,'

.

,~2~:~-~

j

Functions and Matrices

R i c h a r d Wesley H a m m i n g (1915-1998) was born in Chicago, graduated from the University of Chicago in 1937, and received an M.S. from the University of Nebraska 2 years later. After receiving his Ph.D. in mathematics in 1942 from the University of Illinois, he began his teaching career at the university and moved to the university of Louisville until 1945. After a year working on the Manhattan project at Los Alamos Science Laboratory, he joined the technical staff at Bell Telephone Labs in 1946; he headed the numerical methods research department from 1964 to 1967, and then the computer science research department until 1977. He left Bell in 1977 and became an adjunct professor in computer science at the Naval Postgraduate School, Monterey, California. Recipient of numerous awards, Hamming made significant contributions to algebraic coding theory, numerical methods, statistics, and digital filters.

5. Let al, a 2 , . . . ,an c IR+. Prove t h a t at least one of t h e m is g r e a t e r t h a n n

or equal to their average n1 (Y~-ai). 1

E v a l u a t e each sum and product. 2

3

6. ~_, Z

1

1

7. Z

Z

1

Z(i+J+

k)

i=0 j=-I

i = 0 j = 1 k=O

8.

3

4

~

~

5

~ 2 i +j - k

3

9.

i=1 j = 2 k = l 2

10.

i=0 j=l

+3k)

k=O

2

l-I l-I 2i +j i=1 j=O

3

1--I [-l ( i - j )

2

Z(i+2j

5

11.

3

[1 1-] 2 i=2 j=0

An a r i t h m e t i c s e q u e n c e is a n u m b e r sequence in which every t e r m except the first is obtained by adding a fixed n u m b e r , called the c o m m o n d i f f e r e n c e , to the preceding term. For example, 1, 3, 5, 7 , . . . is an a r i t h m e t i c sequence with c o m m o n difference 2. Let an denote the n t h t e r m of the a r i t h m e t i c sequence with first t e r m a and c o m m o n difference d. 12. Find a formula for an. 13. Let Sn denote the s u m of the first n t e r m s of the sequence. Prove t h a t n Sn - ~12a + ( n - 1)d]. A g e o m e t r i c s e q u e n c e is a n u m b e r sequence in which every t e r m except the first is obtained by m u l t i p l y i n g the previous t e r m by a constant, called the c o m m o n r a t i o . For example, 2, 6, 18, 5 4 , . . . is a geometric sequence with c o m m o n ratio 3. Let an denote the n t h t e r m of the geometric sequence with first t e r m a and c o m m o n ratio r.

Chapter Summary

181

14. Find a formula for an. 15. Let Sn denote the sum of the first n t e r m s of the sequence. Prove t h a t S n -- a ( r n - 1 ) r-1

(r ~= 1).

* 16. Let f: X ~ Y be injective and A __ B _ X. Prove t h a t f(A • B) f (A) n f (B). "17. L e t f : X ~ Y a n d A _ B c X such t h a t f ( A ~ B ) = f ( A ) A f ( B ) . Give a counterexample to show t h a t f need not be injective. "18. Suppose a l + a2 + . . . § an - n + 1 pigeons occupy n pigeonholes Hi, 1 a l pigeons, or H2 contains > a2 p i g e o n s , . . . , or Hn contains > an pigeons. Use Exercise 18, prove each. 19. The pigeonhole principle. 20. The generalized pigeonhole principle. oo

21. The sequence {an }~ satisfies the property t h a t an -

ai for every i=n+l

n > 1. Show t h a t an+l - 89

> 1. (T. Fletcher, 1978).

Computer Exercises Write a program to perform each task. 1. Read in the a m o u n t of water used by a household for 6 m o n t h s and compute the water bill, using the rate in Example 3.6. 2. Read in a positive integer n < 1000 and print all perfect n u m b e r s < n. (There are three perfect n u m b e r s < 1000. See Exercises 3.1.) 3. Read in a year y > 1600 and determine each: 9 W h e t h e r or not it is a leap year. 9 The n u m b e r of leap years > 1600 and < y. 4. Read in a year y and find the following (see Exercises 3.2): 9 The day of J a n u a r y 1 in year y and year y + 1. 9 The n u m b e r of F r i d a y - t h e - t h i r t e e n t h s in year y. 5. J a n u a r y 1, 2000, fell on a Saturday. Determine the day of the week of J a n u a r y 1, 1776, and J a n u a r y 1, 3000. P r i n t the calendar for J a n u a r y in each year. 6. Read in a series of years greater t h a n 2000 and determine the Easter date in each year. (See Exercises 3.2.)

182

Chapter3 Functions and Matrices 7. The discrete probability p ( r ) t h a t two people in a g r o u p of r people selected at r a n d o m have the same b i r t h d a y is given by p(r)-

1-

365.364 ... (365 - r + 1) 365 r

a s s u m i n g 365 days in a year. C o m p u t e the probability for each value of r: 10, 20, 3 0 , . . . , 100, including 23. (You will see t h a t i f 2 3 people are selected at r a n d o m , t h e r e is a b e t t e r t h a n 50% chance t h a t two have the same birthday. This is k n o w n as the b i r t h d a y p a r a d o x . ) 8. Assign the n u m b e r s 0-51 in order to the 52 playing cards in a standard deck. Read in a n u m b e r x, w h e r e 0 _< x < 51. I d e n t i f y t h e card n u m b e r e d x. Use the suit labels 0 - clubs, 1 - d i a m o n d s , 2 - h e a r t s , and 3 - spades, and the card labels 0 - ace, 1 - deuce, 2 - t h r e e , . . . , in each suit. 9. Assign the n u m b e r s 0-63, row by row, to the various s q u a r e s on an 8 x 8 chessboard. Read in two n u m b e r s x and y, w h e r e 0 < x , y < 63. D e t e r m i n e if the queen at square x can c a p t u r e the q u e e n at s q u a r e y. 10. Read in n c u s t o m e r s ' nine-digit account n u m b e r s at a b a n k , w h e r e n is a positive integer _< 100. Store t h e m in a hash table u s i n g t h e h a s h i n g function h ( x ) - x mod 113, w h e r e x denotes an account n u m b e r . P r i n t the h a s h table. 11. Read in n s t u d e n t s ' social security n u m b e r s and store t h e m in a h a s h table using the h a s h i n g function h ( x ) - x mod 109, w h e r e x d e n o t e s a social security n u m b e r and n is a positive integer _< 100. P r i n t the hash table. 12. Read in the two-letter a b b r e v i a t i o n s of all states in the U n i t e d States. Store t h e m in a hash table of 26 cells, using the hash f u n c t i o n h ( x ) first letter in x. 13. Read in a positive integer n < 15 and a square m a t r i x A of o r d e r n. D e t e r m i n e if it is symmetric. 14. Read in an m x n m a t r i x A and a p x q m a t r i x B. Find A + B, A - B, and AB if they are defined. n

15. The t r a c e of a m a t r i x ( a i j ) n x n is ~ aii. Read in a positive i n t e g e r i=l

n < 20 and an n x n matrix, and p r i n t the trace of the m a t r i x .

Exploratory Writing Projects Using library and I n t e r n e t resources, write a t e a m r e p o r t on each of the following in y o u r own words. Provide a w e l l - d o c u m e n t e d bibliography. 1. Discuss the d e v e l o p m e n t of the concept of a function.

Chapter Summary

183

2. Collect the various rates for water and electricity consumption from neighboring towns and cities. Write each as a word problem and then define each as a function. 3. Describe the history of perfect numbers and their relationship to Mersenne primes. Comment on the existence of odd perfect numbers. 4. Give a number of applications of the floor and ceiling functions to everyday life. 5. Investigate the two-queens puzzle on an n x n chessboard, where 2

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