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1 ca_main March, 08 3:5 Page i College Algebra Preparing for Calculus J. B. T

2 ca_main March, 08 3:5 Page ii College Algebra: Preparing for Calculus Copyright c 08 by J. B. Thoo. All rights reserved. No part of this book may be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Author. Composed using the L A TEX Document Preparation System. TEX is a trademark of the American Mathematical Society. The Author bears no responsibility for the persistence or accuracy of URLs for external or third-party Internet Web sites referred to in this book, and the Author does not guarantee that any content on such Web sites is, or will remain, accurate or appropriate. ISBN INFO ISBN 3:

3 ca_main March, 08 3:5 Page iii Contents Preface Logic, Sets, Real Numbers, Complex Numbers. Logic Conditional statement; negation Conjunction, disjunction Biconditional statement Quantifiers Methods of Proof Sets What Is a Set? Existence, Equality, Subset Union, Intersection, and Complement The Natural Numbers and Induction Power, Ordered Pair, and Cartesian Product Functions and Relations Countable Sets and Cardinality Real Numbers Properties of Real Numbers The Extended Real Number System Rational Numbers Cardinality of Q and R Other Ways to Define R Complex Numbers Supernumbers Supernumbers Are Complex Numbers Polar Form Absolute Value, Product, Power, Root Exercises Functions, Graphs, and Equations 95. Functions Definition Domain, Range One-to-One, Onto Composition Graphs Exercises v Index 7 iii

4 ca_main March, 08 3:5 Page Chapter Logic, Sets, Real Numbers, Complex Numbers. LOGIC Superior thinking has always overwhelmed superior force... Conditional statement; negation United States Marine Corps recruiting poster Think About It. Johnny and his parents are sitting down to dinner, and a conversation goes like this. Johnny: Mom, may I have ice cream tonight? Mom: You may have ice cream if you finish your broccoli. Johnny: Thanks, Mom. Johnny s mom leaves the room, and Johnny turns to his dad. Johnny: Dad, I m full. May I have ice cream now? Dad: You haven t finished your broccoli. Johnny: But, Dad, I don t like broccoli; and I ate half of it already. Dad: Well, alright, go and get the ice cream from the freezer. You be the judge: Did Johnny s dad contradict the mom because he let Johnny have ice cream even though Johnny had not finished his broccoli? When you study calculus, you will encounter statements such as the following: A function f is continuous at a point x [a, b] if f is di erentiable at x. When we are presented with a statement such as this, in addition to understanding the notation and the terminology, it will also be important for us to know what it asserts and what it does not assert; in other words, it will also be important for us to understand the logical structure of statements like this. As another example, the statement that Johnny s mom made to him,

5 ca_main March, 08 3:5 Page CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS You may have ice cream if you finish your broccoli, has the same logical structure as the mathematical statement above. Certainly, the latter statement tells us that Johnny may have ice cream if he finishes his broccoli; however, it does not tell us that Johnny may not have ice cream if he does not finish his broccoli; that is, in the dinner scene above, Johnny s dad did not contradict his mom by letting Johnny have ice cream even though Johnny had not finished his broccoli. Is that surprising? Think About It. In the dinner scene above, would it have made a di erence if Johnny s mom had said instead, You may have ice cream only if you finish your broccoli? Explain. The statement, You may have ice cream if you finish your broccoli, is an example of a conditional statement; that is, it is a statement that is in the form Q if P, where here Q stands for you may have ice cream and P stands for you finish your broccoli. There are di erent equivalent ways to express a conditional statement; here are a few examples. Q if P You may have ice cream if you finish your broccoli. If P, then Q If you finish your broccoli, then you may have ice cream. P only if Q You finished your broccoli only if you may have ice cream. P implies Q Your finishing your broccoli implies that you may have ice cream. P is su cient for Q Your finishing your broccoli is su Q is necessary for P cient for you to have ice cream. Your being allowed to have ice cream is necessary for you to have finished your broccoli. Think About It.3 What is the di erence between Your finishing your broccoli is su cient for you to have ice cream and Your finishing your broccoli is necessary for you to have ice cream? Which do you think Johnny s mom meant to say in the dinner scene above? Explain. In a conditional statement, If P, then Q, P is called the hypothesis or antecedent, and Q is called the conclusion or consequent. Note that, in all of the di erent ways of stating, If you finish your broccoli, then you may have ice cream, above, P = you finish your broccoli is the hypothesis and is the conclusion. Q = you may have ice cream

6 ca_main March, 08 3:5 Page 3.. LOGIC 3 Now You Try. For each of the following conditional statements, identify the hypothesis and the conclusion, and then rephrase each statement in several di erent ways.. If it is raining, then I will take my umbrella.. A function f is continuous at a point x if f is di erentiable at x. Now, consider the statements, Every turkey is a bird and All turkeys are birds. Both statements are actually the same conditional statement in disguise: If it is a turkey, then it is a bird, where it is a turkey is the hypothesis and it is a bird is the conclusion. Think About It.4 Convince yourself that both statements Every turkey is a bird and All turkeys are birds are actually the same conditional statement: If it is a turkey, then it is a bird. Now You Try. For each of the following conditional statements, identify the hypothesis and the conclusion, and then rephrase each statement in several di erent ways.. All Marines have courage.. Every infinite subset of a countable set A is countable. When is a conditional statement true or a rmed, and when is it false or contradicted? We answer this question using the truth table below. In this table, the arrow! denotes implies so that P! Q is P implies Q or, equivalently, If P, then Q, and so on. Also, in the table, T denotes true, and F denotes false. Conditional P Q P! Q T T T T F F F T T F F T We go through this table line by line. To begin, says that the conditional P! Q is a (both hold), and P Q P! Q T T T rmed when both its hypothesis P and conclusion Q are satisfied P Q P! Q T F F says that P! Q is contradicted when its hypothesis P is satisfied (holds), but its conclusion Q is not satisfied (does not hold). The next two lines,

7 ca_main March, 08 3:5 Page 4 4 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS P Q P! Q F T T F F T say that P! Q is always a rmed whenever its hypothesis is not satisfied (does not hold), regardless of whether its conclusion is satisfied (holds). A conditional statement is contradicted (is shown to be false) when and only when one finds an example where the hypothesis is satisfied, but the conclusion does not hold. Of note, a conditional statement cannot be contradicted by an example where the hypothesis is not satisfied; in this case, we say that the statement is vacuously true. Let us examine the dinner scene above, in which Johnny s mom said to him, You may have ice cream if you finish your broccoli. Again, this is Q if P or P! Q, where the hypothesis P stands for you finish your broccoli and the conclusion Q stands for you may have ice cream. Since Johnny did not finish his broccoli, the hypothesis was not satisfied; hence, Johnny s dad did not contradict the mom by allowing Johnny to have ice cream even though Johnny had not finished his broccoli because the conditional was vacuously true. Remark. Note that we observe. the principle of the excluded middle in which every sentence under consideration is assigned a truth value of true T or false F, and. the principle of the excluded contradiction in which no sentence under consideration is assigned both truth values T and F. Example.. Consider the statement, If it is raining, then I will take my umbrella. The hypothesis is it is raining, and the conclusion is I will take my umbrella. To contradict this statement (to show that it is false or that it is not always true), one must find a specific occasion of when it is raining, but I have not taken my umbrella; that is, when the hypothesis is satisfied, but the conclusion does not hold. Such an example in which the hypothesis is satisfied, but the conclusion does not hold is called a counterexample. Note that the inability to find a counterexample is not su cient to conclude that a statement is false or is not always true.. Consider the statement, A function f is continuous at a point x if f is di erentiable at x. A counterexample is a specific example that disproves a statement (shows that the statement is not always true).

8 ca_main March, 08 3:5 Page 5.. LOGIC 5 The hypothesis is a function f is di erentiable at a point x, and the conclusion is f is continuous at x. To find a counterexample (that shows that the statement is false or is not always true), one must find a specific function f and a specific real number x such that f is di erentiable at x, but f is not continuous at x. Again, the inability to find a counterexample is not su cient to conclude that a statement is false or is not always true. Now You Try.3 What would constitute a counterexample for each of the following conditionals?. You may have ice cream if you finish your broccoli.. If it is a turkey, then it is a bird. 3. Every infinite subset of a countable set A is countable. Remark. The conditional If it is raining, then I will take my umbrella may be translated as R! U, where R stands for it is raining and U stands for I will take my umbrella. It is important that we do not confuse an English sentence, such as It is raining, with its translation, such as R. While the English sentence is presumably either true or false (it really is raining or it really is not raining), the translation into a logical expression might have a di erent interpretation (true or false) as an English sentence depending on the context. In other words, the truth values T and F might or might not have anything to do with reality; indeed, instead of T and F, we could have used and 0, say. Now, compare the following two conditionals.. You may have ice cream if you finish your broccoli.. You may have ice cream only if you finish your broccoli. If we let P stand for you finish your broccoli and Q stand for you may have ice cream, then we see that. you may have ice cream if you finish your broccoli is P! Q, and that. you may have ice cream only if you finish your broccoli is Q! P. We use a truth table to see how the two related conditionals compare. <Verify.> P Q P! Q Q! P T T T T T F F T F T T F F F T T Observe that P! Q and Q! P are not both true (T) or both false (F) together in all of the cases; in particular, when P is true and Q is false, P! Q is false and Q! P is true, and when P is false and Q is true, P! Q is true and Q! P is false.

9 ca_main March, 08 3:5 Page 6 6 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS Hence, P! Q and Q! P are not two ways of saying the same thing. In fact, if we return to the dinner scene from the beginning of this section, because Johnny s mom had said, You may have ice cream if you finish your broccoli, Johnny s dad did not contradict the mom when he let Johnny have ice cream even though Johnny had not finished his broccoli. However, had Johnny s mom said instead, You may have ice cream only if you finish your broccoli, then Johnny s dad would have contradicted the mom by allowing Johnny to have ice cream even though Johnny had not finished his broccoli. Think About It.5 Convince yourself of what is said above, namely, that Johnny s dad would have contradicted the mom if she had said, You may have ice cream only if you finish your broccoli. The conditional statement Q! P is the converse of P! Q. Likewise, P! Q is the converse of Q! P. When two statements say the same thing that is, they are both true or both false together in all cases then the two statements are said to be tautologically equivalent or simply equivalent. Hence, as we have seen, a conditional and its converse are not tautologically equivalent (not equivalent). To a conditional P! Q, besides its converse Q! P, are related two other conditionals; to state them, however, we need first the notion of negation, which we give in the following table. In this table, P, read not P, denotes the negation of P. Thus, for example, if then As another example, if then Negation P P T F F T R = it is raining, R = it is not raining. C = candy is not good for your teeth, C = candy is not not good for your teeth; that is, is good for your teeth. Moreover, as we see in the truth table, when a sentence is true, its negation is false; and when a sentence is false, its negation is true. Think About It.6 How is negation for logic similar to negative for real numbers? We now state the three related conditionals to any given conditional statement. The related conditionals to P! Q are called the converse (that we have already introduced), the contrapositive, and the inverse of P! Q.

10 ca_main March, 08 3:5 Page 7.. LOGIC 7 Example. Related conditionals Given conditional Converse Contrapositive Inverse P! Q Q! P ( Q)! ( P) ( P)! ( Q). Consider again the conditional statement, You may have ice cream if you finish your broccoli or P! Q, where P stands for you finish your broccoli and Q stands for you may have ice cream. Note that P = you do not (or did not) finish your broccoli, Q = you may not have ice cream. Thus, the converse, contrapositive, and inverse statements here are converse: Q! P: If you may have ice cream, then you finished your broccoli; contrapositive: ( Q)! ( P): If you may not have ice cream, then you did not finish your broccoli; inverse: ( P)! ( Q): If you do not finish your broccoli, then you may not have ice cream.. Recall that the statements, Every turkey is a bird and All turkeys are birds, are both the conditional If it is a turkey, then it is a bird or P! Q, where P stands for it is a turkey and Q stands for it is a bird. Thus, P = it is not a turkey, Q = it is not a bird, and the converse, contrapositive, and inverse statements are converse: Q! P: If it is a bird, then it is a turkey; contrapositive: ( Q)! ( P): If it is not a bird, then it is not a turkey; inverse: ( P)! ( Q): If it is not a turkey, then it is not a bird. Now You Try.4 Write the converse, contrapositive, and inverse of the given conditional statement.. If it is raining, then I will take my umbrella.

11 ca_main March, 08 3:5 Page 8 8 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS. A function f is continuous at a point x if f is di erentiable at x. 3. All Marines have courage. 4. Every infinite subset of a countable set A is countable. Now, we have already seen that a conditional statement and its converse are not tautologically equivalent. It turns out, however, that a conditional and its contrapositive are tautologically equivalent; and so are the converse and the inverse of the conditional. We use a truth table to show that a conditional and its contrapositive are equivalent. <Verify.> A conditional and its contrapositive P Q Q P P! Q ( Q)! ( P) T T F F T T T F T F F F F T F T T T F F T T T T Since P! Q and ( Q)! ( P) are both true (T) or both false (F) together in all of the cases of when P and Q are true or false, w conclude that P! Q and ( Q)! ( P) both say the same thing; that is, P! Q and ( Q)! ( P) are equivalent. Thus, the conditional, and its contrapositive <verify>, You may have ice cream if you finish your broccoli, You did not finish your broccoli if you may not have ice cream, both say the same thing: they are tautologically equivalent statements. Now You Try.5 Use a truth table to show that the converse and the inverse of P! Q are tautologically equivalent. A conditional statement P! Q and its contrapositive ( Q)! ( P) are tautologically equivalent. Likewise, its converse Q! P and inverse ( P)! ( Q) are tautologically equivalent. A conditional and its converse are not tautologically equivalent. Remark.3 Sometimes it is easier to prove the contrapositive of a given conditional statement than it is to prove the conditional directly.

12 ca_main March, 08 3:5 Page 9.. LOGIC 9.. Conjunction, disjunction It might be that the hypothesis or the conclusion of a conditional statement has more than one part. For example, consider the following statements in which m and n are whole numbers. 3. If m is even and n is odd, then m + n is odd.. If mn is even, then either m is even or n is even. The hypothesis in the first statement is the conjunction of two sentences m is even and n is odd and the conclusion in the second statement is the disjunction of two sentences m is even or n is even. When is a conjunction or a disjunction true or false? We answer this question using the truth tables below. In the tables, the wedge ^ denotes a conjunction (and) and the vee _ denotes a disjunction (or). Conjunction P Q P ^ Q T T T T F F F T F F F F Disjunction P Q P _ Q T T T T F T F T T F F F Besides using truth tables, it might help to think of P and Q as doors. In this way, P ^ Q means that we must pass through both doors, and P _ Q means that we must pass through either one or both of the doors. P P ^ Q Q P P _ Q Thus, we see that a conjunction is false in all but one case (because it is false when either or both doors are closed), and a disjunction is true in all but one case (because it is true when either or both doors are open). A conjunction P ^ Q (read, P and Q ) is true when and only when both P and Q are true. On the other hand, the disjunction P _ Q (read, P or Q ) is true as long as at least one of P or Q is true. Q Now You Try.6. Complete the following truth tables. What can you conclude about (P! Q) and P ^ ( Q)? 3 The whole numbers are commonly considered to be the numbers 0,, 3,... The natural numbers are commonly considered to be,, 3,... In set theory and computer science, however, the natural numbers often are taken to be the whole numbers: 0,,, and so on.

13 ca_main March, 08 3:5 Page 0 0 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS P Q P! Q (P! Q) T T F F T F T F P Q Q P ^ ( Q). Is the negation of a conditional statement equivalent to any of the conditional s converse, contrapositive, or inverse? Explain. 3. Negate each of the following conditional statements. a) If it is raining, then I will take my umbrella. b) Every infinite subset of a countable set A is countable. Example.3 In the following, m and n are whole numbers.. Let P stand for m is even and Q stand for n is odd. Then, the conjunction P ^ Q = m is even and n is odd is satisfied (is true) when and only when both m is even and n is odd. <Why?>. Let P stand for m is even and Q stand for n is even. Then, the disjunction P _ Q = m is even or n is even holds (is true) when either m is even or n is even, or both m and n are even. <Why?> Remark.4 A disjunction P or Q is said to be inclusive if it is true when both P and Q are true. On the other hand, P or Q is said to be exclusive if it is false when both P and Q are true. Unless we state otherwise, we shall assume that a disjunction is inclusive. Now You Try.7 Find a counterexample for each of the following conditional statements. Explain why they are counterexamples. Here, m and n are whole numbers.. The sum m + n is even only if m is even and n is even.. If m is even or n is odd, then mn + n is odd. Think About It.7 When is the negation of the conjunction P and Q true, and when is the negation of the disjunction P or Q true? Now, consider the following statement in which p, a, and b are natural numbers. If p is a prime number and p divides ab, then either p divides a or p divides b. T T F F T F T F This is the conditional U! V, where U stands for p is a prime number and p divides ab and V stands for either p divides a or p divides b. How do we state the contrapositive, Literally, the contrapositive is, ( V)! ( U)?

14 ca_main March, 08 3:5 Page.. LOGIC If not either p divides a or p divides b, then not p is a prime number and p divides ab, in which we see that we have to negate a disjunction and a conjunction. How do we do this? The negation of a conjunction P ^ Q is (P ^ Q). Now, seeing the expression (P ^ Q), it is understandable that we feel the urge to distribute the negation; that is, (P ^ Q)? $ ( P) ( Q), where a reasonable guess for is either ^ or _. Which is it, if either at all? We use truth tables to help us to answer the question. <Verify the two truth tables below.> Negation of a conjunction P Q P ^ Q (P ^ Q) T T T F T F F T F T F T F F F T Distributed negation possibilities P Q P Q ( P) ^ ( Q) ( P) _ ( Q) T T F F F F T F F T F T F T T F F T F F T T T T Comparing the two tables, we see that (P ^ Q) and ( P) _ ( Q) are both true or both false together in all cases; that is, (P ^ Q) and ( P) _ ( Q) are tautologically equivalent. Now You Try.8 Determine the negation of a disjunction P _ Q in terms of P and Q. Thus, we have what are called De Morgan s laws for negating a conjunction and a disjunction. In the summary below, the double arrow $ denotes tautologically equivalent. De Morgan s laws The negation of P and Q is not P or not Q : (P ^ Q) $ ( P) _ ( Q); and the negation of P or Q is not both P and Q : (P _ Q) $ ( P) ^ ( Q). Think About It.8 Is neither P nor Q tautologically equivalent to not P or not Q? Now, let us recall why we wanted to negate a conjunction and a disjunction: we determined that the contrapositive of the statement, is literally If p is a prime number and p divides ab, then either p divides a or p divides b,

15 ca_main March, 08 3:5 Page CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS If not either p divides a or p divides b, then not p is a prime number and p divides ab, in which we see that we have to negate a disjunction, and to negate a conjunction, p divides a or p divides b, p is a prime number and p divides ab. We negate the disjunction using De Morgan s laws. To negate the disjunction p divides a or p divides b, we let Then, by De Morgan s laws, P = p divides a, Q = p divides b. (P _ Q) $ ( P) ^ ( Q) or not p divides a and not p divides b ; that is, the negation of either p divides a or p divides b is p does not divide a and p does not divide b. Now You Try.9 Use De Morgan s laws to negate p is a prime number and p divides ab. Finally, then, we conclude that the contrapositive of the conditional statement, is the statement, If p is a prime number and p divides ab, then either p divides a or p divides b, If p does not divide a and p does not divide b, then either p is not a prime number or p does not divide ab, Now You Try.0 For each of the following conditional statements, (i) write the converse, contrapositive, and inverse of the given conditional statement; (ii) negate the given conditional.. If m is even and n is odd, then m + n is odd.. Either m is even or n is even if mn is even...3 Biconditional statement Recall the dinner scene on page in which Johnny s mom said to him, You may have ice cream if you finish your broccoli, which is the conditional statement Q if P or P! Q, where P stands for you finish your broccoli and Q stands for you may have ice cream. What if the scene were to play out in this way, with Johnny s mom saying to him the converse instead? Think About It.9 Johnny: Mom, may I have ice cream tonight? Mom: You may have ice cream only if you finish your broccoli. Johnny: Thanks, Mom.

16 ca_main March, 08 3:5 Page 3.. LOGIC 3 Johnny s mom leaves the room, and Johnny turns to his dad. Johnny: Dad, I ve finished my broccoli. May I have ice cream now? Dad: No, Son, not until you also finish washing the dinner dishes. Johnny: But, Dad, Mom said that I could have ice cream after I finish my broccoli. Dad: No, Son, not until you also finish washing the dinner dishes. You be the judge: Did Johnny s dad contradict the mom because he will not let Johnny have ice cream until after Johnny also finishes washing the dinner dishes? Let us analyze the mom s statement, This statement is in the form You may have ice cream only if you finish your broccoli. Q only if P or Q! P, where, again, Q stands for you may have ice cream and P stands for you finish your broccoli. To ease the analysis, we consider the contrapositive of this statement, namely, ( P)! ( Q): If you do not finish your broccoli, then you may not have ice cream. In the contrapositive, the hypothesis P is you do not finish your broccoli and the conclusion Q is you may not have ice cream. Now, recall that a conditional is contradicted in only one case, namely, when the hypothesis is satisfied, but the conclusion does not hold. (See page 4.) In this alternative dinner scene, however, the hypothesis of the contrapositive is not satisfied because Johnny finished his broccoli <explain>; hence, the statement cannot be contradicted because it is vacuously true; in other words, Johnny s dad did not in fact contradict the mom by not letting let Johnny have ice cream until after Johnny also finishes washing the dinner dishes. Now, consider the conjunction of here, Q if P and Q only if P ; You may have ice cream if you finish your broccoli, and you may have ice cream only if you finish your broccoli. Since the second statement can be stated equivalently in the contrapositive, You may not have ice cream if you do not finish your broccoli <verify>, the conjunction can be stated equivalently as You may have ice cream if you finish your broccoli, and you may not have ice cream only if you do not finish your broccoli ; that is, which we can state more simply as (P! Q) and (Q! P), You may have ice cream if and only if you finish your broccoli.

17 ca_main March, 08 3:5 Page 4 4 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS Had Johnny s mom said this to him, then Johnny s dad would not be able to have Johnny finish washing the dishes before Johnny could have ice cream even though he had already finished his broccoli; and Johnny s dad would not be able to let Johnny have ice cream even though Johnny had not finished his broccoli at least not without contradicting the mom. The conjunction of a conditional P! Q and its converse Q! P is called a biconditional statement. When is a biconditional true or false? We answer this using a truth table; here, the double arrow denotes if and only if. <Verify the truth table for the conjunction of a conditional and its converse.> Conjunction of conditional and its converse P Q P! Q Q! P (P! Q) ^ (Q! P) T T T T T T F F T F F T T F F F F T T T Biconditional P Q P $ Q T T T T F F F T F F F T The truth values for a biconditional are taken to agree with the truth values of the conjunction of its component conditional and the converse. Moreover, when proving a biconditional statement P $ Q, we prove each of its component conditionals, P! Q and Q! P. As with a conditional statement, there are di erent equivalent ways to express a biconditional statement; here are a few examples. P if and only if Q. P only if Q, and conversely. P implies Q, and conversely. Example.4 Consider the statement, P implies and is implied by Q. Q is necessary and su cient for P. A necessary and su cient condition for P is Q. A series of nonnegative terms converges if and only if its partial sums form a bounded sequence. The conditional statement here is If the partial sums of a series of nonnegative terms form a bounded sequence, then the series converges and its converse is If a series of nonnegative terms converges, then its partial sums form a bounded sequence. Here are a few ways of rephrasing the statement. A series of nonnegative terms converges only if its partial sums form a bounded sequence, and conversely. That a series of nonnegative terms converges implies that its partial sums form a bounded sequence, and conversely. That a series of nonnegative terms converges implies and is implied by the fact that its partial sums form a bounded sequence. That the partial sums of a series of nonnegative terms form a bounded sequence is necessary and su cient for the series to converge. A necessary and su cient condition for a series of nonnegative terms to converge is that its partial sums form a bounded sequence.

18 ca_main March, 08 3:5 Page 5.. METHODS OF PROOF 5 Now You Try. For each of the following biconditional statements, identify the conditional statement and its converse, and then rephrase the biconditional in several di erent ways.. You may have ice cream if and only if you finish your broccoli.. A set E is open if and only if its complement is closed. Remark.5 Of note is that definitions are always biconditional statements, even if they may not be written that way. For example, consider the following definition. Let A and B be sets. The set A is called a subset of B, denoted A B or B A, if every element of A is an element of B. Although the definition appears to be a conditional statement, it is actually a biconditional statement: Let A and B be sets. The set A is called a subset of B, denoted A B or B A, if and only if every element of A is an element of B. This will always be so in definitions, and in definitions only. It is convention to omit the and only if part in definitions for brevity...4 Quantifiers Finally, we turn to quantifiers ever so briefly. There are two quantifiers that you will encounter: the universal quantifier and the existential quantifier. 8 universal English: for all 9 existential English: there exists For all such and such (8x) means for every such and such or for each such and such or for any such and such, and so on; and there exists such and such (9x) means there is at least one such and such or there is some (one or more) such and such, and so on. For example, consider the statement, If p(x) is a polynomial and p(a)p(b) < 0, then there exists a number c between a and b such that p(c) = 0. Applying this statement to the polynomial p(x) = x 3 6x + x 6 with a = 0 and b = 4, we find that p(0)p(4) = ( 6)(6) < 0; thus, there is at least one number c between 0 and 4 such that p(c) = 0; and, in fact, there are three such numbers:,, and 3. Think About It.0 How would we negate a quantified statement? For example, what is the negation of Every student in this class likes dessert ; what is the negation of There is at least one student in this class who likes dessert?. METHODS OF PROOF Think About It. What are some everyday events or situations that might require proof? (Think of some simple events or situations, and also some more complex or elaborate ones.) What would constitute acceptable proof in these events or situations? Be specific. A mathematical proof is an explanation or an argument that is intended to convince the reader or listener of the correctness of a statement. The amount of details that a proof needs to contain depends very much on who is the presenter and who is the reader or listener. There are several standard methods for proving statements. We give examples of the main ones. The statements that we shall prove shall be either conditional statements or biconditional statements. Since a

19 ca_main March, 08 3:5 Page 6 6 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS biconditional statement is the conjunction of a conditional statement and its converse, let us say a few words about conditional statements in general first. Consider the conditional statement If P, then Q. In this statement, as we noted earlier, P is called the hypothesis or antecedent, and Q is called the conclusion or consequent. Further, recall that a conditional statement is contradicted (is shown to be false) in only one case: when the hypothesis is satisfied (is made to hold), but the conclusion does not hold. Moreover, if the hypothesis is not satisfied, then the conditional is not contradicted regardless of whether or not the conclusion holds; we call this the vacuous case, and we say that the conditional is vacuously true. Example.5 Recall that the natural numbers or counting numbers are,, 3, and so on. Now, consider the statement, Let p, a, and b be natural numbers. If p is a prime number 4 and p divides ab, then either p divides a or p divides b. This is a conditional statement: if..., then... The hypothesis is p is a prime number and p divides ab and the conclusion is p divides a or p divides b. The statement would be contradicted if only if there is a choice of p, a, and b such that the hypothesis is true, but the conclusion is false. Let us test the statement by trying di erent choices of p, a, and b. p =, a = 9, b = 6 First, p = is prime and p divides ab = 54; thus, the hypothesis is satisfied (is true). Now, does not divide 9, but divides 6 so that p divides b; thus, the conclusion holds (is true). Hence, the statement is not contradicted (is true). p = 3, a = 9, b = 6 First, p = 3 is prime and p divides ab = 54; thus, the hypothesis is satisfied. Now, 3 divides 9 and 3 divides 6 so that either p divides a or p divides b (recall that or is inclusive, so it permits both); thus, the conclusion holds. Hence, the statement is not contradicted. p = 3, a = 5, b = 4 First, p = 3 is prime, but ab = 70 so that p does not divide ab; thus, the hypothesis is not satisfied (is not true). Now, 3 does not divide 5 and 3 does not divide 4 so that p does not divide a and p does not divide b; thus, the conclusion does not hold (is false). Nevertheless, the statement is not contradicted because the hypothesis is not satisfied: the statement is vacuously true (even though the conclusion does not hold). p = 6, a = 4, b = 9 First, p = 6 is not prime so that the hypothesis is not satisfied even though p divides ab = 36. Thus, the statement is not contradicted regardless of whether or not the conclusion holds; indeed, the conclusion does not hold because 6 does not divide 4 and 6 does not divide 9 so that p does not divide a and p does not divide b. The statement is vacuously true. p = 6, a =, b = 9 As above, the hypothesis is not satisfied because p = 6 is not prime. Thus, the statement is not contradicted regardless of whether or not the conclusion holds; in this case, however, the conclusion does hold because 6 divides so that p divides a. Nevertheless, the statement is vacuously true. So far, we have not found a choice of p, a, and b that contradicts the statement. Note that we have neither proved nor disproved the statement at this point. 4 Recall that a prime number is a natural number that has exactly two distinct factors. For example, the only factors of 3 are and 3, and so 3 is a prime number; on the other hand, the factors of 9 are, 3, and 9, and so 9 is not a prime number. The first few prime numbers are, 3, 5, 7,, 3, 7, 9, 3, 9, and 3.

20 ca_main March, 08 3:5 Page 7.. METHODS OF PROOF 7 Example.6 Consider the statement, If n is a natural number that is less than 40, then n + n + 4 is a prime number. This is an if..., then... statement. We can test this statement by checking it for every natural number from to 39. If we were to find some number n from to 39 for which n + n + 4 is not a prime number, then we would have disproved the statement, for we would have found a counterexample that contradicts the statement: we would have found an example where the hypothesis is satisfied, but the conclusion does not hold. On the other hand, if, after checking every natural number n from to 39, we were to find that n + n + 4 is indeed a prime, then we would have proved the statement because we would not have found a counterexample after having exhausted all of the possibilities. This would be an example of a direct proof we directly exhausted all of the possibilities. We leave it to you to check if n +n+4 is a prime number for each natural number n =,, 3,..., 39. Think About It. Do you think that the statement in If not, then why not? Example.7 Consider the statement, If n is a natural number, then n + n + 4 is a prime number..5 can be proved exhausting all of the possibilities? This is a conditional statement statement. <Identify the hypothesis and the conclusion.> We note that 40 is a counterexample for the statement, which disproves the statement. For n = 40 implies n + n + 4 = 68 = 4 <check>, so that 68 is not prime because it has more than two distinct factors (at least the factors, 4, and 68). This contradicts the statement because the hypothesis is true, but the conclusion is false. Now You Try. Prove or disprove the statement: If n is a natural number, then n is even. Example.8 Consider the statement, Let a and b be natural numbers. Then a + b is even if both a and b are even. This is a conditional statement where the hypothesis is a and b are even natural numbers and the conclusion is a + b is even. We shall employ a direct proof, but not by exhausting all of the possibilities <why?>; instead, we shall assume that the hypothesis is satisfied, and then show that it follows that the conclusion necessarily holds. Let a and b be natural numbers. Assume that a and b are even; then a = m and b = n for some natural numbers m and n because an even number is a multiple of. Thus, a + b = m + n = (m + n). Since m + n is a natural number <why?>, we see that a + b = (m + n) is even <why?>. This proves the statement. The beauty of this proof is that, by assuming that a and b are even natural numbers without specifying which they are, we have in a sense tried all possible even natural numbers at once without having tried any specific even natural numbers. Think About It.3 Would it be possible to prove the statement in the example above by exhausting all of the possibilities, that is, by trying specific even numbers? If it would, do so; if it would not, explain why not.

21 ca_main March, 08 3:5 Page 8 8 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS Now You Try.3 Prove or disprove the statement: Let a and b be natural numbers. If both a and b are odd, then a + b is even. We now turn to biconditional statements. That a biconditional statement is the conjunction of a conditional statement and its converse gives us a recipe for proving biconditional statements. To prove a biconditional statement P $ Q, prove the two component conditionals P! Q and Q! P. On the other hand, to disprove a biconditional P $ Q, disprove either P! Q or Q! P. Example.9 Consider the statement, Let a and b be natural numbers. Then a + b is even if and only if both a and b are even. This is a biconditional statement, for it is the conjunction of a conditional statement and its converse: the if or backward direction: If both a and b are even natural numbers, then a + b is even ; and the only if or forward direction: If a + b is an even natural number, then both a and b are even natural numbers. Consequently, to prove the biconditional statement, we prove each of its component conditionals. On the other hand, we would disprove the biconditional statement if we were to disprove either one of its component conditionals, for a biconditional is true when and only when both of its component conditionals are true. Recall that we proved the backward (if) direction in.8 above: If both a and b are even natural numbers, then a+b is even. So, it remains for us to prove or disprove the forward (only if) direction: If a + b is an even natural number, then both a and b are even natural numbers. A little thought and our experience, however, quickly reveals that the forward direction is not true it is easy to come up with a counterexample that contradicts the statement; that is, it is easy to come up with an example where the hypothesis a + b is an even natural number is satisfied, but the conclusion both a and b are even natural numbers does not hold. This disproves the biconditional because a biconditional is true in only one case: when both of it component conditionals are true. We remark that what we have shown in the course of our proof is that a + b is even if both a and b are even, but not conversely. Now You Try.4. Find a counterexample that contradicts the statement: Let a and b be natural numbers. Then a + b is even only if both a and b are even.. Prove or disprove the statement: Let a and b be natural numbers. Then a + b is even if and only if both a and b are odd. Example.0 Consider the statement, Let a be a natural number. Then a is even if and only if a is even. This is a biconditional statement, for it is a conjunction of a conditional statement and its converse: the if or backward direction: If a is an even natural number, then a is even ; and

22 ca_main March, 08 3:5 Page 9.. METHODS OF PROOF 9 the only if or forward direction: If a is an even natural number, then a is even. Consequently, to prove the biconditional statement, we prove each of its component conditionals; or to disprove the biconditional, we disprove one of its component conditionals. Now, you might want to try a few numerical examples first to test the statement, just to see if you think that it might be true. We leave it to you to try some numerical examples (for instance, square 3, 4, 0, 3, 5, and 8, and see if their squares are even or odd). Having done that, we hope that you believe that the statement might be true; hence, we should attempt to prove it. We begin with the backward (if) direction: If a is even, then a is even. We attempt a direct proof; that is, we shall assume that a is even, and then attempt to show that it follows that a is even. Let a be a natural number. Assume that a is even; then a = m for some natural number m. <Why?> Thus, a = (m) = 4m = (m ). Since m is a natural number <why?>, we see that a = (m ) is even <why?>. This proves the backward direction. We turn next to the forward (only if) direction: If a is even, then a is even. A direct proof seems to be di cult in this case, for a = m if a is even, but then a = p m, and it would seem to be di cult to conclude that p m is even (is a multiple of ). Hence, we shall attempt an indirect proof of the backward direction; in this case, we shall attempt a proof by contraposition, that is to say, we shall attempt to prove the contrapositive statement: If a is not even, then a is not even. (Recall that a conditional statement and its contrapositive are tautologically equivalent; see 8.) We shall attempt a direct proof of the contrapositive; that is, we shall assume that a is not even (a is odd), and then attempt to show that it follows that a is not even (a is odd). Assume that a is odd; then a = m for some natural number m because an odd natural number is one less than an even natural number. <May we assume, alternatively, that a = m +? Why?> Thus, a = (m ) = 4m 4m + = (m m) +. Since m m is a natural number, (m m) is even and, hence, a = (m m)+ is odd. <Why?> This proves the forward direction. Therefore, having proved both the forward and the backward directions, we have proved the biconditional statement: a is even if and only if a is even. Now You Try.5 Prove or disprove the statements.. Let a be a natural number. Then a 3 is even if and only if a is even.. Let a be a natural number. Then 3 divides a (that is, a is a multiple of 3) if and only if 3 divides a. Hint: 3 divides n if and only if n = 3m; 3 does not divide n if and only if either n = 3m or n = 3m. Example. If r = 0.3 = , where 3 repeats indefinitely, then r is a rational number. 5 We prove this directly. To begin, we note that 00r = ; thus, 00r r =. <why?> =) 99r =. =) r =. 99 = 990. Therefore, r = 0.3 is a rational number; specifically, r = / We shall discuss the real numbers, rational and irrational, in.4. A real number is a rational number if it can be expressed as a common fraction a/b, where a and b are integers with b 6= 0; and it is an irrational number if it cannot be expressed as a/b. Every real number is either rational or irrational, but not both. The integers, which are rational, are the numbers 0,,,...and,, 3,...

23 ca_main March, 08 3:5 Page 0 0 CHAPTER. LOGIC, SETS, REAL NUMBERS, COMPLEX NUMBERS Example. If =, then is an irrational number; that is, the real number p is an irrational number. (The symbol is the Greek small letter alpha. See. for a list of Greek letters.) A direct proof seems to be hard in this case. <Why?> Hence, we shall attempt an indirect proof; in this case, we shall attempt a proof by contradiction. (Indeed, the proof of this statement is a classic example of a proof by contradiction.) What does this mean? It is certainly true that is either an irrational number or a rational number, and not both. Hence, if we assume that is a rational number and this assumption were to lead to a contradiction (an absurd statement), then we must conclude that is in fact an irrational number after all. So, to begin, we assume that = a/b in lowest terms. <Why are we allowed to assume that a/b is in lowest terms?> = a =) p = a b b Then, squaring, we find that = a b, so that a = b. Consequently, we see that a is even (a multiple of ) and, hence, a is even. (See.0.) Since a is even, a = m for some natural number m. Substituting this into the equation above, we find that (m) = b, so that 4m = b or b = m. Consequently, we see that b is even and, hence, b is even. Since b is even, b = n for some natural number n. Hence, a and b have a common factor of, and so a/b is not in lowest terms because a and b have a common factor greater than. Hence, we have shown that if p = a b in lowest terms, then a is not in lowest term. b Since a/b cannot both be in lowest terms and not be in lowest terms, we have arrived at a contradiction (an absurd statement); hence, the assumption that is a rational number must be false. Therefore, if =, then is an irrational number; in other words, the real number p is an irrational number. Letter Capital Small Letter Capital Small Letter Capital Small Alpha A Beta B Gamma Delta Epsilon E or " Zeta Z Eta H Theta or # Iota I Kappa K apple Lambda Mu M µ Nu N Xi Omicron O o Pi Rho P Sigma Tau T Upsilon Phi or ' Chi X Psi Omega! Table.: The letters of the Greek alphabet. Think About It.4 Show that If P, then Q is false only if If P, then not Q is true, but not conversely. Now You Try.6 Prove or disprove the statement.. If a real number is irrational and nonzero, then / is irrational.

24 ca_main March, 08 3:5 Page.3. SETS. The real number p 3 is an irrational number. There is one more method of proof that you will find useful, and that is proof by induction that uses the principle of mathematical induction. We shall discuss proof by induction in SETS If you study mathematics long enough, you will come to a point when it will be useful for you to learn something about sets. How much about sets you should learn will depend on where you are in your studies and on what direction you take in your studies. You are now at a point when it will be useful for you to learn a little something about sets. Set theory is a relatively new branch of mathematics, even though today we consider set theory to be foundational, meaning that all of mathematics may be formally defined within set theory. Moreover, unlike much of mathematics, set theory was not developed by many people over a long period of time; indeed, set theory was essentially born out of the researches of one man, Georg Cantor (845 98), in the late nineteenth century. Since then, many people have contributed to the theory of sets. The axioms of set theory that we commonly use today are called the Zermelo-Fraenkel (ZF) axioms of set theory. We will scratch the surface only, and it will be a whirlwind tour. After that, we may forget all this and just do mathematics; that is to say, we can just do mathematics and not think about the underlying set theory actively..3. What Is a Set? Intuitively, a set is a collection of objects that is not too large. The primary relation is that of membership or belonging, that is, given a set A and an object x, either x is a member of A (x belongs to A) or x is not a member of A (x does not belong to A). The objects or members of a set are also called the elements of a set and, consequently, we may also say that an object is an element of a set or is not an element of a set. For example, if A is the set of all prime numbers, then 5 is a member of A or is an element of A, but 6 is not a member of A or is not an element of A. At this point, we introduce a little notation with which you might already be familiar. We use curly brackets, { and }, to enclose the members of a set. Within the curly brackets, we may list the objects of the set, or a description or condition that specifies the objects of the set. For example, if A is the set of prime numbers, then we may write A = {, 3, 5, 7,,...}, where the dots mean and so on, or A = {prime numbers} or A = {n: n is a prime number}. In the last case, the colon is read such that ; thus, A = {n: n is a prime number} may be read, A is the set of (numbers) n such that n is a prime number. To indicate membership in a set, we use the Greek small letter epsilon, which may be written variously as, ", and. In this book, we choose to use for membership in a set, and reserve the other two for use as parameters or variables. Thus, if A is the set of prime numbers, then 5 A, but 6 6 A, where the slash / has the usual meaning of not so that 6 means is not a member of or is not an element of. It might surprise you to hear that not every collection of objects forms a set. For example, suppose that C is the collection of all sets that are not members of themselves: C = {A: A is a set and A 6 A}.