Bridge To Abstract Mathematics

Bridge to Abstract Mathematics Ralph W. Oberste-Vorth and Aristides Mouzakitis

1

c Copyright
2003 by Ralph W. Oberste-Vorth and Aristides Mouzakitis

i

Table of Contents Contents

i

Some Notes on Notation

iii

Preface

iv

PART I.

THE AXIOMATIC METHOD

1

Chapter 1. Section 1. Section 2. Section 3. Section 4.

Introduction The History of Numbers The Algebra of Numbers The Axiomatic Method Parallel Mathematical Universes

2

Chapter 2. Section 1. Section 2. Section 3. Section 4. Section 5.

Statements in Mathematics Mathematical Statements Mathematical Connectives Symbolic Logic Compound Statements in English Predicates and Quantifiers Supplemental Exercises

8

Chapter 3. Section 1. Section 2. Section 3. Section 4. Section 5.

Proofs in Mathematics What is Mathematics? Direct Proof Contraposition and Proof by Contradiction Proof by Induction Examples and Counterexamples Supplemental Exercises

8

2 2 3 5

8 8 8 8 11 74

12 13 15 16 22 74

How to THINK about mathematics: A Summary How to COMMUNICATE mathematics: A Summary

24 24

PART II.

SET THEORY

25

Chapter 4. Section 1. Section 2. Section 3. Section 4.

Basic Set Operations Introduction Subsets Intersections and Unions Differences and Complements

26 26 27 28 31

ii Section 5. Section 6.

Power Sets Russell’s Paradox Supplemental Exercises

32 33 74

Chapter 5. Functions Section 1. Functions as Rules Section 2. Cartesian Products, Relations and Functions Section 3. Injective, Surjective and Bijective Functions Section 4. Compositions of Functions Section 5. Inverse Functions and Inverse Images of Functions Section 6. Another Approach to Compositions Supplemental Exercises

34

Chapter 6. Relations on a Set Section 1. Properties of Relations Section 2. Order Relations Section 3. Equivalence Relations Supplemental Exercises

56

Chapter 7. Cardinality of Sets Section 1. Introduction Section 2. Finite Sets Section 3. Infinite Sets Section 4. Countable Sets Section 5. Uncountable Sets Supplemental Exercises

46

PART III.

64

ALGEBRA OF NUMBER SYSTEMS

34 34 38 40 42 44 74

56 57 61 74

46 46 49 51 52 74

Chapter 8. Algebra of Number Systems 65 Section 1. Primary Properties of Number Systems 65 Section 2. Secondary Properties Involving Addition and Multiplication 66 Section 3. Secondary Properties Involving Order 66 Section 4. Isomorphisms and Embeddings 68 Supplemental Exercises 74 Chapter 9. Archimedean Ordered Fields Section 1. The Main Goal and Archimedean Principle Supplemental Exercises

65

Chapter 10. The Natural Numbers Section 1. Introduction Section 2. Zero, the Natural Numbers and Addition Section 3. Multiplication Supplemental Exercises

71

69 74

71 71 74 74

Summary of the Properties of Z+ Chapter 11. The Integers Section 1. Introduction: Integers as Equivalence Classes

74 75 75

iii Section 2. Section 3. Section 4. Section 5.

A Total Ordering of the Integers Addition of Integers Multiplication of Integers Embedding the Natural Numbers in the Integers Supplemental Exercises Summary of the Properties of Z

Chapter 12. The Rational Numbers Section 1. Introduction Section 2. A Total Ordering of the Rationals Section 3. Addition of Rationals Section 4. Multiplication of Rationals Section 5. An Ordered Field Containing the Integers Supplemental Exercises

75 75 75 78 74 74 79 79 75 75 75 79 74

Summary of the Properties of Z+

74

Chapter 13. The Real Numbers Section 1. Dedekind Cuts Section 2. Order and Addition of Real Numbers Section 3. Multiplication of Real Numbers Section 4. Embedding the Rationals in the Reals Section 5. Uniqueness of the set of Real Numbers Supplemental Exercises

82

Chapter 14. The Complex Numbers Section 1. Introduction Section 2. Algebra of Complex Numbers Section 3. Order on the Complex Field Section 4. Embedding the Real Numbers in the Complex Numbers Supplemental Exercises

89

Chapter 15. The Real Numbers According to Cantor Section 1. Convergence of Sequences of Rational Numbers Section 2. Cauchy Sequences of Rational Numbers Section 3. Cantor’s Set of Real Numbers Section 4. The Isomorphism from Cantor’s to Dedekind’s Reals Supplemental Exercises

92

PART IV.

HINTS

98

Chapter 16.

Hints for the Exercises Hints for Chapter 1 Hints for Chapter 3 Hints for Chapter 4 Hints for Chapter 5 Hints for Chapter 7 Hints for Chapter 6 Hints for Chapter 8 Hints for Chapter 10

99

82 82 84 85 85 74

89 89 89 90 74

92 93 95 97 74

99 99 99 100 100 102 103 104

iv Hints for Chapter 11 Hints for Chapter 12 Hints for Chapter 13 Hints for Chapter 14 Hints for Chapter 15 Postscript and Selected References

104 105 106 107 108 109

Index

110

v

Some Notes on Notation We will use the following sets of numbers: N

denotes the set of natural numbers: { 1, 2, 3, 4, 5, . . . }

Zn

denotes the set of the first n integers: { 1, 2, 3, . . . , n }∗

Z

denotes the set of integers

Z+

denotes the set of nonnegative integers: { 0, 1, 2, 3, 4, . . . }

Z∗

denotes the set of nonzero integers

Q

denotes the set of rational numbers

Q+

denotes the set of nonnegative rational numbers

Q+∗

denotes the set of positive rational numbers

R

denotes the set of real numbers

R+

denotes the set of nonnegative real numbers

R+∗

denotes the set of positive real numbers

C

denotes the set of complex numbers denotes the open interval: { x ∈ R | a < x < b } denotes the closed interval: { x ∈ R | a ≤ x ≤ b }

(a, b) [a, b] (a, b] [a, b)

denotes the interval: { x ∈ R | a < x ≤ b } denotes the interval: { x ∈ R | a ≤ x < b }

Beware of our use of the following notation for subsets: A⊂B

A is a subset of B (A may equal B)†

A
B

A is a proper subset of B (A = B)

*Most authors use Zn = { 0, 1, 2, . . . , n − 1 }. This is especially true in number theory and abstract algebra, where 0 and n are really the same. The choice here is more natural in our discussion of cardinality in Chapter 7. †We never use the notation A ⊆ B. Many authors use A ⊆ B for subsets and A ⊂ B for proper subsets.

vi

Notation

We use the following notation for relations and functions: f :X →Y
f
S

f is a function with domain X and codomain Y restriction of f to a subset S of its original domain

The symbol
will denote the end of a proof; purists may read it as “Q.E.D.” abbreviating the Latin quod erat demonstrandum, which translates the ancient Greek of Euclid:

˜περ ¶δει δε›ξαι. Similarly, we use the symbol to mark the end of an example or remark.

vii

Preface ’Mathematicians,’ [Uncle Petros] continued, ’find the same enjoyment in their studies that chess players find in chess. In fact, the psychological make-up of the true mathematician is closer to that of the poet or the musical composer, in other words of someone concerned with the creation of Beauty and the search for Harmony and Perfection. He is the polar opposite of the practical man, the engineer, the politician or the ...’ – he paused for a moment seeking something even more abhorred in his scale of values – ’... indeed, the businessman.’ — Apostolos Doxiadis, Uncle Petros & Goldbach’s Conjecture (2000)

A Historical Perspective. We, the authors of this text, were undergraduates together at the end of the 1970’s. We learned how to construct proofs essentially by osmosis. Also, we learned most of the material contained in this text in much the same way. That is, we learned the art of proof mostly by imitating the experts, after listening to and watching their presentations, reading their textbooks and thinking these things through. For us, the process of learning how to prove theorems began in the calculus sequence and continued through linear algebra and beyond. Undergraduate mathematics curricula in the U.S. have undergone some changes since that time. The most visible change may be the end product of Calculus Reform. With an emphasis on the integration of the numerical, analytical, and geometrical aspects of the subject as well as the integration of computing technology, some topics have been removed or minimized in the curricula. We saw proofs of most everything in our calculus courses, even if we were not generally held responsible for them on the examinations. This is not always the case today. In fact, at many institutions a variety of calculus sequences exists. Some of the calculus textbooks used in some of those courses do not give any proofs. Calculus Reform is certainly not the only factor for the changing curricula. The growing number of high school graduates who continue their educations at post-secondary institutions has also put some pressure on the calculus curricula by requiring increasingly extensive reviews of algebra and trigonometry. Also, the fluid nature of the liberal arts education often leads a future major in mathematics to take a calculus course designed for other majors.

viii

Preface

As a result of these fundamental changes in calculus curricula, students have been enrolling in linear algebra courses with weaker backgrounds in constructing and understanding proofs. Recently, many institutions have inserted new (or, sometimes, recycled) courses into their curricula to address this problem. These courses go by many names. The choice of mathematical content is not universal, but the prime objective is universal. That objective is to prepare students to deal with proofs in their later courses. At the University of South Florida, the faculty gave this course the somewhat pretentious title “Bridge to Abstract Mathematics.” (Amazingly, the title was more contentious than the course itself!) An interesting aside lies in the fact that the usual content for these new “Bridge” courses is set theory and related topics. Courses in set theory were a standard part of the major earlier in the twentieth century. Between the 1950’s and 1970’s most of these types of courses were squeezed out of the requirements for the mathematics major by a combination of courses in linear algebra and abstract algebra as well as a relaxation of requirements in favor of elective courses; many of the elective courses were new to the undergraduate curriculum. We are curious how our own mathematical development would have been altered had we taken such a course during our own mathematical training. We certainly hope that you—today’s majors—find this type of course useful. About This Text and How to Use It. This text has a dual purpose. As suggested above, the goal is both to present appropriate mathematical content and to guide students on how to do proofs. While the content is, we hope, interesting, the process of transforming the thought processes from a passive computational orientation to an active creative orientation is at least as important. This text consists of four parts. Part I is an introduction (with a few historical comments). The second and third chapters should be impossible to absorb on a cursory reading. You may want to read it carefully and refer to it as you read the remaining parts. (If you wish to try to ignore it completely, we suggest you at least look at Section 5 of Chapter 2 on the language of mathematics.) Parts II and III present the mathematical content. Part II presents topics from set theory and Part III presents the constructions of various number systems in the context of some of their properties. The definitions are presented with examples and comments. The proofs of propositions and theorems are almost all left as exercises. Yes, you can learn mathematics— how to prove theorems—passively, by watching, listening, and reading. However, we strongly believe that you will be better off learning actively, that is, by supplying the proofs yourself. You have a number of resources

Preface

available to you. The text takes small steps forward, with some proofs provided where they involve a larger leap forward. Of course, you can ask the experts. Lastly, do not overlook this resource: your fellow students. It is difficult to do mathematics without discussing it with someone, even a non-expert. We should mention here that the chapters on the constructions of the natural numbers and the real numbers, Chapters 10 and 13, respectively, are much more difficult than the other chapters of Part III. They can be skipped as long as their properties are understood for use in the other chapters. Part IV constitutes another resource for completing proofs. It contains hints and comments for many of the exercises. Some of the hints will tell you how to start a proof. Other hints may only be helpful once you have already thought about how to approach the proof. In any case, we recommend that you attempt each exercise without consulting the hints. If you get stuck, then consult the hints (or a fellow student or an expert). Once you have completed an exercise, read the hint and tell someone about your solution. By the way, proper etiquette is to challenge everything your fellow students tell you about their proofs. Passive listeners are useless in the pursuit of knowledge. It is not rude to call someone an idiot (at least with respect to a faulty part of their proof), but you should be ready for someone else to call you an idiot (with respect to your own faulty proofs). Perhaps this is a bit overblown: calling a nonsensical statement made by a friend idiotic will hopefully be taken (and given) constructively, while such behavior with a total stranger could be quite dangerous. We are merely supporting a ruthless attack on knowledge through free discourse and not an attack on your neighbors! Acknowledgments. We would like to thank the Fall 2000, Spring 2001 and Summer 2001 classes of MGF 3301 taught by the first author at the University of South Florida for “testing” drafts of this text. Their corrections and suggestions have been invaluable. In particular, we would like to thank Paul Anderson, Ray Burrus, Christie Burton, Leon Calleja, Thuc Cao, Nathan Chau, Teresa Chung, Jason Copenhaver, Mindy Eason, Adam Francis, Alynne Frewin, Russell Gerbers, Bridget Giroux, Erika Johnson, Kristy Kazemfar, Sarah Lahlou-Amine, Christopher Ledwith, Carson McCoy, Rose Nestor, Cheryl Ng, Ryan Parrish, Michelle Richardson, Patrick Robbins, Emily Roberts, Cheryl Scilex, Anthony Upchurch, Adrianne Waltz, Jeanne Waser and Aimee Yates. We would also like to thank professors Edwin Clark and Boris Shekhtman for using drafts of this text at the University of South Florida in 2001 to 2003.

ix

1

PART I THE AXIOMATIC METHOD

2

Chapter 1 Introduction 1. The History of Numbers. The seemingly easy concept of number is actually an abstraction which came quite late in the history of our intellectual evolution. The ancient shepherd had no method of counting the sheep in his herd. He used the more primitive concept of a one-to-one correspondence instead, perhaps by notching a bone as many times as the number of his sheep. The counting numbers or natural numbers 1, 2, 3, 4, . . . were the first to appear on the scene. Pythagoras, a mystic who lived in the sixth century B.C. and who is famous for the theorem which bears his name, considered the natural numbers to be the very stuff out of which the universe is made. The positive rational numbers, which come up when we divide a whole thing into smaller pieces, were invented second. They constitute a natural extension of the natural numbers. A number is positive rational if and only if it can be expressed as a quotient of natural numbers. Such quotients are called fractions. Pythagoras and his disciples knew about fractions. However, the discovery of numbers that are not fractions shook the foundations of the Pythagorean school and caused the first big crisis in mathematics. Eudoxos devised an ingenious theory of irrational quantities, which anticipated the rigorous treatment given by G. Cantor and R. Dedekind in their constructions of the real number system almost two and one half millennia later. The number zero was introduced by the Hindus in the ninth century A.D., but some historians conjecture that it was known to the Babylonians. The negative and complex numbers go hand in hand because of the definition of imaginary numbers as the square roots of negative numbers. Negative and complex numbers were recognized at this time as well. However, the existence of negative and complex numbers was viewed as absurd for another millennium, despite their increased formal use. Economic developments—for example, an international banking system spreading out of Italy in the fourteenth century just as Europe was

Section 2. The Algebra of Numbers

moving from the Middle Ages to the Renaissance—provided favorable circumstances for wider acceptance of negative numbers. The use of negative numbers was perhaps influenced by Fibonacci in the thirteenth century; he noted that a negative number could be regarded as a loss whereas a positive number could represent a gain. The universal acceptance of the complex numbers and, hence, the negative numbers, did not occur until the 1830’s. The early part of the nineteenth century saw developments in algebra, particularly the extensions of the naturals to the integers and rationals and of the reals to the complex numbers. The notion of equivalence classes stood at the heart of these developments, though this language was not developed until the twentieth century. The real numbers, at least the positive real numbers, were certainly accepted since the time that Eudoxus developed the irrationals numbers. However, the axiomatic construction of the real numbers did not occur until the nineteenth century. Many mathematicians had attempted to put the reals on a solid foundation; this was finally achieved by R. Dedekind and G. Cantor 1870’s. This brings us full circle back to the naturals. The easiest of the numbers were the most difficult, requiring the very axioms of set theory which were taking form at the beginning of the twentieth century. Today, we give B. Russell and A. Whitehead the credit for formalizing what was known and used for ages. 2. The Algebra of Numbers. Apart from any social conditions which may have played a role in the development of number systems, there also exist intrinsic reasons for their development that peculiar to the inner dynamics of mathematical activity. The step from natural numbers to rational numbers derives easily from human activity where an apple and yet another make two apples and where two people sharing an apple make for one half of an apple for each. Irrational numbers are at a deeper level; in the time of Eudoxus, it centered on geometrical issues such as the lengths of diagonals of a rectangle whose sides had lengths expressed by natural numbers. Negative numbers are understandable, as by Fibonacci above, if not as easily realizable. Where do complex numbers come from? Pause for a moment to reflect on the interesting names given to the expanding sets of numbers. They surely reflect the prejudices of the namers or, at the very least, the harsh criticism their introduction produced! Consider the positive versus the negative, the rational versus the irrational and the real versus the imaginary. There is a purely algebraic way of explaining the need for most of these numbers and it is this need which first pointed to the complex numbers.

3

4

Chapter 1. Introduction

We will give successive elementary equations in terms of successive sets of known numbers whose solutions are not in those sets, but in successively expanded sets. Consider the equations m + x = n, for natural numbers m and n. It is true that exactly one of the following holds: n > m or n = m or n < m. If n > m, then the equation has a solution in N, which is called n − m. If n = m, then the existence of 0 is suggested. Finally, if n < m, then the negative integers are needed for the equation to have a solution. The equations qn = m, for integers m and n, with n = 0, lead us to the rational numbers. Polynomial equations with integral coefficients such as an x n + an−1 x n−1 + an−2 x n−2 + · · · + a1 x + a0 = 0 lead us to irrational numbers such as the solutions of x 2 − 2 = 0 and complex numbers such as the solutions of x 2 + 1 = 0. The set that is missing from the above discussion is the set of real numbers. The set of all real solutions of polynomial equations with integer coefficients is called the set of algebraic numbers. Some transcendental numbers (real numbers, such as π, which are not algebraic) were known before the real numbers were rigorously defined. The actual set of real numbers—not just the rationals and some algebraic irrationals—was not fully explained until the nineteenth century. One has to view the real numbers as some sort of completion of the rational numbers. Consider the rational numbers. All can be written as repeating decimals such as 1 1

= 1.0 . . . ,

1 2

= 0.50 . . . ,

and

1 3

= 0.3 . . . .

We can now imagine all other possible decimals, where the digits need not form any pattern, and call this the set of real numbers. Making this precise is not as easy as it may appear! 3. The Axiomatic Method. Mathematics is highly respected in our culture. To most people it inspires awe or fear, while to a few it is a source of joy and admiration, an artistic game of the highest quality, the very language in which the book of nature is written.

Section 3. The Axiomatic Method

Whence mathematics derives its power? Is there a royal road to it? Is it accessible to all mortals? It certainly requires hard work, discipline, concentration, self-respect and self-confidence. Armed with these qualities, you have to plunge into this book and be actively engaged in deciphering its content. That means that you have to ask yourself questions and, by trying to answer them, ask more questions; you have to try to discover the real path that the intellect has followed to derive a proof, a path that is now hidden either for aesthetic purposes or for the sake of precision and simplicity. Mathematics started as an empirical discipline in Mesopotamia and Egypt to serve everyday practical needs. Rules were invented to measure distances, areas, and volumes, thus constituting an embryonic form of the body of knowledge called geometry. It is with the Greeks that geometry acquired a drastically different character. With Thales as a pioneer and then with the school of Pythagoras, geometry was constructed as an axiomatic, deductive discipline. Since, in fact, this point of view is still prevailing as a way to sanction the results of the mathematician’s creative imagination, let us describe and explain its main features. In order to establish human discourse, the meaning of the terms employed has to be specified and universally accepted. To determine the meaning of a word, this word has to be explained in terms of other words and this process will never terminate unless we reach a word whose meaning is generally accepted. A word whose meaning is generally accepted is called a primitive term or an undefined term. The terms point, straight line, and plane are examples of primitive terms used in geometry. Other terms of the system in question are defined in terms of previously defined terms and the primitive terms. The following example gives a definition phrased in three different ways. Example 1. The following definitions are exactly the same: • The midpoint of a line segment is the point that divides the line segment into two equal parts. • If a point divides a line segment into two equal parts, then that point is called the midpoint of that line segment. • A point is the midpoint of a line segment if and only if it divides the line segment into two equal parts. All definitions are really if and only if statements,* even if we are not careful to state them in that way. The context should tell you when a statement is really a definition. *We will discuss if and only if statements more in Section 2 of Chapter 2.

5

6

Chapter 1. Introduction

When starting from scratch—from the level of undefined terms—it is not so easy to give definitions of even simple objects. However, this is necessary since we cannot assume that terms are understood the same way by different people. Take the basic geometric notion of a triangle. See Figure 2.

Figure 2: Triangle vs. triangular region You may think everyone knows what a triangle is. Try this experiment: ask a few non-mathematicians “What is a triangle?” You will probably get some different answers, perhaps including drawings. The most popular answer may describe a triangular region, which a geometer would call a region bounded by a triangle. Clearly, we need definitions, even for simple mathematics objects. Exercise 3. Define the concept of a triangle. Do you need more primitive terms than the ones mentioned above? There follows another basic term to define. Exercise 4. Define the concept of an angle. How might addition of two angles be defined? A key concept in mathematics is that of a proof. Suppose you state an opinion on an issue using a single sentence, for example “freedom is preferable to happiness” and your interlocutor asks you to justify it. Just after you have pronounced your next sentence he asks you to justify this sentence also. It is clear that this process will never end, unless you and your interlocutor agree upon the truth of some sentence. Such a sentence is called an axiom. Exercise 5. Are any axioms involved in the definition of a triangle? We list a few axioms from “The Elements,” a book written around 300 B.C. by Euclid and preserving the geometric achievements of the ancients in their final form. Axiom 1. If two straight lines intersected by a transversal make the sum of the measures of the interior angles on the same side less than 180◦ , then the two straight lines meet on that same side.

Section 4. Parallel Mathematical Universes

Axiom 2. Things which are equal to the same thing are equal to one another. Axiom 3. If equals are added to the same thing, the sums are equal. Axiom 4. If equals are subtracted from equals, the differences are equal. Axiom 5. All right angles are equal to one another. A proof consists of a series of logical steps by which we deduce the truth of a statement from the axioms that are explicitly stated. A statement which can be proved is called a theorem. When we present a proof of a theorem, we present only a skeleton; this skeleton indicates the ideas necessary to give a complete proof. Such skeletal proofs rarely go back to the level of axioms, rather they usually depend on theorems which are already proved. Examples of theorems are: (1) The diagonals of any rectangle are congruent. (2) The square of an even number is an even number. Frequently, a theorem is called a proposition or a lemma or a corollary. Logically, all of these are the same. A statement which is called a proposition is usually a less important result than a statement which is called a theorem. A lemma is usually only important as a logical step in the proof of some theorem.* A corollary is a theorem which follows, usually easily, from the preceding theorem. But what do we mean by logical steps? Mathematicians have contrived various methods of deducing a desired result, that is, they have devised different kinds of proof. We illustrate the methods with specific examples in the next chapter. However, there is more to the methodology of doing proofs than the types of proofs. This involves the very thought processes themselves. The only way to develop and refine these processes is to use them frequently. 4. Parallel Mathematical Universes. You may think that there is only one mathematics, that is, one “correct” set of axioms from which to start. Surprisingly, this is quite wrong, in two different ways! The removal, addition or other change of just one axiom can dramatically change the mathematics. Also, two very different collections of axioms can lead to the same mathematics. You may already know the story of Euclid’s “Parallel Postulate” from your high school geometry class. (By the way, we regard the words axiom and postulate as synonyms, though Euclid and others regard them as technically different.) In plane geometry, the Parallel Postulate is equivalent to *It is interesting to note that posterity may judge—sometimes has judged—a lemma to be a theorem. For example, Zorn’s Lemma and the Urysohn Lemma are famous theorems in set theory and topology, respectively, that were considered lemmas by their original authors.

7

8

Chapter 1. Introduction

the statement that there is a unique line that is parallel to a given line passing through a given point not on the original line. One may instead assume that no parallel line may exist or that more than one parallel line may exist. Such changes in the axioms lead to new and different geometries; these are called non-Euclidean geometries. In fact, non-Euclidean geometries are as legitimate as Euclidean geometry. Despite the fact that Euclid and the ancient Greeks tried to model the physical universe with their geometry, in physics, for example in relativity theory, it is a non-Euclidean geometry which is the correct model. Recall your high school geometry class. You may remember that there were quite a few axioms. Perhaps the notions of betweenness and congruence were left undefined, while the ideas of measurement were developed from the axioms. You may remember developing the ideas of segment length and angle measure from the axioms. It is quite possible to develop Euclidean geometry by starting from different collections of undefined terms and axioms. The following lengthy example gives a different set of axioms with the opposite approach. Example 6. In 1932, G. D. Birkhoff published a paper* giving a collection of four axioms for Euclidean plane geometry. Birkhoff assumes you have constructed the set of real numbers. The terms set, point, line, on, line measure (or distance or length), and angle measure are undefined. Birkhoff gave his axioms the following names: I. Postulate of Line Measure II. Point-Line Postulate III. Postulate of Angle Measure IV. Postulate of Similarity Let us consider the statements of these four axioms (or postulates) individually; we will state each one and examine its content. You should remember the big picture as you read this; ask yourself if this really gives you the same plane geometry that you already know. (We follow the usual custom of using the term “line” to mean “straight line.”) Postulate of Line Measure. Given any line, there is a one-to-one correspondence from the set of points of this line onto the set of real numbers, where every point A on this line corresponds to the real number x A , so that the length of the line segment AB (or distance between A and B) is given by d(A, B) = |x B − x A | for all points A, B. *A Set of Postulates for Plane Geometry (based on scale and protractor) appeared in volume 33 of the journal Annals of Mathematics.

Section 4. Parallel Mathematical Universes

The idea of this postulate is that every line comes with coordinates. Consider Figure 7. xA

| xB – xA |

xB

xC

| xC – xB |

B

A

C

Figure 7: Birkhoff’s Postulate of Line Measure You are familiar with thinking of the set of real numbers as a line. The Postulate of Line Measure indicates that every line can be thought of as a number line. The length of a segment is then the distance between its endpoints. The absolute value is necessary to avoid negative length since we do not know, a priori, which endpoint will have the larger coordinate; of course, |x A − x B | = |x B − x A |. Point-Line Postulate. One and only one line contains two given distinct points. The Point-line Postulate should be familiar to you; it can be restated as “Two points determine a line.” The first two postulates allow us to define the distance between any two distinct points, A and B, of the plane. By the Point-Line Postulate, there exists a line containing A and B. The Postulate of Line Measure then gives the distance between the points. This is well-defined since, by the Point-Line Postulate, there is only one line containing the two points. Before proceeding with a postulate on measuring angles, consider Figure 8.

as s B as-ar (mod 2π)

O A

r ar

Figure 8: Birkhoff’s Postulate of Angle Measure

9

10

Chapter 1. Introduction

Postulate of Angle Measure. Given any point O, there is a one-to-one correspondence from the set of rays with endpoint O onto the set of real numbers (mod 2π),* where every ray r with endpoint O corresponds to a real number ar (mod 2π), so that, if A = O and B = O are points on rays r and s, respectively, then the measure of angle AO B is given by mAO B = as − ar (mod 2π). Furthermore, if the point B on ray s varies continuously in a line  not containing the vertex O, the number as varies continuously also. Figure 8 may help you to understand the Postulate of Angle Measure better. The angle measure defined here is a little different than you may think. In general, mAO B = mB O A; in fact, the sum of mAO B and mB O A is 0 (mod 2π), which equals 2π (mod 2π). See Figure 9.

B

B

AOB

BOA AOB

O

BOA O

A

A

Figure 9: mAO B + mB O A = 0 = 2π (mod 2π) Before proceeding to the last of Birkhoff’s postulates, which is about triangles, we should make sure we understand how triangles are defined. Exercise 10. Define triangle using only Birkhoff’s first three postulates. Postulate of Similarity. Given two triangles, ABC and A B  C  , and a constant k > 0, if d(A , B  ) = k · d(A, B), and

d(A , C  ) = k · d(A, C)

mB  A C  = ±mB AC,

*You may not be familiar with the phrase “mod 2π .” However, you are probably familiar with the concept from trigonometry. In trigonometry, one may consider angles with negative measure or with measure larger than 2π . The sine or cosine of these angles only depend on the equivalent angle between 0 and 2π or, more precisely, in the interval [0, 1). For example, sin(−π ) = sin π = sin(3π ) and cos(−π ) = cos π = cos(3π ). Therefore, we write −π = π = 3π (mod 2π ); actually we usually read the equal sign as either “is congruent to” or “is equivalent to” rather than “is equal to.” You may recall that the symbol ≡ is usually used instead of = in this context. We consider this idea more carefully in Section 3 of Chapter 6.

11

Section 4. Parallel Mathematical Universes

then d(B  , C  ) = k · d(B, C), and

mC  B  A = ±mC B A

mA C  B  = ±mAC B. A


A C

B

C


B


Figure 11: Birkhoff’s Postulate of Similarity Figure 11 may help you to understand the Postulate of Similarity better. This is the familiar Side-Angle-Side statement (usually called SAS) for similar triangles. In Figure 11, k = 2; that is, A B  is twice as long as AB, A C  is twice as long as AC and B  A C  is congruent to B AC. The triangles are congruent if k = 1. Of course, the statement at the beginning of this example, that these are axioms for Euclidean plane geometry, must be proved. We will not do this here. Exercise 12. Using Birkhoff’s axiomatic system in Example 6, define the concepts of betweenness and congruence (for line segments and angles). We will see other axioms in Chapter 7 that play similar roles. These are the Axiom of Choice and the Continuum Hypothesis. Mathematicians are free to accept or reject these as axioms of their mathematics; a theorem of one mathematics may be false in another, which is based on a different collection of axioms.

12

Chapter 2 Statements in Mathematics 1. Mathematical Statements. Whatever mathematics may be as a mental activity, it is communicated as a language. Therefore, it has its specific syntax, its own technical terms, and its own conventions. Mathematics is also an exact science, which means that we are obliged to express our mathematical thoughts with high precision. A deviation from the norm may easily lead to a complete distortion of the intended meaning. In mathematics, we assert the truth of certain statements. Other statements are to be proved or disproved. This is a fundamental dichotomy: No mathematical statement is both true and false. Viewed axiomatically, you can take true and false to be undefined terms; the above dichotomy can be taken as an axiom. We should clarify what is meant by a mathematical statement. Definition 1. A declarative sentence is a logical statement iff, according to its logical content, it is unambiguously either true or false. A mathematical statement is a logical statement used in mathematical discourse. By statement, without the adjective logical or mathematical, we will always mean a logical or, usually, a mathematical statement. Let us consider this carefully with some examples. The sentence The number 2 is positive. is a true mathematical statement and the sentence The number 2000 is divisible by 3. is a false mathematical statement. Note that the previous two sentences are both declarative.

Section 1. Mathematical Statements

By definition, sentences that are not declarative cannot be logical statements; such sentences are neither true nor false. An interrogative sentence such as Who is there? is not a logical statement. An imperative sentence such as Go away. is not a logical statement. An exclamatory sentence such as Wow! is not a logical statement. Of course, a sentence like Hello. is not a logical statement either. A declarative sentence may be a logical statement even though its truth value is not known. This is very common in the most advanced areas of mathematics; it is the fuel of mathematical research. Consider the following sentence: The 1010 digit in the decimal expansion of the number π is 3. This is a mathematical statement. Even if no one knows whether this is true or false, it must be exactly one or the other. This is what we mean by unambiguous in Definition 1. Not every declarative sentence is a statement. Opinions are not logical statements. For example, the declarative sentence Vanilla is the best ice cream flavor. is not a logical statement since it does not have a truth value; it is true that a person may have an opinion on this subject, but this does not make the statement true or false. On the other hand, Vanilla is Jon’s favorite ice cream flavor. may be a statement for a specific person named Jon. The sentence Socrates was a good philosopher.

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is a less clear example. We should ask what constitutes a good philosopher; the sentence about the famous ancient Greek philosopher, Socrates, is a logical statement if we give precise criteria for the adjective good. The examples given above are usually called simple statements since they cannot be broken into pieces that are themselves statements. Some authors call simple statements atoms or atomic statements. You can probably think of examples that are not so simple, such as compound statements involving words like not, and, or, if, then, implies, and equivalent. We will consider these terms, called connectives, in Section 2. The same authors who call simple statements either atoms or atomic statements call compound statements molecules or molecular statements. You might also imagine statements involving variables. Consider the sentences The square of every real number is nonnegative. and There exists a real number whose square is not positive. Though these are true statements, they are rather different from our simple statements. Both contain variables. The former states a fact about every real number and the latter states a fact about at least one real number (specifically, 0). Predicates—sentences involving variables—that are quantified, as in the examples above, will be studied in Section 5. You may say that you know exactly what the words not, and, or, etc. mean and how their use will change the meaning of a statement. However, the colloquial usage of these terms does not entirely agree with the mathematical usage. For this reason, we will take a rather pedantic approach and define these words in terms of truth values—the truth values of compound statements formed by connectives are functions of the truth values of the simple statements from which they are formed. Linguistics also comes into play. Translating between mathematical, or logical, language and a natural language, such as English, is not always straight forward. We will start by using statements in symbolic logic and their English translations, side by side. After this chapter, we will abandon the logical notation completely; however, we recommend that you adopt some of it as a type of short hand for your own writing of mathematics. We will remind you of this in Remark 37 at the end of this chapter. 2. Mathematical Connectives. We are now ready to study the different connectives used in creating compound statements. Remember that these are being defined in a formal way, based on truth values, rather than a linguistic way, based on the colloquial usage of English.

Section 2. Mathematical Connectives

Negation (not) You may recall that an integer is called an even number when it equals twice some integer and is called an odd number when it equals one plus twice some integer. The statement “3 is an odd number.” is a true mathematical statement; the statement “3 is not an odd number.” is a false mathematical statement. How are these two statements related? You might say that the second is the negation of the first. This is correct. However, we will define negation not in colloquial terms, but in terms of truth values. Definition 2. For a mathematical statement p, the negation of p is a mathematical statement, denoted ¬ p, whose truth value is the opposite of the truth value of p. In general, we denote the negation of a statement, p, by not p; some denote it symbolically as either ∼ p or − p, rather than ¬ p. We can illustrate Definition 2 using the following truth table: p T F

¬p F T

In a truth table, one reads the truth values—T for true and F for false— of the column headings across a given row. In the truth table above for not p, the first row indicates that whenever p is true, then not p is false while the second row indicates that whenever p is false, then not p is true. Truth tables give us a nice shorthand for expressing the various truth values of related statements, starting with simple statements like p. Colloquially, we generally negate a statement by using the word not. Of course, it isn’t* as simple as changing p to not p. For example, suppose p is the statement “3 is an odd number.” For not p, we write “3 is not an odd number.” rather than “Not 3 is an odd number.” Conjunction (and) Two mathematical statements can be combined to form new, more complicated statements. One way to do that is to connect two statements with the word and. Let us consider this connective, defining it by a truth table. Definition 3. For mathematical statements p and q, the conjunction of p and q is the mathematical statement, denoted p ∧ q, whose truth value varies according to the following truth table: *Sorry for the contraction; we didn’t want the word not to appear here.

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p T T F F

q T F T F

p∧q T F F F

We will generally write p and q for the conjunction p ∧ q. Some authors use the notation p · q or p & q instead of p ∧ q. An example of conjunction is the statement “2 is positive and 3 is negative.” Of course, this conjunction is false since “3 is negative.” is a false statement. The conjunction of two statements is a true statement exactly when both statements are true and is a false statement when at least one statement is false. This agrees with the truth table in Definition 3. Disjunction (or) You might think we were excessively pedantic in defining negation and conjunction of mathematical statements. The next two definitions may change your mind. Consider, for example, the statement 2 is positive or 3 is positive. Is this statement true or false? You might want to say this is false because it is not true that only one or the other is true. For a less mathematical example, consider the statement I will take a bath or I will take a shower. Would you expect that the speaker may take both a bath and a shower? Definition 4. For mathematical statements p and q, the disjunction of p and q is a mathematical statement, denoted p ∨ q, whose truth value varies according to the following truth table: p q p∨q T T T T F T F T T F F F We will generally write p or q for the disjunction p ∨q. You should be extremely careful here, since the word or is not used with its most common meaning. When someone says I will give my ticket to Alex or to Jamie.

Section 2. Mathematical Connectives

it is implied that only one person will receive the ticket: either Alex or Jamie, but not both Alex and Jamie. In mathematics, we use the word or in the sense of at least one. The disjunction of two statements is true if at least one of the statements is true. In mathematics, the word or is equivalent to the hideous word and/or. As we use it in mathematics, or is referred to as the mathematical or and the inclusive or (since the possibility of both is included). Similarly, the or of everyday language is referred to as the exclusive or (since the possibility of both is excluded).* Example 5. It is usually easy to recognize negation, conjunction, and disjunction by use of keywords: the word not in negations, the word and in conjunctions, and the word or in disjunctions. However, notation can hide these constructions. Consider the following three true statements. • 1≤1 • 2<3<4 • 5≤6≤6 Reading the symbol ≤ as “is less than or equal to” shows that the first statement is a disjunction; it can be interpreted as 1 < 1 or 1 = 1. The second statement is the conjunction 2 < 3 and 3 < 4. The third statement is a conjunction of two disjunctions. Implication (implies/if, then) We will consider two additional types of compound statements, implications and equivalences. These are also called conditionals and biconditionals, respectively, by some authors. The next construct, implication, also disagrees with the common colloquial interpretation. Definition 6. For mathematical statements p and q, the implication (or conditional statement) denoted p ⇒ q is a mathematical statement whose truth value varies according to the following truth table: p q p⇒q T T T T F F F T T F F T We will generally write either if p, then q or p implies q for the implication p ⇒ q; some authors use the notaion p → q or p ⊃ q instead of p ⇒ q. For the implication p implies q, the statement p is called the *Some people go to the extreme of using the new word xor for the exclusive or. We do not like it and you may (or may not) like it. Is xor any better (or worse) than and/or?

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hypothesis or antecedent and the statement q is called the conclusion or consequent of the implication. Usually, in ordinary language, it is understood, for a statement of the form if p, then q, that there exists a causal connection between p and q. We cite as an example the following statement: If you leave me, then I will be badly hurt. In mathematical language, a conditional statement is false only when the hypothesis is true and the conclusion is false. Since a false hypothesis always yields a true implication, no sense of causality can be interpreted. Example 7. Cinsider the following four examples: • “If 2 is positive, then 3 is positive.” is a true statement since both the hypothesis and the conclusion are true statements. • “If 2 is negative, then 3 is positive.” is a true statement since the hypothesis is false. • “If 2 is negative, then 3 is negative.” is a true statement since the hypothesis is false. • “If 2 is positive, then 3 is negative.” is a false statement since the hypothesis is true, but the conclusion is false. Make sure you read the middle two examples carefully and understand why they are true! Do not let the trivial nature of the hypotheses and conclusions fool you: once the truth values of the hypothesis and the conclusion of any given implication are known, the truth value of that implication is trivially known as well. An implication if p, then q is called vacuously true if its hypothesis, p, is false. For a vacuously true implication, it does not matter whether its conclusion is true or false. Look again at Example 7 and notice that we did not indicate whether the conclusion was true or false in the middle two examples since it did not matter. We have already said that if p, then q and p implies q are used interchangeably to mean p ⇒ q. In fact, all of the following are equivalent English sentences: (1) If p, then q. (2) p implies q. (3) p only if q. (4) q if p. (5) p is sufficient for q. (6) q is necessary for p. Forms (3) and (4) may seem peculiar to you; think of them, for a true implication p implies q, as “ p is true only if q is true” and “q is true if

Section 2. Mathematical Connectives

p is true.” (You should compare these with the truth table for p ⇒ q.) The forms (5) and (6) are more esoteric. We have a prejudice against this language unless used together, as explained after the following example. Example 8. Consider the sentence For all real numbers a and b, if a = 0, then ab = 0. This is a statement of a different type since it contains variables. Notice that the truth value of “a = 0” depends on the value of the variable a. We will discuss such statements further in Section 5; we use it here because it better illustrates the different forms of implications mentioned above. The following forms should strike you as equivalent. • For all real numbers a and b, a = 0 only if ab = 0. • For all real numbers a and b, ab = 0 if a = 0. • For all real numbers a and b, a = 0 is sufficient for ab = 0. • For all real numbers a and b, ab = 0 is necessary for a = 0. Equivalence (equivalent/if and only if/necessary and sufficient) Our final construct, equivalence, probably agrees with your colloquial use of that word. Definition 9. For mathematical statements p and q, the equivalence of p and q, denoted p ⇔ q, is a mathematical statement whose truth value varies according to the following truth table: p q p⇔q T T T T F F F T F F F T Instead of writing p ⇔ q for an equivalence, we will generally write one of the following: • p is equivalent to q, • p if and only if q, • p is necessary and sufficient for q. For brevity, many people write p iff q in place of p if and only if q. Some authors use p ↔ q or p ≡ q instead of p ⇔ q. Remark 10. You may recall that, in Section 3 of Chapter 1, we explained that definitions are often stated in the form of if and only if statements. Of course, definitions are not logical statements whose truth value can be determined.* Rather, for a definition in the form of an if and only if *You may choose to think of all mathematical terms as undefined, in which case definitions become axioms. This make the biconditional appearance of definitions more reasonable.

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statement, the first half of the statement names the term being defined by the other half. While we try to use the word iff in our definitions, most authors simply write if. For example, we could define even integers as follows: an integer is even iff it is two times some integer. 3. Symbolic Logic. We could make an entire course out of symbolic logic. Rather than do that, we will discuss some of the more important compound statements, those we will use from time to time. Before doing this, let us discuss how statements of symbolic logic are parsed. Note that we will use the symbolic notation throughout this section and return to English statements in the next section. Remark 11. You may remember rules for order of operations in arithmetic. For example 1 + 2 × 3 = 7 = 9 since multiplications are performed before additions. There are also rules for order of operations in logic. Reading from left to right, do things in the following order: (1) (2) (3) (4) (5)

parentheses (or brackets, etc.) negation conjunctions and disjunctions (in any order) implication equivalence

Groupings by parentheses, brackets, etc. are treated hierarchically. That is, nested groupings are evaluated from the inside towards the outside (just as for algebraic expressions). We remember this as working from the inside out. Therefore, ¬ p∨q is the same as (¬ p)∨q and different from ¬( p∨q). Remember that between matched pairs of parentheses, you must apply the same rules. For a more complicated example, the following two compound statements based on simple statements a, b, c, . . . , i are equivalent: a ∨ ¬b ∧ c ⇒ ¬d ∧ e ⇔ ¬ f ∧ g ∨ h ⇒ i,            (¬ f ) ∧ g ∨ h ⇒ i . a ∨ (¬b) ∧ c ⇒ (¬d) ∧ e ⇔



Notice that, in order to evaluate this, we must start with the innermost pairs of parentheses, then the square brackets, then the curly braces, and then, finally, the angle brackets.

Section 3. Symbolic Logic

Example 12. Consider the disjunction p ∨ ¬ p. If p is true, then the disjunction is true.* On the other hand, if p is false, then ¬ p is true and the disjunction is true. This can be expressed by the following truth table: p ¬p p ∨ ¬p T F T F T T Notice how the column for p ∨ ¬ p has all T’s in its column; that is, p or not p is always true.† Definition 13. A compound statement is a tautology iff it is true for all possible truth values of its component statements. Tautologies are easily recognized in truth tables: their columns contains only T’s. Next we examine the other extreme. Example 14. Consider the conjunction p ∧ ¬ p. If p is false, then the conjunction is false.‡ On the other hand, if p is true, then ¬ p is false and the conjunction is false. This can be expressed by the following truth table with only F’s in the column for p ∧ ¬ p. p ¬p p ∧ ¬p T F F F T F Definition 15. A compound statement is a contradiction iff it is false for all possible truth values of its component statements. Contradictions are easily recognized in truth tables: their columns contains only F’s. Contradictions are actually quite important in mathematics. In fact, we will see in the next chapter that contradictions can actually be useful in proving theorems. Let us look at a couple additional examples. Example 16. Consider the implication p ⇒ p. The following truth table shows that this is a tautology. p p⇒p T T F T *The fact that ¬ p is false in the case when p is true does not matter. †The moral of this example is that you should never ask your mathematics instructor a question like “Will this topic be covered on the exam or not?” because you are likely to receive the perfectly correct, yet uninformative, response “Yes.” ‡The fact that ¬ p is true in the case when p is false does not matter.

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Example 17. Consider the equivalence ¬¬ p ⇔ p. This is a tautology: p ¬ p ¬¬ p ¬¬ p ⇔ p T F T T F T F T The following shows that some data can be extraneous. Exercise 18. Use truth tables to show that the following are tautologies: (a) p ∧ q ⇒ p (b) p ⇒ p ∨ q The following gives us an alternative definition of implication. Exercise 19. Use a truth table to show that p ⇒ q ⇔ ¬ p ∨ q. The following exercise may remind you of the concept of equality: x = x, x = y if and only if y = x, and x = y = z implies x = z. Exercise 20. Use truth tables to show that the following are tautologies: (a) p ⇔ p (b) ( p ⇔ q) ⇔ (q ⇔ p) (c) ( p ⇔ q) ∧ (q ⇔ r ) ⇒ ( p ⇔ r ) In the language of Chapter 6, the previous three exercises say that the relation defined on any set of statements by equivalence is, indeed, an equivalence relation. Recall that p ⇒ q can be read as p only if q and q ⇒ p can be read as p if q. It seems reasonable that p if and only if q should be equivalent to p if q and p only if q. This suggests the following important (though, perhaps, obvious) result; we will use this very often. Exercise 21. Use a truth table to show that ( p ⇔ q) ⇔ ( p ⇒ q)∧(q ⇒ p) is a tautology. One can use truth tables in this way to prove many such statements in logic. Remember that this is just a shorthand. To write out the proof of Exercise 21 in words would go something like this: Proof. We consider four cases: (1) (2) (3) (4)

p is true and q is true, p is true and q is false, p is false and q is true, and p is false and q is false.

Section 3. Symbolic Logic

First, suppose p is true and q is true. Hence, if p, then q is true and if q, then p is true. Therefore, the conjunction if p, then q, and if q, then p is true. Moreover, since p and q are true, p is equivalent to q is true. This completes the first case. Second, . . . .
Exercise 22. Complete the proof, started above, of Exercise 21 using words. Let us return to conditional statements. Given statements p and q, we can make implications p implies q as well as q implies p. We can also take the negations of p and q and form other implications. Three of these constructions are important enough to have names. Definition 23. The converse of the implication p ⇒ q is the implication q ⇒ p. That is, the converse of the statement if p, then q is the statement if q, then p. If a statement is true, then its converse may be either true or false; you can examine the possibilities in the following exercise. Exercise 24. Give examples of each of the following, if possible: (a) a true implication whose converse is true, (b) a true implication whose converse is false, (c) a false implication whose converse is true, (d) a false implication whose converse is false. Let p and q be simple statements. Remember that the implication if p, then q can only be false when p is true and q is false. Similarly, if q, then p can only be false when q is true and p is false. So, as you should have discovered in the previous exercise, it is impossible for both such an implication—between simple statements—and its converse to be false. If you thought otherwise, you probably introduced a variable in your statements. Soon, we will introduce variables and quantifiers; this will make things more difficult, but far more interesting. Definition 25. The contrapositive of the implication p ⇒ q is the implication ¬q ⇒ ¬ p. So, the contrapositive of the statement if p, then q is the statement if not q, then not p. Example 26. Consider the statement If the moon is made of cheese, then 1 = 2.

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The contrapositive of this statement is If 1 = 2, then the moon is not made of cheese. You should recognize that both of these statements are true!



The truth value of a contrapositive statement is always the same as the truth value of the original statement. Exercise 27. Prove that any implication is equivalent to its contrapositive. That is, p ⇒ q ⇔ ¬q ⇒ ¬ p. Definition 28. The inverse of the implication p ⇒ q is the implication ¬ p ⇒ ¬q. The inverse of the statement if p, then q is the statement if not p, then not q; it is the contrapositive of the converse of if p, then q. Therefore, the converse and the inverse of a given statement are equivalent: q ⇒ p ⇔ ¬ p ⇒ ¬q. Exercise 29. (a) Use a truth table to show that, given an implication, its converse and its inverse are equivalent; that is, q ⇒ p ⇔ ¬ p ⇒ ¬q. (b) Are an implication and its inverse equivalent? Prove your answer. So, from every given implication we can derive three implications called the converse, the contrapositive, and the inverse of the original implication: Original:

p ⇒

q,

if p, then q

Converse:

q ⇒

p,

if q, then p

Contrapositive: ¬q ⇒ ¬ p,

if not q, then not p

¬ p ⇒ ¬q,

if not p, then not q

Inverse:

4. Compound Statements in English. Of course, statements can get quite complicated and writing statements in English is further complicated by the fact that we do not use parentheses to indicate order of thoughts; commas can play an important role. You may have noticed that we used italics and quotation marks to try to clarify sentences in this chapter; this is a luxury upon which we should not depend. For example, consider the sentence: 1 is negative and 2 is negative or 3 is positive.

Section 5. Predicates and Quantifiers

If we let p denote the statement 1 is negative, let q denote the statement 2 is negative and let r denote the statement 3 is positive, then our compound statement looks like p ∧ q ∨ r. For symbolic logic, order of operations dictates that this is equivalent to ( p ∧ q) ∨ r. Since p and q are false and r is true, this statement is true. Notice that p ∧ q ∨ r is different from p ∧ (q ∨ r ); in fact, p ∧ (q ∨ r ) is false. Now how do we make this distinction in written English? What do you think of the following two statements? 1 is negative and 2 is negative, or 3 is positive. 1 is negative, and 2 is negative or 3 is positive. There certainly is a lot of weight riding on those two little commas. Notice that the former should be interpreted as ( p ∧ q) ∨ r while the latter should be interpreted as p ∧ (q ∨ r ). Now consider spoken English. It is very difficult to hear those little commas! An alternative is to rephrase the sentences completely. This is not so simple. For example, The conjunction of the statement 1 is negative and the disjunction of the statements 2 is negative and 3 is positive. While this may be clearer, it is not a very satisfying solution. Look at the symbolic statement in Exercise 21. In words, we could say something like the following: p if and only if q is equivalent to the conjunction of the implications if p, then q and if q, then p. 5. Predicates and Quantifiers. Take another look at Example 8. Also, did you have any difficulties with Exercise 24? These were tricky since they got you thinking (perhaps, in the case of Exercise 24) about introducing variables into statements. In this section, we introduce the use of variables in both logical and mathematical statements. The statements of most mathematical theorems involve

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variables. By the way, variables are sometimes called free variables, to emphasize that they are free to change. Consider the statement The square of −2 is positive. We know that this statement is true. Now consider the sentence The square of a negative number is positive. Is this a logical statement? You might think that it is, since you see that it is true for every real number. However, this sentence is not a logical statement since there is an unresolved variable involved. Now, let us consider the following sentence: The number x is positive. Is that a logical statement? No, again, since there is an unresolved variable involved. Again, no answer can be given as to the truth value of the sentence, unless x is given a specific value. For example, if x = 2, the sentence becomes a true statement, and, if x = −4, the sentence becomes a false statement. Any declarative sentence that contains one or more unresolved variables is called a predicate iff it is a statement whenever specific values are substituted for the variables. The word unresolved means that no values, or possible set of values, are given for the variables. We will return to the business of resolving, or substituting for, the variables shortly. Two other examples of predicates are x ≥ 0 and x + y = 0; the former has one variable and the latter has two variables. Using functional notation, we could denote these predicates by P(x) and Q(x, y), respectively. Just as in the case of logical statements, we can negate a predicate. For instance, if P(x) is the predicate x ≥ 0, then not P(x) is the predicate x ≥ 0.* Predicates also can be combined to form new predicates as is illustrated in the following exercise. Exercise 30. Equations and inequalities involving variables are predicates. (a) Determine all real numbers x for which the conjunction x + 3 > 6 and x 2 < 3 is a true statement. (b) Similarly, determine all real numbers x and y for which the disjunction x y = 0 or 2x − 3y = 0 is a true statement. *Thinking of the real numbers, we could have said x < 0, but, technically, that requires proof!

Section 5. Predicates and Quantifiers

Substituting a single value for a variable is not generally of interest since it can be done directly, thereby turning a predicate into a statement. For example, instead of writing a predicate with a substitution such as x 2 > 3 for x = 2. we could just write the statement 22 > 3. However, there are two very important ways of creating statements out of predicates; this is called quantifying a predicate. Consider the sentence: For all real numbers, x 2 = 4. This sentence should be a false statement since 52 = 4. This is one kind of quantifier. Definition 31. Let S be a set and let P(x) be a predicate. The universal quantifier is a phrase of the form for all x ∈ S, denoted ∀x ∈ S. The universally quantified predicate sentence ∀x ∈ S, P(x) is a mathematical statement which is true iff P(x) becomes a true statement whenever an element of S is substituted for x in the predicate P(x). We will generally write for all x ∈ S, P(x) for ∀x ∈ S, P(x). The phrases for every x ∈ S and for any x ∈ S are also used for universal quantifiers. We recommend that you be careful with the use of the word any since it may be misinterpreted as some rather than every. The phrase for all real numbers x is, of course, equivalent to for every x ∈ R. Example 32. Consider the following statements. Which are true and which are false? (a) For all real numbers x, √ (x + 3)2 + (x − 2)2 > 0. (b) For all real numbers x, x 2 = x. For (a), the squares are nonnegative. (x + 3)2 = 0 only if x = −3 and (x − 2)2 = 0 only if x = 2. So they cannot both be equal to 0 for the same x. Therefore, the statement in (a) is true. For (b), the square denotes the nonnegative square root.  root notation √ 2 Since, for x = −1, (−1) = 1 = 1 = −1, statement (b) is false.

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Remark 33. Sometimes a universal quantifier is hidden in a statement. It is sometimes preferable, for purposes of logical manipulation, to restate such a statement, as is done in the next example. Sometimes the set defining the universal quantifier is hidden. Consider the statement For all x, there exists a positive square root. This is true for the set of positive real numbers, but is false for the set of real numbers. When the set is omitted, we will assume that it is the set of real numbers. Example 34. In geometry, the Isosceles Triangle Theorem may be stated: The angles at the base of an isosceles triangle are congruent. This could be restated as follows: For every triangle T , if T is isosceles, then the angles at the base are congruent. You might even wonder exactly what set of triangles we are talking about! Next, consider the sentence: There exists x, such that x 2 = 4. This sentence is a true statement since 22 = 4; it is also true since (−2)2 = 4. The expression “there exists” is a new type of quantifier. Definition 35. Let S be a set and let P(x) be a predicate. The existential quantifier is a phrase of the form there exists x ∈ S, denoted ∃x ∈ S. The existentially quantified predicate sentence ∃x ∈ S, P(x) is a mathematical statement which is true iff P(x) becomes a true statement for at least one element of S substituted for x in the predicate P(x). We will generally write there exists x ∈ S such that P(x) for ∃x ∈ S, P(x). The phrase for some x ∈ S is also used for existential quantifiers. Again, we remind you to be careful with the use of the word any.

Section 5. Predicates and Quantifiers

Example 36. Which of the following statements about sets of real numbers are true and which are false? (a) There exists x such that (x + 5)2 = 0. (b) There exists x such that |x − 1| = x. (c) There exist x, y such that x 2 + y 2 = 1. Statement (a) is true since x = −5 satisfies the equation. Statement (b) is true since x = 12 satisfies the equation. Statement (c) is true since, √ for example, x = y = 12 2, satisfies the equation. Notice that the set of values for x is important: if the word real were changed to natural, then all three statements above would be false. Let us see how we negate a sentence which is introduced by a universal quantifier. Consider the statement: “For all x and y, x + y = 0.” In fact, this statement is false. What is the negation of this sentence? If not all pairs of numbers add up to zero, there exist numbers x and y, whose sum is not zero, so the negation of our original sentence is “There exist x and y such that x + y = 0.” which is a true statement. In general, the negation of the statement For all x, P(x). is the statement There exists an x such that not P(x). The negation of predicates with more than one variable is defined in an analogous way. To see how we negate a sentence which is introduced by the existential quantifier, we look at the following example. Consider the statement: “There exists a number x, such that x 2 < 0.” which is a false statement. If there does not exist a number whose square is negative, then the square of all numbers will be nonnegative, so the negation of the original sentence is “For all numbers x, x 2 ≥ 0.” In general, the negation of the statement There exists x such that P(x). is the statement For all x, not P(x). Remark 37: Logical notation. Although we have mentioned the symbols ¬, ∧, ∨, ⇒, ⇔, ∀, and ∃, we will not use them in this text after this chapter. Many people use them in writing mathematics by hand. The symbols for implications, equivalences, and quantifiers are especially useful. Some people also use  or s.t. for “such that,” ∵ for “since” or “because” and ∴ for “therefore” or “thus” or “hence.”

29

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Chapter 2. Statements in Mathematics

We prefer to use double barred arrows (⇒, ⇐, ⇔) for logical operations since single barred arrows, especially →, are used frequently in the notation for functions. We recommend that you use whatever shorthand you feel comfortable with. Moreover, we want to remind you to keep in mind your audience, those who will read what you write!

Supplemental Exercises

Supplemental Exercises Definition Review. There were 13 definitions in this chapter. You can take the truth values true and false to be undefined. Let p and q be logical or mathematical statements (or predicates). Define each of the following and give an example of each: (1) p is a logical statement (1) p is a mathematical statement (2) ¬ p is negation of a statement p (3) p ∧ q (or p and q) is conjunction of statement p and q (4) p ∨ q (or p or q) is disjunction of statement p and q (6) the implication p ⇒ q (or p implies q or if p, then q) (9) the equivalence p ⇔ q of statements p and q (13) a statement is a tautology (15) a statement is a contradiction (23) the converse of an implication p ⇒ q (25) the contrapositive of an implication p ⇒ q (28) the inverse of an implication p ⇒ q (31) the universal quantifier (31) an universally quantified predicate sentence (35) the existential quantifier (35) an existentially quantified predicate sentence In Exercises S1—S35 below, use truth tables to show that each statement is a tautology involving statements p, q, and r . Exercise S1. p ∧ ( p ⇒ q) ⇒ q Exercise S2. ¬ p ∧ (q ⇒ p) ⇒ ¬q Exercise S3. ¬ p ∧ ( p ∨ q) ⇒ q Exercise S4. p ⇒ (q ⇒ p ∧ q) Exercise S5. ( p ⇒ q) ∧ (q ⇒ r ) ⇒ ( p ⇒ r ) Exercise S6. ( p ∧ q ⇒ r ) ⇔ ( p ⇒ [q ⇒ r ]) Exercise S7. ( p ∧ q ⇒ r ) ⇔ ( p ⇒ [q ⇒ r ]) Exercise S8. ( p ⇒ [q ∧ ¬q]) ⇒ ¬ p Exercise S9. ( p ⇒ q) ⇒ ( p ∨ r ⇒ q ∨ r ) Exercise S10. ( p ⇒ q) ⇒ ([q ⇒ r ] ⇒ [ p ⇒ r ]) Exercise S11. ( p ⇔ q) ∧ (q ⇔ r ) ⇒ ( p ⇔ r ) Exercise S12. ( p ⇔ q) ⇔ (q ⇔ p)

31

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Chapter 2. Statements in Mathematics

Exercise S13. ( p ⇒ q) ∧ (r ⇒ q) ⇔ ( p ∨ r ⇒ q) Exercise S14. ( p ⇒ q) ∧ ( p ⇒ r ) ⇔ ( p ⇒ q ∧ r ) Exercise S15. ( p ∨ q) ⇔ (q ∨ p) Exercise S16. ( p ∧ q) ⇔ (q ∧ p) Exercise S17. ( p ∨ q) ∨ r ⇔ p ∨ (q ∨ p) Exercise S18. ( p ∧ q) ∧ r ⇔ p ∧ (q ∧ p) Exercise S19. p ∨ (q ∧ r ) ⇔ ( p ∨ q) ∧ ( p ∨ r ) Exercise S20. p ∧ (q ∨ r ) ⇔ ( p ∧ q) ∨ ( p ∧ r ) Exercise S21. p ∨ p ⇔ p Exercise S22. p ∧ p ⇔ p Exercise S23. ¬( p ∨ q) ⇔ ¬ p ∧ ¬q Exercise S24. ¬( p ∧ q) ⇔ ¬ p ∨ ¬q Exercise S25. p ⇒ q ⇔ ¬ p ∨ q Exercise S26. p ⇒ q ⇔ ¬( p ∧ ¬q) Exercise S27. p ∨ q ⇔ ¬ p ⇒ q Exercise S28. p ∨ q ⇔ ¬(¬ p ∧ ¬q) Exercise S29. p ∧ q ⇔ ¬( p ⇒ ¬q) Exercise S30. p ∧ q ⇔ ¬(¬ p ∨ ¬q) Exercise S31. ( p ⇔ q) ⇔ ( p ⇒ q) ∧ (q ⇒ p) In Exercises S32—S35, let F and T represent statements that are always false or true, respectively. Exercise S32. p ∨ F ⇔ p Exercise S33. p ∧ F ⇔ F Exercise S34. p ∨ T ⇔ T Exercise S35. p ∧ T ⇔ p Exercise S36. Which of the following statements are true and which are false? Explain. (a) For all x, √ (x + 3)2 + (x − 2)2 > 0. (b) For all x, x 2 = x. (c) For all x and y, |x − y| = |y − x|. Exercise S37. Show that q ∧ ( p ⇒ q) ⇒ p is not a tautology. This is sometimes referred to as the fallacy of asserting the conclusion.

33

Chapter 3 Proofs in Mathematics 1. What is Mathematics? Using the ideas of the previous section, we are now in a position where we can understand a given statement. The next step would be to prove or disprove that statement. Is this mathematics? Well, not quite. Mathematics is, foremost, a creative activity. Theorems are not handed out by God (or professors) to the faithful (students?) with the expectation that the supplicants (students??) will supply the appropriate proof before awaiting the next theorem from on high.* Theorems are a product of human endeavor, dependent on the human experience and knowledge. To do mathematics is to create mathematics, theorems along with proofs. Yes, perhaps the scholastic experience is slightly different. Most professors and their texts play the roles of gods and kings, expecting their slaves (students!) to do the grungy work. Before you proceed you should give the preceding some thought. This text deals with finding proofs, not theorems. From elementary school through the master’s degree, students of mathematics rarely encounter the other side, the creation of the theorems themselves. Finding theorems is the primary goal of a doctoral dissertation and of mathematical research. Most times you are given a proof immediately after the theorem. Your job is then to comprehend both the theorem and its proof. In order to learn how to do proofs, you should think about how you would prove the theorem first. To start thinking like a mathematician you must read slowly, pausing to think often. If you pause now and then to ask the question “What can I try to prove next?” you will be taking a large step towards becoming a mathematician and you will be better able to construct proofs. Two things that you should do frequently, which are helpful both in finding theorems and proofs (not surprising since they go hand in hand), are to think of examples and draw pictures. In fact, these are not unrelated activities. Often, working out an example will help you to better understand *Some philosophers certainly disagree with this point of view.

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Chapter 3. Proofs in Mathematics

what makes a theorem true, that is, how to prove it. Pictures are usually an extension of that process. Just as important is to change the hypotheses and consider examples where the conclusion fails. These “counterexamples” again can help you to isolate the missing ideas which make up the proof. You may be thinking that you have seen lots of pictures in mathematics texts that you did not find the least bit illuminating. That is quite understandable. The important part of the picture is usually in its construction, not in its final presentation. It is like looking up the answer to an exercise in the back of a text. Usually, the answer alone does not illuminate the path from the problem to the answer. If you get nothing else out of this chapter, we hope you will understand the following two things. First, you must spend far more time thinking than reading. (It is not uncommon for a mathematician to take hours, even days, to read and comprehend a single page!) Second, you should always think of examples (and counterexamples), drawing pictures as a guide. (Yes, you can draw mental images, but it is surprising how different things can look on paper.) In the remainder of this chapter, we will examine various techniques of proof. 2. Direct Proof. The first type of proof we will consider is the direct proof. When someone says that a proof is “by brute force” or by “following your nose,” they are referring to a direct proof. While “brute force” generally implies some sort of explicit computation, all of these descriptions describe the basic nature of a direct proof. Many mathematicians will say that they “construct” a proof. We warn you that this na¨ıve use of the word “construct” may be misunderstood.* The basic idea of direct proof is simple: start from the hypotheses, make a sequence of implications, and arrive at the conclusion. An Example of Direct Proof Consider the following proposition from geometry. You may refer back to Section 3 of Chapter 1 for our discussion of the axioms of Euclid; the axiom numbers used below refer to the list of axioms in that section. Proposition 1. If two lines intersect, then they make the vertical angles equal to one another. Before writing down a proof, let us start with a picture. Consider Figure 2, which shows two intersecting lines. *“Constructive proof” may refer to the constructivist philosophy of mathematics. Not everyone subscribes to this philosophy. In this philosophy, the existence of a mathematical object is accepted only when it can be constructed (for instance, only when an explicit example is known).

Section 2. Direct Proof

C

B A

D

E

Figure 2: Vertical angles If nothing else, the figure gives us a shorthand notation for our proof by giving names to the angles. By the way, if you do not remember the definition of vertical angles, you should be able to figure it out from the proposition and the figure! Proof. Let B AD and C AE be two lines intersecting at the point A. Then the angles B AD and C AE are both equal to two right angles.† If we subtract from both angles the angle C AD, by Euclid’s Axiom 4, we conclude that B AC = D AE, which is what we wanted to prove. Using the same line of reasoning, we can prove that C AD = E AB.
Exercise 3. Complete the proof of Proposition 1: prove that C AD = E AB. If you prefer to think about the measures of angles (perhaps following Birkhoff’s axioms in Example 6 of Chapter 1), then you might rewrite the above proof as follows: Proof. Let B AD and C AE be two lines intersecting at the point A. Then mB AD = π

and

mC AE = π.

Therefore, mB AD = mC AE and mB AD − mC AD = mC AE − mC AD. Since mB AC = mB AD − mC AD and mD AE = mC AE − mC AD, †By Axiom 5 and the familiar construction of perpendicular lines, the angle at a point on a line is equal to two right angles. In more modern language, we can say these angles measure π or 180◦ .

35

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Chapter 3. Proofs in Mathematics

we have proved that mB AC = mD AE. Using the same line of reasoning, he reader can prove that mC AD = mE AB.
Proof by Brute Force: Another Example of Direct Proof Let A = { 1, 3, 5, 7, . . . } be the set of odd natural numbers. We know, by definition, that every odd natural number can be expressed in the form 2n + 1 for some nonnegative integer n. Proposition 4. The square of any odd integer is odd. Proof. Consider the odd natural number x = 2n + 1 for some nonnegative integer n. Then x 2 = (2n + 1)2 = 4n 2 + 4n + 1 = 2(2n 2 + 2n) + 1 = 2k + 1, where k is the natural number 2n 2 + 2n. Hence, x 2 is odd.




Exercise 5. Prove that the square of an even integer is even. Exercise 6. (a) Prove that the sum of two even integers is even. (b) Prove that the sum of an even integer and an odd integer is odd. Exercise 7. Prove that the cube of an odd integer is odd. Exercise 8. Prove that the sum, the difference, the product, and the quotient (when it is defined) of two rational numbers is a rational number. Proof by cases. In proving a proposition, we sometimes have to distinguish cases. The following example is suggestive. Example 9. To prove that (−1)n + (−1)n+1 = 0 for all integers n, we distinguish two cases: n is either even or odd. Case 1: suppose n is even. Then n + 1 is odd. Hence, (−1)n = 1 and (−1)n+1 = −1. Therefore, the sum equals 0. Case 2: suppose n is odd. Then n + 1 is even. Hence, (−1)n = −1 and (−1)n+1 = 1. Therefore, the sum equals 0. We give another proof of (−1)n + (−1)n+1 = 0 for all integers n: (−1)n + (−1)n+1 = (−1)n + (−1)n · (−1) = (−1)n · (1 + (−1)) = 0. The moral of this example is that there are generally many ways to prove a theorem. The usual problem is to find some proof. Some mathematicians

Section 3. Contraposition and Proof by Contradiction

like to find “the best” proof of every theorem; of course, this is subjective. The twentieth century mathematician P. Erd¨os liked to say that God has “The Book,” which contains all of the theorems, along with their best proofs. Recall the definition of the absolute √ value, |x|, of a real number, x. Of course, we can define |x| to equal x 2 . However, the first definition you saw was probably something like  x , if x ≥ 0 |x| = −x , if x < 0 where −x plays the role of “dropping the negative sign” when x is negative. Since this is a definition by cases, proofs of statements involving absolute values often consider cases. Example 10. Let us prove that |x| ≥ x for all x ∈ R. We consider two cases according to the definition of absolute value. If x ≥ 0, then |x| = x by definition; hence |x| ≥ x. On the other hand, if x < 0, then |x| = −x is positive; hence |x| ≥ x. Exercise 11. Prove that for all x ∈ R, |2x + 1| > x. We would like to point out that “proof by cases” is not a proof technique at all. Instead, “proof by cases” is a method of reducing a problem to easier, smaller problems. In fact, the methods of proof used to prove each of the individual cases may be different. Exercise 12. Prove that, for all real numbers a, if a = 0, then a 2 > 0. 3. Contraposition and Proof by Contradiction. We noted that a statement is equivalent to its contrapositive. Therefore, to prove a statement, we can prove its contrapositive. We call this proof by contraposition. An Example of Proof by Contraposition Consider the converse of Exercise 5. Proposition 13. If the square of an integer is even, then the integer is even. Proof. By contraposition, this statement is equivalent to saying “If an integer is odd, then its square is odd,” which is Proposition 4.
The following exercises are examples of statements whose proofs are most easily done by contraposition. Exercise 14. Prove that if

2n 1+n 2

is irrational, then n is irrational.

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Chapter 3. Proofs in Mathematics

Exercise 15. Prove that, for all real numbers a, if a 2 > 0, then a = 0. Look at Exercise 7. The converse can be proved by contraposition. Exercise 16. Prove that the cube of an integer n is odd if and only if n is odd. A similar idea to contraposition is proof by contradiction. In proof by contraposition, we used the fact that an implication and its contrapositive are logically equivalent. Mechanically, this amounts to the following: to prove that p implies q, we assume that the negation of q is true (i.e., q is false) and proceed o show that the negation of p is true (i.e., p is false). Seen as a direct proof, the hypothesis p has been contradicted since both p and its negation are apparantly true. We use this idea to generalize this proof technique. In proving an implication, p implies q, by contradiction, if we first suppose the hypotheses of a statement, p, and we assume the negation of its conclusion, q, and use these to derive a contradiction (to a hypothesis or to our assumption or to any other true statement), then we have proved the original implication. Note that the case of deriving the negation of the hypotheses is exactly proof by contraposition. To understand that this makes sense, consider that a contradiction should never arise logically unless there is a false assumption and the only assumption we made was the negation of the conclusion of our implication. Two Examples of Proof by Contradiction Let us consider two examples of proof by contradiction. First, Proposition 17 shows another way to approach Exercise 6(b). Proposition 17. The sum of an odd integer and an even integer is odd. Proof. Let x be odd and y be an even integer. Assume that x + y is not odd, that is, x + y is even. Then x = 2n + 1 for some integer n, y = 2m, for some integer m and x + y = 2 p for some integer p. Hence, x + y = 2n + 1 + 2m = 2 p. This is equivalent to 2( p − m − n) = 1. But this is a contradiction since the number on the left-hand side is even, but 1 is odd. Therefore, x + y is odd.
This proof would not make Erd¨os’ The Book (see Example 9). The direct proof is shorter and neater. The second example of a proof by contradiction is given in Theorem 18. This is a famous theorem. √ Theorem 18. 2 is irrational. Before giving a proof of this proposition, you might wonder how we √ know that there exists a real number called 2. This is actually proved by

Section 3. Contraposition and Proof by Contradiction

constructing the set of real numbers in Part III. The following proof shows that this real number is indeed irrational. √ Proof. Assume that 2 is rational. Then there exist natural numbers m and n such that √ m 2= . n Without loss of generality, we assume that the fraction mn is reduced to lowest terms; hence, m and n are not both even. Squaring both sides we get m2 = 2. n2 This is equivalent to the equality 2n 2 = m 2 . Since the left-hand side is an even number, m 2 is even. By Proposition 13, m is even. This means that m = 2k for some natural number k. Hence 2n 2 = (2k)2 = 4k 2 , which implies that n 2 = 2k 2 . a contradiction since we Therefore, n 2 is even and so n is even. This is √ assumed that m and n are not both even. Hence, 2 is irrational.
Remark 19. The following remarks concern the style of writing proofs by contradiction. (1) Many mathematicians start their proofs by indicating how they plan to proceed. A good way to start a proof by contradiction is a statement such as “This proof is by contradiction.” (2) Although the words “assume” and “suppose” are synonyms, you may find it helpful to use “assume” when trying to construct a proof by contradiction and “suppose” for all hypotheses. In this way the word “assume” is a reminder that you are trying to find a contradiction. We are divided on this advice; one of us tries to follow it regularly while the other is indifferent. (3) It is common to leave out the punch line. What is the punch line? It is the declaration of a contradiction and the negation of the original assumption. Sometimes such proofs end with the simple declarative statement

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Chapter 3. Proofs in Mathematics

“Contradiction.” This may leave the reader with a little bit to think about! For example, a much terser proof of Proposition 17 may look as follows. Proof. Suppose x = 2n+1 is odd and y = 2m is even. Assume x +y = 2 p is even. Hence, x + y = 2(n + m) + 1 = 2 p. Contradiction.
We strongly suggest that you write as much as is necessary so that your fellow students could understand your proofs. We will need the following important theorem to generalize Proposition 13 and Theorem 18; its proof is postponed until the Exercise 42. Recall that a integer p is prime iff p is neither 0 nor ±1, and its only divisors are ±1 and ± p. Theorem 20 (Fundamental Theorem of Arithmetic). Each natural number other than 1 is the product of a uniquely determined finite collection of positive primes. For example, 12 = (2)(2)(3) = (2)(3)(2) = (3)(2)(2); the factorization is unique in that each includes two 2’s and one 3. We think of 2 as a “product” of only one prime, namely 2. That is, every integer m greater than 1 can be written uniquely in the form m = p1n 1 p2n 2 · · · pkn k , where p1 < p2 < · · · < pk are k distinct positive prime numbers and n 1 , n 2 , . . . , n k are natural numbers. Consider the following generalization of Proposition 13; they are the same when n = 2 and p = 2. Theorem 21. Let n be a natural number, m be an integer greater than 1, and p be a positive prime number. If the m n is divible by p, then m is also divisible by p. Proof. By the Fundamental Theorem of Arithmetic, there exists a factorization m = p1n 1 p2n 2 · · · pkn k as a product of positive prime numbers where p1 < p2 < · · · < pk . By the definition of prime, if m is divisible by p, then there exists j (between 1 and k inclusive) such that p j is divisible by p and, hence, p = p j . Now m n = ( p1n 1 p2n 2 · · · pkn k )n = p1n 1 n p2n 2 n · · · pkn k n .

Section 4. Proof by Induction

By the Fundamental Theorem of Arithmetic, the uniqueness of the factor
ization of m n guarantees that m n is divisible by p = pk . Use these ideas for the following exercise. Exercise 22. Prove that

√ √ 3 and 5 are irrational.

The following exercises are two more examples of statements whose proofs are best done by contradiction (or contraposition). Exercise 23. Let a and b be real numbers. Prove that a 2 + b2 = 0 if and only if a = b = 0. Exercise 24. Let x be a real number. Prove that if x 2 < x, then x < 1. The next exercise is a type of problem in combinatorics. Be careful with counting problems as they are notoriously difficult. Exercise 25. Prove that if at least two people are present at a party, at least two of them know the same number of participants. (Assume that either everyone or no one counts themselves as someone they know and that if one person knows another, then the latter also knows the former.) 4. Proof by Induction. Before we discuss the next example we need some preliminaries. Suppose we want to prove that n 2 ≥ n for all natural numbers n. In fact, we have to prove infinitely many statements: 12 ≥ 1 22 ≥ 2 32 ≥ 3 .. . We can denote these statements by P(1), P(2), P(3), etc. The following tells us one way to prove such a sequence of statements. Theorem 26 (Principle of Induction). Statements P(n) are true for all n ∈ N if (1) the first statement, P(1), is true and (2) whenever a statement in the sequence, P(n), is true, then the succeeding statement, P(n + 1), is also true.

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Chapter 3. Proofs in Mathematics

Notice that we called the preceding a theorem. In fact, it is very close to being an axiom, usually called the Axiom of Infinity. We state this axiom below in a somewhat different (also more readable and less precise) form than you would encounter in an axiomatic approach to set theory. Axiom 27 (Axiom of Infinity). There exists a set S of natural numbers such that (1) 1 ∈ S and (2) if n ∈ S, then n + 1 ∈ S. If we have a sequence of statements P(1), P(2), P(3), P(4), . . . , then we can define a set S = { n | P(n) is true } . This axiom characterizes the set of natural numbers: S = N. For our purposes, we may as well assume that the Principle of Induction is itself an axiom. Let us see how this gives us an important proof technique. The second step in the Principle of Induction can be restated as P(n) implies P(n + 1) for all n ∈ N. This step is called the inductive step and P(n) is called the inductive hypothesis (or induction hypothesis). Sometimes, it is more advantageous to use the implication P(n − 1) implies P(n); in this case, P(n − 1) is called the inductive hypothesis. The idea is actually quite natural: if P(1) is true and if P(n) → P(n + 1) for all n ∈ N, then P(1) is true → P(2) is true → P(3) is true → P(4) is true → . . . . Remark 28. We like to think of a proof by induction as building a machine: you need an “ON” switch to get it going and then the machine needs to crank out the desired product. The switch is part (1) above and the machine mechanism, or crank, is part (2). We are now ready to apply the Principle of Induction to prove the assertion that n 2 ≥ n, for all n ∈ N. Actually we will consider three examples of proof by induction in Propositions 29, 30 and 31. Proposition 29. n 2 ≥ n for all n ∈ N. Proof. The proof uses the Principle of Induction. Let P(n) denote the statement “n 2 ≥ n.” We first prove P(1). For n = 1, we have 12 ≥ 1, which is obvious. Next we suppose that P(n) is true for some fixed n, that is, n 2 ≥ n; this is the inductive hypothesis. Next we prove that P(n + 1) is true, that is, (n + 1)2 ≥ n + 1. We compute (n + 1)2 = n 2 + 2n + 1 and, by the

43

Section 4. Proof by Induction

inductive hypothesis, n 2 + 2n + 1 ≥ n + 2n + 1 ≥ n + 1 since 2n ≥ 0.*
Therefore, (n + 1)2 ≥ n + 1. You should take a close look at the proof of Proposition 29. Do you see that the work in the second paragraph is necessary? Many students who are seeing induction proofs for the first time want to simply assert the truth of P(n + 1) by substituting n + 1 into the inductive hypothesis. Of course, this is wrong. Proposition 30. 1 + 2 + 3 + · · · + n =

n(n+1) 2

for all n ∈ N.

You may already be familiar with summation notation. In general, b

f (k)

k=a

indicates a sum of terms of the form f (k), for some function f , where we substitute sequential values from a to b in place of k. The statement in Proposition 30 can be rewritten as n k=1

k=

n(n + 1) . 2

Proof of Proposition 30. The proof is by induction on n. First we prove the statement for n = 1: we have 1 = 1·(1+1) which is obviously true. 2 . Add n + 1 to Next, we suppose that is 1 + 2 + 3 + · · · + n = n(n+1) 2 both sides of our inductive hypothesis and simplify the right-hand side to get n(n + 1) + (n + 1) 2 n(n + 1) + 2(n + 1) = 2 (n + 1)(n + 2) . = 2

1 + 2 + 3 + · · · + n + (n + 1) =




The above proof would probably not make Erd¨os’ book. Never be afraid to try different techniques in proving a result. The following proof *In fact, 2n ≥ 2 since n ≥ 1, but we do not need this in the proof. Of course, we could say that n 2 > n for n > 1. Since 12 = 1, this would slightly improve the statement of Proposition 29.

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Chapter 3. Proofs in Mathematics

of Proposition 30 seems more intuitive. It often seems that inducive proofs obscure the motivation of mathematical statements. Another Proof of Proposition 30. Let x = 1 + 2 + 3 + · · · + n. Then x = n + (n − 1) + (n − 2) + · · · + 1 also. Therefore, 2x = x + x = (1 + 2 + 3 + · · · + n) + (n + (n − 1) + (n − 2) + · · · + 1) = (1 + n) + (2 + n − 1) + (3 + n − 2) + · · · + (1 + n) = n(n + 1). Hence, x=

n(n + 1) . 2




Now there is actually nothing special about starting an induction proof from n = 1; an induction can start with any integer and continue with all integers greater than the original integer. (Even wilder possibilities exist!) To understand this you should realize that it is just a matter of relabeling which integer is “the first.” Consider the following proposition. n

1 − r n+1 for n = 0, 1, 2, . . . and r = 1. [For 1−r k=0 r = 0, use the convention r 0 = 1.] Proposition 31.

rk =

Proof. The proof is by induction. Let r be a fixed real number different n

1 − r n+1 . rk = from 1. For n = 0, 1, 2, . . . , let P(n) be the statement 1−r k=0 0

We want to prove P(0); that is, we start with n = 0. rk = r0 = k=0

1−r 1 − r 0+1 1= = . 1−r 1−r Next, we want to show, for every n = 0, 1, 2, . . . , that P(n) implies n

P(n + 1). Let n be a fixed nonnegative integer. Suppose that rk = k=0

1 − r n+1 ; this is the inductive hypothesis. We must show that the inductive 1−r n+1

k 1 − r (n+1)+1 1 − r n+2 hypothesis implies r = = . 1−r 1−r k=0  n n+1

k

k 1 − r n+1 Now r = r + r n+1 = + r n+1 by the inductive 1−r k=0 k=0 1 − r n+1 1 − r n+1 + (1 − r )r n+1 hypothesis. Moreover, + r n+1 = = 1−r 1−r

45

Section 4. Proof by Induction

1 − r n+2 1 − r n+1 + r n+1 − r n+2 = . This is what we needed to show and 1−r 1−r completes the proof.
Remark 32. The following remarks concern the style of writing proofs by induction. (1) Since a proof by induction has two steps, two paragraphs make it more readable. (2) Usually, the first step is much easier than the inductive step. (3) Many authors like to use a new variable when considering the inductive step in the proof. For example, for the inductive step in the proof of n 2 ≥ n for all n ∈ N they will use k 2 ≥ k to show that (k + 1)2 ≥ k + 1. We are not convinced that this makes proofs by induction any easier to understand. (4) During a lecture, a mathematician may give a shorthand proof which looks like the following. Short proof of Proposition 29. n = 1: 12 ≥ 1. n + 1: (n + 1)2 = n 2 + 2n + 1 ≥ n + 2n + 1 ≥ n + 1.




Short proof of Proposition 30. n=1: 1 = n + 1: (1 + 2 + 3 + · · · + n) + (n + 1) (n+1)(n+2) . 2

1·(1+1) . 2 n(n+1) = 2 + (n + 1)

=

n 2 +3n+2 2

=


0

r k = r 0 = 1 = 1−r . Short proof of Proposition 31. n=0: 1−r k=0  n n+1

k

k n+1 n+1 )r n+1 r = r +r n+1 = 1−r +r n+1 = 1−r +(1−r = n+1: 1−r 1−r k=0

k=0

1−r n+2 . 1−r




Can you follow the three proofs above? Do you find the original, wordy proofs more readable? Remember that written words do help the reader to understand your proofs. We do not support shorthand for its own sake, but it is acceptable as long as there is no loss of understanding such as in a lecture where the writer is there to answer questions! The following statement and proof are wrong. [Recall that, for any natural number n, n factorial is the number n! = n(n − 1)(n − 2) · · · (1).] Absurdum. 2n < n! for all n ∈ N. Proof? The proof is by mathematical induction. Suppose that 2k < k! is true; this is our inductive hypothesis. We need to show this implies that 2k+1 < (k + 1)!. We see that (k + 1)! = (k + 1)(k!) > (k + 1)(2k ) ≥ (1 + 1)(2k ) = (2)(2k ) = 2k+1 .


46

Chapter 3. Proofs in Mathematics

Exercise 33. (a) Explain why the statement above is absurd and why its proof is wrong. (b) Prove that 2n < n! for all n = 4, 5, 6, 7, 8, . . . . Once more, the following statement and proof are wrong. Absurdum. For n = 1, 2, 3, . . . , every bag containing n solid-colored balls contains balls of only one color. Proof? n=1: A bag with one ball clearly has balls of only one color. n+1: Take a bag with n + 1 balls. Remove one ball. By hypothesis, the bag now has balls of only one color. Replace the first ball and remove a different ball. Again, the bag now has balls of only one color. Since the other balls which never left the bag did not change color, all the balls must be the same color.
Exercise 34. Explain why the statement above is absurd and why its proof is wrong. Now it’s time for you to try some proofs by induction. Exercise 35. Prove that 1 + 3 + 5 + · · · + (2n − 1) = n 2 for all n ∈ N.

Exercise 36. Prove that, for all natural numbers n,

n k=1

Exercise 37. Prove that, for all natural numbers n,

n

n 1 = . k(k + 1) n+1

k2 =

n(n+1)(2n+1) . 6

k=1

In the following exercise, there are two variables, p and n. Remember that induction is on one variable which varies through integer values. Exercise 38. Prove that for every real number p > −1 and any natural number n, (1 + p)n ≥ 1 + np. We now state the Principle of Complete Induction, which is an equivalent version of the Principle of Induction. Theorem 39 (Principle of Complete Induction). Statements P(n) are true for all n ∈ N if (1) P(1) is true and (2) for every n ∈ N if P(k) is true for k = 1, 2, 3, . . . , n, then P(n + 1) is also true.

Section 4. Proof by Induction

The next result is quite amazing at first glance; we prove it by complete induction. The Fibonacci sequence is defined inductively as follows: set  an =

1,

if n = 1, 2

an−1 + an−2 ,

if n = 3, 4, 5, . . .

That is, the Fibonacci sequence is { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, . . . }. You probably do not see a formula to determine the thousandth term in this sequence as you would for an arithmetic or geometric sequence. However, there is the following formula. Proposition 40. The n th term of the Fibonacci sequence is

1+

an =

√ n √ n 5 − 1− 5 . √ 2n 5

What makes this statement amazing is the appearance of irrational numbers—square roots of 5—in this sequence of sums of pairs of natural numbers! The number √ 1+ 5 2 is a very special number called the golden ratio. It has interesting properties √ and comes up in surprising situations, like this one. The number 1−2 5 is referred to as the conjugate of the golden ratio. Before giving the proof, let us consider what must be done. It is a simple computation to check the formula for a1 = 1. Consider the inductive step using complete induction: we want to use the inductive hypothesis

ak =

1+

√ k √ k 5 − 1− 5 , √ 2k 5

for all k = 1, 2, 3, . . . , n − 1, to prove that

an =

1+

√ n √ n 5 − 1− 5 . √ 2n 5

We know that an = an−1 + an−2 .

47

48

Chapter 3. Proofs in Mathematics

By the inductive hypothesis for k = n − 1 and k = n − 2, we have

an =

1+

√ n−2 √ n−1 √ n−2 √ n−1 1+ 5 5 − 1− 5 − 1− 5 + . √ √ 2n−1 5 2n−2 5

So we wish to verify that

1+

√ n √ n √ n−1 √ n−1 5 − 1− 5 5 − 1 − 5 1 + ? = √ √ 2n 5 2n−1 5

√ n−2 √ n−2 1+ 5 − 1− 5 . + √ 2n−2 5

Using algebra, you might derive the following equivalent equations:   √ n √ n ? √ n−1 √ n−1 1+ 5 − 1− 5 =2 1+ 5 − 1− 5   √ n−2 √ n−2 − 1− 5 +4 1+ 5





 √ n √ n−1 √ n−2 1+ 5 −2 1+ 5 −4 1+ 5

√ n √ n−1 √ n−2 ? = 1− 5 −2 1− 5 −4 1− 5  

√ n−2 √ 2 √  1+ 5 1+ 5 −2 1+ 5 −4  

√ 2 √ n−2 √  ? 1− 5 −2 1− 5 −4 = 1− 5

 √ n−2  √ √ 1+ 5 1+2 5+5−2−2 5−4

 √ √ n−2  √ ? 1−2 5+5−2+2 5−4 = 1− 5

(1 +

√ √ n−2 ? 5) [0] = (1 − 5)n−2 [0] ?

0=0 You know that since the last of these equations is true, all of the previous, equivalent equations are true also. Of course, you never see a computation with question marks like this in a textbook and you should be careful in doing so even as a side computation. It is important that every step

Section 4. Proof by Induction

is reversible! You can rewrite this computation in a better form, going forwards rather than backwards. We will give a less messy proof below. Does this complete the proof? No, since our proposed inductive step never settles the case of n = 2. This is because we used the fact that an = an−1 + an−2 which is only valid for n ≥ 3. The proof is completed by checking this case. Let us now rewrite the proof, following the ideas just discussed. Before doing so, let us consider the golden ratio and its conjugate again. Note that these are exactly the two zeros of the quadratic polynomial x 2 − x − 1. Let √ 1+ 5 ϕ= 2

and

√ 1− 5 ϕˆ = . 2

So ϕ 2 − ϕ − 1 = 0 and ϕˆ 2 − ϕˆ − 1 = 0. That is, ϕ2 = ϕ + 1

and

ϕˆ 2 = ϕˆ + 1.

Proof of Proposition 40. We wish to show that

an =

1+

√ n √ n 5 − 1− 5  1  = √ ϕ n − ϕˆ n . √ 2n 5 5

For n = 1 and n = 2, we compute

 1  √ ϕ 1 − ϕˆ 1 = 5

1+

√  √  5 − 1− 5 = 1 = a1 √ 2 5

and    1  1  1  √ ϕ 2 − ϕˆ 2 = √ (ϕ + 1) − (ϕˆ + 1) = √ ϕ − ϕˆ = 1 = a2 . 5 5 5 For an integer n ≥ 3, suppose that the induction hypothesis that

ak =

1+

√ k √ k 5 − 1− 5  1  = √ ϕ k − ϕˆ k √ 2k 5 5

49

50

Chapter 3. Proofs in Mathematics

holds for all k = 1, 2, 3, . . . , n − 1. By the definition of the Fibonacci sequence and the induction hypotheses for k = n − 1 and k = n − 2,   1  1  an = an−1 + an−2 = √ ϕ n−1 − ϕˆ n−1 + √ ϕ n−2 − ϕˆ n−2 . 5 5 To show that an =

√1 5



 ϕ n − ϕˆ n , we need to show that

   1  1  1  √ ϕ n − ϕˆ n = √ ϕ n−1 − ϕˆ n−1 + √ ϕ n−2 − ϕˆ n−2 . 5 5 5 We compute that (ϕ n−1 − ϕˆ n−1 ) + (ϕ n−2 − ϕˆ n−2 ) = ϕ n−1 (ϕ + 1) − ϕˆ n−1 (ϕˆ + 1) = ϕ n−1 (ϕ 2 ) − ϕˆ n−1 (ϕˆ 2 ) = ϕ n − ϕˆ n . Dividing both sides by

√

5 yields the desired result.




Let’s look at two erroneous attempts at using complete induction. Absurdum. For all n ∈ Z+ , n = 0. Proof? The proof is by complete induction. Let P(n) be the predicate n = 0. Since the least number in Z+ is 0, the first step is for n = 0. This already proves the first step. For the inductive step, we hypothesize that, for k = 0, 1, 2, . . . , n, P(k) is true. That is, for k = 0, 1, 2, . . . , n, k = 0. We want to use this to show that n + 1 = 0. Since, by hypothesis, 1 = 0 and n = 0, then n + 1 = 0 + 0 = 0. This proves the inductive step and completes the proof of the statement.
Of course, there is a problem in the proof, but where is it? You may notice that the hypothesis 1 = 0 is nonsense. However, that alone does not explain how the logic of the proof is faulty. You may also notice that we proved only one starting step (n = 0), but use two hypothses (1 = 0 and n = 0) in proving the inductive step. This brings us closer to the error. The very first application of the inductive step is when n + 1 = 1 or, equivalently, when n = 0. The induction hypothesis is k = 0 for k = 0, 1, 2, . . . , n, which is just 0 = 0 when n = 0. That is, 1 = 0 is not part of the induction hypothesis when n = 0. The next proof attempt, in short-hand form, suffers a similar fate.

Section 5. Examples and Counterexamples

Absurdum. For all n ∈ Z+ , 10n = 1. Proof? n=0: 100 = 1. n+1: 10n+1 = 102n−(n−1) = 10n = 10n−1 = 1.

102n (10n )2 12 = 1 since = = 10n−1 10n−1 1


Exercise 41. Explain why the statement above is absurd and why its proof is wrong. You can use the Principle of Complete Induction to prove the following, which is the existence part of Theorem 20, the Fundamental Theorem of Arithmetic. Exercise 42. Prove that every integer n greater than 1 equals a product of one or more positive prime numbers. The uniqueness part of the Fundamental Theorem of Arithmetic is not so easy to write down. Think about why it is true and compare your thoughts with the following. Proof of the Fundamental Theorem of Arithmetic. In Exercise 42, we proved the existence of such a factorizaion. Now, suppose there are two distinct postive prime factorizations of an integer n with n ≥ 2: n = p1 p2 p3 · · · pa = q1 q2 q3 · · · qb ; that is, n is factored into a primes on one hand and b primes on the other. Since p1 is a divisor of n = q1 q2 · · · qb , p1 must be a divisor of a prime q j ; in fact, p1 = q j . Next rearrange the factorization into qk ’s as q1 q2 · · · qb with q1 = q j . Cancelling p1 = q1 yields p2 p3 · · · pa = q2 q3 · · · qb . Continue to cancel primes in this way until one side of the equation equals 1. Since 1 is neither prime nor the product of primes greater than 1, both sides must equal 1. Therefore, a = b. This shows that the factorization is unique up to the order of the factors. The Fundamental Theorem of Arithmetic is proved.
5. Examples and Counterexamples. Thus far we have considered methods of proof. We will now consider a method for disproving statements. Let us consider the following claim: There do not exist integers a, b, and c, such that a 2 + b2 = c2 .

51

52

Chapter 3. Proofs in Mathematics

Obviously this claim is false, since for a = 3, b = 4, and c = 5 the above relation holds. In this case, we say that we have provided a counterexample. Exercise 43. Find a counterexample to the assertion that the absolute value of every real number is positive. Exercise 44. Find a counterexample to the assertion that every continuous function is differentiable. As we mentioned in Section 1, we consider examples and counterexamples to be extremely important. You should always consider them; often they will help you to find a proof of a statement. You should keep in mind that you can never prove or disprove a statement simply by giving an example unless • you examine all possible examples (this is a proof by cases), or • the statement is an existence theorem, or • the statement is false in at least one case (this is a counterexample). A counterexample to a statement is also a proof of the negation of the statement. Remember that the negation of a statement of the form “for all x, P(x)” is a statement of the form “there exists an x such that not P(x).” So a counterexample for such a statement gives an example that proves a corresponding existence theorem. For example, consider the following existence theorem. Proposition 45. There exists a continuous function f which is not differentiable. Proof. Your counterexample in Exercise 44 proves this proposition.




Supplemental Exercises

Supplemental Exercises Exercise S1. Prove that 2n + 2 ≤ 2n+1 for all n ∈ N. Use this result to prove that 2n + 1 ≤ 2n for n = 3, 4, 5, . . . .

53

54 How to THINK about mathematics: A Summary • Before trying to prove a statement: • Construct examples (under the same and modified hypotheses) • Construct counterexamples (under the same and modified hypotheses) • Construct pictures • Consider different proof techniques: • Direct Proof (including Brute Force) • Proof by Contraposition (or proof by proof of the contrapositive) • Proof by Contradiction • Proof by Induction (including Complete Induction) • Consider proof by cases and breaking a proof into smaller pieces

How to COMMUNICATE mathematics: A Summary • Discuss mathematics with people—all parties should be honest and active • Organize your ideas (rewrite your proofs several times, if necessary) • Remember to use enough words (a reader of your written proofs cannot ask you questions) • Use notation only as a shorthand or to clarify your proofs

55

PART II SET THEORY

56

Chapter 4 Basic Set Operations 1. Introduction. Perhaps our ability to distinguish and to classify objects constitutes the essential part of our intellectual armor as human beings. These objects may be tangible—inanimate objects, distant galaxies, living beings, our beloved persons—or they may even be ideas, feelings or figments of our imagination. Very often we view a certain collection of these objects as a unity. When Aristotle claimed that man is a rational animal, he meant to say that all human beings are rational. In fact, of all the animals, he isolated those, and only those, who shared a common characteristic. Of course, categorization is not always an easy process. Very often, it stumbles over insurmountable ambiguities, especially due to the fact that universal agreement on the meaning of terms is not always achieved. If we restrict reality to the mathematical realm, it turns out that we can come up with an elegant and powerful theory. The key concept is that of a set. To cite Cantor’s so-called na¨ıve definition, a set is any collection of definite, distinguishable objects of our intuition or of our intellect to be conceived as a whole. You are probably already familiar with the notation used to describe sets. We may list the elements of sets explicitly as in T = { earth, wind, fire, water }

and

E = { 2, 4, 6, . . . }.

The set T has exactly four elements. E, the set of all even natural numbers, has infinitely many; the three dots indicate that the list continues indefinitely in the same pattern. We may specify E in different ways as follows: E = { n ∈ N | 2 divides n } = { 2n | n ∈ N } . This kind of notation is called set-builder notation. It is common to read this notation by substituting “such that” for the vertical bar. Some authors

Section 2. Subsets

prefer to use a colon in place of the vertical bar. Notice that the set N appears on different sides of the vertical bar. We must specify a set for every variable. 2. Subsets. To study set theory, as with any new mathematics, we start with some undefined terms and some definitions. Undefinition* 1. The terms set, element, and is an element of (represented by the symbol ∈) are taken as undefined. For example, the notation x ∈ S means that x is an element of a set, S. We will be sloppy with this notation when we write something like x, y ∈ S to mean x ∈ S and y ∈ S. To make set theory interesting, we need an axiom which tells us that some set actually exists. The existence of the following set is assumed axiomatically. Definition 2. A set is called empty iff the set contains no elements. The empty set is denoted in two ways, ∅ and { }. You should be very careful about notation! The sets { {} } and { ∅ } are not empty. Both have exactly one element, namely, the empty set. You may have noticed that we wrote “the” empty set instead of “an” empty set in Definition 2. This will be justified in Proposition 12; there is only one empty set. Definition 3. A set S is called a subset of a set X iff every element of S is also an element of X . We write either S ⊂ X or X ⊃ S. That is, S is a subset of X iff s ∈ S implies s ∈ X . Important Remark 4. You should be careful with the notation used in this text. Many authors, especially of more elementary texts, use the notation S ⊆ X to describe subsets. Those authors use S ⊂ X to mean that S is a proper subset of X , that is, S is a subset of X , but is not equal to X †. We will use the notation S
X for proper subsets. It is more common to discuss subsets than proper subsets; we prefer to use the simpler notation for the more common usage. In more advanced mathematical texts, it is normal to use ⊂ for all subsets. *Since a sentence introducing defined terms is called a definition, we will call a sentence introducing undefined terms an undefinition. Or perhaps we should call it a primativum since primitive terms are introduced? †You may take the concept of set equality for granted. However, we will define this in Definition 9 later in this section.

57

58

Chapter 4. Basic Set Operations

There is a moral to this story: whenever you pick up a new mathematics text, you should be sure that you understand the notation and the terminology since usage may differ among different authors. In constructing proofs, it is often helpful, as we noted in Section 1 of Chapter 3, to draw pictures. For proofs about sets, Venn diagrams, such as in Figure 5, may help you to construct proofs. U A x B

Figure 5: Example of a Venn diagram There is no geometry in Figure 5; the circular nature of the actual shapes is irrelevant. The sets A and B are subsets of some set U . Also, A is a subset of B. Sometimes it is useful to indicate elements explicitly. The element x is an element of B. However, x is not an element of A, which we denote by x ∈ / A. Remember this tool when you construct proofs in this chapter. Absurdum. Let X be a set. The empty set is not a subset of X . Proof? For the empty set to be a subset of X , every element of the empty set must belong to X . Since the empty set has no elements, this condition is not satisfied. Therefore, the empty set is not a subset of X .
Exercise 6. Find the fallacy in the “Proof?” of the Absurdum and prove that ∅ ⊂ X . Remember that the notation A ⊂ B does not imply that A = B (not that we have even defined the equality—or inequality—of sets yet). Keeping this in mind, prove the following: Exercise 7. Let X be a set. Prove that X ⊂ X . The following is an important property called the transitive property of subsets. Exercise 8. Let A, B, and C be sets. Prove that if A ⊂ B and B ⊂ C, then A ⊂ C. If you ask yourself what it means for two sets A and B to be equal, you might say that A and B have exactly the same elements. We make this precise in the following definition.

Section 3. Intersections and Unions

Definition 9. Two sets A and B are equal iff the following equivalence holds: x ∈ A if and only if x ∈ B. Of course, when A equals B we write A = B. Proving that two sets are equal can be quite easy, as in the following example. Example 10. Let A = { 2, 4, 6 } and B = { 2n | n ∈ Z3 }. Since Z3 = { 1, 2, 3 }, we see that the elements of B are 2(1) = 2, 2(2) = 4, and 2(3) = 6. But these are exactly the elements of A! Hence, A = B. Of course, proving that two sets are equal is generally much more difficult than in Example 10! The following is a useful way of thinking about equality of sets. Exercise 11. Let A and B be sets. Prove that A = B if and only if A ⊂ B and A ⊃ B. In other words, to prove that two sets A and B are equal, we must show that a ∈ A implies a ∈ B, and b ∈ B implies b ∈ A. You may have wondered why we wrote “the empty set” (and not “an empty set”) in and after Definition 2. The following proposition shows that there is a unique empty set. Proposition 12. If A and B are empty sets, then A = B. Proof. By Exercise 11, to show that A = B, it suffices to show that A ⊂ B and B ⊂ A. By Exercise 6, since A is empty, A ⊂ B. Again, by Exercise 6, since B is empty, B ⊂ A.
In the following exercise, we examine the important properties of equality, called the reflexive, symmetric, and transitive properties. These three properties virtually characterize equality; this will be explored in detail in Chapter 6. Exercise 13. Let A, B, and C be sets. Prove that (a) A = A, (b) if A = B, then B = A, and (c) if A = B and B = C, then A = C. 3. Intersections and Unions. Next we consider two basic set operations, intersection and union. We will do this first for two sets and then for arbitrarily many sets. The intersection of two sets A and B should be the set consisting of all elements common to both of the sets A and B.

59

60

Chapter 4. Basic Set Operations

Definition 14. Let A and B be sets. The intersection of A and B is the set A ∩ B = { x | x ∈ A and x ∈ B } . That is, x ∈ A ∩ B if and only if x ∈ A and x ∈ B. Consider how you see intersections in Venn diagrams, such as in Figure 15. U A

U

C

B

x D

y E

Figure 15: Intersections in Venn diagrams The left figure shows two sets, A and B with the intersection A ∩ B shaded. The right figure shows three sets intersecting each other in the most generic way possible. Note that x, y ∈ C ∩ E, x ∈ (C ∩ E) ∩ D but y∈ / (C ∩ E) ∩ D. Let us consider two easy intersection properties. Exercise 16. Let A be a set. Prove that A ∩ ∅ = ∅. Exercise 17. Let A be a set. Prove that A ∩ A = A. Definition 18. Two sets A and B are disjoint iff A ∩ B = ∅. By Exercise 16, every set is disjoint from the empty set. By Exercises 12 and 17, only the empty set is disjoint from itself. Next, we consider unions. The union of A and B should be the set consisting of all elements of the sets A and B. Definition 19. Let A and B be sets. The union of A and B is the set A ∪ B = { x | x ∈ A or x ∈ B } . That is, x ∈ A ∪ B if and only if x ∈ A or x ∈ B. Remember that or is not exclusive; x ∈ A or x ∈ B is true when x ∈ A and x ∈ B. An element may appear only once in a set. So {1, 2} ∪ {2, 3} = {1, 2, 3}; in fact, {1, 2, 2, 3} is not a set. The Venn diagrams in Figure 20 show the unions of two and three sets, respectively.

61

Section 3. Intersections and Unions

U A

U

C

B D

E

Figure 20: Unions in Venn diagrams Here are some easy exercises. Exercise 21. Let A be a set. Prove that A ∪ ∅ = A. Exercise 22. Let A be a set. Prove that A ∪ A = A. The following exercise relates intersections and unions. Exercise 23. Let A and B be sets. Prove* A ∩ B ⊂ A ⊂ A ∪ B. Prove that this implies that A ∩ B ⊂ A ∪ B. Remember that A ∩ B ⊂ A ∪ B does not necessarily mean that A ∩ B is a proper subset of A∪ B; this should remind you of the non-exclusiveness of the logical “or” operation. Exercises 16 and 17 are generalized by Exercise 24; Exercises 21 and 22 are generalized by Exercise 25. Exercise 24. Let A and B be sets. Prove that A ∩ B ⊃ A if and only if A ⊂ B. Conclude that A ∩ B = A if and only if A ⊂ B. Exercise 25. Let A and B be sets. Prove that A ∪ B ⊂ B if and only if A ⊂ B. Conclude that A ∪ B = B if and only if A ⊂ B. The following shows that the equality of subsets is truly exceptional. Exercise 26. Let A and B be sets. Prove that A ∩ B = A ∪ B if and only if A = B. In the following exercises, you will prove the commutative, associative, and distributive properties of intersections and unions. Exercise 27. Prove the commutative properties for intersections and unions. That is, for sets A and B show that A∩B = B∩ A

and

A ∪ B = B ∪ A.

62

Chapter 4. Basic Set Operations

Exercise 28. Prove the associative properties for intersections and unions. That is, for sets A, B, and C show that (A ∩ B) ∩ C = A ∩ (B ∩ C)

and

(A ∪ B) ∪ C = A ∪ (B ∪ C).

Exercise 29. Prove the distributive properties. That is, for sets A, B, and C show that A ∩(B ∪C) = (A ∩ B)∪(A ∩C)

and

A ∪(B ∩C) = (A ∪ B)∩(A ∪C).

Notice that, by Exercises 27 and 29, we also have the following distributive properties: (A∪ B)∩C = (A∩C)∪(B ∩C)

and

(A∩ B)∪C = (A∪C)∩(B ∪C).

Some people use terms like “left” or “right” distributive properties; we do not use these terms since we can never remember which is which! Let us now return to the definitions of intersection and union, this time for a collection of sets. The intersection of a collection of sets should be the set consisting of the common elements of all of the sets in the collection. Definition 30. Let A be a collection of sets. The intersection of the sets in A is the set { a | a ∈ A for all A ∈ A } . Some authors simply say the intersection of A. The notation is usually 

A

or



A

A∈A

Recall Definition 18 of a disjoint pair of sets. Before generalizing that concept to collections of sets, consider Figure 31. Figure 31 shows two different collections, A and B, of subsets of R2 which exhibit some kind of “disjointness” property; in both A and B, the circular and triangular regions are disjoint as in Definition 18. A

Figure 31: Disjointness properties of collections of sets

B

63

Section 3. Intersections and Unions

Definition 32. Let A be a collection of sets. The sets in A are pairwise disjoint iff A1 and A2 are disjoint for all A1 , A2 ∈ A such that A1 = A2 . Definition 33. Let A be a collection of sets. The sets in A are disjoint iff  A = ∅. The above definition of a disjoint collection is not universally used since pairwise disjointness is usually desired. Hence, some authors use disjoint to mean pairwise disjoint. Consider Figure 31 again; which collection(s) is (are) pairwise disjoint and which collection(s) is (are) disjoint? Often a collection of sets is indexed by some parameter. For example, the following is such a collection of intervals: 

      
 (0, 1), 0, 12 , 0, 13 , . . . = 0, n1
n ∈ N .

This family is indexed by N. For operations such as intersections of indexed sets the following notations are equivalent: ∞   n=1

       1 0, n1 = 0, n = (0, 1) ∩ 0, 12 ∩ 0, 13 ∩ . . . . n∈N

Example  1   34.  Consider Figure 35, showing the nested intervals (0, 1), 0, 2 , 0, 13 , . . . .

0

x

1/4

1/3

1/2

1

Figure 35: Nested intervals with empty intersection ∞   n=1

 0, n1 = ∅ since 1/n can be made arbitrarily small by choosing

a sufficiently large n. To be more precise, start by assuming that there ∞    exists x ∈ 0, n1 . Hence, x > 0 and, for all n ∈ N, x < 1/n. But n=1

this is clearly false since there are certainly natural numbers greater than 1/x. This “obvious” statement is called the Archimedean Principle; we will study it more carefully in Part III.

Exercise 36. Find

∞   n=1

∞     1 0, n1 and ,1 . n n=2

64

Chapter 4. Basic Set Operations

Exercise 37. Let A be a collection of sets containing at least two sets. (a) Prove that if A is a pairwise disjoint collection, then A is a disjoint collection. (b) Give a counterexample for the converse of the statement in (a). The definition of the union of a collection of sets is similar to that of the intersection. The union of the sets in a collection is the set consisting of all of the elements of all of the sets in the collection. Definition 38. Let A be a collection of sets. The union of the sets in A is the set { a | there exists A ∈ A such that a ∈ A } . In analogy with intersections, some authors simply say the union of A. The notation is usually 

A

or



A.

A∈A

We can rephrase this definition to say that something is an element of the union of A if and only if there is some set in A that contains it. Exercise 39. Find

∞   n=1

∞     1 0, n1 and , 1 . n n=2

The distributive properties can be generalized for collections of sets. Exercise 40. Let A be a collection of sets and let B be a set. Prove:  B∩

 A∈A

A =



 (B ∩ A)

and

A∈A

B∪



A =

A∈A



(B ∪ A).

A∈A

What happened to the associative and commutative properties? You can think of these properties as coming for free in the definitions of intersection and union for collections. Remark 41. Let us consider the definitions of intersection and union for small collections A. For example, suppose A = { A, B }. In this case, we are back to our first definition of intersection and union: ∩A = A ∩ B

and

∪ A = A ∪ B.

Suppose we have a collection A = { A } of only one set. Then ∩A = A

and

∪ A = A.

Section 4. Differences and Complements

The situations for collections with two sets and with one set should be clear to you. Now consider what happens when A = ∅. Since there does not exist a set in the empty collection, no element can be contained in a set in the empty collection A. Therefore, ∪A = ∅. What about ∩A? Which elements are contained in every set of the empty collection A? Every element satisfies this condition vacuously! That is, the statement if A ∈ A, then x ∈ A is true for all elements x. So what is this set of all possible elements? If we assume that every set we are considering is a subset of some universal set U , then ∩A = U. The necessity of universal sets comes up often, e.g., Definition 49 of the complement of a set; it is also useful for avoiding paradoxes such as we dscuss in Section 6. Also, you may have wondered whence the elements in Definitions 30 and 38 came. 4. Differences and Complements. You have probably seen the operation of taking complements of sets. Before defining this concept we will first look at the operation of set difference. Definition 42. Let A and B be two sets. The set difference A − B is the set of all elements of A that are not elements of B. That is, A − B = { x | x ∈ A and x ∈ / B }. Set difference is characterized by the following property. Exercise 43. For all sets A and B, prove that A − B is the largest subset of A that is disjoint from B. (Being the largest such set means that if C ⊂ A and C ∩ B = ∅, then C ⊂ A − B.) Set difference is not a commutative operation. Exercise 44. (a) Give a counterexample to A − B = B − A. (b) Give an example where A − B = B − A. The following exercise relates intersection, union, and difference. Exercise 45. Prove that A−B and A∩B are disjoint and (A−B)∪(A∩B) = A for all sets A and B.

65

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Chapter 4. Basic Set Operations

Set difference is not an associative operation either. Exercise 46. Give a counterexample to A − (B − C) = (A − B) − C. The following exercise concerns properties which are referred to as De Morgan’s Laws. Exercise 47. Let A, B, and C be sets. Prove that A−(B ∩C) = (A− B)∪(A−C)

and

A−(B ∪C) = (A− B)∩(A−C).

This can be generalized to collections of sets; the following are also called De Morgan’s Laws. Exercise 48. Let A be a set and let B be a collection of sets. Prove that  A−

 B∈B

B

=

 B∈B

 (A − B)

and

A−

 B∈B

B

=



(A − B).

B∈B

So what is the complement of a set? This is not so obvious. We rely on the existence of a universal set, of which all sets are subsets. Definition 49. Suppose that U is the universal set. The complement of A is U − A. The complement is denoted differently by many authors; A , Ac , A and, of course, U − A are all commonly used to denote the complement. We will usually use U − A and occasionally use A when U is clear from the context. Exercise 50. Suppose that A and B are subsets of a universal set U . Prove that A − B = A ∩ B. Exercise 51. Suppose that A and B are subsets of a universal set U . Prove that (B  ) = B and A − B  = B − A. Exercise 52. Suppose that A and B are subsets of a universal set U . Prove that (A − B) ∪ (B − A) = (A ∩ B) − (A ∪ B) . The set (A− B)∪(B − A) is sometimes called the symmetric difference of A and B.

Section 5. Power Sets

5. Power Sets. A construction with which you may not be familiar is the power set of a set. It is an elementary construction and it is useful in more advanced topics in mathematics. We will use it again when we discuss different types of infinities in Chapter 7. Definition 53. Let X be a set. The power set of X , denoted P(X ), is the set of all subsets of X . For example, and

    P { 1 } = ∅, { 1 }     P { 1, 2 } = ∅, {1}, {2}, { 1, 2 } .

Of course, since ∅ and X are always subsets of a set X , the power set of X always contains the sets ∅ and X . Exercise 54. What is P(∅)? Prove your result. We have noted that P(Z1 ) = P({ 1 }) = { ∅, { 1 } } has two elements and that P(Z2 ) = P({ 1, 2 }) = { ∅, {1}, {2}, { 1, 2 } } has four elements. This can be generalized. Exercise 55. How many elements does P(Zn ) have? Prove your result. Exercise 55 can be proved by induction and by a direct counting argument. Try to give both proofs. Exercise 56. Prove: if S ⊂ X , then P(S) ⊂ P(X ). It follows from Exercise 56 that if S = X , then P(S) = P(X ). Exercise 57. Let A and B be sets. Prove, or disprove by a counterexample, each of the following: P(A) ∩ P(B) ⊂ P(A ∩ B)

and

P(A) ∩ P(B) ⊃ P(A ∩ B).

67

68

Chapter 4. Basic Set Operations

Exercise 58. Let A and B be sets. Prove, or disprove by a counterexample, each of the following: P(A) ∪ P(B) ⊂ P(A ∪ B)

and

P(A) ∪ P(B) ⊃ P(A ∪ B).

Exercise 59. Let A and B be sets. Prove, or disprove by a counterexample, each of the following: P(A) − P(B) ⊂ P(A − B)

and

P(A) − P(B) ⊃ P(A − B).

6. Russell’s Paradox. We have mentioned the term “collections” of sets in this chapter. We should be careful here. If we were to develop the theory of sets axiomatically, we would need to define this term. Alternatively, we might want to replace the term “collection” with “set.” The use of sets of sets (or collections of sets) has to be dealt with very carefully, however. To see that this is so, let us consider the following paradox, which is usually attributed to Bertrand Russell. Let us consider the set of all sets. Some sets may contain themselves as elements while others may not. Think about this for a moment. Most sets do not contain themselves. For example, the set of natural numbers does not contain itself; N is not an element of N since every element of N is a number and not a set of numbers.* For another example, consider the set of all sets. This set must contain itself. There is no paradox here, yet. Now consider the set R of all sets which do not contain themselves as elements. Now we ask the question, is R contained in R? If R is an element of R, then, by definition of the set R, R cannot contain R. This is a contradiction. In case you think we were unwise to start with the assumption that R is an element of R, consider the other case. If R is not an element of R, then, again by the definition of R, R must be an element of R. Again this is a contradiction. Much effort was spent by mathematicians in trying to remedy this situation. The usual remedy is to use axioms which do not allow the existence of the set of all sets. One axiom allows for the specification of subsets as we described in the introduction. For example, set difference is allowed by this axiom. Other axioms allow for unions, intersections, and power sets. That is, such a formal system of axioms allows us to do what we have done in this chapter. If this all strikes you as a bit abstract, consider the following amusing version of Russell’s paradox. If the barber in a certain village shaves all *You may ask what a number is; indeed Chapter 10 will blur the distinction between numbers and sets.

Section 6. Russell’s Paradox

those, and only those, who do not shave themselves, then who shaves the barber?

69

70

Chapter 4. Basic Set Operations

Supplemental Exercises Definition Review. There were 13 definitions in this chapter. Recall that the terms set, element, and is an element of are undefined. Let A and B be sets and let C be a collection of sets; assume all of these sets are subsets of a universal set U . Define each of the following and give an example of each: (2) A is empty (3) A is a subset of B (9) A equals B (14) the intersection of A and B (18) A and B are disjoint (19) the union of A and B (30) the intersection of C (32) C is a pairwise disjoint collection (33) C is a disjoint collection (38) the union of C (42) the difference of B from A (49) the complement of A (53) the power set of A Exercise S1. List the elements of each of the following four sets.  A=



12
∈N n∈N
n




B = (x, y)
x ∈ N, y ∈ N, x + y = 5
 
2
C = (x, y) ∈ R
|x + y| ≤ 0
 
x +2
<0 D= x ∈Z
x −3 

Exercise S2. After Definition 2, we remarked that { {} } and { ∅ } are not empty. Verify this by finding an element of each. Are these two sets equal or not? Explain! Exercise S3. Let A and B be sets. Can A ∩ B  A? Can A ∩ B
B? Explain!   Exercise S4. Prove that x 3 | x ∈ R = R.

Supplemental Exercises

Exercise S5. Determine which of the following sets are equal and which are proper subsets of which.
√ 
A = x ∈ R
x2 = x
√  
1+ x
B= x ∈R
√ ∈R 1− x
 

C = x ∈ R
x ≥ 0 



x3
D= x ∈R
(x − 1)2  

2
E = x
x ∈ R 

Exercise S6. Prove or disprove that if A = B and B = C, then A = C. Exercise S7. Prove that if A ⊂ B, B ⊂ C, C ⊂ D and D ⊂ A, then A = B = C = D. Exercise S8. Let

b a ∗
A = x ∈ R
x = + , for some a, b ∈ R b a
 

and B = y ∈ R
y = sin t, for some t ∈ R . 

Prove that A ∩ B = ∅. Exercise S9. Find a condition such that A ∩ C = B ∩ C implies A = B. Also, find an example where A ∩ C = B ∩ C, but A =  B. Exercise S10. Let A = { 1, 2, 3, 4, 5, 6 } and B = { 2, 4, 6, 8 }. Find all the subsets of A ∩ B. Also, find all the subsets of A ∪ B, containing exactly three elements. Exercise S11. Let A = { 4, 5 }. Suppose that A ∪ B = { 1, 2, 3, 4, 5, 8, 9, 10 }

and

B ∪ C = { 4, 5, 6, 7, 8, 9, 10 }.

Can we determine uniquely the sets B and C? If not, what is the minimal additional information needed, concerning unions or intersections of the sets, for these sets to be determined uniquely?

71

72

Chapter 4. Basic Set Operations

Exercise S12. Let
  A = n ∈ N
n 2 − 8n + 7 > 0 , B = { x ∈ N | x is a prime number }   and C = n 2 | n ∈ N . Determine the sets A ∩ B, (A ∩ B) ∪ (A ∩ C) and (A ∩ C) − (A ∪ B ∪ C). Exercise S13. Find a condition such that A ∪ C = B ∪ C implies A = B. Prove it! Also, find an example where A ∪ C = B ∪ C, but A = B. Exercise S14. Prove that A ∪ B = ∅ if and only if A = B = ∅. Exercise S15. Prove that A ∪ (A ∩ B) = A Exercise S16. Prove that if C ⊂ A, then (A ∩ B) ∪ C = A ∩ (B ∪ C). Is the converse true? Exercise S17. Prove or disprove that if C ⊂ B, then (A ∪ B) ∩ C = A ∪ (B ∩ C). Exercise S18. Let A1 , · · · , An be n sets. Prove that, for all 1 ≤ i, j ≤ n, Ai ∩ A j ⊃ A1 ∩ · · · ∪ An and Ai ∪ A j ⊂ A1 ∪ · · · ∪ An . Exercise S19. Consider the following three collections of sets. Which are disjoint? Which are pairwise disjoint? A = { {0, 1}, {1, 2}, {2, 3} } B = { (0, 1), (1, 2), (2, 3) } C = { [0, 2], [1, 3], [2, 4] } Exercise S20. Let C be a collection of sets. Prove that if C contains exactly two sets, then C is disjoint if and only if C is pairwise disjoint. Exercise S21. Let C be a collection of sets. Prove that if S ∈ C, then ∩C ⊂ S ⊂ ∪C. Exercise S22. Let C be a collection of sets. Prove that ∩C = ∪C if and only if S1 = S2 for all S1 , S2 ∈ C. Exercise S23. Prove that A − (A − B) = A ∩ B. Exercise S24. Prove that A ∩ (B − C) = (A ∩ B) − (A ∩ C). Exercise S25. Prove that (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).

Supplemental Exercises

Exercise S26. If A − B = B − A, what can you conclude about the two sets? Prove your answer. Also, answer the same question if A− B ⊂ B − A. Exercise S27. Prove that if A ∪ B = A ∩ B, then A − B = ∅. Exercise S28. Prove that (A − B) ∩ (C − A) = ∅. Exercise S29. Prove that the following are equivalent conditions on sets A and B: (1) A − B = B − A; (2) A = B; (3) A − B = B − A = ∅. Exercise S30. Prove that (A ) = A.
  Exercise S31. If A = x ∈ R
x 2 > x , express A as an interval. Exercise S32. Prove that B ⊂ (A ∩ B  ) . Exercise S33. Let A and B be sets such that P(A) = P(B). Prove that A = B. Exercise S34. Show that (A ∩ B ∩ C) = A ∪ B  ∪ C  . Is it also true that (A ∪ B ∪ C) = A ∩ B  ∩ C  ? Exercise S35. Prove that if (A ∪ B) = A ∪ B  , then A = B. Does (A ∩ B) = A ∩ B  imply that A = B? Exercise S36. Prove that (A ∪ B  ) ∪ (A ∪ B) = A. Exercise S37. Can there exist a set whose power set has 4k + 1 elements, where k ∈ N? Exercise S38. Prove that (A − B) ∩ (B − A) = ∅ for all sets A and B. Exercise S39. Prove that P(A − B) ⊂ (P(A) − P(B)) ∪ { ∅ }. Project. We have taken a relatively “na¨ıve” approach to set theory. It is possible to be much more “axiomatic.” Find a text on axiomatic set theory and outline the development of the topics of this chapter in an axiomatic form, giving undefined terms, axioms, definitions, etc. Pay particular attention to the axioms that allow new sets to be created from known sets, such as by unions and intersections. Is a universal set needed in your axiomatic set theory? How is Russell’s Paradox avoided in your axiomatic set theory?

73

74

Chapter 5 Functions 1. Functions as Rules. Much of mathematics concerns the study of functions; the notion of “function” is perhaps the single most important concept in all of mathematics. Let us start with an informal definition of the term function. Probably you have seen (but perhaps forgotten?) a definition of the word function which goes something like this: a function is a pair of sets and a rule which assigns a unique element of the second set to each element of the first set. You may also recall that the first set is called the domain of the function and, again, you may recall (somewhat incorrectly!) that the second set is called the range of the function. Actually, the second set should be called the codomain of the function. Probably your facility with functions is better than your recollection of the actual definition of the term function. In the next section, we will define functions as certain types of sets. From a rigorous (logical, axiomatic) point of view, this is preferable. However, you may (should?!) continue to think of a function as a rule that assigns to each value of its domain a unique value in its range. Before you continue reading, think for a moment about a function and its graph. If, for a graph, you are thinking of a pictoral representation, then what does that picture represent? Is there a difference between that object, called the graph of a function, and the function itself? 2. Cartesian Products, Relations, and Functions. Before giving our definition of the word function, we will remind you of the cartesian product of two sets. Often, cartesian products are simply called products. Definition 1. Let A and B be sets. The cartesian product of A and B, denoted by A × B, is the set of all ordered pairs* (a, b) such that a ∈ A and b ∈ B. *While it is possible to define the term ordered pair by something like (a, b) = { a, {a, b} }, you may accept this as undefined. Since the notation (a, b) also represents an open interval, some authors use the notation a × b for an ordered pair.

Section 2. Cartesian Products, Relations, and Functions

Of course, you should have in mind the usual example and its model. R × R, which is usually denoted as R2 , is the setting for most functions you have studied; the model of R2 is the plane which you usually draw by indicating the two coordinate axes. A point P in this plane corresponds to an ordered pair (x, y); the vertical line through P intersects the horizontal axis at x while the horizontal line through P intersects the vertical axis at y. See Figure 2. 3 2 1 -3

-2

-1

0 -1

1

y

x

2

3

P(x,y)

-2 -3

Figure 2: A point of the plane models an ordered pair in R2 It is possible to construct the products of other sets. Using the kind of geometric intuition that we used for R2 , we may model various products. For example, R3 = R × R × R is modeled by (three dimensional) space.† Similarly, N2 = N × N is modeled by a lattice of isolated points and R × {1, 2} and {1, 2} × R are modeled by pairs of lines. We draw pictures of these models as if drawing the sets themselves. Exercise 3. Draw pictures of (the models of) R3 , N2 , R × {1, 2} and {1, 2} × R. We did not require non-empty sets in Definition 1; of course the product of empty sets is again empty. Empty products are not very important, but may come up on occasion. Exercise 4. Let A and B be sets. Prove that A × B = ∅ if and only if A = ∅ or B = ∅. †You may wonder if R×R×R means (R×R)×R or R×(R×R); it turns out that it does not matter however. The reason for this has to do with the concept of isomorphism, which we will introduce in Chapter 8. The general idea is that every element ((x, y), z) ∈ (R × R) × R corresponds to exactly one element (x, (y, z)) ∈ R × (R × R); we can simply think of both of these points as (x, y, z) ∈ R3 . In fact, we will give a different definition of products of collections of sets in Definition 52 of Chapter 7.

75

76

Chapter 5. Functions

Of course, the operation of taking products is not commutative. For example, if A = { 1 } and B = { 2 }, then A × B = { (1, 2) } while B × A = { (2, 1) }. The following exercise relates to this. Exercise 5. Let A and B be sets. Prove that A × B and B × A are disjoint if and only if A and B are disjoint. Before giving a definition of the term function, we define a simpler term: relation. Definition 6. Let A and B be sets. A relation from A to B is a subset R ⊂ A × B. This relation is denoted as R(A, B) or, when A and B are understood, simply as R. If A = B, then R ⊂ A × A = A2 and we call R a relation on A. An ordered pair (a, b) in a relation R is usually denoted by a R b which is read “a is R-related to b” or, more simply, as “a is related to b.” That is, if we write a R b, we mean (a, b) ∈ R; similarly, a R/ b will mean that (a, b) ∈ / R. y

y 3

1

2 1

-1

0

1

x

-3

-2

-1

0

1

2

3

x

-1 -1

-2 -3

Figure 7: The circle x 2 + y 2 = 1 in R2 and < on R Let us consider two examples of relations that you have seen before. They are represented by Figure 7. Example 8. The equation x 2 + y 2 = 1 determines a subset of the product R2 , that is, a relation on R. Similarly, any equation in two variables determines a relation. Example 9. We define a relation <(R, R), in the notation of the definition. This is the usual relation “less than” on the set of real numbers. We use the notation x < y to mean that (x, y) is an element of this relation. We will refer to this as the relation < on R and never use the hideous notation <(R, R) again!

Section 2. Cartesian Products, Relations, and Functions

In Chapter 6, we will study relations on a set more closely. For the remainder of this chapter, we will consider functions. You may already know that every function is a relation. Also, the converse is false: you may recall that the circle of Example 8 is not the equation of a function. No doubt you would suspect that the relation of Example 9, as depicted in Figure 7, is not a function either. We are now ready to give a formal definition of the term function. This definition will be rather different from the informal definition in Section 1. In fact, the object we will define is often called the graph of a function rather than a function. Definition 10. Let X and Y be sets. A function from X to Y is a relation f from X to Y with the property that for every x ∈ X there exists a unique y ∈ Y such that the ordered pair (x, y) is in f . X is called the domain of the function and Y is called the codomain of the function. That is, a function f from X to Y is a subset of X × Y such that for every x ∈ X there exists a unique y ∈ Y such that (x, y) ∈ f . When we sketch the graph of a function, the domain is usually on the horizontal axis while the codomain is on the vertical axis. Consider again the circle of Example 8; this is a relation on R. It fails to define a function from R to R in two ways. First, the domain is not R since x must be between −1 and 1, inclusive. Second, domain values  do not, in general, correspond to unique codomain values; for example, 35 , 45   and 35 , − 45 are both on the circle. So, this circle is not a function with domain [−1, 1] either. Remark 11. Here are some important words about notation. A function, as in Definition 10, is denoted by f : X → Y . When X and Y are understood, we may simply denote the function by f . If (x, y) ∈ f , then we denote the element y by f (x); that is, f (x) = y and (x, y) = (x, f (x)). Be careful, while most mathematicians are happy to call a function f , many frown at calling a function f (x) since f (x) is an element of the codomain. The notation y = f (x) is better and is very commonly used, as in the function y = x 2 or f (x) = x 2 . You may have noticed that we used a lower case f to represent a set; of course, there is no rule which requires the name of a set to be an upper case Latin letter. Some authors substitute the words source and target for domain and codomain, respectively. When specifying a function f without explicitly giving the domain and codomain, we will use the notations D f and C f for the domain and codomain, respectively. This formal definition certainly looks very different from the informal definition, but it is not so very different at all. The domain and codomain are actually exactly the same in both the informal and the formal definitions.

77

78

Chapter 5. Functions

The difference is in the interpretation of f , which is either a rule or a set. The primary difference may be in the way one thinks of a function. In fact, we ignore this ambiguity and think of a function in whatever way is convenient for us at the time; consider Figure 12. R

R

1

f(x)

1

-1

3

4

2

2

.5

.25

f(x) = x2

1 -2

-1

0

1

2

x

Figure 12: A function thought of as a rule and as a subset of R2 So what happened to the range, which you may remember from earlier days? Definition 13. Let f : X → Y be a function and S ⊂ X . The image of S under f is the set   f (S) = y ∈ Y | (x, y) ∈ f for some x ∈ S . Definition 14. The range of a function f : X → Y is the set f (X ), the image of the domain. Exercise 15. Let f : X → Y be a function and S ⊂ X . Prove that f (S) = { f (x) | x ∈ S }. The following examples should help to clarify these matters. Example 16. Consider again the function f (x) = x 2 . Figure 12 shows a picture of this function. What are the domain and codomain for this function? Of course, the domain is R. By convention, when the domain is not given explicitly, then the domain is the largest subset of real numbers (or whatever set fits the situation) for which the rule defining the function makes sense. The range of this function is [0, ∞). Since the codomain must contain the range, any set containing the set [0, ∞) would suffice. In the case of real-valued functions, that is, functions whose range is a subset of R, we use the convention that the codomain is R. That is, we can write f : R → R. Now what is f ? Since the domain of f is R and the codomain is R, the function f is a subset of the product R2 = R × R. We can write it as   f = (x, x 2 ) | x ∈ R .

79

Section 2. Cartesian Products, Relations, and Functions

√ Example 17. Consider the familiar function f (x) = x. The domain and range of this function is [0, ∞) and the codomain of this function is R; so we can write f : [0, ∞) → R. The range of this function is [0, ∞). f is the following subset of [0, ∞) × R:   √ f = (x, x) | x ∈ [0, ∞) .



Exercise 18. Draw a picture of f from the preceding example. In general, if we know the rule defining a function f and its domain X , then we know that   f = (x, f (x)) | x ∈ X . Notice that we did not specify the codomain of the function f . Hence, we do not know the product, in the definition of function, that is supposed to contain the set f . However, this is generally good enough. In practice, we may be given a formula defining a function where the domain and range are implied and the codomain may be inferred. (Perhaps this is why you remembered the range rather than the codomain at the beginning of this chapter?) Exercise 19. Determine the domain, range, and function defined by the rule y = x 2x−1 . Draw a picture of f . When are two functions equal? Since functions depend on the domain X , the codomain Y , and the set of ordered pairs in X × Y , it makes sense to say two functions are equal if their domains, codomains, and sets of ordered pairs are equal. The equality of the sets of ordered pairs implies the equality of the domains. [You should check this!] We therefore give the following definition. Definition 20. The functions f : X → Y and g : U → V are equal if Y = V and f = g as sets of ordered pairs. In practice, to check that two functions f and g are equal, we check that f (x) = g(x) for all x. You should be asking yourself “For all x in what set?” In fact, one must first check that the domains of f and g are equal and then that f (x) = g(x) for all x ∈ D f = Dg ; that is, we must check that X = U in the notation of Definition 20.

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Chapter 5. Functions

Example 21. Let f : R → R be defined by f (x) = x and let g : R+ → R be defined by g(x) = x. Of course, these are very different functions; they are not equal. However, note that f (x) = g(x) for all x ∈ R+ = Dg . On the other hand, f (−1) = −1 while g(−1) is not defined since −1 ∈ / Dg . Another way of saying this is that (−1, −1) ∈ f but (−1, −1) ∈ / g. In Example 21, the functions f and g are very similar. You can think of g as being f with a restricted domain. This idea is important enough to merit a name and some notation. Definition 22. Let f : X → Y be a function and let S ⊂ X . The function g : S → Y defined by g(x) = f (x) for all x ∈ S is called the restriction of f to S and is denoted f | S . We continue with exercises and an example about the image of sets under a function. Exercise 23. Let f be a function and let A ⊂ B ⊂ D f . Prove that f (A) ⊂ f (B). Exercise 24. Let f be a function and let A and B be subsets of D f . Prove that f (A ∪ B) = f (A) ∪ f (B). The situation for intersections is more complicated. Example 25. Let f : R → R be defined by f (x) = |x| and consider the intervals A = (−2, −1) and B = (1, 2). Since A ∩ B = ∅, f (A ∩ B) = ∅. Also f (A) ∩ f (B) = (1, 2). Therefore, f (A ∩ B) = f (A) ∩ f (B). Another example where f (A ∩ B) = f (A) ∩ f (B) is f : R → R defined by f (x) = x 2 with A = { −1 } and B = { 1 }. Then A ∩ B = ∅, but f (A) = f (B) = { 1 }. In both cases, f (A ∩ B)
f (A) ∩ f (B). However, containment in one direction always holds for intersections. Exercise 26. Let f be a function and let A and B be subsets of D f . Prove that f (A ∩ B) ⊂ f (A) ∩ f (B). 3. Injective, Surjective, and Bijective Functions. In this section we will look at some properties of functions which you may find familiar. Definition 27. A function f : X → Y is one-to-one if and only if x1 and x2 are distinct elements in X implies f (x1 ) and f (x2 ) are distinct elements in Y .

Section 3. Injective, Surjective, and Bijective Functions

That is, f : X → Y is one-to-one if, for all x1 , x2 ∈ X , x1 = x2 implies f (x1 ) = f (x2 ). For a one-to-one function we often write that it is “1-1.” A one-to-one function is also called an injective function or an injection. The following exercise gives us an equivalent definition of one-to-one. Exercise 28. Prove that a function f : X → Y is one-to-one if and only if, for all x1 and x2 in X , f (x1 ) = f (x2 ) implies x1 = x2 . One-to-one functions are better behaved than generic functions. Compare Exercises 24, 26, and the following exercise. Exercise 29. Let f be a function. Prove that f is one-to-one if and only if f (A ∩ B) = f (A) ∩ f (B) for all subsets A and B of D f . Definition 30. A function f : X → Y is onto if and only if f (X ) = Y , that is, the range of f is the codomain of f . An onto function is also called a surjective function or a surjection. The following exercise gives us an equivalent definition of onto. Exercise 31. Prove that a function f : X → Y is onto if and only if, for each y ∈ Y , there exists x ∈ X such that f (x) = y. Definition 32. A function f : X → Y is bijective if and only if it is one-to-one and onto. Bijective functions are also called one-to-one correspondences. The authors generally prefer bijective function or bijection to one-to-one correspondence; however, we prefer the latter in connection with cardinality, the topic of Chapter 7. The following will remind you of the tenuous role of the codomain as compared with the range. Exercise 33. Let f : X → Y be a function. (a) Prove that there exists a surjective function from X onto f (X ). (b) Prove that if f is one-to-one, then there exists a bijective function from X onto f (X ). The following exercise gives us an equivalent definition of bijective. Exercise 34. Prove that a function f : X → Y is bijective if and only if, for each y ∈ Y , there exists a unique x ∈ X such that f (x) = y.

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It is possible for a function to be bijective or to have exactly one of the properties of one-to-one or onto or to have neither property. For example, f (x) = tan x, with the conventional codomain of R is onto, but not one-to-one since tan 0 = tan π. Exercise 35. Find three examples of functions with codomain R such that the first is bijective, the second is one-to-one but not onto, and the third is neither one-to-one nor onto. 4. Compositions of Functions. An important operation we perform on functions is that of composition. The most common definition that mathematicians give for the composition of functions requires that the codomain of the first function equals the domain of the second function as in f :X →Y

and

g : Y → Z.

Since the range of a function is more essential than the codomain of that function, the situation for composition can be relaxed to allow the range of the first function to be a subset of the domain of the second function. You will recall defining the function g ◦ f : X → Z by (g ◦ f )(x) = g( f (x)). Below, we define the composition of two functions as a subset of a cartesian product. A more general definition, which does not require the range of the first function to be contained in the domain of the second function, is given in Section 6; it evolves from the definition of composition of relations. Definition 36. Let f : X → Y and g : W → Z be functions with f (X ) ⊂ W . The composition of f followed by g is the function defined by  g ◦ f = (x, z) ∈ X × Z | there exists y ∈ f (X )

 such that f (x) = y and g(y) = z .

The domain of g ◦ f is X , the domain of f , and the codomain of g ◦ f is Z , the codomain of g. Definition 36 is actually quite reasonable if you consider things. The idea is that f sends an x ∈ X to some y ∈ Y . In order to be able to apply g, we must have y ∈ W also; since f (X ) ⊂ W this condition is automatically satisfied. Finally, g sends y to some z ∈ Z . Notice that if Y ⊂ W (for example, when Y = W ), then f (X ) ⊂ W .

Section 4. Compositions of Functions

Exercise 37. For f : X → Y and g : W → Z with f (X ) ⊂ W , verify that g ◦ f = { (x, g( f (x))) | x ∈ X } . You may already have noticed that given two functions f and g, there are often two composite functions: f ◦ g and g ◦ f . In general, f ◦ g = g ◦ f , as the following examples demonstrate. This is true even when both composite functions exist. Example 38. Define f : R → R and g : R → R by f (x) = 2x and g(x) = 2x + 1 for all x ∈ R. Then f ◦ g and g ◦ f are both functions from R to R. However, ( f ◦ g)(x) = f (g(x)) = f (2x + 1) = 2(2x + 1) = 4x + 2 while (g ◦ f )(x) = g( f (x)) = g(2x) = 2(2x) + 1 = 4x + 1. So f ◦ g = g ◦ f . Therefore, composition is not a commutative operation. The following example is much more peculiar, yet more typical. √ √ Example 39. Let f (x) = 2 + x and g(x) = 1 − x. Then, f (g(x)) = √ √ 2+ 1 − x = 2 + 4 1 − x and hence   √ 4 f ◦ g = (x, 2 + 1 − x) | x ∈ (−∞, 1] . Note that the range of g equals the domain of f : the interval [0, ∞). This is consistent with Definition 36.   √ √ On the other hand, g( f (x)) = 1 − (2 + x) = −1 − x is not defined for any real number x. Hence, g ◦ f does not exist. Another way of saying this is that g ◦ f does not exist since the domain of g, which equals (−∞, 1], does not contain the range of f , which equals [2, +∞). In fact, the domain of g and the range of f are disjoint sets. So, certainly f ◦ g = g ◦ f . Examples 38 and 39 show that composition depends on the order of the functions, that is, composition is not commutative. The second example also shows that one of the compositions may be more interesting than the other. In fact, usually we will have the following situation: given f : X → Y and g : W → Z , if Y ⊂ W (or Y = W ), then we have the composite g ◦ f : X → Z .

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Given the noncommutative nature of compositions, you may be surprised to learn that composition is associative. Exercise 40. Show that composition of functions is associative. Compositions of one-to-one and onto functions are well-behaved. Exercise 41. Prove that the composition of two one-to-one functions is again one-to-one. Exercise 42. Prove that if f : X → Y and g : Y → Z are onto functions, then the composite function g ◦ f is onto. Recall that in Definition 36, g ◦ f was defined whenever the range of f was a subset of the domain of g. In Exercise 42, we strengthened this requirement to equality: Y = R f = Dg . The following example shows that this stronger requirement is necessary. Example 43. Define f : R+ → R+ and g : R → R be defined by f (x) = x and g(x) = x for all x in their respective domains. Both functions are onto. By Definition 36, g ◦ f has domain R+ and codomain R; g ◦ f is not onto. From Exercises 41 and 42 we see that the composite g ◦ f of two bijections f : X → Y and g : Y → Z is a bijection. The converses of Exercises 41 and 42 are interesting and are considered in Exercises 44 and 45. Let us consider one-to-one composite functions first. Exercise 44. Suppose that g ◦ f is the composition of f : X → Y and g : Y → Z . Suppose that g ◦ f is one-to-one. (a) Prove that f is one-to-one. (b) Find a counterexample to show that g need not be one-to-one. (c) Prove that if f is onto, then f is bijective and g is one-to-one. Now let us consider onto composite functions. Exercise 45. Suppose that g ◦ f is the composition of f : X → Y and g : Y → Z . Suppose that g ◦ f is onto. (a) Prove that g is onto. (b) Find a counterexample to show that f need not be onto. (c) Prove that if g is one-to-one, then g is bijective and f is onto.

Section 5. Inverse Functions and Inverse Images of Functions

5. Inverse Functions and Inverse Images of Functions. The study of inverses is common in mathematics. It requires identities and composites. Definition 46. A function f is the identity function on X iff D f = C f = X and f = { (x, x) | x ∈ X } ⊂ X × X. That is, f : X → X , defined by f (x) = x for all x ∈ X , is the identity function on X . We use I X to denote the identity function on X . Exercise 47. Prove that I X is the unique function from X to X which satisfies f ◦ I X = f and I X ◦ g = g for all functions f : X → Y and g : W → X . (That is, show that if i : X → X satisfies f ◦ i = f and i ◦ g = g for all functions f : X → Y and g : W → X , then i = I X .) Definition 48. Let f and g be functions. f and g are inverse functions iff f ◦ g = IDg and g ◦ f = ID f . We also say that g is an inverse of f . A function f is invertible iff an inverse function g exists. Notice that this definition implies a relationship between the domains and codomains of the inverse functions, as given in the following exercise. Exercise 49. Prove that if f is an invertible function and g is an inverse of f , then D f = Cg and Dg = C f . It is important to include both f ◦ g = IDg and g ◦ f = ID f in the definition of inverse functions, as Example 50 will show. Example 50. Let f : R+ → R be defined by f (x) = x and let g : R → R+ be defined by g(x) = |x|. Then (g ◦ f )(x) = g(x) = x for all x ∈ R+ ; that is, g ◦ f = IR+ . However, ( f ◦ g)(x) = |x| for all x ∈ R, which is not IR . You may notice that f is one-to-one, but not onto, and that g is onto, but not one-to-one. Exercise 51. Prove that a function is invertible if and only if it is bijective. If it sounds strange to you to read “an inverse” rather than “the inverse” in the definition of inverse functions, the following exercise will help. Of course, we will always write “the inverse” after the following exercise. Exercise 52. Prove that every invertible function has a unique inverse function.

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Since inverse functions are unique, when they exist, we have a dedicated notation for them: the inverse of f is denoted by f −1 . When f is invertible, we have an inverse function f −1 and it makes sense to consider images of the inverse function such as f −1 (S). We call this the inverse image of S. This concept can be generalized to noninvertible functions. Definition 53. Let f : X → Y be a function and T ⊂ Y . The inverse image of T under f is f −1 (T ) = { x ∈ X | (x, y) ∈ f for some y ∈ T } . Remark 54. Be very careful here! The use of the superscript on f −1 (T ) should not make you think the function is invertible. Also, when T is a singleton set { y }, we write f −1 (y) instead of f −1 ({ y }). The use of inverse images gives us a nice way of describing one-to-one and onto functions. Exercise 55. Let f : X → Y be a function. Prove the following: (a) f is one-to-one if and only if f −1 (y) contains at most one element for every y ∈ Y . (b) f is onto if and only if f −1 (y) contains at least one element for every y ∈ Y. (c) f is bijective if and only if f −1 (y) contains exactly one element for every y ∈ Y . Inverse images behave nicely with respect to set operations. Exercise 56. Let f the following: (a) f −1 (A ∩ B) = (b) f −1 (A ∪ B) = (c) f −1 (A − B) =

: X → Y be a function with A ⊂ Y and B ⊂ Y . Prove f −1 (A) ∩ f −1 (B), f −1 (A) ∪ f −1 (B), f −1 (A) − f −1 (B).

Forward and inverse images can be combined. In general, the new set is different from the original. Exercise 57. Let f : X → Y be a function with S ⊂ X . (a) Prove that f −1 ( f (S)) ⊃ S. (b) Give a counterexample to f −1 ( f (S)) = S. (c) Prove that if f is one-to-one, then f −1 ( f (S)) = S. (d) Is the converse of the statement in (c) true? Prove it or find a counterexample.

Section 6. Another Approach to Compositions

Exercise 58. Let f : X → Y be a function with T ⊂ Y . (a) Prove that f ( f −1 (T )) ⊂ T . (b) Give a counterexample to f ( f −1 (T )) = T . (c) Prove that if f is onto, then f ( f −1 (T )) = T . (d) Is the converse of the statement in (c) true? Prove it or find a counterexample. Compare parts (d) of Exercises 57 and 58 with Supplemental Exercise S17. 6. Another Approach to Compositions. In Section 4, we looked at compositions of functions where the range of the first function was contained in the domain of the second function. In this section, we will remove this requirement. The moral of this section is that the easier (i.e., more general) the definitions are, the harder (or less general) the theorems are. We consider first the composition of relations. Definition 59. Let f be a relation from X to Y and let g be a relation from W to Z . The composition of f followed by g is the relation from X to Z defined by  g ◦ f = (x, z) ∈ X × Z | there exists y ∈ Y ∩ W

 such that f (x) = y and g(y) = z .

It is more common to compose functions than relations. However, we will use this idea to generalize composition of functions. Definition 60. Let f : X → Y and g : W → Z be functions. The composition of f followed by g is the function defined by  g ◦ f = (x, z) ∈ X × Z | there exists y ∈ Y ∩ W

 such that f (x) = y and g(y) = z .

The codomain of g ◦ f is Z , the codomain of g. The domain of g ◦ f is the largest subset of X for which it makes sense. Definition 60 differs from Definition 36 in two ways: first, the requirement that f (X ) ⊂ W was removed and, second, the domain is a subset of X , but not necessarily all of X . In fact, the domain may turn out to be the empty subset of X ! You could complain that the phrase “the largest subset of X for which it makes sense” is not very precise. An example should help.

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√ Example 61. Let us consider Example 39 again. Let f (x) = 2 + x and   √ √ √ g(x) = 1 − x. Since (g ◦ f )(x) = 1 − (2 + x) = −1 − x is not defined for any real number x, the domain of g ◦ f is empty. Hence, g ◦ f = ∅! This is OK; the definition of a function does not require a nonempty domain and ∅ × Z = ∅ for every set Z . The following is similar. Exercise 62. Suppose that w, x, y, and z are distinct; in particular, y = z. Let X = { x }, Y = { y }, W = { w }, and Z = { z }. Define f : X → Y and g : W → Z by f (x) = y and g(w) = z. Describe g ◦ f and f ◦ g. Do f and g commute under composition? (That is, does g ◦ f equal f ◦ g?) In general, what is the domain of the composite function g ◦ f ? The composition of f followed by g may easily be rewritten as
  g ◦ f = (x, g( f (x)))
x ∈ D f and f (x) ∈ Dg . [You should check this!] This tells us that the domain is the set
  Dg◦ f = x ∈ D f
f (x) ∈ Dg . We have already seen (Exercise 38, Exercise 39, Example 61, and Exercise 62) that composition is not commutative. Composition is still associative in this setting, but the proof requires you to consider the domains carefully. Here we have an example where our more general definition makes a theorem more difficult to prove. Exercise 63. Show that composition of functions is associative. Next, consider the relationship of compositions with the properties of being one-to-one and onto. The following exercise gives us examples of how generalizing a definition can turn a true theorem into a false statement. Exercise 64. For each of the following, give two functions f and g satisfying the conditions: (a) f and g are onto, but the composite function g ◦ f is not onto. (b) neither f nor g is one-to-one, but the composite function g ◦ f is oneto-one. (c) f is a bijection, g is not one-to-one, but the composite function g ◦ f is one-to-one. (d) g is a bijection, f is not onto, but the composite function g ◦ f is onto. Recall that the definition of inverses is dependent on composition of functions. You may wish to return to the exercises in Sections 4 and 5, this time using Definition 60 for the definition of composition of functions.

Supplemental Exercises

Supplemental Exercises Definition Review. There were 16 definitions in this chapter. Let A, B, X , Y , S, and T be sets with S ⊂ X and T ⊂ Y . Let f : X → Y and g : W → Z be functions (or relations). Assume all of these sets are subsets of a universal set U . Define each of the following and give an example of each: (1) the cartesian product of A and B (6) a relation from A to B (6) a relation on A (10) a function f from X to Y (10) the domain of a function f (10) the codomain of a function f (13) the image of S under f (14) the range of a function f (20) functions f and g are equal (22) the restriction of f to S (27) f is one-to-one (or injective) (27) f is an injection (30) f is onto (or surjective) (30) f is a surjection (32) f is bijective (or a one-to-one correspondence) (32) f is a bijection (36) composition of a function f followed by a function g ( f (X ) ⊂ W ) (36) the domain of g ◦ f ( f (X ) ⊂ W ) (36) the codomain of g ◦ f ( f (X ) ⊂ W ) (46) f is the identity function on X (48) f and g are inverse functions (48) f is an invertible function (53) the inverse image of T under f (59) composition of a relation f followed by a relation g (60) composition of a function f followed by a function g (60) the domain of g ◦ f (60) the codomain of g ◦ f Exercise S1. Find the range of the following functions 2

x (1) f (x) = x 2 +x−2 (2) g(x) = |x 3 + x + 1| 2 (3) h(x) = 2 sin x 2x+1 .

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Exercise S2. How many 1 − 1 functions can be defined from {1, 2, 3} to {1, 2, 3, 4}? Exercise S3. Determine which of the following functions is 1 − 1 (1) f (x) =

3x+1

2x−3  (2) g(x) = 1 + x1 (3) h(x) = √ |2x + 1| (4) t (x) = −x 6

Exercise S4. Determine which of the following functions is a bijection from R to R (1) f (x) = ax + b (2) g(x) = |x| (3) h(x) = x 3 + 1 Exercise S5. Let f : R → R be a function, with f (x) = 2x + 1. Prove that f (N) ⊂ f (R). Find a function g : R → R, such that g(N) = g(R). Exercise S6. If ( f ◦ g)(x) = x 2 + 3 and g(x) = 2x − 1, find the function f. Exercise S7. If ( f ◦ g)(x) = 3x + 2 and f (x) = 4x − 1, find the function g. Exercise S8. Determine the values of a and b, for which the function f (x) = ax + b is invertible and find its inverse. Exercise S9. Determine the values of a, b, and c, for which the function f (x) = ax 2 + bx + c is invertible and find its inverse. Exercise S10. Find a restriction f | S of the function f (x) = 2x 2 − 5x + 1 which is invertible. Prove or disprove that for each function f , there exists a restriction of f which is invertible. Exercise S11. Let f (x) = ax + b and g(x) = 2x − 3. Determine a and b, such that f ◦ g = g ◦ f . Exercise S12. Let f (x) = 5x − 4. Find f −1 ([2, 3]) and f −1 (R). Exercise S13. Let f : R − {−2} → R be a function, with f (x) = 2x−7 . x+2 Prove that f is not onto. Restrict the codomain so that f is onto, prove that with the restricted codomain is invetrible and find its inverse. Exercise S14. Let f : A → A and g : A → A be invertible functions. Show that ( f ◦ g)−1 = g −1 ◦ f −1 .

91

Supplemental Exercises

 Exercise S15. Let f (x) = and find its inverse. Exercise S16. Let f (x) =



x,

x ∈Q

x + 1,

x∈ /Q

x 2,

x ∈Q

x + 1,

x∈ /Q

. Prove that f is invertible

. Is f invertible?

Exercise S17. Let f : X → Y be a function. (a) Prove that if f −1 ( f (S)) = S for every subset S of X , then f is one-toone. (b) Prove that if f ( f −1 (T )) = T for every subset T of Y , then f is onto. (Compare with Exercises 57 and 58.)

92

Chapter 6 Relations on a Set 1. Properties of Relations. As we saw in Chapter 5, a function is a specific kind of relation. Recall that a relation from a set A to a set B is a subset R ⊂ A × B and if A = B, we call R a relation on A. You may have noticed that the examples of relations we looked at in Chapter 5 are relations on a single set. In fact, we are more interested in relations on a set and consider various properties that a relation may or may not have. Definition 1. Let R be a relation on a set S. (a) R is reflexive iff s R s for all s ∈ S. (b) R is nonreflexive iff s R/ s for all s ∈ S. (c) R is symmetric iff s R t implies t R s for all s, t ∈ S. (d) R is asymmetric iff s R t implies t R/ s for all s, t ∈ S. (e) R is antisymmetric iff s R t and t R s implies s = t for all s, t ∈ S. (f) R is transitive iff s R t and t R u implies s R u for all s, t, u ∈ S. (g) R is connected iff s R t or t R s or s = t for all s, t ∈ S. (h) R is trichotomous iff exactly one of s R t, t R s, s = t is true for all s, t ∈ S. In the definition of connected, we say that s and t are comparable iff s R t or t R s or s = t. There are many relationships between the properties in Definition 1; the most obvious is that if a relation is trichotomous, then it must be connected. Example 2. Let us consider the relation in Example 8, given by x 2 + y 2 = 1. It is not reflexive since, for example, 1 is not related to 1, that is, 12 + 12 = 2 = 1. It is not nonreflexive either since ( √12 )2 + ( √12 )2 = 1. It is symmetric by the commutative property of addition. It is not antisymmetric since 02 + 12 = 12 + 02 = 1, but 0 = 1. It is not transitive since 1 is related to 0 which is related to 1 but 1 is not related to 1. It is neither connected nor trichotomous since 1 and 12 satisfy none of the conditions.

Section 2. Order Relations

Exercise 3. Consider whether the relation < on R satisfies or fails to satisfy each of the eight properties of relations given above. Prove your answers. Exercise 4. Consider whether the relation ⊂ on P(U ) for some nonempty U satisfies or fails to satisfy each of the seven properties of relations given above. Prove your answers. Exercise 5. Prove that a relation on a nonempty set cannot be both reflexive and nonreflexive but can be both symmetric and antisymmetric. Exercise 6. Give an example of a relation on a nonempty set which is connected but not trichotomous. Exercise 7. Give examples of relations which have any combination of the properties given above. 2. Order Relations. One natural kind of relation is of the type which puts an order on the set. The relations < on R and ⊂ on P(U ) (which were considered in Exercises 3 and 4) are examples of such order relations. We start with two definitions that model these two examples, but are more general. Definition 8. A relation ≺ on a set S which is nonreflexive and transitive is called a strict partial ordering. A strictly partially ordered set is a set S together with a strict partial ordering ≺ on S. Definition 9. A relation " on a set S which is reflexive, antisymmetric and transitive is called a partial ordering. A partially ordered set is a set S together with a partial ordering " on S. Using the highlighted characters, a partially ordered set is sometimes called a poset. Exercise 10. Consider = as a relation on R. Does it define a strict partial ordering? Does it define a partial ordering? Prove your answers. The example in Exercise 10 is rather trivial since most elements are not comparable. It is sometimes called a totally disordered set. Exercise 11. Consider R with < and P(U ) with ⊂ again. Which is a strictly partially ordered set? Which is a partially ordered set? Prove your answers.

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Sometimes, especially for finite partially ordered sets in which not all elements are comparable, it is useful to make an illustration, called a Hasse diagram, of the set with its relation as Figure 12 illustrates. {1,2}

{1}

{2}

∅

Figure 12: The Hasse diagram for P({1, 2}) with ⊂ Figure 12 shows the elements of the P({1, 2}) with the relation ⊂. The elements of P({1, 2}) are shown as the vertices of a graph. Arrows show the order relation; if there is a directed path from one vertex to another using one or more arrows, then the vertex at the tail of the (first) arrow is related to the vertex at the head of the (last) arrow. For example, the arrow from ∅ to {1} indicates that ∅ ⊂ {1} and the arrow from ∅ to {2} together with the arrow from {2} to {1, 2} indicate that ∅ ⊂ {1, 2}. It is possible to give a precise definition of a Hasse diagram in terms of directed graphs, which are discussed in Chapter ???. Using the convention that arrows always go up simplifies the presentation. Figure 13 shows the more complicated Hasse diagram for P({1, 2, 3}); you should see it as a cube. {1,2,3}

{1,3}

{1}

{2,3}

{1,2} {3}

{2}

∅

Figure 13: The Hasse diagram for P({1, 2, 3}) with ⊂ The next exercise shows that strict partial orderings and partial orderings are intimately related.

Section 2. Order Relations

Exercise 14. (a) Suppose that " is a partial ordering on S and define the relation ≺ on S by s ≺ t if and only if s " t and s = t. Show that ≺ is a strict partial ordering on S. (b) Suppose that ≺ is a strict partial ordering on S and define the relation " on S by s " t if and only if s ≺ t or s = t. Show that " is a partial ordering on S. Remark 15. According to Exercise 14, it does not matter whether the order on a partially ordered set S is derived from a partial ordering " on S or a strict partial ordering ≺ on S. Put another way, whenever either type of order relation is present, both types are actually present. Hence, one can use whichever is more convenient. A mathematician may be sloppy and write “a partial ordering ≺” when “a strict partial ordering ≺” is meant. Definition 16. A strict partial ordering ≺ on S which is connected is called a strict total ordering. A strictly totally ordered set is a set S together with a strict total ordering ≺ on S. Definition 17. A partial ordering " on S which is connected is called a total ordering. A totally ordered set is a set S together with a total ordering " on S. Exercise 18. (a) Prove that every strict total ordering ≺ on a set S is trichotomous. (b) Suppose that R is a relation on S which is transitive and trichotomous. Prove that R is a strict total ordering. That is, a relation is a strict total ordering if and only if it is transitive and trichotomous. In our notation, we will always assume that we can go back and forth between ≺ or ", as in Exercise 14. Perhaps you were wondering if we can use # or $ when we have ≺ or ", respectively? The answer is yes, of course. That is, x # y is defined as y ≺ x and x $ y is defined as y " x. The following definitions are related, but not the same. Definition 19. Let S be a partially ordered set with partial ordering ". An element m ∈ S is maximal iff there exists no s ∈ S such that m ≺ s. Definition 20. Let S be a partially ordered set with partial ordering ". An element M ∈ S is greatest iff s " M for all s ∈ S. Definitions 19 and 20 are different. The key thing to remember is that, in a partially ordered set, all elements need not be related. The following example should help you to see the difference between maximal and greatest elements.

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Example 21. Let A = { 1, 2 }. Consider S = P(A) and T = S − { A } with the partial ordering ⊂. S has A as its only maximal element and its only greatest element. T has two maximal elements, { {1} } and { {2} }, but no greatest element. Example 22. Consider N with the partial ordering ≤. N has neither a maximal element nor a greatest element. [Though this is intuitively obvious, its proof is surprisingly difficult. The proof depends on the Principle of Archimedes, which we postpone until Chapter 10.] Of course, maximal and greatest elements are intimately related as we will see from the following four exercises. Exercise 23. Let S be a partially ordered set with partial ordering ". Prove that if S has a greatest element, then this greatest element is unique. Exercise 24. Let S be a partially ordered set with partial ordering ". Prove that the greatest element of S, if it exists, is a maximal element. Exercise 25. Let S be a partially ordered set with partial ordering ". Prove that if S has more than one maximal element, then S does not have a greatest element. Exercise 26. Let S be a totally ordered set with total ordering ". Prove that a maximal element of S, if it exists, is the greatest element. In a totally ordered set, by Exercises 24 and 26, maximal and greatest elements are identical and, by Exercise 23, if a maximal element exists, then it is unique. We can define minimal elements and the least element of a partially ordered set in analogy with the definitions of maximal and greatest elements. Analogous results can be proved about these. Next we consider subsets in a partially ordered set. Of course, the partial ordering on the ambient set induces a partial ordering on the subset. Definition 27. Let A be a subset of S, a partially ordered set with partial ordering ". An element m ∈ S is an upper bound on A iff a " m for all a ∈ A. A is bounded above iff A has an upper bound. Definition 28. Let A be a subset of S, a partially ordered set with partial ordering ". An element M ∈ S is a least upper bound on A iff M is an upper bound on A and M ≤ s for all upper bounds s on A. A least upper bound is also called a supremum.

Section 2. Order Relations


 Example 29. Let A = x
0 < x 2 < 2 . Clearly, 2 is an upper bound on A. What is the least upper bound of A? It turns out that we have not been precise enough to answer this question! If we consider A as a subset √ of R, then 2 is the least upper bound. What if we consider A as a subset of Q? In that case, there is no least upper bound! You may recall that √ 2 is irrational (see Proposition 18 of Chapter 3). You can think of the decreasing sequence of rational upper bounds 2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . . √ This sequence converges to 2 in R. Since this limit is unique, it can be shown that there is no rational least upper bound. We will not go into the details. Example 30. Consider N with the partial ordering ≤. N has no upper bounds. Also the set E of all even natural numbers has no upper bounds. [Again, while this is intuitively obvious, its proof is surprisingly difficult and is postponed until Exercise 3 of Chapter 8.] Exercise 31. Let S be a partially ordered set with partial ordering " and A ⊂ S. Prove that if A has a least upper bound, then A has a unique least upper bound. Exercise 32. For both of the following, give an example of a nonempty subset A of a partially ordered set S such that: (a) the least upper bound of A exists and equals the greatest element of A, (b) the least upper bound of A exists but is not the greatest element of A. Prove your answers. Exercise 33. Let A be a subset of a partially ordered set S with partial ordering ". Prove that if A has a greatest element, then the least upper bound of A equals the greatest element of A. We can define lower bounds and the greatest lower bound of a subset of a partially ordered set in analogy with the definitions of upper bound and least upper bound. Analogous results can be proved about these. Definition 34. Let S be a partially ordered set with partial ordering ". S has the least upper bound property iff every nonempty subset which is bounded above has a least upper bound. The least upper bound property is important. It distinguishes the rational numbers from the real numbers and will be discussed more in Chapters 8 and 13.

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  Example 35. Let us again consider A = x ∈ Q
0 < x 2 < 2 . We √ indicated that 2 is the least upper bound of A as a subset of R and that A has no least upper bound as a subset of Q. Hence Q does not have the least upper bound property. We mention here one more definition related to order. Definition 36. A totally ordered set S is well-ordered iff every nonempty subset of S has a least element. The ordering is called a well-ordering. The fact that N is well-ordered is proved in Exercise 7 of Chapter 10. The question of well-ordering a set in general is quite difficult. The statement that every set has a well-ordering is actually equivalent to the Axiom of Choice. 3. Equivalence Relations. Definition 37. A relation ≈ on a set S which is reflexive, symmetric and transitive is called an equivalence relation. The prototypical example is equality on a set. There are other, more interesting, examples. Example 38. Congruence and similarity are equivalence relations on the set of triangles. Example 39. Let k ∈ N. Define the relation ≡ on N by n ≡ m if and only if n−m ∈ Z. We usually say that n is congruent to m modulo k and we k write “n ≡ m (mod k).” This equivalence relation is well known to small children, who learned it in the form that 3 hours after 11 o’clock comes 2 o’clock; in this case k = 12 and the statement is that 11 + 3 ≡ 2 (mod 12). In Chapter 7, given two sets A and B, we will say that A has the same cardinality as B if and only if there exists a bijection f : A → B. Exercise 40. Prove that “has the same cardinality as” is an equivalence relation on the power set of a given set. Partitioning the set of clothes in your house into categories such as coats, shirts, pants, underwear, socks and shoes is an example of an equivalence relation at work. Rather than giving the relation in terms of pairs of objects, it is given in terms of the subsets which supply the entries of these pairs. This leads us to the following definition. Definition 41. Let ≈ be an equivalence relation on S and let x ∈ S. The set [x]≈ = { s ∈ S | x ≈ s }

Section 3. Equivalence Relations

is called the equivalence class of x under ≈. The notation for the equivalence class of x is sometimes written simply as [x] when the equivalence relation is understood. Exercise 42. Let ≈ be an equivalence relation on a set S and let s, t ∈ S. Prove that [s] = [t] if and only if s ≈ t. Exercise 43. Let ≈ be an equivalence relation on a set S and let s, t ∈ S. Prove that if [s] = [t], then [s] ∩ [t] = ∅, that is, [s] and [t] are disjoint. We will rephrase the previous exercises using the following language. Definition 44. Let S be a nonempty set. A partition of S is a set of nonempty, pairwise disjoint subsets of S whose union is S. Definition 45. Let ≈ be an equivalence relation on a nonempty set S. Denote by S/≈ denote the set of all equivalence classes of S under ≈. Assuming we read ≈ as “equivalence,” we read S/ ≈ as “S mod equivalence.” After the following exercise, we call S/≈ the partition of S induced by ≈. Exercise 46. Prove that if ≈ is an equivalence relation on a nonempty set S, then S/≈ is a partition of S. We have gone from equivalence relations to partitions. Next we go in the other direction. Definition 47. Let C be a partition of a nonempty set S. Define a relation R on S by s R t if and only if there exists C ∈ C such that s, t ∈ C. R is called the equivalence relation induced by the partition C and is denoted by S/C. We read S/C as “S mod C” and s S/C t as “s is S mod C related to t.” Of course, it must be verified that S/C actually is an equivalence relation. Exercise 48. Let C be a partition of a nonempty set S. Prove that S/C is an equivalence relation. Of course, since we can pass from an equivalence relation to a partition and then back to an equivalence relation, we can ask how the original relation and the induced relation are related. Actually they are the same. One can do this starting with a partition as well.

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Theorem 49. If ≈ is an equivalence relation on a nonempty set S, then S/(S/≈) = ≈. Proof. By Exercise 46, S/≈ is a partition of S and, by Exercise 48, S/(S/≈ ) is an equivalence relation on S. Since a relation on S is a subset of S × S, it suffices to show that S/(S/≈) ⊂ ≈ and ≈ ⊂ S/(S/≈). Suppose (s, t) ∈ S/(S/ ≈); that is, s S/(S/ ≈) t. By Definition 47, there exists C ∈ S/≈ such that s, t ∈ C. By Definition 45, there exists u ∈ S such that C = [u]. Since s, t ∈ [u], u ≈ s and u ≈ t. Hence, s ≈ t. Therefore, S/(S/≈) ⊂ ≈. The remainder of the proof is left for Exercise 50.
Exercise 50. Complete the proof of Theorem 49: prove that ≈ ⊂ S/(S/≈). Theorem 51. If C is a partition of a nonempty set S, then S/(S/C) = C. Proof. By Exercise 48, S/C is an equivalence relation on S, and, by Exercise 46, S/(S/C) is a partition of S. Since a partition of S is a subset of P(S), it suffices to show that S/(S/C) ⊂ C and C ⊂ S/(S/C). Suppose C ∈ C. Choose an element s ∈ C. Since S/(S/C) is a partition of S, there exists C  ∈ S/(S/C) such that s ∈ C. C  = [s] with respect to the equivalence relation S/C. We must show that C = [s]. First, fix t ∈ C. By Definition 47, s S/C t. By Exercise 42, t ∈ [t] = [s]. So C ⊂ [s]. Next, fix t ∈ [s]. So s S/C t. By Definition 47, there exists C  ∈ S/(S/C) such that s, t ∈ C  . Since S/(S/C) is a partition and s ∈ C ∩ C  , C  = C. So t ∈ C and [s] ⊂ C. Hence, C ∈ S/(S/C). Therefore, C ⊂ S/(S/C). The remainder of the proof is left for Exercise 52.
Exercise 52. Complete the proof of Theorem 51: prove that S/(S/C) ⊂ C.

Supplemental Exercises

Supplemental Exercises Definition Review. There were 12 definitions in this chapter. Let S be a set with subset A and let R, ≺, ", and ≈ be relations on S. Define each of the following and give an example of each: (1) R is reflexive (1) R is nonreflexive (1) R is symmetric (1) R is asymmetric (1) R is antisymmetric (1) R is transitive (1) R is connected (1) R is trichotomous (8) ≺ is a strict partial ordering (8) S is a strictly partially ordered set (9) " is a partial ordering (9) S is a partially ordered set (16) ≺ is a strict total ordering (16) S is a strictly totally ordered set (17) " is a total ordering (17) S is a totally ordered set (19) m ∈ S is a maximal element (20) m ∈ S is a greatest element For 27–34, let S be a partially ordered set with partial ordering ". (27) m ∈ S is an upper bound on A (27) A is bounded above (28) m ∈ S is a least upper bound on A (34) S has the least upper bound property For 36, let S be a totally ordered set with total ordering ". (36) S is well-ordered (36) " is a well-ordering (37) ≈ is an equivalence relation (41) the equivalence class of x under ≈, an equivalence relation (44) a partition of a nonempty set, S For 45, let ≈ be an equivalence relation on S. (45) S/≈, the set of all equivalence classes of S under ≈ For 47, let C be a partition of S. (47) S/C, the equivalence relation induced by C

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Exercise S1. Let X be the set of all 2 × 2 matrices with real entries.  a b Recall that the determinant of a matrix equals ad − bc; denote c d the determinant of a matrix M by |M|. Define a relation < on X by A < B iff |A| < |B|. Prove that < is a strict partial ordering on X . Is < a total ordering? Explain your answer.

103

Chapter 7 Cardinality of Sets 1. Introduction. If we consider the two sets A = { a, b, c }

and

B = { δ, ε, ζ },

it is clear that they have the same “size.” Perhaps we should say what “size” means exactly. If you think about this for a moment, you may say something like “Of course, both sets have exactly three elements.” You would be showing your heritage and knowledge. The idea of counting is a very advanced concept compared to what is needed here. a

δ

b

 ζ

c

Figure 1: A one-to-one correspondence A more primitive idea would be to compare the two sets by matching their elements as in the following: a↔δ

b↔ε

c ↔ ζ.

Perhaps you recognize that this is just the concept of producing a one-toone correspondence, or bijective function, between the two sets. We think this is a fancy description for a fairly simple idea. The comparison of infinite sets such as N, Z, Q, R and C is trickier. In fact, many people have been intimidated by the notion of infinity. The ideas of one-to-one correspondence and counting, which we considered for finite sets, lead to analogous ideas about infinite sets. In the infinite case, the notion of one-to-one correspondence is virtually no more difficult, while

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Chapter 7. Cardinality of Sets

the notion of “infinite numbers” to represent the “size” of infinite sets is far more difficult. To see why people have been intimidated by infinity for a long time, consider the paradox of Zeno, an ancient Greek philosopher. According to Zeno, you can never reach the end of a racecourse since you must first cover the first half before you can cover the whole course, and half of the remainder before covering the second half of the course, and so on, ad infinitum. That is, you cannot start from 0, go through 12 , 34 , 78 , 15 , etc., and 16 arrive at 1 in finite time. This paradox, though often forgotten or ignored, has been around for nearly 2500 years. It points to the very foundations of mathematics and issues unresolved until the end of the nineteenth century. 2. Finite Sets. We would like to formalize the definition of the “size” of a set, but rather we will do something a little different. Definition 2. Sets A and B have the same cardinality iff there exists a one-to-one correspondence f : A → B. As indicated above, the sets A = { a, b, c } and B = { δ, ε, ζ } have the same cardinality since f : A → B defined by f (a) = δ

f (b) = ε

f (c) = ζ

is a one-to-one correspondence. But you should notice that we did not say what “cardinality” is, only what “same cardinality” is. That is, we are saying that A and B have the same number of elements without ever saying what that number is! Example 3. Infinite sets are more interesting. Consider N = { 1, 2, 3, 4, . . . }

E = { 2, 4, 6, 8, . . . }.

and

9 10 ...

N

2 4 6 8 10 12 14 16 18 20 ...

E

1 2 3 4 5

6 7

8

Figure 4: A one-to-one correspondence from N to E Of course, E
N. If you think that E and N are not the same “size,” then you are not thinking of the cardinality of the sets! In fact, it is easy to see that f : N → E given by f (n) = 2n is a one-to-one correspondence. Hence E and N have the same cardinality. Let us turn our attention to finite sets now.

Section 2. Finite Sets

Exercise 5. Let A be a set. Prove that A and ∅ have the same cardinality if and only if A = ∅. The next three exercises are easy and important consequences of facts about bijective functions. These properties are called, respectively, reflexivity, symmetry, and transitivity; these are the properties of equivalence relations, which we studied in Section 3 of Chapter 6. Exercise 6. Let A be a set. Prove that A and A have the same cardinality. Exercise 7. Let A and B be sets. Prove that A and B have the same cardinality if and only if B and A have the same cardinality. Exercise 8. Let A, B and C be sets. Prove that if A and B have the same cardinality and B and C have the same cardinality, then A and C have the same cardinality. For n ∈ N, consider the sets Zn = { 1, 2, 3, . . . , n }. We will use these to further our understanding of finite sets. For example, if A and Zn have the same cardinality, then we can think of Zn as an index for A, that is, A must look like { a1 , a2 , a3 , . . . , an }. So how do we know if { a1 , a2 , a3 , . . . , am } and { b1 , b2 , b3 , . . . , bn } have the same cardinality? Clearly, the answer should be whenever m = n. You may try to prove this by taking one element at a time away from each set until one is reduced to the empty set. Then you check if the other is also reduced to the empty set. We will turn this into an induction argument. We do this first for the following special case. Theorem 9. Zm and Zn have the same cardinality if and only if m = n. Proof. That m = n implies Zm and Zn have the same cardinality follows from the fact that the identity function on Zm is a bijection. The proof in the other direction is by induction on n; that is, n is fixed while m is variable. Consider the step n = 1. Then there exists a one-toone correspondence f : Z1 → Zm . Assume m = n. Hence, m ≥ 2 and 1, 2 ∈ Zm . Since f is onto, f (1) = 1 and f (1) = 2. This contradiction shows that m = n. To prove the inductive step, we suppose the induction hypothesis that Zm and Zn have the same cardinality implies m = n. Suppose that Zm and Zn+1 have the same cardinality. Thus, there exists a one-to-one correspondence f : Zn+1 → Zm . Since 1 = n + 1, f (1) and f (n + 1) are distinct elements of Zm , m − 1 ∈ N. Let g be the permutation of Zm which

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exchanges f (n + 1) and m while leaving everything else fixed; that is, g : Zm → Zm is defined by ⎧ ⎨ f (n + 1), if x = m g(x) = m, if x = f (n + 1) . ⎩ x, otherwise (If f (n + 1) = m, then g = IZm .) Since f and g are bijections, g ◦ f : Zn+1 → Zm is a one-to-one correspondence. Since g( f (n + 1)) = m, we can define h : Zn → Zm−1 by h(x) = g( f (x)). This is a one-toone correspondence. By the induction hypothesis, m − 1 = n. Hence, m = n + 1.
We are now in a position to define what a finite set is. Definition 10. A set S is finite iff either S = ∅ or there exists some n ∈ N such that S and Zn have the same cardinality. We say that S has cardinality 0 or n, respectively, and denote this by |S| = 0 or |S| = n. Exercise 11. Suppose A and B are finite sets with cardinalities n and m, respectively. Prove that A and B have the same cardinality if and only if n = m. Exercise 12. Let A be a set with |A| = n and suppose x ∈ / A and y ∈ A. (a) Prove that |A ∪ { x }| = n + 1. (b) Prove that |A − { y }| = n − 1. Exercise 13. Let B be a finite set. Prove that if A ⊂ B, then A is finite and |A| ≤ |B|. The results of this section now enable us to prove the following theorem. Before looking at the proof, you should read the statement and try to prove it for yourself. While this elementary statement may appear obvious, it is not so easy to prove; in fact, the “counting elements”-type argument you might envision has been isolated in Exercise 13. Theorem 14. Suppose A
B. If B is finite, then A and B do not have the same cardinality. Proof. Pick x ∈ B − A. Now A ⊂ B − { x }. By Exercise 12(b), |B − { x }| = |B| − 1.

Section 3. Infinite Sets

By Exercise 13, |A| ≤ |B − { x }| = |B| − 1 < |B|. Hence, |A| = |B|. By Exercise 11, A and B do not have the same cardinality.
This theorem and its proof reflect one of the most important ideas which we presented in Chapter 3. It is an example of breaking a proof into smaller, easier pieces. These pieces, usually called lemmas, are easier to work on in isolation from the remainder of the theorem. Here the lemmas are the earlier exercises in this section: Exercises 12, 13 and 11. Now that we understand finite sets, we can prove things like the following, which seem “obvious.” For the moment, you can suppose that the familiar infinite sets, such as N and R, really are infinite. Exercise 15. Let f : A → B be a one-to-one function. (a) Prove that if B is finite, then A is finite. (b) Give a counterexample to: if A is finite, then B is finite. Exercise 16. Let f : A → B be an onto function. (a) Prove that if A is finite, then B is finite. (b) Give a counterexample to: if B is finite, then A is finite. 3. Infinite Sets. The definition of infinite sets is obvious. Definition 17. A set S is infinite iff S is not finite. The next exercise establishes the existence of infinite sets. Proposition 18. N is an infinite set. We will consider three different proofs of this statement. The first two are similar, while the third is quite different and will come later (Exercise 27). Proof #1. Assume that N is a finite set. Then we can write N in the form { n 1 , n 2 , n 3 , . . . , n k }. Hence, N has a largest element, n j for some j. Now n j + 1 > n j and, therefore, n j + 1 ∈ / N. But, since n j is a natural number, n j + 1 is also a natural number. Contradiction!


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You may object to the proof above, asking how we know that the set { n1, n2, n3, . . . , nk } has a largest element or how we know that n j + 1 > n j or, for that matter, how we know that n j +1 is a natural number. These questions are answered by the construction of the natural numbers in Chapter 10. If you find this proof reasonable now, that is OK. Here is a different, but similar proof. Proof #2. Assume that N is a finite set. Then we can write N in the form { n 1 , n 2 , n 3 , . . . , n k }. We know that 1, 2 ∈ N. Consider the number N = n 1 +n 2 +n 3 +· · ·+n k . N is a natural number. However, N > n j for all j since n j + 2 > n j + 1 > n j and 1, 2 and n j are at least two different elements of N. So N ∈ / N. Contradiction!
You may have similar objections to Proof #2 as to Proof #1. A different proof is contemplated in Exercise 27 based on the following important characterization theorem for finite and infinite sets. Theorem 19. Let S be a set. (a) S is infinite if and only if it has a proper subset of the same cardinality. (b) S is finite if and only if it has no proper subset of the same cardinality. As in the proof of Theorem 14, the proof of Theorem 19 is also completed in steps. Theorem 14 provided half of part (b); half of part (a) is to be proved in the following exercise. Exercise 20. Prove that if a set B has a proper subset with the same cardinality, then B is infinite. Our next goal is to prove the converses of Theorem 14 and Exercise 20, thereby completing the proof of Theorem 19. Definition 21. A set S is denumerable iff S and N have the same cardinality. Sets which are denumerable are also called countably infinite. Definition 22. A set S is countable iff S is either finite or denumerable. Since a denumerable set is in one-to-one correspondence with N, we can can think of it as an infinite list indexed by the natural numbers such as { a1 , a2 , a3 , . . . }.

109

Section 3. Infinite Sets

Theorem 23. Every infinite set contains a denumerable subset. Proof. Let A be an infinite set. We wish to construct a denumerable subset D = { a1 , a2 , a3 , . . . } by identifying elements a1 , a2 , etc., in that order. The proof is by induction on the index of the elements. Choose a1 ∈ A; this is the first step, n = 1, of the induction. Next we wish to prove the inductive step: if Dk = { a1 , a2 , a3 , . . . , ak } ⊂ A, then there exists ak+1 ∈ A − Dk . The induction hypothesis is that this is true for k = n; we need to show that it is then true for k = n + 1. Assume that it is false for k = n + 1. That means that A − Dn = ∅. Hence A = Dn+1 is finite. This is a contradiction. That is, we have constructed a sequence of distinct elements of A: { a1 , a2 , a3 , . . . }, as desired.




We should pause here to consider the last sentence of the proof of Theorem 23. It looks deceptively simple. The induction actually provides a sequence of finite sets D1 , D2 , D3 , . . . and not a denumerable set. In fact, mathematicians isolated an axiom in the eighteenth century that is required in circumstances exactly like this one. It is called the Axiom of Choice and most, but not all, mathematicians use and accept this axiom. We do not wish to delve into the intricacies of the Axiom of Choice. Suffice it to say that whenever an infinite number of choices are made, the Axiom of Choice is being implemented, unless a universal rule exists to make our infinitely many choices clear in one fell swoop. For example, to choose one sock from each of an infinite number of pairs of socks requires the Axiom of Choice; to choose one shoe from each of an infinite number of pairs of shoes does not require the Axiom of Choice since we can choose all of the left shoes. Having said all of this, we now present a new proof of Theorem 23, written as most mathematicians might write it. Second Proof of Theorem 23. Let A be an infinite set. Since A is infinite, A is nonempty. Choose a1 ∈ A. Now choose a2 ∈ A different from a1 ; this can be done since A − { a1 } is nonempty. Continue in this way: having chosen a1 , a2 , a3 , . . . , ak we choose ak+1 ∈ A−{ a1 , a2 , a3 , . . . , ak } which can be done since the difference of a finite set from an infinite set is always nonempty. Continuing this process indefinitely, we construct a denumerable subset { a1 , a2 , a3 , . . . } of A.
The next exercise requests a proof of a useful lemma. You will then use it to prove the converses of Exercise 20 and Theorem 14.

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Chapter 7. Cardinality of Sets

Exercise 24. Suppose D is denumerable and x ∈ D. Prove that D − { x } is denumerable. Exercise 25. Prove that every infinite set has a proper subset of the same cardinality. Exercise 26. Let S be a set. Prove that if no proper subset of S has the same cardinality as S, then S is finite. This completes the proof of Theorem 19. If it interests you, you can keep in mind that the equivalent definitions of finite and infinite sets, as given in Theorem 19, depend on the Axiom of Choice. We now return to the proof that N is infinite, Proposition 18. The equivalent definition of infinite in Exercise 25 gives us a new proof of this fact. Exercise 27. Use Theorem 19 to prove that N is infinite. The fact that R is infinite now follows. Exercise 28. Prove that R is infinite. 4. Countable Sets. As we noted, there are two kinds of infinite sets: those which are countable and those which are not countable, that is, uncountable. We consider countable sets next, remembering that countable sets are either finite or denumerable. Theorem 29. Every subset of a countable set is countable. Proof. Let A be a countable set. We have already settled the case where A is finite in Exercise 13. So suppose that A is denumerable. Then we can list the elements of A as, say, A = { a1 , a2 , a3 , . . . }. Let B be a subset of A. Then, using the indices of the list of the elements of A, we let S = { n ∈ N | an ∈ B } . Then k &→ ak defines a one-to-one correspondence from S to B. Either S is finite or infinite. If S is finite, then B is finite and, therefore, countable.

Section 4. Countable Sets

Suppose S is infinite. Since we can order the elements of S starting with the smallest, we can call the elements of S n1, n2, n3, . . . in increasing order. Now k &→ n k defines a one-to-one correspondence from N to S. Hence S and, therefore, B are denumerable.
We next ask the question whether all infinite sets are countable? The answer is negative. However, we will see that the sets N, Z and Q are all countable. Exercise 30. Prove that all denumerable sets have the same cardinality. Despite the facts that N is a proper subset of Z and that Z appears to have, roughly, twice as many elements as N, they are shown to have the same cardinality by the following exercise. Exercise 31. Prove that Z is denumerable. Now consider Q, the set of rational numbers. An amazing theorem proved by Georg Cantor in 1874 shows that Q also has the same cardinality as N. The proof depends on a diagonal argument that we present next. Theorem 32. The set of positive rational numbers, Q+∗ , is denumerable. Proof. Consider the following table of the elements of Q+∗ : 1 1 1 2 1 3 1 4 1 5 1 6

.. .

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

.. .

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

.. .

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

.. .

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

.. .

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

.. .

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

We use the pattern in Figure 33 on the table above to create a list of the positive rationals, skipping over duplicate rationals in the table as we go along.

111

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Chapter 7. Cardinality of Sets

... ...

... ...

... ...

... ...

... ... ...

...

...

...

...

...

...

...

...

...

...

...

...

Figure 33: Cantor’s scheme That list is   Q+∗ = 1, 2, 12 , 13 , 3, 4, 32 , 23 , 14 , 15 , 5, 6, 52 , . . . . Hence, Q+∗ is denumerable.




Exercise 34. Prove that Q is denumerable. Exercise 35. Prove that a countable union of countable sets is countable. Exercise 36. Prove that a finite product of countable sets is countable. It turns out that countable products of finite sets need not be countable; this will be shown in Exercises 54 and 55. Definition 37. The set of algebraic numbers is the union of the real solution sets of all polynomial equations (in one variable) with integer coefficients. Exercise 38. Prove that the set of algebraic numbers is denumerable. 5. Uncountable Sets. To complete this chapter we will consider uncountable sets. Yes, they do exist; that is, there are different orders of infinity! Definition 39. A set S is uncountable iff S is not countable. So, every set is exactly one of the following properties: finite or denumerable or uncountable. The existence of uncountable sets is shown next. The proof below is again due to Georg Cantor in 1891.

113

Section 5. Uncountable Sets

Theorem 40. The subset [0, 1] ⊂ R is uncountable. Proof. Assume that [0, 1] is countable. Then we can list the elements of [0, 1] as [0, 1] = { r1 , r2 , r3 , . . . }. Every real number can be written as an infinite decimal; those in [0, 1] can be written with 0 to the left of the decimal point. Therefore, the elements of [0, 1] can be written in the form r1 = 0.a1,1 a1,2 a1,3 a1,4 a1,5 . . . r2 = 0.a2,1 a2,2 a2,3 a2,4 a2,5 . . . r3 = 0.a3,1 a3,2 a3,3 a3,4 a3,5 . . . r4 = 0.a4,1 a4,2 a4,3 a4,4 a4,5 . . . r5 = 0.a5,1 a5,2 a5,3 a5,4 a5,5 . . . .. . where the ai, j ’s are the j th decimal places of ri . Define the number  r = 0.b1 b2 b3 b4 b5 . . .

where bn =

3, if an,n = 3 4,

if an,n = 3

.

Clearly, r = rn for all n ∈ N since they differ in the n th decimal place. Therefore, r ∈ [0, 1], but r ∈ / { r1 , r2 , r3 , . . . }. This is a contradiction.
In the sense of cardinality, there are more real numbers between any two given rational numbers than there are rational numbers! Exercise 41. Prove that R is uncountable. There are lots of examples of uncountable sets. With the following six exercises, we will show that all nonempty open intervals (including R) have the same cardinality, and are therefore uncountable. Exercise 42. Prove that R and (0, 1) have the same cardinality. Exercise 43. Prove that R and (0, +∞) have the same cardinality. Exercise 44. Prove that, for all real numbers a, (a, +∞) and (0, +∞) have the same cardinality.

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Exercise 45. Prove that, for all real numbers a, (−∞, −a) and (a, +∞) have the same cardinality. Exercise 46. Prove that, for all real numbers a and b with a < b, (a, b) and (0, 1) have the same cardinality. Exercise 47. Prove that all nonempty open intervals have the same cardinality. In fact, all nonempty and non-singleton intervals are uncountable and have the same cardinality. Exercise 48. Prove that (0, 1), [0, 1), (0, 1] and [0, 1] have the same cardinality. Exercise 49. Prove that all nonempty and non-singleton intervals have the same cardinality. Recall that a number is irrational if it is not rational. That is, the set of irrational numbers is R − Q. Exercise 50. Prove that the set of irrational numbers is uncountable. Recall that a number is transcendental if it is not algebraic. Exercise 51. Prove that the set of transcendental numbers is uncountable. The next three exercises show that uncountable sets naturally arise from countable processes. First, we must define products of infinite collections of sets. In Section 2 of Chapter 5, we defined finite products. Actually, we only considered products of two sets and R 3 , a product of three sets; this is easily generalized to finite products. A collection, S, of sets is indexed by a set K if S = { Sk | k ∈ K } . In an indexed collection, the elements need not be distinct; that is, an indexed collection may not be a set. For example, {Z |k ∈ N } is an indexed collection consisting of denumerably many copies of Z.

115

Section 5. Uncountable Sets

Definition 52. Suppose S = { Sk | k ∈ K } is a collection of sets indexed by a set K . The product of the sets in S is the set of all functions f : K → ∪S with f (k) ∈ Sk for all k ∈ K . Suppose S is a set of sets; every set in S appears exactly once. In this case, we can think of the product of the sets in S as the set of all functions f : S → ∪S with f (S) ∈ S for all S ∈ S. You may think of this as S being indexed by S since S is a set. As long as you distinguish multiple occurrences of a set S in S, you can do this. As for notation, we use   Sk or S k∈K

S∈S

to denote the product of the sets in S. Some mathematicians call this a direct product and some use the notation

×S

k

×S.

or

k∈K

S∈S

It is one of the most amazing facts in mathematics that the product of infinitely many non-empty sets cannot generally be proven non-empty except by using the Axiom of Choice. The notation YX =



Y

x∈X

is also commonly used for the set of all functions from X to Y . Example 53. The collection S = { S1 , S2 } is indexed by Z2 . By definition, ! k∈Z2 Sk is the set of all functions f : { 1, 2 } → S1 ∪ S2 with f (k) ∈ Sk for both k ∈ Z2 . That is, for k = 1, 2, an element of the product depends on elements s1 = f (1) ∈ S1 and s2 = f (2) ∈ S2 . We can think of this as the ordered pair (s1 , s2 ). Of course, this is a generic element of S1 × S2 according to Definition 1 of Chapter 5. In light of Example 53, though this is not technically correct, when K is a finite set such as K = Zn = { 1, 2, . . . , n }, then we write S1 × S2 × · · · × Sn =



Sk .

k∈Zn

We ask you to believe that this is not circular: products of collections of sets are defined in terms of functions that are defined in terms of products of two sets.

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! Exercise 54. Let S = { 0, 1 } and let P = n∈N S = S × S × S × . . . . (P is a denumerable product of copies of S.) Prove that P is uncountable. Exercise 55. Prove that the denumerable product of finite sets, each with two or more elements, is uncountable. Exercise 56. Prove that the power set of N is uncountable. Moreover, P(N) and R have the same cardinality. The proof of this depends on showing that every element of (0, 1) has a (unique) binary representation, corresponding to a sequence of 0’s and 1’s (which corresponds to an element of P(N) as suggested by the hint for Exercise 56). The following result is often called Cantor’s Theorem. Exercise 57. Let S be a nonempty set. Prove that the power set P(S) and S do not have the same cardinality. Remark. It is interesting to note that it is not clear, from what we have done here, whether there exists an uncountable subset of R that does not have the same cardinality as R. The rejection of this possibility is called the Continuum Hypothesis. It turns out that, as is true for the Axiom of Choice, the Continuum Hypothesis is an independent axiom; that is, both its acceptance and its rejection are consistent with our usual set of axioms.

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PART III CONSTRUCTION OF NUMBER SYSTEMS

118

Chapter 8 Algebra of Number Systems 1. Primary Properties of Number Systems. The goal of Part III is to construct the familiar sets of numbers, together with the usual arithmetic operations and the usual ordering. We do this by starting with known concepts from set theory. The empty set will give us our starting number, zero. From zero, we use set theoretic tools to construct the natural numbers. Zero and the natural numbers lead us to the integers, then to the rational numbers, then to the real numbers and, finally, to the complex numbers. This succession of constructions creates each new set such that it extends the previous set. In each construction, except the complex numbers, a total ordering will be given. Indeed, we will see that there is no ordering of the complex numbers that extends the properties of the ordering of the real numbers. In each construction, we will also have two operations, addition and multiplication. We have already made precise what it means for ≤ to be a total ordering (or, equivalently, what it means for < to be a strict total ordering) in Section 2 of Chapter 6. The following definition makes precise what it means to have an operation like addition or multiplication. Definition 1. A binary operation on a set X is a function  : X × X → X . For x1 , x2 ∈ X , we denote (x1 , x2 ) by x1  x2 . We will call a binary operation, more simply, an operation. Some mathematicians prefer binary operator or operator. You should recognize addition and multiplication as operations. Notice the comment on notation in the definition of operation: the functional notation for addition, +(x, y), looks strange compared with the more natural x + y. Remark 2. We always use the symbol + to denote addition. We also will follow the usual conventions by writing multiplication in numerous ways: x · y = x × y = (x)(y) = x y. The usual order of operations applies: multiplications have priority over additions. Hence, x + yz = x + (yz) and (usually!) x + yz = (x + y)z.

Section 1. Primary Properties of Number Systems

Remark 3. Consider the following properties on a set X with addition and multiplication operations: (A0) Closure of Addition: for all x, y ∈ X , x + y ∈ X (M0) Closure of Multiplication: for all x, y ∈ X , x y ∈ X The closure properties are often listed as special properties of addition and multiplication. Since we have defined operations as functions from X × X to X , the closure properties are redundant. That is, +:X×X →X

and

·:X×X →X

are functions. Therefore, if x, y ∈ X , then (x, y) ∈ X × X and, hence, + (x, y) ∈ X

and

· (x, y) ∈ X.

This notation is clumsy and we revert to the usual notation for addition and multiplication: x + y = +(x, y)

and

x y = ·(x, y).

In this chapter and throughout Part III of this text, we will concern ourselves with the algebraic properties of addition and multiplication operations on totally ordered sets of numbers. Definitions 4 and 8, given below, mimic the familiar properties of the real numbers. You should pay careful attention to the properties given in these definitions. We will not give the most general definitions for the terms being defined, which are field and ordered field, since we are only interested in whether our constructions satisfy the individual properties. Definition 4. A set X with two operations, + and ·, is a field iff the following properties are satisfied, for all x, y, z ∈ X : (AC) Commutativity of Addition: x+y = y+x (AA) Associativity of Addition: (x + y) + z = x + (y + z) (AZ) Existence of a Unique Additive Identity: there exists a unique 0 ∈ X such that 0 + x = x + 0 = x (AI) Existence of Unique Additive Inverses: there exists a unique −x ∈ X such that x + (−x) = (−x) + x = 0 (MC) Commutativity of Multiplication: x y = yx (MA) Associativity of Multiplication: (x y)z = x(yz) (MU) Existence of a Unique Multiplicative Identity:

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there exists a unique 1 ∈ X , with 1 = 0, such that 1x = x1 = x (MI) Existence of Unique Multiplicative Inverses: if x = 0, there exists a unique x −1 ∈ X such that x x −1 = x −1 x = 1 (DP) Distributive Properties: (x + y)z = x z + yz and x(y + z) = x y + x z The two letters between parentheses before each property name should be suggestive of the name of the property. For example, (AC) stands for Addition is Commutative. The additive identity of a field is often called a zero and the multiplicative identity of a field is often called a unity: (AZ) refers to Zero and (MU) refers to Unity. Remark 5. According to Definition 4, 0 = 1 in a field. Suppose all of the other field properties hold and 0 = 1; it will follow from Exercise 20, which shows that multiplication by 0 yields 0, that X = { 0 } since, for all x ∈ X, x = 1x = 0x = 0. To avoid this trivial case, most authors include the property that 0 = 1 in their definitions of field, as we have done in (MU). Remark 6. Notice that the two Distributive Properties, if written as follows, might be called Factoring Properties: x z + yz = (x + y)z

and

x y + x z = x(y + z).

Remark 7. Notice that if X has a commutative multiplication, then the two Distributive Properties are equivalent. These are often called the left and right distributive properties. Such choices are arbitrary and are not made consistently. In our constructions of N, Z, etc., we will always have a commutative multiplication. Similar remarks are in order for identity elements and inverses. One can define things like left identity and right inverse; for a commutative operation, lefts and rights are equal and we simply call them identity and inverse, respectively. Next, we add an ordering to the mix. Definition 8. A field X with a total ordering < is an ordered field iff the following properties are satisfied, for all x, y, z ∈ X : (PP) Product of Positives: if x > 0 and y > 0, then x y > 0 (IX) Additive Cancellation in Inequalities: if x + y < x + z, then y < z Remark 9. Consider property (IX). That y < z implies x + y < x + z should strike you as obvious. But how do we know this? Perhaps you

Section 1. Primary Properties of Number Systems

would say that this is just “substitution.” We need not appeal to an Axiom of Substitution; in Exercise 25, we prove this type of substitution is implied by two kinds of cancellation properties. Remark 10. Any element z that satisfies z + x = x + z = x for all x ∈ X is called an additive identity of X . Any element y that satisfies y + x = x + y = 0 for all x ∈ X is called an additive inverse of x in X . Property (AZ) of a field guarantees the existence of a unique additive identity and property (AI) guarantees the existence of a unique additive inverse for every element of the field. The uniqueness actually follows from the existence according to the following two propositions. Proposition 11. If a set X with addition has an additive identity, then this identity is unique. Proof. Assume that there exist two distinct additive identities 0 and 0 . Then 0 = 0 + 0 = 0. Hence, 0 = 0. Contradiction! Therefore, 0 is the unique additive identity.
The proof of Proposition 11 is a standard proof for many kinds of identity elements. It is sometimes described as “letting the two identities fight it out.” The uniqueness of inverses is similarly proved by letting two inverses fight it out. You should compare this with Exercise 47 and 52 of Chapter 5. Proposition 12. Suppose X is a set with addition satisfying properties (AA) and (AZ). If x ∈ X has an additive inverse, then this inverse is unique. Proof. Assume that −x and x  are distinct additive inverses of x. Then x  = x  + 0 = x  + (x + (−x)) = (x  + x) + (−x) = 0 + (−x) = −x. So x  = −x. Contradiction! Therefore, −x is the unique additive inverse of x.
The existence of multiplicative identity and multiplicative inverses (of nonzero elements) similarly implies their uniqueness. Exercise 13. Prove that if a set X with multiplication has an multiplicative identity, then this identity is unique.

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Exercise 14. Suppose X is a set with multiplication satisfying properties (MA) and (MU). Prove that if x ∈ X has a multiplicative inverse, then this inverse is unique. As we indicated above, it is not our purpose to study fields or ordered fields in their full generality. You may already have realized that N and Z are not ordered fields. So we will really study sets with addition and multiplication operations and, possibly, a total ordering that satisfy some, but not necessarily all, of the properties listed in this section. 2. Secondary Properties Involving Addition and Multiplication. One of the most useful aspects of algebra is that the properties characterize the structures, as suggested by the attempt, in the definition of ordered field, to characterize R. Moreover, the properties themselves may be used to prove other properties, without ever using any direct knowledge about the sets or operations involved. We will consider a few secondary properties. At first, we will assume that X has only those properties—mostly primary properties—indicated. Later we will simply let X be a field or an ordered field. We could give a minimal list of conditions that imply the desired conclusion; however, there may exist two or more distinct minimal lists. That is, a sufficient condition may not be necessary. (See the comments after Exercise 16.) The following three properties depend only on addition. Remark 15. Suppose that X has an addition operation and that x, y, z ∈ X . The following seems reasonable: if y = z, then x + y = x + z. You may regard this as a kind of substitution: z is substituted for y. However, you can also reason as follows: x = x and y = z imply (x, y) = (x, z), which implies, since addition is a function, that x + y = x + z. Similarly, if y = z, then y + x = z + x. These properties are sometimes referred to as Adding Equals to Equals. The converse of Adding Equals to Equals, for example, that x + y = x + z implies y = z, is called Cancellation; compare it with property (IX). This property requires proof. Exercise 16: (AX) Additive Cancellation in Equations. Suppose X is a set with addition satisfying properties (AA), (AZ), and (AI). (a) Prove that x + y = x + z if and only if y = z for all x, y, z ∈ X . (b) Prove that y + x = z + x if and only if y = z for all x, y, z ∈ X . In proving Exercise 16, you used additive inverses. Consider for a moment whether additive cancellation holds in N. Of course, for y ∈ N,

Section 2. Secondary Properties Involving Addition and Multiplication

5 + y = 5 + z seems to imply that y = z. We will prove this is true in Chapter 10. The proof will be very different from your proof of Exercise 16 since there are no additive inverses in N: for example, −5 ∈ / N. As we noted at the beginning of this section, the conditions we give are not, in general, necessary and sufficient. The next exercise gives a familiar property about double negatives. Exercise 17: Double Additive Inverses. Suppose X is a set with addition satisfying properties (AA), (AZ), and (AI). Prove that, for all x ∈ X , −(−x) = x. If your proof of Exercise 17 followed the technique of making two inverses fight it out, then you gave the expected proof. However, there is another proof that does not require (AA)! Notice that (AI) says that −x is the unique inverse of x in X if and only if x + (−x) = (−x) + x = 0. This same condition indicates that x is the unique inverse of −x in X . What is −(−x)? The following two properties depend only on multiplication and are similar to the previous properties of addition except for the avoidance of multiplication by 0. Exercise 18: (MX) Multiplicative Cancellation in Equations. Suppose X is a set with multiplication satisfying properties (MA), (MU), and (MI). (a) Prove that x y = x z if and only if y = z for all x, y, z ∈ X with x = 0. (b) Prove that yx = zx if and only if y = z for all x, y, z ∈ X with x = 0. In the proof of Exercise 18, you should have noticed that one direction of the equivalences is easy; these are often referred to as Multiplying Equals by Equals. Exercise 19: Double Multiplicative Inverses. Suppose X is a set with multiplication satisfying properties (MA), (MU), and (MI). Prove that, for all x ∈ X with x = 0, (x −1 )−1 = x. Try to prove this without using (MA).

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The remaining properties in this section use both addition and multiplication. The next property is the last one that we list depending on only some of the properties of a field. In (MI) we consider multiplicative inverses of nonzero elements. Does 0 have a multiplicative inverse? Can x −1 = 0? The next exercise addresses these issues. Exercise 20: Multiplication by 0. Suppose that X is a set with addition and multiplication satisfying properties (AZ), (AX) and (DP). Prove that, for all x ∈ X , 0x = x0 = 0. Since 1 = 0, 0 has no multiplicative inverse. Also, a multiplicative inverse cannot equal 0. In a field X , since all of the assumed properties hold, 0x = x0 = 0 for all x ∈ X . Consider the following statement, the converse of Exercise 20. Exercise 21: 0 Product Property. Suppose that X is a field. Prove that, for all x, y ∈ X , if x y = 0, then x = 0 or y = 0. In a field X , the following are equivalent for all x, y ∈ X : • if x = 0 and y = 0, then x y = 0 , • if x = 0 and x y = 0, then y = 0 for all x, y ∈ X , • if y =  0 and x y = 0, then x = 0 for all x, y ∈ X . Combining the Multiplication by 0 and the 0 Product Properties, we see that in a field X , for all x, y ∈ X , x y = 0 if and only if x = 0 or y = 0. The following two properties depend on the Uniqueness of Additive Inverses. Exercise 22: Factoring a Minus Property. Suppose that X is a field. Prove that, for all x, y ∈ X , (−x)y = x(−y) = −(x y).

Section 3. Secondary Properties Involving Order

Exercise 23: Product of Negatives Property. Suppose that X is a field. Prove that, for all x, y ∈ X , (−x)(−y) = x y. 3. Secondary Properties Involving Order. The properties in this section involve operations, addition and multiplication, as well as an order relation. Given an ordering, < or ≤, on a set X , we have three transitive relations on X : <, ≤, and =. (Recall, from Exercise 14 of Chapter 6, that we can always derive < from ≤ or vice versa.) The following proposition may seem obvious by “substitution,” but we can prove it. Proposition 24. Suppose X is a set with a strict partial ordering < and x, y, z ∈ X . If x < y and y = z, then x < z. Proof. By the definition of ≤, x < y implies x ≤ y and y = z implies y ≤ z. Hence, by transitivity of ≤, x ≤ z. Assume that x = z. By symmetry of =, z = y. By transitivity of =, x = y. However, x < y, contradicting the trichotomy of <. Hence, x = z. Therefore, x < z.
Since the preceding proposition involves both an order relation (<) and an equivalence relation (=), it is not simply a transitive property. Similar properties, involving other permutations of <, ≤, and =, are given in Exercise 51. The following property is the converse of property (L2) Additive Cancellation in Inequalities. It is more complicated than the property of Adding Equals to Both Sides of an Equation. Exercise 25: Adding Equals to Both Sides of an Inequality. Suppose that X is an ordered field. Prove that, for all x, y, z ∈ X , if y < z, then x + y < x + z. In the following exercises, we explore the dualities between less than and negative and between greater than and positive. Exercise 26: Negation Reverses Order Property. Suppose that X is an ordered field. Prove that x > 0 if and only if −x < 0 for all x ∈ X .

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Exercise 27: Product by Positive Preserves Order Property. Suppose that X is an ordered field. Prove that if x > 0 and y < z, then x y < x z for all x, y, z ∈ X . Exercise 28: Product by Negative Reverses Order Property. Suppose that X is an ordered field. Prove that if x < 0 and y < z, then x y > x z for all x, y, z ∈ X . Exercise 29: Squares of Nonzero Numbers are Positive. Suppose that X is an ordered field. Prove that if x = 0, then x 2 = x x > 0 for all x ∈ X . Exercise 30: Multiplicative Inverses Reverse Order Property. Suppose that X is an ordered field. Prove that if y > x > 0, then 0 < y −1 < x −1 for all x, y ∈ X . 4. Isomorphisms and Embeddings. In mathematics, one often studies structures which are the same in all respects except for the names of the elements, relations, operations, etc. which define them. Such structures are called isomorphic structures. To better understand what kind of structure we are referring to, consider R as the underlying set of our structure with the order relation and operations of addition and multiplication providing the structure. Consider the following definition. Definition 31. Let X and Y be totally ordered sets together with addition and multiplication. We denote the strict orderings by < and the operations by + and × for both X and Y . X and Y are isomorphic structures iff there exists a bijective function f : X → Y , called an isomorphism, satisfying the following three conditions for all x1 , x2 ∈ X : x1 < x2 if and only if f (x1 ) < f (x2 ), f (x1 + x2 ) = f (x1 ) + f (x2 ), f (x1 × x2 ) = f (x1 ) × f (x2 ). We call the three conditions in the definition of an isomorphism order-preserving, addition-preserving, and multiplication-preserving, respectively. Taken collectively, we say that an isomorphism is structurepreserving.

Section 4. Isomorphisms and Embeddings

The following example and the discussion and exercises following it should shed some light on the nature of isomorphisms. Example 32. Let E be the set of even integers. Define a function f : Z → E by f (n) = 2n. f is clearly bijective. f is order- and addition-preserving since m < n if and only if f (m) = 2m < 2n = f (n) and f (m + n) = 2(m + n) = 2m + 2n = f (m) + f (n). However, f is not multiplication-preserving since f (1 · 1) = f (1) = 2 while f (1) · f (1) = 2 · 2 = 4. Note that this shows that f is not an isomorphism; it does not show that Z is not isomorphic to E using some other function as an isomorphism. The following exercise lists some of the many ways that isomorphisms preserve structure. Exercise 33. Suppose that X and Y are sets together with addition and multiplication (no order relations are assumed here). Suppose that f : X → Y is an isomorphism, that is, a bijection which preserves addition and multiplication. Prove each of the following: (a) If 0 is the additive identity in X , then f (0) is the additive identity in Y . (b) If 1 is the multiplicative identity in X , then f (1) is the multiplicative identity in Y . (c) If −x is the additive inverse of x in X , then f (−x) is the additive inverse of f (x) in Y . (d) If x −1 is the multiplicative inverse of x in X , then f (x −1 ) is the multiplicative inverse of f (x) in Y . Results such as those of the previous exercise are useful in showing that structures are not isomorphic. Example 34. We know that 1 is the multiplicative identity of Z. Does E have a multiplicative identity? Since 1 ∈ / E, the answer appears to be no. However, this does not prove it since another element of E could be the multiplicative identity. Assume n ∈ E is the multiplicative identity in E. Therefore, 2n = 2. Since 2, n ∈ E and E ⊂ Z, 2, n ∈ Z. In Z, 2 = 2(1). Hence, 2n = 2(1) and, by Multiplicative Cancellation, n = 1. This implies that 1 ∈ E, a contradiction. (We will check that all necessary properties of Z hold in Chapter 11.) Exercise 35. Prove that Z is not isomorphic to E.

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Since we think of isomorphic structures as the same, the following idea is natural. Exercise 36. Consider “is isomorphic to” as a relation on a collection of sets with addition and multiplication. Prove that this is an equivalence relation. The equivalence classes for the relation in Exercise 36 are called isomorphism classes. Consider the following definition. Definition 37. Let X and Y be totally ordered sets together with addition and multiplication. X is embedded in Y if there exists Z ⊂ Y and an isomorphism f : X → Z . The function g : X → Y defined by g(x) = f (x) is called an embedding. Notice that an embedding is a one-to-one function that is order-preserving, addition-preserving, and multiplication-preserving. We could have done this differently by defining an embedding this way. Then the existence of the isomorphism f : X → g(X ) is assured. (You should verify this.)

Supplemental Exercises

Supplemental Exercises Exercise 38. Give reasons for every step in the proof of Proposition 11. Exercise 39. Give reasons for every step in the proof of Proposition 12. Exercise 40. In Exercise 20, we essentially showed that the distributive properties imply 0x = x0 = 0. Consider the two distributive properties separately. Which one implies that 0x = 0 and which one implies x0 = 0? Exercise 41. Suppose that X = { 0 }. How many different operations can you find on X ? Define addition and multiplication on X in the obvious way. Verify that X with this addition and multiplication is a field. Exercise 42. Before continuing beyond this chapter, use your knowledge of the sets N, Z+ , Z, Q, R, and C to complete the following table. For each set, with the usual addition, multiplication, and ordering, write Yes or No for each property of an ordered field. For every No, try to give a reason or counterexample. (For example, N does not satisfy property A3, the existence of an additive identity, since 0 ∈ / N and Z does not satisfy property M4, the existence of multiplicative inverses, since 2 has no multiplicative inverse in Z.) Property N Z+ Z Q R C (AC) Commutative Addition (AA) Associative Addition (AZ) Additive Identity 0 (AI) Additive Inverses (MC) Commutative Multiplication (MA) Associative Multiplication (MU) Multiplicative Identity 1 (MI) Multiplicative Inverses (DP) Distributive Properties (IX) Cancellation in Inequalities (PP) Product of Positives Exercise 43. Let X be the set of all 2 × 2 matrices with real entries. Using ordinary addition and multiplication of matrices, determine which of the properties of a field hold. (Check each property for addition, (AC), (AA), (AZ), and (AI), for multiplication, (MC), (MA), (MU), and (MI), and the distributive properties (DP).) Exercise 44. Suppose X has addition and multiplication operations. Prove that if multiplication is commutative, then the First Distributive Property is equivalent to the Second Distributive Property.

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Exercise 45. Explain why a multiplication operation is closed. Explain why any binary operation is closed. Exercise 46. Prove that in a field X , (−1)x = x(−1) = −x for all x ∈ X . Exercise 47. Suppose that Z3 = { 1, 2, 3 } and addition and multiplication are defined by the following tables. + 1 2 3 · 1 2 3 1 2 3 1 1 1 2 3 2 3 1 2 2 2 1 3 3 1 2 3 3 3 3 3 Prove that Z3 is a field. Exercise 48. Consider the field Z3 from Exercise 47. Prove that the only field automorphism of Z3 is the identity function, IZ3 . Exercise 49. Suppose that V = { 0, 1, a, b } and addition and multiplication are defined by the following tables. + 0 1 a b × 0 1 a b 0 0 1 a b 0 0 0 0 0 1 1 0 b a 1 0 1 a b a a b 0 1 a 0 a a 0 b b a 1 0 b 0 b 0 b Is V a field? If not, indicate all properties in the definition of a field that fail. Exercise 50. Consider V = { 0, 1, a, b } with the addition and multiplication operations given in Exercise 49. Prove that the function f : V → V defined by f (0) = 0, f (1) = 1, f (a) = b and f (b) = a is an automorphism; i.e., the two operations are preserved. Exercise 51. Suppose X is a set with a partial ordering ≤ and x, y, z ∈ X . (a) Prove that if x = y and y < z, then x < z. (b) Prove that if x < y and y ≤ z, then x < z. (c) Prove that if x ≤ y and y < z, then x < z. (d) Prove that if x ≤ y and y = z, then x ≤ z. (e) Prove that if x = y and y ≤ z, then x ≤ z. Exercise 52. Suppose F is an ordered field. Prove that, for all x, y ∈ F, x < y if and only if −y < −x. Exercise 53. Suppose X is an ordered field. Prove that if 0 = 1, then 1 > 0. Exercise 54. Suppose X is a field. Prove that −(x + y) = (−x) + (−y).

131

Chapter 9 Archimedean Ordered Fields 1. The Main Goal and Archimedean Principle. In constructing the number systems, we will check which of the properties of an ordered field are satisfied and whether the least upper bound property is satisfied. As we progress towards the real numbers, the new constructs will possess more of these properties than their predecessors. This culminates in the construction of the real numbers; the complex numbers form an interesting digression or continuation, depending on your personal point of view. The main goal of Part III of this text is to prove the following theorem, which is proved in Section 5 of Chapter 13. Main Theorem (Theorem 28 of Chapter 13): The Uniqeness of R. There exists, up to isomorphism, a unique ordered field with the least upper bound property. By “unique up to isomorphism” we mean unique isomorphism class. From the constructions, we will see that R, with the usual ordering and operations, is the ordered field with the least upper bound property. Let us now prove two important properties of ordered fields with the least upper bound property. Let F be an ordered field with the least upper bound property. We denote the sets of natural, integral and rational field elements of F, respectively, by FN = { m ∈ F | m = 1 + 1 + · · · + 1 for finitely many 1’s } , FZ = { m ∈ F | m = 0 or m ∈ FN or − m ∈ FN } , and   FQ = mn −1 | m ∈ FZ and n ∈ FN . The following result is usually called the Archimedean Principle. Theorem 1. Let F be an ordered field with the least upper bound property. If x, y ∈ F and x > 0, then there exists n ∈ FN such that nx > y.

132

Chapter 9. Archimedean Ordered Fields

x 0

2x 3x

mx (m+1)x s-x

s

y

Figure 2: The set S in the proof of Theorem 1 Proof. Let S = { nx | n ∈ FN }. Since x ∈ S, S is nonempty. Assume that nx ≤ y for all n ∈ FN . Then y is an upper bound of S. Since F has the least upper bound property, S has a least upper bound, s. Now x > 0 implies s − x < s. Thus, s − x is not an upper bound of S. Hence, there exists m ∈ FN such that s − x < mx. Therefore, s < (m + 1)x. This contradicts the fact that s is an upper bound of S.
The Archimedean Principle is intimately related to the fact that the natural numbers are not bounded. Exercise 3. Let F be an ordered field with the least upper bound property. Prove that FN is not bounded. The following corollary of the Archimedean Principle is a useful, stronger version of the Archimedean Principle. Corollary 4. Let F be an ordered field with the least upper bound property. If x, y ∈ F and x > 0, then there exists m ∈ FZ such that mx ≤ y < (m + 1)x. Proof. By the Archimedean Principle, there exists q ∈ FN such that q x > y. Next, we show that there exists p ∈ FZ such that px < y. Consider two cases: either y ≥ 0 or y < 0. Case 1. If y ≥ 0, then −x < 0 ≤ y. Set p = −1 so that px < y. Case 2. If y ≥ 0, then, by the Archimedean Principle, there exists n ∈ FN such that nx > −y. Hence, (−n)x < y. Set p = −n so that px < y. So, there exist p, q ∈ FZ with p < 0 and q > 0 such that px < y < q x. Consider the set S = { n ∈ FZ | n ≥ p and nx ≤ y } . S is nonempty since p ∈ S. If n ∈ S, then nx < y < q x and, hence, n < q. So q is an upper bound of S and S is a finite set. Since a finite, totally ordered set has a greatest element, let m be the greatest element of

133

Section 1. The Main Goal and Archimedean Principle

S. Since m ∈ S, mx ≤ y. Since m + 1 > m, m + 1 ∈ / S. So (m + 1)x ≤ y. Therefore, mx ≤ y < (m + 1)x.
Exercise 5. Let F be an ordered field with the least upper bound property. If x, y ∈ F and x < 0, then there exists m ∈ FN such that mx ≤ y < (m − 1)x. For the following theorem, we sometimes say “the rational field set FQ is dense in the field F.” Theorem 6. Let F be an ordered field with the least upper bound property. If x, y ∈ F and x < y, then there exists r ∈ FQ such that x < r < y. Proof. We will use the Archimedean Principle and Corollary 4 to produce m, n ∈ FN so that mn −1 is the desired element r of FQ . Since x < y, y − x > 0. By the Archimedean Principle, there exists n ∈ FN such that n(y − x) > 1. Thus, nx + 1 < ny.

(7)

By Corollary 4, there exists m ∈ FZ such that m − 1 ≤ nx < m. Hence, m ≤ nx + 1

and

nx < m.

(8)

Putting together the inequalities in (7) and (8) yields nx < m ≤ nx + 1 < ny. Therefore nx < m < ny and multiplying all three terms by n −1 gives the desired result: x < mn −1 < y.


134

Chapter 10 The Natural Numbers 1. Introduction. We now begin a long journey. We have used numbers (natural, integer, rational, and real) in the previous chapters. A fundamental question is to ask if these objects that we have been using even exist. One must go to the very foundation of our axiomatic system to answer this fundamental question. In our journey we will not quite go back to the beginning; instead we will take a more na¨ıve approach and assume certain constructions are already known to be realizable in our axiomatic development. The first such na¨ıve construction is the immediate successor of a set. The following collection of “axioms” will be helpful. Depending on where we start in our development of set theory, these statements can be proved. Theorem 1(Peano Axioms). (1) ∅ exists. (2) For every n ∈ Z+ , there exists Succ(n) = n ∪ { n } ∈ Z+ . (3) For every n ∈ Z+ , 0 = Succ(n). (4) For every m, n ∈ Z+ , if m = n, then Succ(m) = Succ(n). (5) If S ⊂ Z+ , 0 ∈ Z+ , and, for every n ∈ Z+ , Succ(n) ∈ S, then S = Z+ . Definition 2. The immediate successor of a set S is the set S ∪ { S }. The immediate successor of S will be denoted by Succ(S). In our construction, we will actually start with 0. While we do not regard 0 as a natural number, it will be more natural for our construction to do this. In effect, we will construct the set { 0 } ∪ N which we have decided to call Z+ . In the developement of numbers, the natural numbers came into use first; the number 0 did not come into use until much later. The notion of constructing the numbers is far more recent yet; it is due to the work in set theory initiated in the nineteenth century by Georg Cantor and others. The second na¨ıve notion will be the principle of induction. Here, since we do not yet have the set N of natural numbers, we will use induction to prove a sequence of statements P(k) where k = n, Succ(n), Succ(Succ(n)), . . . .

Section 2. Zero, the Natural Numbers and Addition

That is, we prove the “first” step P(n) and prove the inductive implication P(k) implies P(Succ(k)). 2. Zero, the Natural Numbers and Addition. First we define 0. Definition 3. Let 0 be ∅. Next we use the immediate successor of a set to define the natural numbers inductively; this is another incidence of our na¨ıvet´e. Definition 4. Let 1 be Succ(0), 2 be Succ(1), 3 be Succ(2), 4 be Succ(3) and so on. The set of all of the numbers constructed in this way is N, the set of natural numbers. So you may have started to compile a list that looks something like the following: 0=∅ 1 = 0 ∪ {0} = {∅} 2 = 1 ∪ { 1 } = { ∅, { ∅ } } 3 = 2 ∪ { 2 } = { ∅, { ∅ }, { ∅, { ∅ } } } ... Unfortunately, having done this will not help us in the development of the natural numbers. Definition 5. Let n ∈ N. The set of successors of n or descendency set of n is the subset Desc(n) = { Succ(n), Succ(Succ(n)), Succ(Succ(Succ(n))), . . . }. For p ∈ N, we will use the notation Succ p (n) to denote the p th successor of n, which is Succ(. . . (Succ(n)) . . . ), where the Succ is taken p times. Note that n ∈ / Desc(n). Otherwise, there would exist p ∈ N such that n = Succ p (n). ............................ Definition 6. We define a relation < on N by n < m if and only if m ∈ Desc(n). Exercise 7. Prove that < is a strict total ordering on N and, in fact, N is well-ordered. Using < and the usual rules for a strict total ordering, we can write n ≤ m, m > n and m ≥ n. Before defining the operation of addition, we consider a type of inverse operation to the immediate successor.

135

136

Chapter 10. The Natural Numbers

Definition 8. Let n ∈ N. The number m ∈ Z+ is the immediate predecessor of n iff the immediate successor of m is n. We denote the immediate predecessor of n by Pred(n). That is, if the immediate predecessor of n exists, then n = Succ(Pred(n)). Lemma 9. For all n ∈ Z+ , n = Pred(Succ(n)). Proof. Consider Pred(Succ(n)). By definition of the immediate predecessor, it is the number whose immediate successor is Succ(n), that is, n.
Next we define addition. The definition is again inductive. Definition 10. Let m, n ∈ Z+ . Define m + n to be  m+n =

m,

if n = 0

Succ(m + Pred(n)), if n > 0

m + n is called the sum of m and n. This defines addition on Z+ . Lemma 11. For all n ∈ Z+ and p ∈ N, n + p = Succ p (n). Moreover, Desc(n) = { n + 1, n + 2, n + 3, . . . }. Proof. We proceed by induction on p. For p = 1, n + 1 = Succ(n + Pred(1)) = Succ(n + 0) = Succ(n). For the inductive step, assume the inductive hypothesis n + p = Succ p (n). Now n + Succ( p) = Succ(n + Pred(Succ( p))) = Succ(n + p) = Succ(Succ p (n)) = Succ p+1 (n).




Exercise 12. Prove that, for all m, n ∈ N, m < n if and only if there exists a unique p ∈ N such that m + p = n. Let us consider some of the properties of addition in Z+ .

Section 2. Zero, the Natural Numbers and Addition

Theorem 13. Addition on Z+ is associative. Proof. We wish to prove that for all m, n, p ∈ Z+ , (m + n) + p = m + (n + p). Let m and n be arbitrary elements of Z+ . We proceed by induction on p. Consider p = 0. By the definition of addition, (m + n) + 0 = m + n and m +(n +0) = m +n. Since = is an equivalence relation, (m +n)+0 = m + (n + 0). For the inductive step, we assume the inductive hypothesis (m + n) + p = m + (n + p). Now, by definition of addition, Lemma 9 and the induction hypothesis,    (m + n) + Succ( p) = Succ (m + n) + Pred Succ( p)   = Succ (m + n) + p   = Succ m + (n + p) . Also, using the definition of addition twice and Lemma 9 twice,    m + (n + Succ( p)) = Succ m + Pred n + Succ( p)      = Succ m + Pred Succ n + Pred Succ( p)   = Succ m + (n + p) . Using the fact that = is an equivalence relation yields the desired result.
We next wish to show that addition on Z+ is commutative. Recall that m + 0 = m for all m ∈ Z+ by definition of addition. The next exercise is the first step in the proof of commutativity. Exercise 14. Prove that 0 + m = m for all m ∈ Z+ . The following is a corollary. Exercise 15. Prove that n = Succn (0) for all n ∈ N. Together with the definition m + 0 = m, Exercise 14 shows that 0 is the additive identity element of Z+ . The following exercise justifies use of the definite article when we say “the” additive identity. Exercise 16. Prove that 0 is the unique additive identity element in Z+ . Moreover, N does not have an additive identity.

137

138

Chapter 10. The Natural Numbers

Exercise 17. Prove that addition on Z+ is commutative. We will refer to the following as the additive cancellation property for equality. Exercise 18. Let m, n, p ∈ Z+ . Prove that m + n = m + p if and only if n = p. Next we consider the relationship between the total ordering < and addition. Exercise 19. Let m, n ∈ Z+ . Prove that m + n = 0 if and only if m = n = 0. Exercise 20. Let m, n, p ∈ Z+ . Prove that m + n < m + p implies n < p. We will refer to the previous as the additive cancellation property for inequality. When we say the cancellation properties for addition, we will mean both for equalities and for inequalities. Of course, the operation may change! 3. Multiplication. Next we tackle multiplication in Z+ . Definition 21. Let m, n ∈ Z+ . Define mn to be  mn =

0,

if n = 0

mPred(n) + m,

if n > 0

mn is called the product of m and n. This defines multiplication on Z+ . We next examine the properties of multiplication. Since the definition of multiplication resembles the distributive property, it seems the natural place to start. Exercise 22. Prove that, for m, n, p ∈ Z+ , (m + n) p = mp + np. Exercise 23. Prove that 1 is the multiplicative identity for Z+ . Exercise 24. Prove that multiplication on Z+ is commutative. Combining the distributive and commutative properties yields the second distributive property: for all m, n, p ∈ Z+ , m(n + p) = mn + mp.

Section 3. Multiplication

Exercise 25. Prove that multiplication on Z+ is associative. Exercise 26. Let m, n ∈ Z+ . Prove that mn = 0 if and only if m = 0 or n = 0. The following is known as the multiplicative cancellation property. Theorem 27. Let m, n, p ∈ Z+ . Prove that if mn = mp and m = 0, then n = p. Proof. Assume n = p. By trichotomy, either n < p or n > p. First, let us suppose that n < p. Hence, there exists q ∈ N such that p = n + q (by Exercise 12). Now mn + 0 = mn = mp = m(n + q) = mn + mq. By additive cancellation, 0 = mq. By Exercise 26, m = 0 or q = 0. Since m = 0, we must have q = 0. This contradicts that q ∈ N. The case n > p also leads to a contradiction. Therefore, n = p.
Exercise 28. Complete the proof of Theorem 27: consider the case of n > p. Next we have the other cancellation property for multiplication. Exercise 29. Let m, n, p ∈ Z+ and m = 0. Prove that n < p if and only if mn < mp.

139

140 Summary of the Properties of the Non-negative Integers The set Z+ = { 0, 1, 2, 3, 4, 5, . . . }, with addition, multiplication and an order relation, satisfies all of the following properties, for all m, n, p ∈ Z+ : m+n =n+m (m + n) + p = m + (n + p) 0 is the unique element such that 0 + m = m + 0 = m m + n = m + p if and only if n = p and n + m = p + m if and only if n = p (MC): mn = nm (MA): (mn) p = m(np) (MU): 1 is the unique element such that 1m = m1 = m (MX): for m = 0, mn = mp if and only if n = p and for m = 0, nm = pm if and only if n = p (DP): (m + n) p = mp + np and m(n + p) = mn + mp (PP): if m > 0 and n > 0 then mn > 0 (IX): n + m < p + m if and only if n < p and n + m < p + m if and only if n < p Despite the fact that there are no additive inverses for nonzero elements of Z+ , the following Difference Property holds: for all m, n ∈ Z+ , there exists a unique d ∈ N such that m + d = n if and only if m < n. This property can be thought of as the definition of the relation <. Recall that m ≤ n is defined by m < n or m = n. (AC): (AA): (AZ): (AX):

141

Chapter 11 The Integers 1. Introduction: Integers as Equivalence Classes. In Chapter 10, we constructed the following set: Z+ = { 0, 1, 2, 3, 4, 5, . . . }. We called this the set of non-negative integers. We constructed two operations, addition and multiplication, on Z+ and a strict order relation <. In this chapter, you will not need to refer back to the construction in Chapter 10; rather, we have listed on the previous page all of the relevant properties of the construction. These properties of Z+ are all that we will need in this chapter. Addition and multiplication were the two operations defined on Z+ . Notice that subtraction and division were never mentioned. In fact neither operation can be defined universally on Z + . The deficiency of Z+ with respect to subtraction vis-`a-vis addition necessitated its extension to the set of the integers which will be discussed in this chapter. For each pair of numbers m and n we wish the difference m − n to have a meaning, that is, we require the existence of a number q such that m = n + q. A natural way to proceed would be to attempt to expand Z+ into Z. One way to do this is to assume that, for each m ∈ Z+ , there exists a number m  such that m + m  = 0; we call m  the (additive) inverse of m. Then the set of all integers Z would be the set of all natural numbers together with their inverses and 0. Setting aside for a moment the problem of defining the new negative elements, it turns out that this approach makes it difficult to give even a cumbersome definition of addition. Such a definition, with several different cases, makes the proofs of the basic properties of the operations in Z quite tedious. Therefore, it is preferable to start with the following notion. We note that, for example, −1 = 0 − 1 = 1 − 2 = 2 − 3 = 3 − 4 = 4 − 5 = . . .

142

Chapter 11. The Integers

Similar statements can be made for all negative integers and, in fact, for all integers. We can therefore formulate a definition of the integers as equivalence classes of pairs of nonnegative integers. This new construction has the obvious disadvantage that Z no longer contains Z+ as we had previously defined it. However, in Section 4, we show that Z+ is embedded in Z; this allows us to think of our previously constructed set Z+ as a “subset” of Z. We start with a definition of a set which we will call Z, the set of integers. Definition 1. Define a relation ∼ on Z+ × Z+ by (m, n) ∼ ( p, q) if and only if m + q = p + n. Exercise 2. Prove that the relation ∼ on Z+ ×Z+ is an equivalence relation. Definition 3. The set of integers is (Z+ × Z+ )/∼. We denote this set as Z. We denote the equivalence class of (m, n) by [m, n]; this is an element of Z. 2. A Total Ordering of the Integers. First, let us define a relation on Z that we will show is a total ordering. Definition 4. Define the relation < on Z by [m, n] < [ p, q] if and only if m + q < p + n. It is important to realize that the symbol < is used in two different ways in the definition above. The first is a new relation which we are defining on Z. The second is our old strict total ordering on Z+ . We could use different symbols for each, e.g., use subscripts as in
and

p + q = p + q  ,

143

Section 2. A Total Ordering of the Integers

respectively. Adding these two equations gives us the following: (m  + n) + ( p + q  ) = (m + n  ) + ( p  + q). Using the commutative and associative properties of addition in Z+ yields: (m  + q  ) + (n + p) = (n  + p  ) + (m + q).

(6)

Since m + q < n + p, (m  + q  ) + (m + q) < (m  + q  ) + (n + p).

(7)

Now, (6) and (7) imply (m  + q  ) + d) + (m + q) < (n  + p  ) + (m + q). By cancellation in Z+ , m  + q  < n  + p .




Notice that we used only properties (A1), (A2) and (L3) for Z+ in the proof above. Exercise 8. Prove that < is a strict total ordering on Z. As for any strict total ordering <, there is an associated total ordering ≤: define [m, n] ≤ [ p, q] iff [m, n] < [ p, q] or [m, n] = [ p, q]. Recall that Z+ is well-ordered. That is, Z+ is totally ordered and every nonempty subset of Z+ has a least element. This total ordering on Z+ extends to (or induces) a total ordering on Z, but ≤ is not a well-ordering on Z. Exercise 9. Prove that < is not a well-ordering on Z. Theorem 10. For all [m, n] ∈ Z, there exists a unique p ∈ Z+ such that  [ p, 0], m ≥ n [m, n] = . [0, p], m ≤ n Proof. If m = n, then m + 0 = n + 0 = 0 + n. Hence, [m, n] = [0, 0] and p = 0. If m > n, then, by the Difference Property for Z+ , there exists a unique p such that m = n + p. Hence, m + 0 = n + p and [m, n] = [ p, 0]. If m < n, then there exists a unique p such that n = m + p. Hence, m + p = n + 0 and [m, n] = [0, p].
At this point in our discussion we tire of writing [m, n] for an integer and go back to the more usual looking form p ∈ Z. We will write p for [ p, 0], 0 for [0, 0] and − p for [0, p]. You should be very careful here. Nothing has changed; elements of Z are still equivalence classes of ordered pairs!

144

Chapter 11. The Integers

Definition 11. Let m ∈ Z. m is positive iff m > 0 and m is negative iff m < 0. Exercise 12. Let m ∈ Z. Prove that exactly one of the following holds: m is positive, m is negative, or m = 0. 3. Addition of Integers. We next define the operation of addition on Z. The idea behind the definition is that we want (m − n) + ( p − q) = (m + p) − (n + q). Definition 13. Define addition in Z by [m, n] + [ p, q] = [m + p, n + q] for all [m, n], [ p, q] ∈ Z. It is important to realize that the symbol + is used in two different ways in the definition of addition. The first is addition in Z, that is, addition of equivalence classes. The second and third are our old addition in Z+ . As with any operation on equivalence classes, we must check that this is well-defined. That is, we must check that the equivalence class of the sum (m + p, n + q) is independent of the choice of representatives of the equivalence classes [m, n] and [ p, q]. Exercise 14. Prove that addition on Z is well-defined. Recall that addition in Z+ is commutative and associative with a unique identity. The corresponding properties for addition in Z follow. Exercise 15. Prove that addition on Z is commutative. Exercise 16. Prove that addition on Z is associative. Exercise 17. Prove that 0 is the unique additive identity for Z. For algebraic structures, we usually denote the additive identity element as 0. Recall that 0 = [0, 0]. Again you should see that there are two meanings of 0 in the previous statement. We come to the punch line of this chapter now: the purpose of defining negative integers was to have additive inverses. Exercise 18. Prove that every element of Z has a unique additive inverse.

Section 4. Multiplication of Integers

As one might expect, the additive inverse of m is denoted by −m. We can now define subtraction: m − n = m + (−n) for all m, n ∈ Z. Exercise 19. Prove that, for every m ∈ Z, −(−m) = m. The proof of the cancellation properties is now much simpler than it was for Z+ . Exercise 20. Prove that, for all m, n, p ∈ Z, if m + n = m + p, then n = p. Commutativity of addition and Exercise 20 imply that, for all m, n, p ∈ Z, if n + m = p + m, then n = p. Exercise 21. Prove that, for all m, n, p ∈ Z, if m + n < m + p, then n < p. 4. Multiplication of Integers. Next, we consider multiplication. Since we want (m − n)( p − q) = mp − mq − np + nq, we define multiplication as follows. Definition 22. Define multiplication in Z by [m, n][ p, q] = [mp + nq, np + mq] for all [m, n], [ p, q] ∈ Z. Proposition 23. Multiplication on Z is well-defined. Proof. Suppose (m  , n  ) ∈ [m, n] ∈ Z and ( p  , q  ) ∈ [ p, q] ∈ Z. Hence, m  + n = n  + m and p  + q = q  + p. We will use properties of addition in Z+ . Multiplying the previous equations by equals yields the following: m(q  + p) = m( p  + q) n( p  + q) = n(q  + p) (n  + m) p  = (m  + n) p  (m  + n)q  = (n  + m)q  By the distributive properties, mq  + mp = mp  + mq np  + nq = nq  + np n  p  + mp  = m  p  + np  m  q  + nq  = n  q  + mq 

145

146

Chapter 11. The Integers

By adding these equations, we obtain (mq  + mp) + (np  + nq) + (n  p  + mp  ) + (m  q  + nq  ) = (mp  + mq) + (nq  + np) + (m  p  + np  ) + (n  q  + mq  ). By the associative and commutative properties, we rewrite this as (mp + nq) + (n  p  + m  q  ) + (mq  + np  + mp  + nq  ) = (np + mq) + (m  p  + n  q  ) + (mq  + np  + mp  + nq  ). By the cancellation property, (mp + nq) + (n  p  + m  q  ) = (np + mq) + (m  p  + n  q  ), which implies [m, n][ p, q] = [mp + nq, np + mq] = [m  p  + n  q  , n  p  + m  q  ] = [m  , n  ][ p  , q  ] Therefore, multiplication is well-defined.




Exercise 24. Prove that multiplication on Z is commutative. Exercise 25. Prove that multiplication on Z is associative. Exercise 26. Prove that 1 is the unique multiplicative identity for Z. Notice that we again have labeled a multiplicative identity 1. However, Z is lacking multiplicative inverses. To prove this, we will use the following exercises. Exercise 27. Prove that the distributive property holds in Z. Exercise 28. Prove that mn > 0 if and only if m > 0 and n > 0, or m < 0 and n < 0. Alas, Z still has deficiencies, as the following indicates.

Section 5. Embedding the Natural Numbers in the Integers

Proposition 29. 2 has no multiplicative inverse in Z. Proof. Assume that there is an m ∈ Z such that 2m = 1. If m ≤ 0, then 2m ≤ 0 and, hence, 2m = 1. If m > 0, then 2m = (1 + 1)m = m + m > m ≥ 1. In any case, 2m = 1 and hence 2 has no multiplicative inverse in Z.
We end this section with two more standard algebraic properties. Exercise 30. Prove that, for all m ∈ Z, 0m = 0. Exercise 31. Prove that, for all m, n ∈ Z, m(−n) = (−m)n = −(mn). 5. Embedding the Natural Numbers in the Integers. It is important to realize that, contrary to what you might have expected, N and Z+ , as constructed in Chapter 10, are not subsets of Z = (Z+ × Z+ )/∼. This is obviously true since an element of Z is defined as the equivalence class of a pair of elements of Z+ . However, this poses no problem, since we can think of Z+ as a subset of Z in a very natural way. Recall the discussion of isomorphic structures in Section 4 of Chapter 8. In our case, we have Z+ and Z as the underlying sets of our structures and the order relations and operations of addition and multiplication providing the organization of these structures. Do you see an isomorphic copy of Z+ in Z? If so, you are now in a position to prove the following exercise. Exercise 32. Prove that there exists an embedding of Z+ into Z. Since isomorphic structures can be thought of as identical, we will simply say, from now on, that Z+ is a subset of Z. To be more precise, we could have called our original Z+ something else and called its isomorphic image Z+ ! That would be OK until we define the rational numbers, when we will have to repeat the process of finding an embedding yet again (and again for real numbers and yet again for complex numbers).

147

148

Chapter 11. The Integers

Supplemental Exercises Exercise 33. Explain what is wrong with the notation [− p, 0] ∈ Z. (The statement [ p, 0] + [− p, 0] = 0 is nonsense!) Exercise 34. Explain what is wrong with the following proof of the existence of additive inverses in Z (Exercise 18): For every [ p, 0] ∈ Z, [ p, 0] + [0, p] = [ p, p] = [0, 0] = 0. Similarly, [0, p] + [ p, 0] = [ p, p] = [0, 0] = 0.

149 Summary of the Properties of the Integers The set of integers, Z, with addition (+), multiplication (·) and a strict total ordering (<), satisfies all of the following properties, for all m, n, p ∈ Z: (AC): m + n = n + m (AA): (m + n) + p = m + (n + p) (AZ): 0 is the unique element such that 0 + m = m + 0 = m (AI): −m is the unique element such that −m + m = m + (−m) = 0 (AX): m + n = m + p if and only if n = p and n + m = p + m if and only if n = p (MC): mn = nm (MA): (mn) p = m(np) (MU): 1 is the unique element such that 1m = m1 = m (MX): for m = 0, mn = mp if and only if n = p and for m = 0, nm = pm if and only if n = p (DP): (m + n) p = mp + np and m(n + p) = mn + mp (PP): if m > 0 and n > 0 then mn > 0 (IX): n + m < p + m if and only if n < p and n + m < p + m if and only if n < p The following properties also hold: for all m, n, p ∈ Z+ , Multiplication by 0: 0m = m0 = 0 Factoring a Minus: (-m) n = m(-n) = -(mn) Negation Reverses Order: m > 0 if and only if −m < 0

150

Chapter 12 The Rational Numbers 1. Introduction: Rationals as Equivalence Classes. We now turn to the construction of the rationals which, from a historical point of view, were actually identified before the integers; it seems easier to understand what one half of a melon means than to understand zero melon or minus one melon. As we commented in the introduction of the previous chapter, to make division between any two integers as universally possible as multiplication, we need an extension of our number system. The construction of Q from Z is quite similar to the construction of Z from Z+ in the previous chapter. For each pair of integers m and n, we require the existence of a unique rational number q (to be viewed as the quotient m/n), such that m = nq. Division by n = 0 is not allowed for the following reasons. Suppose n = 0. Since any product with 0 in Z equals 0 we would get m = 0q = 0. By the transitivity of equality on Z, m would have to be 0. This is no good either since 0 = 0q has infinitely many solutions in Z.* Given a rational number q, the numbers m and n satisfying the equality m = nq, i.e., mn = q, are not uniquely determined. For example 2 3 4 5 1 = = = = = .... 2 4 6 8 10 Therefore, we define the rationals to be equivalence classes of pairs of integers. Recall that Z ∗ = Z − { 0 }, the set of all nonzero integers. *So if anyone asks you “What is 0 divided by 0?” you can answer with any number you like! More seriously, this may remind you of limits like lim kx x = k for every k ∈ R. x→0

Section 2. A Total Ordering of the Rationals

Definition 1. Define a relation ∼ on Z × Z∗ by (m, n) ∼ ( p, q) if and only if mq = pn. Exercise 2. Prove that ∼ is an equivalence relation on Z × Z∗ . Definition 3. The set of rational numbers is the set (Z×Z∗ )/∼. We denote this set by Q. The equivalence class of (m, n) is denoted by [m, n]; that is, [m, n] is an element of Q. We could have chosen to define rational numbers differently, only allowing positive n (for the “denominator”). The following two exercises show that this is possible. Exercise 4. Prove that, for all (m, n) ∈ Z × Z∗ , [m, n] = [−m, −n]. Exercise 5. Prove that, for all [m, n] ∈ Q, there exists ( p, q) ∈ [m, n] with q > 0. Remark 6: Important! For the remainder of this chapter, when we consider a rational number [m, n], we will assume that n > 0. It does not much matter whether you think that Q = (Z × N)/∼ or you think we only choose representatives (m, n) of rational numbers with n ∈ N. 2. A Total Ordering of the Rationals. We now define a total ordering on Q. The definition is suggested by the following statement: m p < n q

implies mq < pn.

Notice that Remark 6 is crucial here: since n and q are positive, this works. On the contrary, −1 1 < 2 −3

does not imply (−1)(−3) < (2)(1).

Definition 7. Define the relation < on Q by [m, n] < [ p, q] if and only if mq < pn. Again, note that the symbol < is used in two different ways in Definition 7. Exercise 8. Prove that the relation < on Q is well-defined.

151

152

Chapter 12. The Rational Numbers

Next, we check that < is a strict total ordering on Q. Exercise 9. Prove that < is a strict total ordering on Q. Remember that a strict total ordering is trichotomous. Of course, we can define a total ordering ≤ in terms of < and =. As was the case for the total ordering on Z, < is not a well-ordering. Exercise 10. Prove that < is not a well-ordering on Q. Moreover, Q has no least element. Let 0 = [0, 1]; we will see in Exercise 17 that 0 is the unique additive identity. Since n > 0, we wish that 0 m > n 1

implies m = (m)(1) > (n)(0) = 0.

Definition 11. Suppose [m, n] ∈ Q. [m, n] is positive if and only if m > 0 and [m, n] is negative if and only if m < 0. Instead of always writing an element in Q as the equivalence class of a pair of integers, we can simply write q ∈ Q. Nothing has changed: there is a pair (m, n) such that q = [m, n]. Exercise 12. Let q ∈ Q. Prove that exactly one of the following holds: q is positive, q is negative, q = 0. 3. Addition of Rationals. You remember that addition of rationals (fractions) is a bit more difficult than multiplication of rationals (fractions). Nevertheless, let us start with addition. Following the idea that m p mq + np + = , n q nq we next define addition. Definition 13. Define addition in Q by [m, n] + [ p, q] = [mq + np, nq] for all [m, n], [ p, q] ∈ Q. In the definition above, the + sign on the first side is understood to denote the operation of addition in Q, which differs from the operation of

Section 4. Multiplication of Rationals

addition in Z that appears on the second side and is denoted by the same symbol. Since Definition 13 depends on the choice of equivalence class representatives, we must check that addition is well-defined. Exercise 14. Prove that addition on Q is well-defined. The following four exercises establish properties (AC), (AA), (AZ) and (AI) for addition in Q. Exercise 15. Prove that addition on Q is commutative. Exercise 16. Prove that addition on Q is associative. Exercise 17. Prove that 0 = [0, 1] is the unique additive identity for Q. Exercise 18. Prove that every element of Q has a unique additive inverse. Again, we denote the additive inverse of a number x ∈ Q by −x. Subtraction is defined as with the integers: for all x, y ∈ Q, x − y = x + (−y). Property (AX) is established by the following exercise. Exercise 19. Prove that the cancellation properties hold for addition in Q: (a) for all x, y, z ∈ Q, x + y = x + z if and only if y = z and (b) for all x, y, z ∈ Q, y + x = z + x if and only if y = z. 4. Multiplication of Rationals. Since we want   m p mp = , n q nq we next define multiplication as follows. Definition 20. Define multiplication in Q by [m, n][ p, q] = [mp, nq] for all [m, n], [ p, q] ∈ Q. Again, since Definition 20 depends on the choice of equivalence class representatives, we must check that multiplication is well-defined.

153

154

Chapter 12. The Rational Numbers

Exercise 21. Prove that multiplication on Q is well-defined. The following four exercises establish properties (MC), (MA) and (MU) for multiplication in Q. Exercise 22. Prove that multiplication on Q is commutative. Exercise 23. Prove that multiplication on Q is associative. Exercise 24. Prove that [1, 1] is the unique multiplicative identity for Q. We set 1 = [1, 1]. By now you should realize the special nature of the rationals of the form [m, 1]. Set Q∗ = Q − { 0 }. The following exercise establishes property (MI) for multiplication in Q. Exercise 25. Prove that, for each x ∈ Q∗ , there exists a unique multiplicative inverse. Again, we denote the multiplicative inverse of a number x ∈ Q by x −1 . Division is defined, for all x ∈ Q and y ∈ Q∗ , by x÷y=

x = x/y = x y −1 . y

The following exercise establishes property (MX) for multiplication in Q. Exercise 26. Prove that the cancellation properties hold for multiplication in Q: (a) for all x ∈ Q and y, z ∈ Q, x y = x z if and only if y = z and (b) for all x ∈ Q and y, z ∈ Q, yx = zx if and only if y = z. 5. An Ordered Field Containing the Integers. To show that Q is a field, all that remains is to verify the distributive property (DP). First, we give an exercise that will be needed in the proof of the distributive property. Exercise 27. Prove that [mp, np] = [m, n] in Q for all m ∈ Z and n, p ∈ Z∗ . While you probably proved Exercise 27 by using the arithmetic of Z, we could argue as follows. We can show that [mp, np] = [m, n][ p, p] and that [ p, p] = 1 (see Exercise S12). The result follows from a property of the

Section 5. An Ordered Field Containing the Integers

multiplicative identity. That is, Exercise 27 simply states that (x)(1) = x for all x ∈ Q. Exercise 28. Prove that the distributive property holds in Q. Exercise 28 completes the proof that Q is a field. We next show that Q is an ordered field. We need to verify that properties (PP) and (IX) hold. Exercise 29. Suppose x, y ∈ Q. Prove that x y > 0 if and only if either x > 0 and y > 0, or x < 0 and y < 0. Exercise 30. Prove that cancellation in inequalities holds in Q: (a) for all x, y, z ∈ Q, x + y < x + z if and only if y < z and (b) for all x, y, z ∈ Q, y + x < z + x if and only if y < z. This shows that Q is an ordered field. Of course, we “think” of Z as a subset of Q even though it is not. Exercise 31. Prove that there exists an embedding of Z into Q. The next two exercises show that Q satisfies the Archimedean Principle and as well as a stronger version of the Archimedean Principle. Exercise 32. Prove that if x, y ∈ Q and x > 0, then there exists n ∈ N such that nx > y. Exercise 33. Prove that if x, y ∈ Q and x > 0, then there exists n ∈ Z such that nx ≤ y < (n + 1)x. The next exercise shows that there are many rational numbers using a “denseness” type of property. Exercise 34. Prove that if x, z ∈ Q and x < z, then there exists y ∈ Q such that x < y < z. As discussed in Chapter 6, Q does not have the least upper bound property. We postpone a proof of this fact until we have constructed the √ real numbers. At that point we will be able to show that 2 ∈ R and use the denseness of Q in R to derive a contradiction from the assumption that the least upper bound property holds for Q. It is interesting to note that Q is an ordered field without the least upper bound property that nevertheless satisfies the Archimedean Principle. You should compare this with Theorem 1 of Chapter 9.

155

156

Chapter 12. The Rational Numbers

Supplemental Exercises Exercise S1. Explain why [1, 0] is not an element of Q. Exercise S2. Verify the following using Definition 3 of elements of Q: (a) [1, 2] = [2, 4] in Q, (b) [−11, −6] = [11, 6] in Q, (c) [−3, 5] = [3, −5] in Q. Exercise S3. Use Definition 3 to do the following: (a) find (m, n) ∈ [5, −7] in Q such that n > 0, (b) find ( p, q) ∈ [−16, 3] in Q such that q < 0, (c) find (r, s) ∈ [−6, −13] in Q such that s > 0. Exercise S4. Show that [0, n] = [0, 1] in Q for all n ∈ Z ∗ . Exercise S5. Show that [2, 3] = [2n, 3n] in Q for all n ∈ Z ∗ . Exercise S6. Verify the following using Definition 7 of order on Q: (a) [1, 2] < [2, 3] in Q, (b) [−10, 3] > [−7, 2] in Q, (c) [−10, 3] > [7, −2] in Q. [Be careful!] Exercise S7. Prove that (m, n) ∈ Z × Z∗ represents a positive rational number if and only if mn > 0 and (m, n) represents a negative rational number if and only if mn < 0. Exercise S8. Suppose that m, n ∈ N. For each of the following pairs of elements of Q, determine which is greater: (a) [m, n] and [m, n + 1] (b) [m, n] and [m + 1, n] Exercise S9. Repeat Exercise S8 with m ∈ Z∗− (i.e., m < 0) and n ∈ N. Exercise S10. Suppose that m, n ∈ N. Under what condition on m and n is it true that [m, n + 1] ≤ [m − 1, n]? Exercise S11. Compute the following using Definition 13 of addition on Q: (a) [2, 3] + [3, 4] in Q, (b) [−2, 5] + [−7, 3] in Q, (c) [−4, 8] + [9, 18] in Q. Exercise S12. Show that [n, n] = [1, 1] in Q for all n ∈ Z ∗ . Exercise S13. Show that 0 = 1 in Q.

Supplemental Exercises

Exercise S14. Compute the following using Definition 20 of multiplication on Q: (a) [2, 3][3, 4] in Q, (b) [−2, 5][−7, 3] in Q, (c) [−4, 8][9, 18] in Q.

157

158 Summary of the Properties of the Rationals The set of rationals, Q, with addition, multiplication and an order relation is an ordered field. That is, all of the following properties are satisfied, for all x, y, z ∈ Q: (AC): x + y = y + x (AA): (x + y) + z = x + (y + z) (AZ): 0 is the unique element such that 0 + x = x + 0 = x (AI): −x is the unique element such that −x + x = x + (−x) = 0 (AX): x + y = x + z if and only if y = z and y + x = z + x if and only if y = z (MC): x y = yx (MA): (x y)z = x(yz) (MU): 1 is the unique element such that 1x = x1 = x (MI): x −1 is the unique element such that x −1 x = x x −1 = 1 (MX): for x = 0, x y = x z if and only if y = z and for x = 0, yx = zx if and only if y = z (DP): (x + y)z = x z + yz and x(y + z) = x y + x z (PP): if x > 0 and y > 0, then x y > 0 (IX): y + x < z + x if and only if y < z and y + x < z + x if and only if y < z

159

Chapter 13 The Real Numbers 1. Dedekind Cuts. The set of rationals, Q, has the defect that it does not have the least upper bound property. Another way of looking at this is to say that the polynomial x 2 − 2 has no rational roots. In this chapter we construct the familiar real numbers. The first constructions were given, independently, by Richard Dedekind and Georg Cantor in 1872. In this chapter, we will present the construction due to Dedekind.
  We saw, in Chapter 6, that x ∈ Q
0 < x 2 < 2 does not have a least upper bound in Q. We want a set like this to represent the real number √ √ 2 for us. In this spirit, what set should represent − 2? We modify our original idea somewhat to let 


 x ∈ Q
x < 0 or x 2 < 2

and




 x ∈ Q
x < 0 and x 2 > 2

√ √ to represent 2 and − 2, respectively. That is, a real number is to be represented by the set of all rational numbers that we wish to be less than the given real number. Then it makes sense to let {x ∈ Q |x < q } represent the rational number q. Let us formalize this. Definition 1. A Dedekind cut is a set D ⊂ Q satisfying (1) (2) (3) (4)

D = ∅, D = Q, if x ∈ D, y ∈ Q and y < x, then y ∈ D, if x ∈ D, then there exists y ∈ D such that x < y.

A Dedekind cut is also called a cut or a real number. The set of all cuts is denoted R.

160

Chapter 13. The Real Numbers

We have defined a cut D to be a nonempty, proper subset of Q with no greatest element containing all rational numbers less than any number in D. Note that another reasonable candidate to represent q is D = { x ∈ Q | x ≤ q } . Since we are trying to connect real numbers with the least upper bounds of sets of rational numbers that are bounded above, sets like D  are not useful. Note that the least upper bound of D  is the greatest element, q. Property (4) precludes cuts from having a greatest element. Throughout this chapter you should keep in mind that real numbers are defined as sets of rational numbers; two real numbers are equal if and only if they are the same subset of Q. Important Remark. It will be useful in this chapter to denote all rational numbers by lower case letters and all cuts by upper case letters. We will assume this to be the case unless otherwise specified. 2. Order and Addition of Real Numbers. Before continuing with the construction, let us consider property (3) of a cut D that x ∈ D and y < x implies y ∈ D. This means that D contains all rational numbers which are less than some rational number in D. Two statements equivalent to this are given in the following exercise. Exercise 2. Suppose D ∈ R and x, y ∈ Q. Prove that the following are equivalent. (a) If x ∈ D and y < x, then y ∈ D. (b) If x ∈ D and y ∈ / D, then y > x. (c) If x ∈ / D and x < y, then y ∈ / D. First, we define a total ordering on R in the natural way. Definition 3. Define the relation < on R by A < B iff A
B. You may find it more appealing to rewrite the condition as A ≤ B if and only if A ⊂ B. Exercise 4. Prove that < is a total ordering on R. Exercise 5. Prove that R has the least upper bound property. Next we define addition, whose definition is also quite natural.

Section 2. Order and Addition of Real Numbers

Definition 6. Define addition on R by X + Y = { x + y | x ∈ X and y ∈ Y } . Unlike our previous constructions, we must prove that X + Y ∈ R, that is, the closure property of addition. Exercise 7. Prove that addition on R is closed. Exercise 8. Prove that addition on R is commutative. Exercise 9. Prove that addition on R is associative. Exercise 10. Prove that { x ∈ Q | x < 0 } is the unique additive identity on R. As always, we denote the additive identity as 0. The existence of additive inverses is more difficult. Intuitively, we might let the additive inverse of X be the set of all x such that −x ∈ / X. This will not quite work since this set may have a greatest element. For example, consider X = {x ∈ Q |x < 0 }. The set { x ∈ Q | −x ∈ / X } = { x ∈ Q | −x < 0 } = { x ∈ Q | x ≤ 0 } has 0 as its greatest element and hence is not a cut. We define the following set: −X = { x ∈ Q | −x ∈ / X and −x is not the least element of Q − X } = { x ∈ Q | there exists y > 0 such that −x − y ∈ / X }. We must show that −X ∈ R, and X + (−X ) = 0 or −X + X = 0. Lemma 11. For all X ∈ R, −X ∈ R. Proof. Recall the 4 properties in Definition 1 of a cut. By property (2), X = Q. So there exists y ∈ Q such that y ∈ / X . Set x = −y − 1. Hence, y = −x − 1 = −(x + 1) ∈ / X . Since −(x + 1) < −x, −x ∈ / X and x ∈ −X . This proves property (1), −X = ∅. If y ∈ X , then −y ∈ / −X . This proves property (2), −X = Q. Choose x ∈ −X and w > 0 such that −x − w ∈ / X . Suppose y < x. Then −y − w > −x − w and hence −y − w ∈ / X . Therefore, y ∈ −X and property (3) is proved. For x and w as in the previous paragraph, z = x + w2 > x and −z − w2 = −x − w ∈ / X . Therefore, z ∈ −X and property (4) is proved.


161

162

Chapter 13. The Real Numbers

Lemma 12. For all X ∈ R, X + (−X ) = 0. Proof. Let x ∈ X and y ∈ −X . Thus, −y ∈ / X . By Exercise 2, −y > x. Therefore, x + y < 0 and X + (−X ) ≤ 0. On the other hand, choose v ∈ 0. Let w = − v2 . Since v < 0, w > 0. By Exercise 33 of Chapter 12, a strong version of the Archimedean Property of Q, there exists n ∈ Z such that nw ∈ X but (n + 1)w ∈ / X . Set x = nw and y = −(n + 2)w. Since −y − w = (n + 2)w − w = (n + 1)w ∈ / X, y ∈ −X . Now x + y = nw − (n + 2)w = −2w = v. Therefore, 0 ≤ X + (−X ). Combining the above results gives us X + (−X ) = 0.
Exercise 13. Complete the proof that every element of R has a unique additive inverse. Having proved the properties for addition, we know from Chapter 8, that several secondary properties hold. Exercise 14. Prove that cancellation in inequalities holds: X + Y < X + Z if and only if Y < Z for all X, Y, Z ∈ R. 3. Multiplication of Real Numbers. Defining multiplication in R is a little less natural. First we consider the product of positive real numbers. We use the desired rules of signs to define products involving negative real numbers. Definition 15. Let X, Y ∈ R. If X > 0 and Y > 0, then X Y = { x y | x ∈ X and y ∈ Y } . Otherwise, define the product as follows: ⎧ 0, ⎪ ⎪ ⎪ ⎨ −[(−X )Y ], XY = ⎪ −[X (−Y )], ⎪ ⎪ ⎩ (−X )(−Y ),

if X = 0 or Y = 0 if X < 0 and Y > 0 if X > 0 and Y < 0 if X < 0 and Y < 0

We will prove the primary properties for multiplication on the set R+∗ = { X ∈ R | X > 0 } of positive real numbers first.

Section 4. Embedding the Rationals in the Reals

Exercise 16. Prove that multiplication on R+∗ is closed. This also proves the product of positives property: if X > 0 and Y > 0, then X Y > 0 for all X, Y ∈ R. Exercise 17. Prove that multiplication on R+∗ is commutative. Exercise 18. Prove that multiplication on R+∗ is associative. Exercise 19. Prove that { x ∈ Q | x < 1 } is the multiplicative identity on R+∗ . As always, we denote the multiplicative identity as 1. Exercise 20. Prove that every X ∈ R+∗ has a multiplicative inverse. Second, prove the primary properties for multiplication on R. Exercise 21. Prove that multiplication on R is closed. Exercise 22. Prove that multiplication on R is commutative. Exercise 23. Prove that multiplication on R is associative. Exercise 24. Prove that 1 is the multiplicative identity on R. Exercise 25. Prove that every X ∈ R with X = 0 has a multiplicative inverse. As always, we denote the multiplicative inverse of X as X −1 . We have now shown that R is an ordered field. 4. Embedding the Rationals in the Reals. Once again, we have not defined R to extend Q. However Q is naturally embedded in R. By renaming things, we can think of Q as a subset of R. Exercise 26. Prove that there exists an embedding of Q into R. Collecting the results of this chapter thus far, we have proved the following theorem. [A subfield of a field is a subset which is also a field under the operations on the ambient set.]

163

164

Chapter 13. The Real Numbers

Theorem 27. R is an ordered field with the least upper bound property and with Q as a subfield of R. From Theorem 1 of Chapter 9, we know that R satisfies the Archimedean Principle and from Theorem 6 of Chapter 9, we know that Q is dense in R. That is: • if x, y ∈ R and x > 0, then there exists n ∈ N such that nx > y and • if x, y ∈ R and x < y, then there exists q ∈ Q such that x < q < y. 5. Uniqueness of the set of Real Numbers. So far we have shown that R has certain properties. Now we will show that those properties completely characterize R; that is, every X with these properties is isomorphic to R. This is the Main Theorem mentioned in Section 1 of Chapter 9. Theorem 28 (Main Theorem). There exists, up to isomorphism, a unique ordered field with the least upper bound property. The proof of the Main Theorem is long; we will break it into eight steps! Suppose F is an ordered field with the least upper bound property. We will write everything having to do with F —its elements, order relation and operations—in bold. We will prove Theorem 28 by constructing an isomorphism f : R → F . Here is an outline of the construction: Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Step 8:

construct a function f Z : Z → F extend f Z to a function f Q : Q → F extend f Q to a function f : R → F show that f preserves order show that f is one-to-one show that f is onto show that f preserves addition show that f preserves multiplication

Step 1. Define f Z inductively by f Z (0) = 0 and ⎧ n times # $% & ⎪ ⎪ ⎨ 1 + . . . + 1, f Z (n) = ⎪ −(1 + . . . + 1), ⎪ ⎩% &# $

if n > 0 if n < 0

|n| times

Let us give some obvious notation for the n-fold sums and its negation: n = 1 + ... + 1

and

− n = −(1 + . . . + 1).

165

Section 5. Uniqueness of the set of Real Numbers

Exercise 29. Prove that f Z : Z → F preserves operations. Step 2. We will write all rational numbers in the form m/n where m ∈ Z and n ∈ N. (Of course, n = n/1.) We extend the function f Z of Step 1 to f Q : Q → F by defining f Q (m/n) = m n−1 . By the following exercise, the right-hand side is defined. Exercise 30. Prove that if n ∈ N, then f Q (n) = f Z (n) = n = 0. The function f Q is well-defined since p m = ⇒ mq = np ⇒ mq = np ⇒ mn−1 = pq −1 , n q by field properties of Q and F and Exercise 29. Note that f Q extends f Z since m ∈ Z implies f Q (m/1) = m1−1 = m1 = m = f Z (m). Exercise 31. Prove that f Q : Q → F preserves operations. Step 3. Let x ∈ R. Thinking of x as a cut, x = { q ∈ Q | q < x }. Extend the function f Q of Step 2 to f : R → F by f (x) = sup f Q ({ q ∈ Q | q < x }) = sup



f Q (q) | q ∈ Q and q < X



.

If the least upper bound exists, it is unique and f is well-defined. Exercise 32. Prove that sup f Q ({ q ∈ Q | q < x }) exists. Recall from Section 1 of Chapter 9, the definitions of the sets FN , FZ and FQ of natural, integral and rational field elements of F , respectively. Exercise 33. Prove that f Q (N) = FN , f Q (Z) = FZ and f Q (Q) = FQ . We next prove that f extends f Q , that is, for all q ∈ Q, f (q) = f Q (q). Since, by Exercise 31, f Q ( p) < f Q (q) for all p < q, we know that f (q) = sup



f Q ( p) | p ∈ Q and p < q



≤ f Q (q).

166

Chapter 13. The Real Numbers

Assume that f (q) < f Q (q). Since FQ is dense in F (Theorem 6 of Chapter 9) and by Exercise 33, there exists r ∈ Q such that f (q) < f (r ) < f Q (q).

Exercise 34. Complete the proof that f extends f Q by deriving a contradiction. Step 4. Exercise 35. Prove that f preserves order. Step 5. Exercise 36. Prove that f is one-to-one. Step 6. Exercise 37. Prove that f is onto. Step 7. Exercise 38. Prove that f preserves addition. Step 8. Exercise 39. Prove that f preserves multiplication. Proof of Theorem 28. Since we have shown that every ordered field with the least upper bound property is isomorphic to R, Theorem 28 follows from the fact that isomorphism is an equivalence relation.
The construction of the isomorphism above may seem a bit contrived to you at first. Nothing could be further from the truth. In fact, it is natural and unique, as the following exercises demonstrate. Exercise 40. Prove that the only isomorphism of R onto itself is the identity function.

Section 5. Uniqueness of the set of Real Numbers

Definition 41. Suppose X is a field (respectively, ordered field). An isomorphism from X onto X is called an automorphism of X . The result of Exercise 40 can be stated that the identity function IR is the unique (ordered) field automorphism of R. Exercise 42. Let F be an ordered field with the least upper bound property. Prove that the function f , constructed above to prove the Main Theorem, is the unique isomorphism from R onto F .

167

168

Chapter 14 The Complex Numbers 1. Introduction. In this chapter, we will consider a rigorous construction of the field C of complex numbers. The addition and multiplication operations on C will be defined to preserve the operations on R. We will show that C is a field containing R as a subfield. Let us start informally. Complex numbers arise when we consider solutions of equations like 2 x = −1. We declare that there is a solution i to this equation. Having defined i, we allow the standard algebraic operations on the real numbers to work on (the closure of) the set containing the real numbers together with the new number i, always obeying the property that i 2 = −1. The complex numbers, defined in this way, are of the form a + bi, where a and b are real numbers. Of course, we should be careful to define the multiplication bi and the addition a + bi. For a moment, let us suppose that we have defined the complex numbers. Using familiar algebraic properties, we see that (−i)2 = (−1)2 (i)2 = (1)(−1) = −1. So, our original equation, x 2 = −1, has two solutions! Actually, this agrees with the notion that a generic quadratic polynomial should have two roots. The choice in the definition of i appears in Definition 14 below. However, this choice is inconsequential as Exercise 20 will show. While addition and multiplication of complex numbers is quite natural, ordering the complex numbers is a different story. Suppose we have a total ordering on C which agrees with the usual ordering on the subset R. Since a total ordering is trichotomous, either i = 0 or i > 0 or i < 0. Of course, i = 0 is just absurd. Suppose for a moment that i > 0. If C is to be an ordered field, then multiplying positives yields a positive. That is, i 2 > 0. But −1 > 0 in R. The other case is no better, as you can check. This also will be made precise below, in Exercises 17 and 18.

Section 2. Algebra of Complex Numbers

2. Algebra of Complex Numbers. We must first define the set, C, of complex numbers as well as the operations of addition and multiplication on C. Definition 1. The set of complex numbers is the set C = R2 . Definition 2. Addition of complex numbers (a, b) and (c, d) is defined by (a, b) + (c, d) = (a + c, b + d). Definition 3. Multiplication of complex numbers (a, b) and (c, d) is defined by (a, b)(c, d) = (ac − bd, ad + bc). These operations are clearly well-defined; unlike the constructions of the integers and the rationals, this construction does not use equivalence classes. We next verify that C is a field. Let us start with the properties of addition. Exercise 4. Prove that addition on C is commutative. Exercise 5. Prove that addition on C is associative. Exercise 6. Prove that there exists a unique additive identity for C. As before, we let 0 denote the additive identity; that is, 0 = (0, 0). Exercise 7. Prove that every element of C has a unique additive inverse. Denote the additive inverse of a number z ∈ C by −z. Now, let us continue with the properties of multiplication. Exercise 8. Prove that multiplication on C is commutative. Exercise 9. Prove that multiplication on C is associative. Exercise 10. Prove that there exists a unique multiplicative identity 1 for C. Once again, we let 1 denote the multiplicative identity (1, 0). Exercise 11. Prove that every nonzero z ∈ C has a unique multiplicative inverse.

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Chapter 14. The Complex Numbers

Denote the multiplicative inverse of a number z ∈ C by z −1 . It remains to prove the distributive property. Exercise 12. Prove the distributive property for C. This completes the proof that C is a field. We will come back to the question of order on C and whether C is an ordered field or not in the next section. Consider the following computation of a product of real numbers: (a, 0)(b, 0) = (ab, 0). Since this looks a lot like the equation a · b = ab for multiplication of real numbers, we will write a for (a, 0) and b for (b, 0). This suggests a candidate for an embedding of the field R in the field C, which is postponed until Exercise 19. Next, we will identify the imaginary number i. Exercise 13. Show that (0, 1)2 = −1. Is there any other complex number whose square equals −1? Definition 14. The element (0, 1) ∈ C is called i. Of course, we have made a choice in Definition 14 since (0, −1)2 = −1 also. The nature of this choice will be explained by Exercise 20. Since (a, b) = (a, 0) + (0, b) = (a, 0)(1, 0) + (b, 0)(0, 1) = (a, 0) · 1 + (b, 0) · i = a · 1 + b · i, we will write a + bi for (a, b) ∈ C. We call a the real part of (a, b) and b the imaginary part of (a, b). For all b ∈ R, the numbers (0, b) ∈ C are called imaginary. 3. Order on the Complex Field. Let us look at order relations on C. Whenever we take products of ordered sets, we can define an order relation on the product called the lexicographic or dictionary order. We will do this specifically for C rather than the general situation. Definition 15. Define the lexicographic ordering ≺ by (a, b) ≺ (c, d) iff a < c, or a = c and b < d. Exercise 16. Prove that ≺ is a total ordering on C and that (a, 0) ≺ (b, 0) if and only if a < b.

Section 4. Embedding the Real Numbers in the Complex Numbers

Exercise 17. Prove that ≺ does not make C an ordered field. Unfortunately, the lexicographic ordering is not a bad choice since no other ordering makes C an ordered field either. To prove this you can consider the both of the following possibilities: −1 < 0 or −1 > 0. Exercise 18. Prove that there is no total ordering on C which makes it an ordered field. 4. Embedding the Real Numbers in the Complex Numbers. Of course, R is not a subset of C as we have defined C. Again, we can find an embedding of R into C which preserves the operations so that we may consider R as a subfield of C. Exercise 19. Prove that there exists an embedding of R into C. Perhaps, we were a bit terse in Exercise 19; let us elaborate. So far, our embeddings have all had the property of preserving order as well as operations. In fact, if you thought of C with the lexicographic ordering, then the ordering on R is quite compatible. However, since C does not have a natural ordering (i.e., one which makes it an ordered field), we really have in mind a weaker kind of embedding which does not include the orderpreserving property, an embedding of a field rather than an ordered field. We would be done with complex numbers now except for one small detail. The definition of i left us some wiggle room as the following exercise explains. Exercise 20. Prove that there are exactly two automorphisms f : C → C such that f (x) = x for all x ∈ R. Since the automorphisms in Exercise 20 induce the identity IR when restricted to R, we say that these automorphisms of C extend the identity on R. By Exercise 40 of Chapter 13, IR is the unique automorphism on R. Suppose f : C → C is an automorphism such that f (R) = R. The induced function on R preserves operations (and order) and is one-to-one since f has these properties, making it an isomorphism onto its range R; the induced function is an automorphism of R, hence, equal to IR . Therefore, the statement in Exercise 20 can be strengthened: there are exactly two automorphisms f : C → C such that f (R) = R. In fact, this result can be strengthened further. This is done in the following exercise.

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Chapter 14. The Complex Numbers

Exercise 21. Prove that if f : C → C is an isomorphism, then f (R) = R. Hence, there are exactly two automorphisms on C.

173

Chapter 15 The Real Numbers According to Cantor 1. Convergence of Sequences of Rational Numbers. In Chapter 13, we constructed the set of real numbers from the set of rational numbers using the method of Dedekind cuts. We saw that these cuts formed a type of “completion” of the rationals with respect to the least upper bound property. In this chapter, we will give a second construction of the reals from the rationals using the method of Cauchy sequences devised by Cantor. This construction will emphasize another way in which the reals “complete” the rationals. If the construction by cuts can be regarded as essentially algebraic, then the construction by Cauchy sequences can be regarded as analytic. Recall that a sequence can be regarded as a denumerable list of numbers, not necessarily distinct. Since we will only be interested in rational numbers and how they define real numbers, we restrict our attention to sequences of rational or real numbers, that is, sequences in Q or R. Definition 1. Let X be a set. A sequence in X is a function from N to X . Of course, you will very rarely see a sequence denoted as, for instance, f : N → Q. Usually a sequence will be denoted as either { qn }

or

{ q1 , q2 , q3 , q4 , . . . }.

The connection between f and this notation is that qn = f (n) for all n ∈ N. You should note that the notation for sequences looks like a set. This is misleading since a term in the sequence may repeat. On the other hand, we use the same notation to represent the range of the sequence, which is a set. Example 2. A = { n } and B = { 1, 1, 1, 1, . . . } denote two sequences in Q. The range of A is N and the range of B is { 1 }. Of course, we could also write the sequences as A = { 1, 2, 3, 4, . . . } and B = { 1 }.

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Chapter 15. The Real Numbers According to Cantor

If you have seen sequences before, you will recall the major issue was the convergence of sequences. This will be our primary concern also, especially since they do not converge in general. To explain, we need a definition and some examples. Definition 3. A sequence { qn } in Q (respectively, R) converges to q iff for every ε > 0 there exists a natural number N such that |qn − q| < ε whenever n ≥ N . A sequence { qn } in Q is convergent if there is some q ∈ Q such that { qn } converges to q. A sequence { qn } in R is convergent if there is some q ∈ R such that { qn } converges to q. The idea is that a sequence { qn } in Q converges to the limit q if the terms qn get closer and closer to q as n gets larger, that is, as you go further into the endless tail of the sequence. Note, a sequence in Q is also a sequence in R. However, a sequence in Q may converge in R but not in Q. Example 4. (a) The sequence { 0 } converges to 0 since the difference from 0 is always 0. (b) The sequence { n1 } also converges to 0 since n1 can be made as small as we wish, simply by choosing n large enough; this is the Archimedean Property of Q: for any positive ε ∈ Q there exists n ∈ N such that nε > 1 (that is, 0 < n1 < ε). (c) The sequence { 2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . } (in which the successive terms are derived by reducing the last decimal place of its predecessor by one and appending, in the next decimal place, one plus the corresponding decimal place in the decimal representation of √ 2) does not converge in Q. The goal of the construction in this chapter is to make sequences like the one in the last example convergent in the larger set; that is, convergent √ to 2 ∈ R. Before continuing with sequences, we must prove an important inequality known as the Triangle Inequality. It will be very useful for us since it deals precisely with expressions like |qn − q| in the definition of convergent sequence. Exercise 5. Prove that, for all p, q ∈ R, | p + q| ≤ | p| + |q|. Moreover, | p + q| ≥ | p| − |q|. Properties like the following about convergent sequences can be useful.

Section 2. Cauchy Sequences of Rational Numbers

Exercise 6. Suppose { xn } is a sequence in R which converges to x and { yn } is a sequence in R which converges to y. Prove that { xn + yn } converges to x + y. Exercise 7. Suppose { xn } is a sequence in R which converges to x and { yn } is a sequence in R which converges to y. Prove that { xn − yn } converges to x − y. Exercise 8. Suppose { xn } is a sequence in R which converges to x and { yn } is a sequence in R which converges to y. Prove that { xn yn } converges to x y. Exercise 9. Suppose { xn } is a sequence in R which converges to x and { yn } is a sequence in R which converges to y. Prove that { xn /yn } converges to x/y provided that yn = 0 for all n ∈ N and y = 0. 2. Cauchy Sequences of Rational Numbers. What is it about convergent sequences which makes them convergent? In some sense, a convergent sequence requires two things. First, it requires the existence of a limit q, in our set, Q or R, for the sequence to converge to. Second, it requires the behavior of the sequence which makes q the number to which it converges. Notice that if the terms qn get closer and closer to q, then they also get closer and closer to each other. The following exercise should help to explain this better. Exercise 10. Suppose that { qn } is a convergent sequence in either Q or R. Prove that for every ε > 0 there exists a natural number N such that |qm − qn | < ε whenever both m ≥ N and n ≥ N . Sequences with the property in the exercise above were first considered by A. Cauchy. Definition 11. A sequence { qn } in Q or in R is called a Cauchy sequence iff for every ε > 0 there exists a natural number N such that |qm − qn | < ε whenever both m ≥ N and n ≥ N . The converse of Exercise 10 is, of course, false. Consider again the following sequence. { 2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214, . . . } given in Example 4(c).This sequence is Cauchy since terms past the k th term differ in, at worst, the k th decimal place, that is, by at most 101k . However, √ this sequence fails to converge in Q; of course, it converges to 2 in R.

175

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Chapter 15. The Real Numbers According to Cantor

By Exercise 10, every sequence in Q or R which converges is Cauchy. By the previous example the converse is false in Q. In the remainder of this section, we will prove that the converse is true in R. The sequence { n } is clearly not Cauchy. In fact, Cauchy sequences never have this kind of unbounded character, as we show in the next exercise. First, we give some definitions. Definition 12. A sequence { qn } in Q or in R is bounded iff there exists q ∈ Q such that |qn | ≤ q for all n ∈ N. { qn } is bounded above iff there exists q ∈ Q such that qn ≤ q for all n ∈ N. { qn } is bounded below iff there exists q ∈ Q such that qn ≥ q for all n ∈ N. Clearly, a sequence is bounded if and only if it is both bounded above and bounded below. On the other hand, a sequence can be bounded above and not bounded, or bounded below and not bounded as the sequences { −n } and { n } demonstrate. Exercise 13. Suppose { qn } is a Cauchy sequence in Q (or R). Prove that { qn } is bounded. Definition 14. A sequence { qn } in Q or in R is increasing iff qn+1 ≥ qn for all n ∈ N. { qn } is decreasing iff qn+1 ≤ qn for all n ∈ N. { qn } is monotone iff it is either increasing or decreasing. Note that a constant sequence { q } is both increasing and decreasing! Of course, you can imagine how a strictly increasing or strictly decreasing sequence should be defined. A subsequence is much like a subset: a subsequence of a sequence { a1 , a2 , a3 , a4 , . . . } is a sequence { an 1 , an 2 , an 3 , an 4 , . . . } where n 1 < n 2 < n 3 < n 4 < . . . . The following is usually called the Bolzano-Weierstrass Theorem. Theorem 15. Every bounded monotone sequence in R is convergent. Proof. Suppose { xn } is an increasing sequence in R which is bounded above. By the least upper bound property, the range of the sequence has a least upper bound x. We wish to show that the sequence converges to x. Since x is an upper bound, xn ≤ x for all n ∈ N. Fix an ε > 0. There exists an N ∈ N such that x − ε < x N ≤ x, since otherwise x − ε would be a smaller upper bound than the least upper bound x. Since the sequence is increasing, x − ε < x N ≤ xn ≤ x

177

Section 3. Cantor’s Set of Real Numbers

for all n ≥ N . This proves the convergence of { xn } to x. The other case is left as an exercise.




Exercise 16. Complete the proof of Theorem 15. Lemma 17. Every sequence in R has a monotone subsequence. Proof. Suppose { xn } is a sequence in R. Consider the set P = { n ∈ N | xn ≥ xm for all m > n } . (If you graph the sequence, the n’s in P indicate “peaks” in the graph which are higher than all later points.) Either P is infinite or finite. In the case when P is infinite, write P = { n 1 , n 2 , n 3 , n 4 , . . . }, where n k+1 > n k for all k ∈ N. Then the subsequence { xn 1 , xn 2 , xn 3 , xn 4 , . . . } is decreasing. In the case when P is finite, set n 1 greater than any n ∈ P, for example, 1 + max P. Since n 1 ∈ / P, there exists n 2 > n 1 such that xn 1 < xn 2 . Continue in this way to produce a subsequence which is strictly increasing.
The following is also sometimes called the Bolzano-Weierstrass Theorem. Exercise 18. Prove that every bounded sequence in R has a convergent subsequence. We can now prove the converse of Exercise 10 in R. Theorem 19. Every Cauchy sequence in R is convergent. Proof. Suppose { xn } is a Cauchy sequence in R. By Exercise 13, { xn } is bounded. By Lemma 17, there is a monotone subsequence { xn k }. By Theorem 15, { xn k } converges; let x be the limit of the subsequence. Let us fix ε > 0. Since { xn } is Cauchy, there exists a natural number N such that |xm − xn | < 12 ε whenever both m ≥ N and n ≥ N . Since { xn k } converges to x, there exists a natural number M such that |xn k − x| < 12 ε whenever k ≥ M. Pick k such that k ≥ M and n k ≥ N . By the Triangle Inequality (Exercise 5), for all n ≥ N , |xn − x| = |(xn − xn k ) + (xn k − x)| ≤ |xn − xn k | + |xn k − x| < Therefore, the Cauchy sequence { xn } also converges to x.

1 1 ε + ε = ε. 2 2


178

Chapter 15. The Real Numbers According to Cantor

3. Cantor’s Set of Real Numbers. For Dedekind cuts, we defined a real number x as the set of all rational numbers less than x. Now, we wish to define a real number x using Cauchy sequences via those sequences which converge to x. Definition 20. Suppose that x = { xn } and y = { yn } are Cauchy sequences in Q. Define a relation ∼ on the set of all Cauchy sequences in Q by x ∼ y if and only if the sequence { xn − yn } converges to 0. To rephrase the above definition, x ∼ y if and only if for every rational number ε > 0 there exists a natural number N such that |xn − yn | < ε whenever n ≥ N . Exercise 21. Prove that the relation ∼ is an equivalence relation on the set of all Cauchy sequences in Q. Definition 22. The set of equivalence classes of ∼ is the set of Cantor’s real numbers which we denote by RC . Of course, we wish to show that R and RC are isomorphic, both being the set of all real numbers. First, we need to define order and operations on RC . It is natural to think that [x] < [y] should mean that [y] − [x] > 0 and, hence, [y] − [x] > ε for some positive rational ε. Definition 23. Define a relation < on RC by [x] < [y], for Cauchy sequences x = { xn } and y = { yn } of rational numbers, if and only if, there exists a rational number ε > 0 and there exists a natural number N such that xn + ε < yn whenever n ≥ N . Notice that Definition 23 of [x] < [y] seems to depend on the choices of representatives of the equivalence classes [x] and [y]. Exercise 24. Prove that the relation < on RC is well-defined. From here on we will assume that { xn } is a representative sequence in [x] and so on for other elements of RC . As always, we define [x] ≤ [y] to mean [x] < [y] or [x] = [y]. Example 25. Let xn = − n1 and yn = n1 for all n ∈ N. Then [x] = [y] = 0 since |xn − yn | = n2 converges to 0. Note that xn < yn for all n ∈ N. That is, there does not exist an N ∈ N such that such that yn ≤ xn whenever n ≥ N . Therefore, we could not define ≤ by comparing the terms xn and yn directly. Exercise 26. Prove that < is a (strict) total ordering on RC . Next we define the operations in the obvious way.

Section 4. The Isomorphism from Cantor’s to Dedekind’s Reals

Definition 27. Define addition on RC by [x] + [y] = [{ xn + yn }]. Exercise 28. Prove that addition on RC is well-defined. Definition 29. Define multiplication on RC by [x][y] = [{ xn yn }]. Exercise 30. Prove that multiplication on RC is well-defined. By the way, RC completes Q in the way desired in Section 1. If q ∈ Q, then it is represented by the constant sequence in RC , that is, q = [{ q }] ∈ RC is the embedded image of q ∈ Q. So a sequence { xn } in Q can be viewed as a sequence { xn } in RC . Exercise 31. If x = { xn } is a Cauchy sequence in Q, then { xn } converges to [x] in RC . Next we could check that RC with this ordering and these operations has all of the properties of an ordered field with the least upper bound property and that Q embedded as a subfield. Only the least upper bound property would give us any serious work. We know all such structures are isomorphic and, in particular, RC is isomorphic to R. However, we will construct the isomorphism directly. 4. The Isomorphism from Cantor’s to Dedekind’s Reals. We next wish to give an isomorphism f : R → RC . Take a cut representing some x in R. If we can extract a sequence from the cut which converges to x, this gives us a way to define a candidate for our isomorphism. By the way, it is much harder to define the isomorphism in the opposite direction. This makes sense since we know all about R but we know almost nothing about RC . You will make frequent use of the properties of R in proving the existence of the isomorphism. Recall that the cut x is the set { q ∈ Q | q < x in R }. Pick q1 ∈ x. Set q2 = q1 + 1 if q1 + 1 < x or q2 = q1 + 12 if q1 + 12 < x < q1 + 1 or etc. Continue in the same way to define q3 , q4 , etc. This defines an increasing sequence { qn } in Q. We will check that this sequence is Cauchy and then define f : R → RC by f (x) = [{ qn }]. Exercise 32. Prove that the process in the previous paragraph actually defines a sequence and that it is Cauchy in Q and converges to x in R. We define f : R → RC by f (x) = [{ qn }]. We must check that this is well-defined, that is, independent of the initial choice q1 .

179

180

Chapter 15. The Real Numbers According to Cantor

Exercise 33. Prove that f is well-defined. It remains to show that f is an isomorphism, that is, a structurepreserving bijection. Exercise 34. Prove that f is order-preserving and, hence, one-to-one. Exercise 35. Prove that f is addition-preserving. Exercise 36. Prove that f is multiplication-preserving. Exercise 37. Prove that f is onto. This completes the proof that the structure RC we constructed in this chapter is really the same set of real numbers R with its order relation and operations.

181

PART IV HINTS

182

Chapter 16 Hints for (and Comments on) the Exercises Hints for Chapter 1. Exercises 3, 4 and 5. These are meant to inspire thought rather than specific answers. Exercise 10. Think trilateral rather than triangle; that is, use line segments, not angles. Exercise 12. Use scale and length of line measure for betweenness and congruence of segments, respectively. Use angle measure similarly. Hints for Chapter 2. Exercise 18. That is, if p and q is true, then p is true and if p is true, then p or q is true. Of course, p ∧ q ⇒ q and q ⇒ p ∨ q are also tautologies. Exercise 19. That is, if p implies q is true, then p is false or q is true. Exercise 20. These are known as the reflexive, symmetric, and transitive properties of equivalence, respectively. Exercise 21. This is why the phrases ”if and only if” and ”necessary and sufficient” make sense. Exercise 22. The point of this exercise is to make sure that you understand the value of truth tables! Every row in a truth table represents an entire paragraph in words. Exercise 24. Be careful! (d) is impossible! Exercise 27. You can do this in words or with truth tables. This is a very important tool for constructing proofs. Exercise 29. (a) is exactly Exercise 27 after a change in variables. You can disprove with a counterexample or with truth tables. Exercise 30. This is the same kind of algebra exercise as you have often seen before.

183

Chapter 16. Hints for (and Comments on) the Exercises

Hints for Chapter 3. Exercise 3. Mimic the first part of the proof of Proposition 1. Exercise 5. Mimic the proof of Proposition 4. Exercise 6. Use proof by brute force. Exercise 7. Use proof by brute force. Exercise 8. Use proof by brute force. Exercise 11. Consider when 2x + 1 is nonnegative and when it is positive. Exercise 12. You may use the facts that products of positive real numbers are positive and (−1)2 = 1. Exercise 14. Try proof by contraposition. You may assume that, for integers p and q, q = 0 implies that p 2 + q 2 = 0; this is proved in Exercise 23. Exercise 15. Since this is the converse of Exercise 12, that does not help to prove it. Try proof by contraposition. Exercise 16. Break the equivalence into two implications: use direct proof for one and contraposition for the other. Exercise 22. For each, mimic the proof of Theorem 18. The phrase “is even” can be replaced by “is divisible by 2;” this becomes “is divisible by 3” Similarly, the condition m = 2k becomes m = 3k. Exercise 23. Use proof by brute force for one direction and proof by contradiction for the other. You can use Exercise 12. Exercise 24. Multiply both sides of an inequality by x. Exercise 25. Suppose there are n ≥ 2 people. Use proof by contradiction. Consider the set of numbers of people that each person knows. Show that someone must know no one else while someone else must know every one. Exercise 33. Where is the first step in the induction? Exercise 34. Examine how the situation is different when n + 1 = 2 from when n + 1 = 3.   Exercise 35. Consider 1 + 3 + 5 + · · · + (2n − 1) + (2(n + 1) − 1) in a proof by induction.  n n+1

1

1  in a = +   1 Exercise 36. Use k(k+1) k(k+1) k=1

k=1

n+1

(n+1)+1

proof by induction. Exercise 37. Use proof by induction. Exercise 38. Use proof by induction (on n). Exercise 41. Examine the first application of the inductive step, going from 100 = 1 to 100+1 = 1. Exercise 42. Given n, n is either prime or the product of two integers strictly between 1 and n.

184

Chapter 16. Hints for (and Comments on) the Exercises

Exercise 43. Consider 0. Exercise 44. Consider f (x) = |x|. Hints for Chapter 4. Remember Venn diagrams and logic! Exercise 6. Is it true that if x ∈ ∅, then x ∈ X ? Recall that statements such as ∅ ⊂ X are sometimes called vacuously true. Exercise 7. Remember that ⊂ does not mean proper subset. Exercise 8. You could use ( p ⇒ q) ∧ (q ⇒ r ) ⇒ ( p ⇒ r ) (Supplemental Exercise S5 of Chapter 2), but this is actually straightforward. Exercise 11. Use ( p ⇔ q) ⇔ ( p ⇒ q) ∧ (q ⇒ p) (Exercise 21 of Chapter 2). Exercise 13. Use Exercises 7, 8, and 11. Exercise 16. Use Exercises 6 and 11; use the tautology p ∧ F ⇔ F, where F is always false (Supplemental Exercise S33 of Chapter 2). Exercise 17. Use the tautology p ∧ p ⇔ p (Supplemental Exercise S22 of Chapter 2). Exercise 21. Use the tautology p ∨ F ⇔ p (Supplemental Exercise S32 of Chapter 2). Exercise 22. Use the tautology p ∨ p ⇔ p (Supplemental Exercise S21 of Chapter 2). Exercise 23. Break the compound statement into A ∩ B ⊂ A and A ⊂ A ∪ B. Use Exercise 8 for the second part. Exercise 24. You could use Exercise 23 (A ∩ B ⊂ B) to show that A ∩ B ⊃ A implies A ⊂ B. Exercise 25. You could use Exercise 23 (A ⊂ A ∪ B) to show that A ∪ B ⊂ B implies A ⊂ B. Exercise 26. Use both parts of Exercises 25 and 24. Exercise 27. Use p ∧ q ⇔ q ∧ p and p ∨ q ⇔ q ∨ p (Supplemental Exercises S15 and S16 of Chapter 2). Exercise 28. Use ( p∧q)∧r ⇔ p∧(q ∧r ) and ( p∨q)∨r ⇔ p∨(q ∨r ) (Supplemental Exercises S17 and S18 of Chapter 2). x ∈ A ∩ (B ∪ C) means x ∈ A and x ∈ (B ∪ C), which means x ∈ A, and x ∈ B or x ∈ C. Notice how the comma plays the role of parentheses. Be careful! It is unclear what the following means: x ∈ A and x ∈ B or x ∈ C. Exercise 29. Use p ∧ (q ∨ r ) ⇔ ( p ∧ q) ∨ ( p ∧ r ) and p ∨ (q ∧ r ) ⇔ ( p ∨ q) ∧ ( p ∨ r ) (Supplemental Exercises S19 and S20 of Chapter 2). Exercise 36. Recall that [a, b) = { x ∈ R | x ≥ a and x < b } and use the definition of intersection. Exercise 37. See Figure 31 and Example 34. Exercise 39. Recall the definitions of [a, b), (a, b), and union.

Chapter 16. Hints for (and Comments on) the Exercises

Exercise 40. Compare with Exercise 29. Exercise 43. Recall the definition of A − B. Exercise 44. Consider B = ∅ and A = B. You have been using Venn diagrams, right? Exercise 45. Recall the definition of A− B; logical and is commutative and associative. Exercise 46. Try ∅, { 1 } and { 1, 2 } Exercise 47. Follow the Venn diagram. Exercise 48. This is similar to Exercise 47. Exercise 50. Recall the definitions of A − B and B  . Exercise 51. (B  ) = B is analogous to ¬(¬ p) ⇔ p. Use Exercise 50. Exercise 52. Follow the Venn diagram. Exercise 54. Remember that ∅ and X are always subsets of X . You must prove that ∅ is the only subset of ∅! Exercise 55. We have seen that |P(∅)| = 1 = 20 , |P(Z1 )| = 2 = 21 , and |P(Z1 )| = 4 = 22 . Power, get it? Exercise 56. Use Exercise 8. Exercise 57. Both are true: P(A) ∩ P(B) = P(A ∩ B). Exercise 58. Only one is true. X ⊂ A ∪ B does not imply that X ⊂ A or X ⊂ B. Exercise 59. Neither one holds! Check ∅; compare with Supplemental Exercise S35. Hints for Chapter 5. Exercise 3. They are, respectively, 3-dimensional space, a 2-dimensional lattice of points, 2 horizontal parallel lines, and 2 vertical parallel lines (given that the plane has the standard orientation). Exercise 4. Assume that A × B = ∅; assume that A = ∅ and B = ∅. Exercise 5. Assume that (A × B) ∩ (B × A) = ∅; assume that A ∩ B = ∅. Exercise 15. This is probably the more familiar definition; it is just a matter of the meaning of f (x). √ Exercise 18. Graph the half-parabola y = x. Exercise 19. Use your knoweldge of precalculus and/or calculus. Exercise 23. If y ∈ f (A), then there exists x ∈ A such that y = f (x). Exercise 24. Suppose y ∈ f (A ∪ B); suppose y ∈ f (A) ∪ f (B). Exercise 26. Suppose y ∈ f (A ∩ B). Exercise 28. Use contraposition. Exercise 29. Compare with Example 25 and Exercise 26. Exercise 31. This is probably the more familiar definition. Exercise 33. Use the same rule and domain with a “smaller” codomain.

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Exercise 34. Compare with Exercise 31; uniqueness comes from oneto-one. Exercise 35. Think about functions you know: polynomial, trigonometric, exponental, etc. Exercise 37. Unravel g( f (x)). Exercise 40. That is, show (( f ◦ g) ◦ h)(x) = ( f ◦ (g ◦ h))(x). Exercise 41. Start with f : X → Y , g : W → Z , Y ⊂ W , and x1 = x2 in X . Exercise 42. Notice that the codomain and range of f equals the domain of g. Determine g( f (X )). Exercise 44. By definition of function, g(y1 ) = g(y2 ) implies y1 = y2 . Consider functions on small finite sets for your counterexample. Exercise 45. By definition of function, f (X ) ⊂ Y . Exercise 47. Assume there is another identity function i : X → X . Show that i ◦ I X = i and i ◦ I X = I X . Exercise 49. What are the domains and codomains of f ◦ g and g ◦ f ? Exercise 51. Compare with Exercises 44 and 45. Exercise 52. Assume f has another inverse function g. Consider g ◦ f ◦ f −1 in two ways using the associative property. Exercise 55. If x1 , x2 ∈ f −1 (y), then f (x1 ) = y = f (x2 ). Exercise 56. x ∈ f −1 (A ∩ B) is equivalent to f (x) ∈ A ∩ B is equivalent to f (x) ∈ A and f (x) ∈ B, etc. Exercise 57. By (c), for (b), f cannot be one-to-one; use small finite sets for your counterexample. (d) is false! Exercise 58. By (c), for (b), f cannot be onto. (d) is false! Exercise 62. No! State the domains and codomains explicitly. Exercise 63. You must show that the domains are equal. Exercise 64. Be sure that the range of f is not a subset of the domain of g. Hints for Chapter 6. Exercise 4. This is different from Exercise 3; ⊂ is different from
and the structure of R is different from that of P(U ). Exercise 5. Consider the relation =. Exercise 6. You may try to consider relations on small finite sets. Exercise 7. Since there are eight properties, this means to give 28 = 256 examples! Since relations cannot be both reflexive and nonreflexive, cannot be both symmetric and asymmetric, and trichotomous implies connected, this eliminates more than half of the cases. Give at least a few, if not all. Exercise 10. Check reflexive and nonreflexive.

Chapter 16. Hints for (and Comments on) the Exercises

Exercise 11. Compare the symbols ≺ and " with < and ≤, respectively; recall that our ⊂ is denoted ⊆ by some authors. Exercise 14. x " y and y " x imply x " x, which implies x = x; x ≺ x implies x = x. Exercise 18. Connected but not trichotomous contradicts nonreflexiveness. Exercise 23. Assume there are two candidates for greatest element. Use antisymmetry to show that they must be equal. Exercise 24. Assume that a greatest element is not maximal. Use nonreflexivity of ≺. Exercise 25. Assume there exists a greatest element and show maximal elements are equal. Exercise 26. Use connectedness. Exercise 31. Assume there are two least upper bounds. Use antisymmetry to show that they must be equal. Exercise 32. Consider intervals in R. Exercise 33. Check the properties. Exercise 40. See Chapter 7, Section 2. Exercise 42. Use s ≈ t to show that [s] ⊂ [t] and [t] ⊂ [s]. Exercise 43. Show that if x ∈ [s] ∩ [t], then [s] = [t]. Exercise 46. Check the properties. Exercise 48. Check the properties. Exercise 50. Exercise 52. Hints for Chapter 7. Exercise 5. Show that any function f : ∅ → ∅ satisfies the conditions of a bijection vacuously. Exercise 6. Consider I A . Exercise 7. Consider the inverse of a one-to-one correspondence. Exercise 8. Consider the composition of two one-to-one correspondences. Exercise 11. Use Exercise 7, Exercise 8 and Theorem 9. Exercise 12. For (a), extend a one-to-one correspondence Zn → A by n + 1 &→ x. (b) follows from (a). Exercise 13. Proceed by induction on |B|. Exercise 15. f induces a one-to-one correspondence onto a subset of B. Exercise 16. There exists a one-to-one correspondence between B and a subset of A.

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Chapter 16. Hints for (and Comments on) the Exercises

Exercise 18. If f : Zn → N (n ≥ 2) is a one-to-one correspondence, show that the sum of the elements of the range is not in the range, contradicting that f is onto. Exercise 20. Compare with Theorem 14. Exercise 24. For a one-to-one correspondence f : N → D, set k = f −1 (x). Map n &→ f (n + 1) for n ≥ k. Exercise 25. Suppose X is an infinite set. By Theorem 23, there is a denumerable subset D of X . Choose x ∈ D. By the proof of Exercise 24, D and D − { x } have the same cardinality. Use the identity function on X − D to complete the construction of a bijection from X onto X − { x }. Exercise 26. Compare with Exercise 25. Exercise 30. By transitivity. Exercise 31. Define a one-to-one correspondence by, for example, alternating positive and negative. Exercise 34. Use the one-to-one correspondence of Theorem 32 to define a one-to-one correspondence by alternating positive and negative. Exercise 35. Make an array of lists of the countable sets and use the “zig-zag” method of the proof of Theorem 32. Exercise 36. Use induction. Exercise 38. For polynomials of a fixed degree there are countably many algebraic numbers. Exercise 41. See Theorem 29. Exercise 42. Consider a tangent function. Exercise 43. Consider “half of” a tangent function. Exercise 44. Consider a translation function (that is, of the form f (x) = x + t). Exercise 45. Consider the negation function. Exercise 46. Consider an affine function (that is, of the form f (x) = mx + b). Exercise 47. Use Exercises 42–46. Exercise 48. For (0, 1) and [0, 1) and for (0, 1] and [0, 1], use the technique of the proof of Exercise 25. Exercise 49. Generalize Exercise 48, considering different cases. (This is similar to the method of Exercises 42–47.) Exercise 50. Assume R − Q is countable. Then Q ∪ (R − Q) is countable. Exercise 51. Similar to Exercise 50. Exercise 54. Assume it is countable. Use the diagonal technique used in the proof that (0, 1) is uncountable (Theorem 40). Exercise 55. Use Exercise 54. Exercise 56. Assume it is countable. A subset S of N, that is, an element of P(N), can be represented by a sequence of 0’s and 1’s where the

Chapter 16. Hints for (and Comments on) the Exercises

n-th entry, sn , in the sequence indicates n ∈ / S if sn = 0 or n ∈ S if sn = 1. Use the diagonal technique on the list of such sequences. Exercise 57. Assume that there is a one-to-one correspondence f : S → P(S). Let A = { s ∈ S | s ∈ / f (s) }. Show that A ∈ P(S) but A∈ / f (S). Hints for Chapter 8. Exercise 13. This is similar to Exercise 11. Exercise 14. This is similar to Exercise 12. Exercise 16. Add −x. Exercise 17. Use Exercise 12. Exercise 18. This is similar to Exercise 16. Exercise 19. This is similar to Exercise 17. Exercise 20. Exercise 21. Exercise 22. Exercise 23. Exercise 25. Exercise 26. Exercise 27. Exercise 28. Exercise 29. Exercise 30. Exercise 33. Exercise 35. Show that E has no multiplicative identity element. [Compare with the proof of Proposition 29 in Chapter 11.] Exercise 36. Hints for Chapter 9. Exercise 3. Exercise 5. Hints for Chapter 10. Exercise 7. Use proof by contradiction to show that N is well ordered: assume S is a nonempty subset with no least element and show by induction that 1 and its successors Succ P (1) are all in the complement of S. Exercise 9. Exercise 11. Exercise 12. Exercise 13.

189

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Chapter 16. Hints for (and Comments on) the Exercises

Exercise 14. Use induction. Exercise 15. Exercise 16. Assume there is another additive identity z. Consider 0 + z. Exercise 17. For the inductive step, it is easiest to convert to powers of Succ. Exercise 18. Exercise 19. Exercise 20. Exercise 22. Use induction on p. Exercise 23. Show 1m = m by induction. Remember to show uniqueness! Exercise 24. Show 0m = 0 by induction. For the inductive step, use Exercise 23. Exercise 25. Exercise 26. Exercise 27. Exercise 29. Hints for Chapter 11. Exercise 2. Use (AC) for Z+ to show that ∼ is reflexive and symmetric on Z+ × Z+ ; use (AC) and (AA) for Z+ to show that ∼ is transitive on Z+ × Z+ . Exercise 8. Exercise 9. Show that, for any [m, n] ∈ Z, [m, n + 1] < [m, n]. Exercise 12. Exercise 14. Exercise 15. Use (AC) for Z+ . Exercise 16. Use (AA) for Z+ . Exercise 17. Remember to show both existence and uniqueness. Also, cf., Proposition 11 of Chapter 8. Exercise 18. It is easier to start with [m, n] than p or − p. Remember to show both existence and uniqueness. Also, cf., Proposition 12 of Chapter 8. Exercise 19. Cf., Exercise 17 of Chapter 8. Exercise 20. Cf., Exercise 16 of Chapter 8. Exercise 21. Use (PP) for Z+ . ????? Exercise 24. Use (MC) and (MA) for Z+ . Exercise 25. Use (MC) and (MA) for Z+ . Exercise 26. Remember to show both existence and uniqueness. Also, cf., Exercise 13 of Chapter 8. Exercise 27. Cf., Exercise 14 of Chapter 8.

Chapter 16. Hints for (and Comments on) the Exercises

Exercise 28. Exercise 30. Exercise 31. Exercise 32. Given p ∈ Z+ , what element of Z resembles p? Hints for Chapter 12. Exercise 2. Use (MC) for Z to show that ∼ is reflexive and symmetric on Z × Z∗ ; use (MC), (MA) and (MX) for Z to show that ∼ is transitive on Z × Z∗ . Exercise 4. Use the Negation Reverses Order and Factoring a Minus Properties for Z. Exercise 5. For n < 0, use Exercise 4. Exercise 8. Use (MC), (MA) and (IX) for Z. Exercise 9. Use (MC), (MA) and (MX) for Z to show that < is transitive; use the trichotomy of < on Z to show that < is trichotomous. Exercise 10. Show that [m − 1, n] < [m, n] for any [m, n] ∈ Z. Exercise 12. Use the trichotomy of <. Exercise 14. Use (MC), (MA), (MX) and (DP) for Z. Exercise 15. Use (AC) and (MC) for Z. Exercise 16. Use (AA), (MA) and (DP) for Z. Exercise 17. Remember to show both existence and uniqueness. Also, cf., Proposition 11 of Chapter 8. Exercise 18. Remember to show both existence and uniqueness. Also, cf., Proposition 12 of Chapter 8. Exercise 19. Cf., Exercise 16 of Chapter 8. Exercise 21. Use (MC) and (MA) for Z. Exercise 22. Use (MC) for Z. Exercise 23. Use (MA) for Z. Exercise 24. Use (MC) and (MA) for Z. Also, cf., Exercise 13 of Chapter 8. Exercise 25. Use (MC) and (MA) for Z. Also, cf., Exercise 14 of Chapter 8. Exercise 26. Cf., Exercise 18 of Chapter 8. Exercise 27. Use (MC) and (MA) for Z. Exercise 28. Use (MC), (MA), (MX) and (DP) for Z. Exercise 29. Use (PP) for Z. Exercise 30. Cf., Exercise 25 of Chapter 8. Exercise 31. You already know that 0 &→ [0, 1] and 1 &→ [1, 1]. Exercise 32. Use the Archimedean Principle on Z. Exercise 33. Cf., Exercise 4 of Chapter 9. Exercise 34. Take the average of p and q.

191

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Chapter 16. Hints for (and Comments on) the Exercises

Hints for Chapter 13. Exercise 2. Exercise 4. Exercise 5. Suppose S ⊂ R is nonempty and bounded above by U ; that is, S ⊂ U for all S ∈ S. Set L = ∪S. Show that S ∈ R and S is the least upper bound of S. Exercise 7. Exercise 8. Exercise 9. Exercise 10. Show that X + 0 ⊂ X and X ⊂ X + 0. Similarly for 0 + X. Exercise 11. Exercise 12. Exercise 13. Exercise 14. Use the Cancellation Property for Addition (Exercise 16) to show strict inequality. Exercise 16. Exercise 17. Exercise 18. Exercise 19. Exercise 20. Exercise 21. Exercise 22. Exercise 23. Exercise 24. Exercise 25. Exercise 26. Exercise 27. Exercise 28. Exercise 29. Exercise 30. Exercise 31. Exercise 32. Exercise 33. Exercise 34. Exercise 35. Thinking of cuts, x < y implies f (x) ≤ f (y). By density of rationals, there exists q, r ∈ Q such that x < q < r < y. Exercise 36. Use Exercise 35. Exercise 37. For q ∈ F , set S = { q ∈ Q | f (q) < a } and x = sup S. Show that f (x) = q.

Chapter 16. Hints for (and Comments on) the Exercises

Exercise 38. If x, y ∈ R and q ∈ Q such that f (x + y) < f (q) < f (x) + f (y), then there exist q1 , q2 ∈ Q such that q1 + q2 = q, x < q1 , and y < q2 . Exercise 39. Start with positive real numbers. Exercise 40. Use the properties of isomorphisms to show that f (x) = x for all x ∈ Z, then for all x ∈ Q, and finally for all x ∈ R. Exercise 42. Consider g −1 ◦ f for isomorphisms f, g : R → F . Hints for Chapter 14. Exercise 4. Exercise 5. Exercise 6. Exercise 7. Exercise 8. Exercise 9. Exercise 10. Exercise 11. Exercise 12. Exercise 13. Exercise 16. Use Definition 3 to prove the properties of a total ordering. To see this is not an ordered field, compare i with 0 and 0 with −1 and consider the product of positives property of ordered fields. Exercise 17. Exercise 18. Either i > 0 or i < 0. (Why?) Use Exercise 26 of Chapter 8. Exercise 19. Exercise 20. Consider the image of i and the preservation of multiplication. Exercise 21. Hints for Chapter 15. Exercise 5. Exercise 6. Exercise 7. Exercise 8. Exercise 9. Exercise 10. Exercise 13. Exercise 15. Exercise 16. Exercise 17.

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Chapter 16. Hints for (and Comments on) the Exercises

Exercise 18. Exercise 19. Exercise 21. Exercise 24. Exercise 26. Exercise 28. Exercise 30. Exercise 31. Exercise 32. Exercise 33. Exercise 34. Use the denseness of Q in R. Exercise 35. Exercise 36. Exercise 37.

195

Postscript and Selected References Completion of Part II of these notes, together with a sequence in calculus, should provide a student with more than adequate preparation for a course in linear algebra or abstract algebra or number theory. Actually, Part II alone should suffice; however, calculus may provide for some examples in all such courses. Completion of Parts II and III of these notes, together with a sequence in calculus, should provide a student with more than adequate preparation for a course in analysis (or advanced calculus) or general (point-set) topology or axiomatic set theory or logic. Conway, Lakatos, Knuth

References [Bell] [Halmos]

Bell, E. T., The Development of Mathematics, Dover, New York, 1945, 1972. Halmos, Paul, Finite Dimensional Vector Spaces, Springer-Verlag, New York, 19??. [Herstein] Herstein, I. N., Topics in Algebra, Ginn, Waltham, MA, 1964. [Rudin] Rudin, Walter, Principles of Mathematical Analysis, Third Edition, McGraw-Hill, New York, 1976. [Spivak] Spivak, Michael, Calculus, Second Edition, Publish or Perish, Wilmington, DE, 1980.

196

Index The numbers after the keywords refer to item numbers; for example, 3.2 refers to Definition 2 (or Example 2, or Remark 2, etc.) in Chapter 3. The notation “iv” refers to page iv, concerning notation, which appears before the Preface; there are no other page numbers yet—sorry! Symbols
, iv ⊂, iv, 4.3 , iv ∅, 4.2 ∈, 4.1 { }, 4.2 (a, b), iv [a, b), iv (a, b], iv [a, b], iv C, iv, 14.1 i, 14.14 N, iv, 10.4 Q, iv, 12.3 Q+ , iv Q∗+ , iv R, iv, 13.1, 15.22 R+ , iv R∗+ , iv Z, iv, 11.3 Zn , iv Z+ , iv, 10.4 Z∗ , iv A Addition on C, 14.1 Addition on Q, 12.13 Addition on R, 13.6 Addition on Z+ , 10.10

Index

Addition on Z, 11.13 Algebraic Nnumbers, 7.37 And, 2.3 Antisymmetric relation, 6.1(d) Archimedean Principle, 8.1 B Bijective function, 5.32, 5.55 Birkhoff, G.D., 1.6 Bolzano-Weierstrass property, 15.15 Bounded sequence, 15.12 C Cardinality of sets, same, 7.2 Cartesian product of sets, 5.1, 7.52 Cauchy sequence, 15.11 Codomain of a function, 5.10 Complex numbers, set of, iv, 14.1 Composition of two functions, 5.36, 5.60 Contradiction, 2.ContradictionDef Contraposition, proof by, 3.13 Connected relation, 6.1(f) Convergent sequence, 15.3 Countable set, 7.22 Cut, 13.1 D Decreasing sequence, 15.14 Dedekind cut, 13.1 Denumerable set, 7.21 Dense, 8.6 Descendency set, 10.5 Difference of sets, 4.42 Disjoint sets, 4.18, 4.33 Domain of a function, 5.10 E Element, 4.1 Embedding, 11.37 Empty set, 4.2 Equal functions, 5.20 Equal sets, 4.9 Equivalence class, 6.41 Equivalence relation, 6.37 F

197

198

Index

Field, 8.4 Finite set, 7.10 Function, 5.10 G Greatest element, 6.20 H I Identity function, 5.46 Iff, 3.10 Image of a set under a function, 5.13 Inclusive or, 2.4 Increasing sequence, 15.14 Induction, 3.26 Infinite set, 7.17 Injective function, 5.27, 5.55 Integers, set of, iv, 11.3 Intersection of two sets, 4.14 Intersection of a collection of sets, 4.30 Inverse image of a set under a function, 5.53 Invertible function, 5.48 Isomorphism, 8.31 J K L Least upper bound, 6.28 Least upper bound property, 6.34 Less than on C, 14.15 Less than on Q, 12.7 Less than on R, 13.3, 15.23 Less than on Z+ , 10.6, 10.12 Less than on Z, 11.4 Lexicographic ordering, 14.15 M Mathematical or, 2.4 Maximal element, 6.19 Mod, for an equivalence relation, 1.6, 6.45 Mod, for a partition, 6.47 Monotone sequence, 15.14 Multiplication on C, 14.1 Multiplication on Q, 12.20 Multiplication on R, 13.15

Index

Multiplication on Z+ , 10.21 Multiplication on Z , 11.22 N Natural numbers, set of, iv, 10.4 Negative integer, 11.11 Nonnegative integers, set of, iv Nonnegative rational numbers, set of, iv Nonnegative real numbers, set of, iv Nonreflexive relation, 6.1(b) Nonzero integers, set of, iv Not, 2.2 O One-to-one function, 5.27, 5.55 Onto function, 5.30, 5.55 Or, 2.4 Ordered field, 8.8 Ordering on C, 14.15 Ordering on Q, 12.7 Ordering on R, 13.3, 15.23 Ordering on Z+ , 10.6 Ordering on Z, 11.4 P Pairwise disjoint collection of sets, 4.32 Partial ordering, 6.9 Partition, 6.44 Positive integer, 11.11 Positive rational numbers, set of, iv Positive real numbers, set of, iv Power set, 4.53 Predecessor, 10.8 Principle of Induction, 3.26 Product of sets (cartesian), 5.1, 7.52 Proper subset, 4.4 Q Quantifier, Hidden, 3.33 R Range of a function, 5.14 Rational numbers, set of, iv, 12.3 Real numbers, set of, iv, 13.1, 15.22 Reflexive relation, 6.1(a) Relation, 6.6 Relation, properties of a, 6.1

199

200

Index

S Same cardinality, 7.2 Sequence, 15.1 Set, 4.1 Set difference, 4.42 Strict partial ordering, 6.8 Strict total ordering, 6.16 Subset, 4.3 Subset, proper, 4.4 Successor, 10.2 Surjective function, 5.30, 5.55 Symmetric relation, 6.1(c) T Tautology, 2.TautologyDef Total ordering, 6.17 Transitive relation, 6.1(e) Trichotomous relation, 6.1(g) U Uncountable set, 7.39 Union of two sets, 4.19 Union of a collection of sets, 4.38 Upper bound, 6.27 V W Well ordered set, 6.36 X Y Z Zero, 10.3