How to Prove It: A Structured Approach 3rd Edition By Daniel J. Velleman SOLUTIONS MANUAL

SOLUTIONS MANUAL

How To Prove It: A Structured Approach Third Edition Solutions Manual Daniel J. Velleman Department of Mathematics and Statistics, Amherst College Department of Mathematics and Statistics, University of Vermont

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SOLUTIONS

Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 Section 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 Section 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Section 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Section 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 Section 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Section 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Section 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Section 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Section 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Section 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Section 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Section 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Section 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Section 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Section 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Section 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Section 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Section 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Section 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Section 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Section 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Section 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Section 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Section 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Section 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Section 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Section 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Section 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Section 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Section 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 iii

iv

SOLUTIONS Section 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Section 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Section 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Section 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Section 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Section 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Section 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Section 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

INTRODUCTION

1

Introduction 1. We use the formula 2ab − 1 = (2b − 1)(1 + 2b + 22b + · · · + 2(a−1)b ) from the proof of Conjecture 2. (a) 215 − 1 = 23·5 − 1 = (25 − 1) · (1 + 25 + 210 ) = 31 · 1057. (b) 232,767 −1 = 21057·31 −1 = (231 −1)(1+231 +· · ·+21056·31 ). The first factor is 231 −1 = 2,147,483,647.

2. When n = 1, 3n − 1 = 2, which is prime. For all n > 1, 3n − 1 is an even integer larger than 2, so it is not prime. If n is not prime, then 3n − 2n is not prime. If n is prime, then 3n − 2n may be either prime or composite.

3. (a) The method gives m = 2 · 3 · 5 · 7 + 1 = 211, which is prime. (b) The method gives m = 2 · 5 · 11 + 1 = 111 = 3 · 37; 3 and 37 are both prime.

4. Using the method of the proof of Theorem 4, with n = 5, we get x = 6! + 2 = 722. The five consecutive composite numbers are 722 = 2 · 361, 723 = 3 · 241, 724 = 2 · 362, 725 = 5 · 145, and 726 = 2 · 363.

5. 25 − 1 = 31 and 27 − 1 = 127 are Mersenne primes. Therefore, by Euclid’s theorem, 24 (25 − 1) = 496 and 26 (27 − 1) = 8128 are both perfect.

6. No. The remainder when any integer n > 3 is divided by 3 will be either 0, 1, or 2. If it is 0, then n is divisible by 3, so it is composite. If it is 1, then n + 2 is divisible by 3 and therefore composite. If it is 2, then n + 4 is divisible by 3 and composite. Thus, the numbers n, n + 2, and n + 4 cannot all be prime.

7. The positive integers smaller than 220 that divide 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110, and 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284. The positive integers smaller than 284 that divide 284 are 1, 2, 4, 71, and 142, and 1 + 2 + 4 + 71 + 142 = 220.

Chapter 1 Section 1.1 1. (a) (R ∨ H) ∧ ¬(H ∧ T ), where R stands for the statement “We’ll have a reading assignment,” H stands for “We’ll have homework problems,” and T stands for “We’ll have a test.” (b) ¬G ∨ (G ∧ ¬S), where G stands for “You’ll go skiing,” and S stands for “There will be snow.” √ √ (c) ¬[( 7 < 2) ∨ ( 7 = 2)].

2. (a) (J ∧ B) ∨ ¬(J ∨ B), where J stands for “John is telling the truth” and B stands for “Bill is telling the truth.” (b) (F ∨ C) ∧ ¬(F ∧ P ), where F stands for “I’ll have fish,” C stands for “I’ll have chicken,” and P stands for “I’ll have mashed potatoes.” (c) S ∧ N ∧ F , where S stands for “6 is divisible by 3,” N stands for “9 is divisible by 3,” and F stands for “15 is divisible by 3.”

3. Let A stand for the statement “Alice is in the room” and B for “Bob is in the room.” (a) ¬(A ∧ B). (b) ¬A ∧ ¬B. (c) ¬A ∨ ¬B; this is equivalent to the formula in (a). (d) ¬(A ∨ B); this is equivalent to the formula in (b).

4. Let P stand for “Ralph is tall,” Q for “Ed is tall,” R for “Ralph is handsome,” and S for “Ed is handsome.” (a) (P ∧ Q) ∨ (R ∧ S).

2

SOLUTIONS (b) (P ∨ R) ∧ (Q ∨ S). (c) ¬(P ∨ R) ∧ ¬(Q ∨ S). (d) ¬[(P ∧ R) ∨ (Q ∧ S)].

5. (a) and (c) are well-formed formulas.

6. (a) I won’t buy the pants without the shirt. (b) I won’t buy the pants and I won’t buy the shirt. (c) Either I won’t buy the pants or I won’t buy the shirt.

7. (a) Either Steve or George is happy, and either Steve or George is not happy. (b) Either Steve is happy, or George is happy and Steve isn’t, or George isn’t happy. (c) Either Steve is happy, or George is happy and either Steve or George isn’t happy.

8. (a) Either taxes will go up or the deficit will go up. (b) Taxes and the deficit will not both go up, but it is not the case that taxes won’t go up and the deficit also won’t go up. (c) Either taxes will go up and the deficit won’t, or the deficit will go up and taxes won’t.

9. See exercise 7 in Section 1.2 for the determination of whether these arguments are valid. (a) Let J stand for “Jane will win the math prize,” P for “Pete will win the math prize,” and C for “Pete will win the chemistry prize.” The premises are: ¬(J ∧ P ), P ∨ C, J. The conclusion is C. (b) Let B stand for “The main course will be beef,” F for “The main course will be fish,” P for “The vegetable will be peas,” and C for “The vegetable will be corn.” The premises are: B ∨ F , P ∨ C, ¬(F ∧ C). The conclusion is ¬(B ∧ P ). (c) Let J stand for “John is telling the truth,” B for “Bill is telling the truth,” and S for “Sam is telling the truth.” The premises are J ∨ B, ¬S ∨ ¬B. The conclusion is J ∨ ¬S. (d) Let S stand for “Sales will go up,” E for “Expenses will go up,” and B for “The boss will be happy.” There is only one premise: (S ∧ B) ∨ (E ∧ ¬B). The conclusion is ¬(S ∧ E).

Section 1.2 1. (a)

P F F T T

Q F T F T

¬P ∨ Q T T F T

(b)

S F F T T

G F T F T

(S ∨ G) ∧ (¬S ∨ ¬G) F T T F

2. (a)

P F F T T

Q F T F T

¬[P ∧ (Q ∨ ¬P )] T T T F

CHAPTER 1 (b)

3 P F F F F T T T T

Q F F T T F F T T

R F T F T F T F T

(P ∨ Q) ∧ (¬P ∨ R) F F T T F T F T

3. (a)

P Q P +Q F F F F T T T F T T T F (b) Here are two possibilities: (P ∧ ¬Q) ∨ (Q ∧ ¬P ), or (P ∨ Q) ∧ ¬(P ∧ Q). The truth table is the same as in part (a).

4. ¬(¬P ∧ ¬Q). The truth table is the same as the truth table for P ∨ Q. P Q P ↓Q F F T F T F T F F T T F (b) ¬(P ∨ Q). (c) ¬P is equivalent to P ↓ P , P ∨ Q is equivalent to (P ↓ Q) ↓ (P ↓ Q), and P ∧ Q is equivalent to (P ↓ P ) ↓ (Q ↓ Q).

5. (a)

P Q P |Q F F T F T T T F T T T F (b) ¬(P ∧ Q). (c) ¬P is equivalent to P | P , P ∨ Q is equivalent to (P | P ) | (Q | Q), and P ∧ Q is equivalent to (P | Q) | (P | Q).

6. (a)

7. We use the premises and conclusions identified in exercise 9 of Section 1.1. (a) Valid: premises are all true only in line 6 of the following truth table, and the conclusion is also true in that line.

Premises Conclusion J P C ¬(J ∧ P ) (P ∨ C) J C F F F T F F F F F T T T F T F T F T T F F F T T T T F T T F F T F T F T F T T T T T T T F F T T F T T T F T T T

4

SOLUTIONS (b) Invalid. Here is a line of the truth table in which all premises are true but the conclusion is false: Premises Conclusion B F P C B ∨ F P ∨ C ¬(F ∧ C) ¬(B ∧ P ) T T T F T T T F (c) Valid: premises are all true in lines 3, 5, 6, and 7 of the following truth table, and the conclusion is also true in all of those lines.

Premises Conclusion J B S J ∨ B ¬S ∨ ¬B J ∨ ¬S F F F F T T F F T F T F F T F T T T F T T T F F T F F T T T T F T T T T T T F T T T T T T T F T (d) Invalid. Here is a line of the truth table in which the premises are all true but the conclusion is false: Premise Conclusion S E B (S ∧ B) ∨ (E ∧ ¬B) ¬(S ∧ E) T T T T F 8. The following truth table shows that (a) and (c) are equivalent, as are (b) and (e): (a) (b) (c) (d) (e) P Q (P ∧ Q) ∨ (¬P ∧ ¬Q) ¬P ∨ Q (P ∨ ¬Q) ∧ (Q ∨ ¬P ) ¬(P ∨ Q) (Q ∧ P ) ∨ ¬P F F T T T T T F T F T F F T T F F F F F F T T T T T F T 9. The following truth table shows that (b) is a contradiction and (c) is a tautology: (a) (b) (c) P Q (P ∨ Q) ∧ (¬P ∨ ¬Q) (P ∨ Q) ∧ (¬P ∧ ¬Q) (P ∨ Q) ∨ (¬P ∨ ¬Q) F F F F T F T T F T T F T F T T T F F T Formula (d) is also a tautology, as the following truth table shows: P Q R [P ∧ (Q ∨ ¬R)] ∨ (¬P ∨ R) F F F T F F T T F T F T F T T T T F F T T F T T T T F T T T T T

CHAPTER 1

5

10. (a)

P F F T T

Q F T F T

¬(P ∨ Q) T F F F

(b)

P F F F F T T T T

Q F F T T F F T T

R F T F T F T F T

11. (a)

(b)

(c)

¬P ∧ ¬Q T F F F

P ∧ (Q ∨ R) F F F F F T T T

(P ∧ Q) ∨ (P ∧ R) F F F F F T T T

P ∨ (Q ∧ R) F F F T T T T T

(P ∨ Q) ∧ (P ∨ R) F F F T T T T T

¬(¬P ∧ ¬Q) is equivalent to ¬¬P ∨ ¬¬Q

(De Morgan’s law),

which is equivalent to P ∨ Q

(double negation law).

(P ∧ Q) ∨ (P ∧ ¬Q) is equivalent to P ∧ (Q ∨ ¬Q)

(distributive law),

which is equivalent to P

(tautology law).

¬(P ∧ ¬Q) ∨ (¬P ∧ Q) is equivalent to (¬P ∨ ¬¬Q) ∨ (¬P ∧ Q)

(De Morgan’s law),

which is equivalent to (¬P ∨ Q) ∨ (¬P ∧ Q)

(double negation law),

which is equivalent to ¬P ∨ (Q ∨ (¬P ∧ Q))

(associative law),

which is equivalent to ¬P ∨ (Q ∨ (Q ∧ ¬P ))

(commutative law),

which is equivalent to ¬P ∨ Q

(absorption law).

12. (a) ¬(¬P ∨ Q) ∨ (P ∧ ¬R) is equivalent to (¬¬P ∧ ¬Q) ∨ (P ∧ ¬R)

(De Morgan’s law),

which is equivalent to (P ∧ ¬Q) ∨ (P ∧ ¬R)

(double negation law),

which is equivalent to P ∧ (¬Q ∨ ¬R) which is equivalent to P ∧ ¬(Q ∧ R)

(distributive law), (De Morgan’s law).

(b) ¬(¬P ∧ Q) ∨ (P ∧ ¬R) is equivalent to (¬¬P ∨ ¬Q) ∨ (P ∧ ¬R)

(De Morgan’s law),

which is equivalent to (P ∨ ¬Q) ∨ (P ∧ ¬R) which is equivalent to (¬Q ∨ P ) ∨ (P ∧ ¬R)

(double negation law), (commutative law),

which is equivalent to ¬Q ∨ (P ∨ (P ∧ ¬R))

(associative law),

which is equivalent to ¬Q ∨ P

(absorption law).

(c) (P ∧ R) ∨ [¬R ∧ (P ∨ Q)] is equivalent to (P ∧ R) ∨ [(¬R ∧ P ) ∨ (¬R ∧ Q)]

(distributive law),

which is equivalent to (P ∧ R) ∨ [(P ∧ ¬R) ∨ (Q ∧ ¬R)]

(commutative law),

which is equivalent to [(P ∧ R) ∨ (P ∧ ¬R)] ∨ (Q ∧ ¬R)

(associative law),

which is equivalent to [P ∧ (R ∨ ¬R)] ∨ (Q ∧ ¬R)

(distributive law),

which is equivalent to P ∨ (Q ∧ ¬R)

(tautology law).

6

SOLUTIONS 13.

¬(P ∨ Q) is equivalent to ¬(¬¬P ∨ ¬¬Q)

(double negation law),

which is equivalent to ¬¬(¬P ∧ ¬Q)

(first De Morgan’s law),

which is equivalent to ¬P ∧ ¬Q

(double negation law).

14. We use the associative law for ∧ twice: [P ∧ (Q ∧ R)] ∧ S is equivalent to [(P ∧ Q) ∧ R] ∧ S which is equivalent to (P ∧ Q) ∧ (R ∧ S).

15. 2n .

16. P ∨ ¬Q. (You can check this by making the truth table.)

17. (P ∧ ¬Q) ∨ (Q ∧ ¬P ). (You can check this by making the truth table.)

18. If the conclusion is a tautology, then the argument is valid. If the conclusion is a contradiction and there is at least one line of the truth table in which all the premises are true, then the argument is not valid. If one of the premises is a tautology, then you can delete that premise without affecting the validity of the argument. If one of the premises is a contradiction, then the argument is valid.

Section 1.3 1. (a) D(6) ∧ D(9) ∧ D(15), where D(x) means “x is divisible by 3.” (b) D(x, 2) ∧ D(x, 3) ∧ ¬D(x, 4), where D(x, y) means “x is divisible by y.” (c) N (x) ∧ N (y) ∧ [(P (x) ∧ ¬P (y)) ∨ (P (y) ∧ ¬P (x))], where N (x) means “x is a natural number” and P (x) means “x is prime.”

2. (a) M (x) ∧ M (y) ∧ [(T (x, y) ∨ T (y, x)], where M (x) means “x is a man” and T (x, y), means “x is taller than y.” (b) (B(x) ∨ B(y)) ∧ (R(x) ∨ R(y)), where B(x) means “x has brown eyes” and R(x) means “x has red hair.” (c) (B(x) ∧ R(x)) ∨ (B(y) ∧ R(y)), where the letters have the same meanings as in part (b).

3. (a) {x | x is a planet}. (b) {x | x is an Ivy League school}. (c) {x | x is a state in the United States}. (d) {x | x is a province or territory in Canada}.

4. (a) {x | x is the square of a positive integer}. (b) {x | x is a power of 2}. (c) {x | x is an integer and 10 ≤ x < 20}.

5. (a) (−3 ∈ R) ∧ (13 − 2(−3) > 1). Bound variables: x; no free variables. This statement is true. (b) (4 ∈ R) ∧ (4 < 0) ∧ (13 − 2(4) > 1). Bound variables: x; no free variables. This statement is false. (c) ¬[(5 ∈ R) ∧ (13 − 2(5) > c)]. Bound variables: x; free variables: c.

6. (a) (w ∈ R) ∧ (13 − 2w > c). Bound variables: x; free variables: w, c. (b) (4 ∈ R) ∧ (13 − 2(4) is a prime number). Bound variables: x, y; no free variables. This statement is true. (c) (4 is a prime number) ∧ (13 − 2(4) > 1). Bound variables: x, y; no free variables. This statement is false.

7. (a) {−1, 1/2}.

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7

(b) {1/2}. (c) {−1}. (d) ∅.

8. (a) {x | Elizabeth Taylor was once married to x} = {Conrad Hilton Jr., Michael Wilding, Michael Todd, Eddie Fisher, Richard Burton, John Warner, Larry Fortensky}. (b) {x | x is a logical connective studied in Section 1.1} = {∧, ∨, ¬}. (c) {x | x is the author of this book} = {Daniel J. Velleman}.

9. (a) {x ∈ R | x2 − 4x + 3 = 0} = {1, 3}. (b) {x ∈ R | x2 − 2x + 3 = 0} = ∅. (c) {x ∈ R | 5 ∈ {y ∈ R | x2 + y 2 < 50}} = {x ∈ R | 5 ∈ R ∧ x2 + 25 < 50} = {x ∈ R | x2 < 25} = {x ∈ R | −5 < x < 5}. Examples of elements: −4, 2, 4.9, π.

Section 1.4 1. (a) {3, 12}. (b) {1, 12, 20, 35}. (c) {1, 3, 12, 20, 35}.

The sets in parts (a) and (b) are both subsets of the set in part (c).

2. (a) {United States, Germany, China, Australia, France, India, Brazil}. (b) ∅. (c) {France}.

The set in part (b) is disjoint from both of the other sets, and also a subset of both other sets. The set in part (c) is a subset of the set in part (a).

3. In the diagram on the left below, the set (A ∪ B) \ (A ∩ B) is represented by the region that is striped but not crosshatched. In the diagram on the right, the set (A \ B) ∪ (B \ A) is represented by the region that striped (in either direction).

U

A

A∪B

U

B

A∩B

A\B

4. (a) Both Venn diagrams look like this:

U

A

A

B

B

B\A

8

SOLUTIONS (b) Both Venn diagrams look like this:

U

B

A

C 5. (a)

(b)

x ∈ A \ (A ∩ B) is equivalent to x ∈ A ∧ ¬(x ∈ A ∧ x ∈ B) which is equivalent to x ∈ A ∧ (x ∈ / A∨x∈ / B)

(definitions of \, ∩), (De Morgan’s law),

which is equivalent to (x ∈ A ∧ x ∈ / A) ∨ (x ∈ A ∧ x ∈ / B)

(distributive law),

which is equivalent to x ∈ A ∧ x ∈ /B

(contradiction law),

which is equivalent to x ∈ A \ B

(definition of \).

x ∈ A ∪ (B ∩ C) is equivalent to x ∈ A ∨ (x ∈ B ∧ x ∈ C) which is equivalent to (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) which is equivalent to x ∈ (A ∪ B) ∩ (A ∪ C)

(definitions of ∪, ∩), (distributive law), (definitions of ∪, ∩).

6. (a) Both Venn diagrams look like this:

U

B

A

C (b) Both Venn diagrams look like this:

U

B

A

C 7. (a)

(b)

x ∈ (A ∪ B) \ C is equivalent to (x ∈ A ∨ x ∈ B) ∧ x ∈ /C

(definitions of ∪, \),

which is equivalent to (x ∈ A ∧ x ∈ / C) ∨ (x ∈ B ∧ x ∈ / C)

(distributive law),

which is equivalent to x ∈ (A \ C) ∪ (B \ C)

(definitions of \, ∪).

x ∈ A ∪ (B \ C) is equivalent to x ∈ A ∨ (x ∈ B ∧ x ∈ / C)

(definitions of ∪, \),

which is equivalent to (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ / C)

(distributive law),

which is equivalent to (x ∈ A ∨ x ∈ B) ∧ ¬(x ∈ / A ∧ x ∈ C)

(De Morgan’s law),

CHAPTER 1

9 which is equivalent to x ∈ (A ∪ B) \ (C \ A)

8. (a)

(b)

x ∈ (A \ B) ∩ C is equivalent to (x ∈ A ∧ x ∈ / B) ∧ x ∈ C

(definitions of ∪, \). (definitions of \, ∩),

which is equivalent to x ∈ A ∧ (x ∈ / B ∧ x ∈ C)

(associative law),

which is equivalent to x ∈ A ∧ (x ∈ C ∧ x ∈ / B)

(commutative law),

which is equivalent to (x ∈ A ∧ x ∈ C) ∧ x ∈ /B

(associative law),

which is equivalent to x ∈ (A ∩ C) \ B

(definitions of ∩, \).

x ∈ (A ∩ B) \ B is equivalent to (x ∈ A ∧ x ∈ B) ∧ x ∈ /B

(definitions of ∩, \).

This is clearly a contradiction, so (A ∩ B) \ B = ∅. (c) x ∈ A \ (A \ B) is equivalent to x ∈ A ∧ ¬(x ∈ A ∧ x ∈ / B)

(definition of \),

which is equivalent to x ∈ A ∧ (x ∈ / A ∨ x ∈ B) which is equivalent to (x ∈ A ∧ x ∈ / A) ∨ (x ∈ A ∧ x ∈ B)

(De Morgan’s law), (distributive law),

which is equivalent to x ∈ A ∧ x ∈ B

(contradiction law),

which is equivalent to x ∈ A ∩ B

(definition of ∩).

9. (a) x ∈ (A \ B) \ C means (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (b) x ∈ A \ (B \ C) means x ∈ A ∧ ¬(x ∈ B ∧ x ∈ / C) which is equivalent to x ∈ A ∧ (x ∈ / B ∨ x ∈ C)

(De Morgan’s law),

which is equivalent to (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ C)

(distributive law).

(c) x ∈ (A \ B) ∪ (A ∩ C) means (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ C). (d) x ∈ (A \ B) ∩ (A \ C) means (x ∈ A ∧ x ∈ / B) ∧ (x ∈ A ∧ x ∈ / C) which is equivalent to ((x ∈ A ∧ x ∈ / B) ∧ x ∈ A) ∧ x ∈ /C

(associative law),

which is equivalent to (x ∈ A ∧ (x ∈ A ∧ x ∈ / B)) ∧ x ∈ /C

(commutative law),

which is equivalent to ((x ∈ A ∧ x ∈ A) ∧ x ∈ / B) ∧ x ∈ /C

(associative law),

which is equivalent to (x ∈ A ∧ x ∈ / B) ∧ x ∈ /C

(idempotent law).

(e) x ∈ A \ (B ∪ C) means x ∈ A ∧ ¬(x ∈ B ∨ x ∈ C) which is equivalent to x ∈ A ∧ (x ∈ / B∧x∈ / C) which is equivalent to (x ∈ A ∧ x ∈ / B) ∧ x ∈ /C

(De Morgan’s law), (associative law).

This shows that sets (a), (d), and (e) are equal, and sets (b) and (c) are equal.

10. (a) A = {1, 2}, B = {2, 3}. (A ∪ B) \ B = {1, 2, 3} \ {2, 3} = {1} = 6 {1, 2} = A. (b) x ∈ (A ∪ B) \ B is equivalent to (x ∈ A ∨ x ∈ B) ∧ x ∈ /B (definitions of ∪, \), which is equivalent to (x ∈ A ∧ x ∈ / B) ∨ (x ∈ B ∧ x ∈ / B)

(distributive law),

which is equivalent to x ∈ A ∧ x ∈ /B

(contradiction law),

which is equivalent to x ∈ A \ B

(definition of \).

11. The Venn diagram for (A\B)∪B looks the same as the Venn diagram for A∪B, so (A\B)∪B = A∪B. Therefore A ⊆ (A\B)∪B, but the two sets need not be equal. For example, if A = {1, 2} and B = {2, 3} then (A \ B) ∪ B = {1} ∪ {2, 3} = {1, 2, 3} = 6 {1, 2} = A. 12. (a) There is no region corresponding to the set (A ∩ D) \ (B ∪ C), but this set could have elements. (b) Here is one possibility:

10

SOLUTIONS

D

U A

B

C 13. (a) The Venn diagrams are given in the solution to exercise 6. Those diagrams show that (A∪B)\C ⊆ A ∪ (B \ C). (b) A = {1, 2, 3}, B = {2, 3, 4}, C = {3, 4, 5}. (A∪B)\C = {1, 2, 3, 4}\{3, 4, 5} = {1, 2}, A∪(B\C) = {1, 2, 3} ∪ {2} = {1, 2, 3}.

14. The Venn diagrams for both sets look like this:

U

B

A

C 15.

(a) (b) (c) (A 4 B) ∪ C = (A ∪ C) 4 (B \ C). (A4B)∩C = (A∩C)4(B ∩C). (A 4 B) \ C = (A \ C) 4 (B \ C).

16.

(a) (b) (c) (A∪B)4C = (A4C)4(B \A). (A∩B)4C = (A4C)4(A\B). (A\B)4C = (A4C)4(A∩B).

17. All of the following answers can be confirmed with Venn diagrams. (a) C \ B. (b) C \ B. (c) A ∪ B.

CHAPTER 1

11

Section 1.5 1. (a) (S ∨ ¬E) → ¬H, where S stands for “This gas has an unpleasant smell,” E stands for “This gas is explosive,” and H stands for “This gas is hydrogen.” (b) (F ∧ H) → D, where F stands for “George has a fever,” H stands for “George has a headache,” and D stands for “George will go to the doctor.” (c) (F → D) ∧ (H → D), where the letters have the same meanings as in part (b). (d) (x 6= 2) → (P (x) → O(x)), where P (x) stands for “x is prime” and O(x) stands for “x is odd.”

2. (a) H → (P ∧ A), where H stands for “Mary will sell her house,” P stands for “Mary can get a good price,” and A stands for “Mary can find a nice apartment.” (b) M → (C ∧ P ), where C stands for “you have a good credit history,” P stands for “you have an adequate down payment,” and M stands for “you will get a mortgage.” (c) ¬S → D, where D stands for “John will drop out of school,” and S stands for “someone will stop him.” (d) (D(x, 4) ∨ D(x, 6)) → ¬P (x), where D(x, y) means “x is divisible by y” and P (x) means “x is prime.”

3. Let R stand for “It is raining,” W for “It is windy,” and S for “It is sunny.” (a) R → (W ∧ ¬S). (b) (W ∧ ¬S) → R. This is the converse of (a). (c) R → (W ∧ ¬S). This is the same as (a). (d) (W ∧ ¬S) → R. This is the converse of (a). (e) (S ∨ ¬W ) → ¬R which is equivalent to R → ¬(S ∨ ¬W ) which is equivalent to R → (¬S ∧ W ) which is equivalent to (a). (f) (R → W ) ∧ (R → ¬S) which is equivalent to (¬R ∨ W ) ∧ (¬R ∨ ¬S)

(contrapositive law), (De Morgan’s law),

(conditional law),

which is equivalent to ¬R ∨ (W ∧ ¬S)

(distributive law),

which is equivalent to R → (W ∧ ¬S) which is the same as (a).

(conditional law),

(g) (W → R) ∨ (¬S → R) which is equivalent to (¬W ∨ R) ∨ (S ∨ R)

(conditional law),

which is equivalent to (¬W ∨ S) ∨ (R ∨ R)

(commutative, associative laws),

which is equivalent to ¬(W ∧ ¬S) ∨ R

(De Morgan’s, idempotent laws),

which is equivalent to (W ∧ ¬S) → R which is the converse of (a).

(conditional law),

4. (a) Let S stand for “Sales will go up,” E for “Expenses will go up,” and B for “The boss will be happy.” Valid.

12

SOLUTIONS Premises Conclusion S E B S ∨ E S → B E → ¬B ¬(S ∧ E) F F F F T T T F F T F T T T F T F T T T T F T T T T F T T F F T F T T T F T T T T T T T F T F T F T T T T T F F (b) Let T stand for “The tax rate will go up,” U for “The unemployment rate will go up,” R for “There will be a recession,” and G for “GDP will go up.” Valid. To shorten the truth table, we take some shortcuts. “-” indicates an entry that could be T or F. Premises Conclusion T U R G (T ∧ U ) → R G → ¬R G ∧ T ¬U F T F T - F T F T F - T T T T T T F T F T T F T T T T T F T F (c) Let W stand for “The warning light will come on,” P for “The pressure is too high,” and C for “The relief valve is clogged.” Invalid. Here is a line of the truth table in which the premises are all true but the conclusion is false: Premises Conclusion W P C W ↔ (P ∧ C) ¬C W ↔P F T F T T F 5. (a) Let J stand for “Jones will be convicted,” P for “Jones will go to prison,” and S for “Smith will testify agains Jones.” Invalid. Here is a line of the truth table in which the premises are all true but the conclusion is false.

Premises Conclusion J P S J → P J → S ¬S → ¬P F T F T T F (b) Let D stand for “The Democrats will have a majority in the Senate,” R for “The Republicans will have a majority in the senate,” and B for “The bill will pass.” Valid.

Premises Conclusion D R B (D ∨ R) ∧ ¬(D ∧ R) B → D R → ¬B F F F F T T F F T F F T F T F T T T F T T T F F T F F T T T T F T T T T T T F F T T T T T F T F 6. (a)

P F F T T

Q F T F T

P ↔Q T F F T

(P ∧ Q) ∨ (¬P ∧ ¬Q) T F F T

CHAPTER 1 (b)

13

(P → Q) ∨ (P → R) is equivalent to (¬P ∨ Q) ∨ (¬P ∨ R)

(conditional law),

which is equivalent to (¬P ∨ ¬P ) ∨ (Q ∨ R)

(commutative, associative laws),

which is equivalent to ¬P ∨ (Q ∨ R)

(idempotent law),

which is equivalent to P → (Q ∨ R)

(conditional law).

7. (a) (P → R) ∧ (Q → R) is equivalent to (¬P ∨ R) ∧ (¬Q ∨ R)

(conditional law),

which is equivalent to (¬P ∧ ¬Q) ∨ R

(distributive law),

which is equivalent to ¬(P ∨ Q) ∨ R

(De Morgan’s law),

which is equivalent to (P ∨ Q) → R

(conditional law).

(b) (P → R) ∨ (Q → R) is equivalent to (¬P ∨ R) ∨ (¬Q ∨ R)

(conditional law),

which is equivalent to (¬P ∨ ¬Q) ∨ (R ∨ R)

(commutative, associative laws),

which is equivalent to ¬(P ∧ Q) ∨ R

(De Morgan’s, idempotent laws),

which is equivalent to (P ∧ Q) → R

(conditional law).

P Q R (P → Q) ∧ (Q → R) (P → R) ∧ [(P ↔ Q) ∨ (R ↔ Q)] F F F T T F F T T T F T F F F F T T T T T F F F F T F T F F T T F F F T T T T T (b) (P → Q) ∨ (Q → R) is equivalent to (¬P ∨ Q) ∨ (¬Q ∨ R) (conditional law),

8. (a)

which is equivalent to (¬P ∨ R) ∨ (Q ∨ ¬Q) which is a tautology

(commutative, associative laws), (tautology law).

9. ¬(P → ¬Q). The equivalence can be checked with a truth table.

10. ¬((P → Q) → ¬(Q → P )). The equivalence can be checked with a truth table.

11. (a)

P F F T T

Q F T F T

(P ∨ Q) ↔ Q T T F T

P →Q T T F T

(b)

P F F T T

Q F T F T

(P ∧ Q) ↔ Q T F T T

Q→P T F T T

12. (a), (b), and (d) are equivalent; (c) and (e) are equivalent. This can be seen in the following truth table.

14

SOLUTIONS

P F F F F T T T T

Q F F T T F F T T

R F T F T F T F T

(a) P → (Q → R) T T T T T T F T

(b) Q → (P → R) T T T T T T F T

(c) (P → Q) ∧ (P → R) T T T T F F F T

(d) (P ∧ Q) →R T T T T T T F T

(e) P → (Q ∧ R) T T T T F F F T

Chapter 2 Section 2.1 1. (a) ∀x[∃yF (x, y) → S(x)], where F (x, y) stands for “x has forgiven y,” and S(x) stands for “x is a saint.” (b) ¬∃x[C(x) ∧ ∀y(D(y) → S(x, y))], where C(x) stands for “x is in the calculus class,” D(y) stands for “y is in the discrete math class,” and S(x, y) stands for “x is smarter than y.” (c) ∀x(¬(x = m) → L(x, m)), where L(x, y) stands for “x likes y,” and m stands for Mary. (d) ∃x(P (x) ∧ S(j, x)) ∧ ∃y(P (y) ∧ S(r, y)), where P (x) stands for “x is a police officer,” S(x, y) stands for “x saw y,” j stands for Jane, and r stands for Roger. (e) ∃x(P (x) ∧ S(j, x) ∧ S(r, x)), where the letters have the same meanings as in part (d).

2. (a) ∀x[B(x) → ∃y(U (y, x) ∧ R(y))], where B(x) stands for “x has bought a Rolls Royce with cash,” U (y, x) stands for “y is an uncle of x,” and R(y) stands for “y is rich.” (b) ∃x(D(x) ∧ M (x)) → ∀x[∃y(F (y, x) ∧ D(y)) → Q(x)], where D(x) stands for “x is in the dorm,” M (x) stands for “x has the measles,” F (y, x) stands for “y is a friend of x,” and Q(x) stands for “x will have to be quarantined.” (c) ¬∃xF (x) → ∀x[A(x) → ∃y(D(y) ∧ T (x, y))], where F (x) stands for “x failed the test,” A(x) stands for “x got an A,” D(x) stands for “x got a D,” and T (x, y) stands for “x will tutor y.” (d) ∃xD(x) → D(j), where D(x) stands for “x can do it” and j stands for Jones. (e) D(j) → ∀xD(x), where the letters have the same meanings as in part (d).

3. (a) ∀z(z > x → z > y). Free variables: x, y. (b) ∀a[∃x(ax2 + 4x − 2 = 0) ↔ (a > −2 ∨ a = −2)]. No free variables. (c) ∀x(x3 − 3x < 3 → x < 10). No free variables. (d) (∃x(x2 + 5x = w) ∧ ∃y(4 − y 2 = w)) → (−10 < w ∧ w < 10). Free variables: w.

4. (a) All unmarried men are unhappy. (b) y is a sister of one of x’s parents; i.e., y is x’s blood aunt.

5. (a) Every prime number other than 2 is odd. (b) There is a largest perfect number.

6. (a) There is no real number that is a solution to both of the equations x2 +2x+3 = 0 and x2 +2x−3 = 0. This is true, because x2 + 2x + 3 and x2 + 2x − 3 cannot both be 0. (b) It is not the case that both of the equations x2 + 2x + 3 = 0 and x2 + 2x − 3 = 0 have real solutions. This is true, because the equation x2 + 2x + 3 = 0 has no real solution. (c) Both of the equations x2 + 2x + 3 = 0 and x2 + 2x − 3 = 0 have no real solutions. This is false, because x = 1 is a solution to the second equation.

CHAPTER 2

15

7. We translate each statement into English before saying whether it is true or false. (a) There is someone who is a parent of everyone. False; no one person is a parent of everyone. (b) Everyone is a parent of someone. False; some people have no children. (c) No one is a parent of anyone. False; some people have children. (d) Someone has no children. True. (e) Someone is not a parent of someone. True; for example, Barack Obama is not a parent of Angelina Jolie.

8. (a), (d), and (e) are true; (b), (c), and (f) are false.

9. (a), (c), (e), and (f) are true; (b) and (d) are false.

10. (a), (d), (e), and (f) are true; (b) and (c) are false.

Section 2.2 1. (a) Let M (x) stand for “x is majoring in math,” F (x, y) for “x and y are friends,” and H(y) for “y needs help with his or her homework.” ¬∀x[M (x) → ∃y(F (x, y) ∧ H(y))] is equivalent to ∃x¬[M (x) → ∃y(F (x, y) ∧ H(y))]

(quantifier negation law),

which is equivalent to ∃x¬[¬M (x) ∨ ∃y(F (x, y) ∧ H(y))]

(conditional law),

which is equivalent to ∃x[M (x) ∧ ¬∃y(F (x, y) ∧ H(y))]

(De Morgan’s law),

which is equivalent to ∃x[M (x) ∧ ∀y¬(F (x, y) ∧ H(y))]

(quantifier negation law),

which is equivalent to ∃x[M (x) ∧ ∀y(¬F (x, y) ∨ ¬H(y))]

(De Morgan’s law),

which is equivalent to ∃x[M (x) ∧ ∀y(F (x, y) → ¬H(y))]

(conditional law).

(b) Let R(x, y) stand for “x and y are roommates” and L(x, y) for “x likes y.” ¬∀x∃y(R(x, y) ∧ ∀z¬L(y, z)) is equivalent to ∃x¬∃y(R(x, y) ∧ ∀z¬L(y, z))

(quantifier negation law),

which is equivalent to ∃x∀y¬(R(x, y) ∧ ∀z¬L(y, z))

(quantifier negation law),

which is equivalent to ∃x∀y(¬R(x, y) ∨ ¬∀z¬L(y, z))

(De Morgan’s law),

which is equivalent to ∃x∀y(¬R(x, y) ∨ ∃z¬¬L(y, z))

(quantifier negation law),

which is equivalent to ∃x∀y(¬R(x, y) ∨ ∃zL(y, z))

(double negation law),

which is equivalent to ∃x∀y(R(x, y) → ∃zL(y, z))

(conditional law).

(c) ¬∀x[(x ∈ A ∨ x ∈ B) → (x ∈ C ∧ x ∈ / D)] is equivalent to ∃x¬[(x ∈ A ∨ x ∈ B) → (x ∈ C ∧ x ∈ / D)]

(quantifier negation law),

which is equivalent to ∃x¬[¬(x ∈ A ∨ x ∈ B) ∨ (x ∈ C ∧ x ∈ / D)]

(conditional law),

which is equivalent to ∃x[(x ∈ A ∨ x ∈ B) ∧ ¬(x ∈ C ∧ x ∈ / D)]

(De Morgan’s law),

which is equivalent to ∃x[(x ∈ A ∨ x ∈ B) ∧ (x ∈ / C ∨ x ∈ D)]

(De Morgan’s law).

(d) ¬∃x∀y[y > x → ∃z(z 2 + 5z = y)] is equivalent to ∀x¬∀y[y > x → ∃z(z 2 + 5z = y)] 2

which is equivalent to ∀x∃y¬[y > x → ∃z(z + 5z = y)] 2

which is equivalent to ∀x∃y¬[¬(y > x) ∨ ∃z(z + 5z = y)]

(quantifier negation law), (quantifier negation law), (conditional law),

16

SOLUTIONS which is equivalent to ∀x∃y[y > x ∧ ¬∃z(z 2 + 5z = y)] 2

which is equivalent to ∀x∃y[y > x ∧ ∀z(z + 5z 6= y)]

(De Morgan’s law), (quantifier negation law).

2. (a) Let F (x) stand for “x is in the freshman class” and R(x, y) for “x and y are roommates.” ¬∃x(F (x) ∧ ¬∃yR(x, y)) is equivalent to ∀x¬(F (x) ∧ ¬∃yR(x, y)) which is equivalent to ∀x(¬F (x) ∨ ∃yR(x, y)) which is equivalent to ∀x(F (x) → ∃yR(x, y))

(quantifier negation law), (De Morgan’s law), (conditional law).

(b) Let L(x, y) stand for “x likes y.” ¬[∀x∃yL(x, y) ∧ ¬∃x∀yL(x, y)] is equivalent to ¬∀x∃yL(x, y) ∨ ∃x∀yL(x, y) which is equivalent to ∃x∀y¬L(x, y) ∨ ∃x∀yL(x, y) (c) ¬∀a ∈ A∃b ∈ B(a ∈ C ↔ b ∈ C) is equivalent to ∃a ∈ A∀b ∈ B¬(a ∈ C ↔ b ∈ C)

(De Morgan’s law), (quantifier negation laws).

(quantifier negation laws),

which is equivalent to ∃a ∈ A∀b ∈ B¬((a ∈ C → b ∈ C) ∧ (b ∈ C → a ∈ C)) (meaning of biconditional), which is equivalent to ∃a ∈ A∀b ∈ B¬(¬(a ∈ C ∧ b ∈ / C) ∧ ¬(b ∈ C ∧ a ∈ / C)) (conditional law), which is equivalent to ∃a ∈ A∀b ∈ B((a ∈ C ∧ b ∈ / C) ∨ (b ∈ C ∧ a ∈ / C)) (De Morgan’s law). (d) ¬∀y > 0∃x(ax2 + bx + c = y) is equivalent to ∃y > 0∀x(ax2 + bx + c 6= y) by quantifier negation laws.

3. (a) True, because 02 + 02 + 02 = 0, 12 + 02 + 02 = 1, 12 + 12 + 02 = 2, 12 + 12 + 12 = 3, 22 + 02 + 02 = 4, 22 + 12 + 02 = 5, and 22 + 12 + 12 = 6. (b) False, because the equation has two solutions, x = 1 and x = 3. (c) True, because x = 5 is the only natural number solution to the equation. (x = −1 is also a solution, but −1 is not a natural number.) (d) True, because x and y can both be 5.

4.

5.

6.

¬∀xP (x) is equivalent to ¬∀x¬¬P (x)

(double negation law),

which is equivalent to ¬¬∃x¬P (x)

(first quantifier negation law),

which is equivalent to ∃x¬P (x)

(double negation law).

¬∃x ∈ A P (x) is equivalent to ¬∃x(x ∈ A ∧ P (x))

(expanding abbreviation),

which is equivalent to ∀x¬(x ∈ A ∧ P (x))

(quantifier negation law),

which is equivalent to ∀x(x ∈ / A ∨ ¬P (x))

(De Morgan’s law),

which is equivalent to ∀x(x ∈ A → ¬P (x))

(conditional law),

which is equivalent to ∀x ∈ A ¬P (x)

(abbreviation).

∃x(P (x) ∨ Q(x)) is equivalent to ¬¬∃x(P (x) ∨ Q(x)) which is equivalent to ¬∀x¬(P (x) ∨ Q(x))

(double negation law), (quantifier negation law),

CHAPTER 2

7.

17 which is equivalent to ¬∀x(¬P (x) ∧ ¬Q(x))

(De Morgan’s law),

which is equivalent to ¬(∀x¬P (x) ∧ ∀x¬Q(x))

(distributive law for ∀ and ∧),

which is equivalent to ¬∀x¬P (x) ∨ ¬∀x¬Q(x)

(De Morgan’s law),

which is equivalent to ¬¬∃xP (x) ∨ ¬¬∃xQ(x)

(quantifier negation law),

which is equivalent to ∃xP (x) ∨ ∃xQ(x)

(double negation law).

∃x(P (x) → Q(x)) is equivalent to ∃x(¬P (x) ∨ Q(x)) which is equivalent to ∃x¬P (x) ∨ ∃xQ(x)

(conditional law), (exercise 6),

which is equivalent to ¬∀xP (x) ∨ ∃xQ(x)

(quantifier negation law),

which is equivalent to ∀xP (x) → ∃xQ(x)

(conditional law).

8. (∀x ∈ A P (x)) ∧ (∀x ∈ B P (x)) is equivalent to ∀x(x ∈ A → P (x)) ∧ ∀x(x ∈ B → P (x))

(expanding abbreviation),

which is equivalent to ∀x[(x ∈ A → P (x)) ∧ (x ∈ B → P (x))]

(distributive law),

which is equivalent to ∀x[(x ∈ / A ∨ P (x)) ∧ (x ∈ / B ∨ P (x))]

(conditional law),

which is equivalent to ∀x[(x ∈ / A∧x∈ / B) ∨ P (x)]

(distributive law),

which is equivalent to ∀x[¬(x ∈ A ∨ x ∈ B) ∨ P (x)]

(De Morgan’s law),

which is equivalent to ∀x[x ∈ / (A ∪ B) ∨ P (x)] which is equivalent to ∀x[x ∈ (A ∪ B) → P (x)]

(definition of ∪), (conditional law),

which is equivalent to ∀x ∈ (A ∪ B) P (x)

(abbreviation).

9. No. For example, if the universe of discourse is N, P (x) stands for “x is even,” and Q(x) stands for “x is odd,” then ∀x(P (x) ∨ Q(x)) means “Every natural number is either even or odd,” which is true, but ∀xP (x) ∨ ∀xQ(x) means “Either every natural number is even or every natural number is odd,” which is false.

10. (a) ∃x ∈ A P (x) ∨ ∃x ∈ B P (x) is equivalent to ∃x(x ∈ A ∧ P (x)) ∨ ∃x(x ∈ B ∧ P (x)) which is equivalent to ∃x[(x ∈ A ∧ P (x)) ∨ (x ∈ B ∧ P (x))]

(expanding abbreviation), (distributive law),

which is equivalent to ∃x[(x ∈ A ∨ x ∈ B) ∧ P (x)]

(distributive law),

which is equivalent to ∃x[x ∈ A ∪ B ∧ P (x)]

(definition of ∪),

which is equivalent to ∃x ∈ A ∪ B P (x)

(abbreviation).

(b) No. Suppose A is the set of even natural numbers, B is the set of odd natural numbers, and P (x) stands for “x is prime.” Then ∃x ∈ A P (x) ∧ ∃x ∈ B P (x) means “There is at least one even prime number and there is at least one odd prime number,” which is true, but ∃x ∈ A ∩ B P (x) means “There is at least one number that is both even and odd and is prime,” which is false.

11. A ⊆ B means ∀x(x ∈ A → x ∈ B) which is equivalent to ∀x(x ∈ / A ∨ x ∈ B)

(conditional law),

which is equivalent to ∀x¬(x ∈ A ∧ x ∈ / B)

(De Morgan’s law),

which is equivalent to ¬∃x(x ∈ A ∧ x ∈ / B)

(quantifier negation law),

which is equivalent to A \ B = ∅

(definitions of \, ∅).

12. C ⊆ A ∪ B means ∀x(x ∈ C → (x ∈ A ∨ x ∈ B)) which is equivalent to ∀x(x ∈ / C ∨ (x ∈ A ∨ x ∈ B))

(conditional law),

which is equivalent to ∀x((x ∈ / C ∨ x ∈ A) ∨ x ∈ B)

(associative law),

18

SOLUTIONS which is equivalent to ∀x(¬(x ∈ C ∧ x ∈ / A) ∨ x ∈ B)

(De Morgan’s law),

which is equivalent to ∀x((x ∈ C ∧ x ∈ / A) → x ∈ B)

(conditional law),

which is equivalent to C \ A ⊆ B

(definitions of \, ⊆).

13. (a) A ⊆ B means ∀x(x ∈ A → x ∈ B) which is equivalent to ∀x((x ∈ A ∨ x ∈ B) ↔ x ∈ B)

(Sec. 1.5, exercise 11(a)),

which is equivalent to A ∪ B = B

(definition of ∪).

(b) A ⊆ B means ∀x(x ∈ A → x ∈ B) which is equivalent to ∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ A)

(Sec. 1.5, exercise 11(b)),

which is equivalent to A ∩ B = A 14. A ∩ B = ∅ means ¬∃x(x ∈ A ∧ x ∈ B) which is equivalent to ∀x¬(x ∈ A ∧ x ∈ B)

(definition of ∩).

(quantifier negation law),

which is equivalent to ∀x(x ∈ / A∨x∈ / B)

(De Morgan’s law),

which is equivalent to ∀x(x ∈ A → x ∈ / B)

(conditional law),

which is equivalent to ∀x((x ∈ A ∧ x ∈ / B) ↔ x ∈ A)

(Sec. 1.5, exercise 11(b)),

which is equivalent to A \ B = A

(definition of \).

15. None of these statements are equivalent to each other. (a) x is a teacher of exactly one person; i.e., x is a teacher who has exactly one student. (b) Someone is a teacher of exactly one person; i.e., some teacher has exactly one student. (c) There is exactly one person who teaches someone; i.e., there is exactly one teacher. (d) There is someone who is taught by exactly one person; i.e., some student has exactly one teacher. (e) There is exactly one person who is a teacher of exactly one person; i.e., there is exactly one person who is a teacher with only one student (there may be other teachers with more students). (f) There is exactly one pair of people, the first of whom is a teacher of the second; this is equivalent to saying that there is exactly one teacher, and that teacher has exactly one student.

Section 2.3 1. (a) ∀x(x ∈ F → ∀y(y ∈ x → y ∈ A)). (b) ∀x(x ∈ A → ∃n ∈ N(x = 2n + 1)). (c) ∀n ∈ N∃m ∈ N(n2 + n + 1 = 2m + 1). (d) We work this out in several steps: S S i. ∃x(x ∈ P( i∈I Ai ) ∧ x ∈ / i∈I P(Ai )).

S ii. ∃x(x ⊆ i∈I Ai ∧ ∀i ∈ I(x ∈ / P(Ai )).

S iii. ∃x(∀y(y ∈ x → y ∈ i∈I Ai ) ∧ ∀i ∈ I(x * Ai )). iv. ∃x(∀y(y ∈ x → ∃i ∈ I(y ∈ Ai )) ∧ ∀i ∈ I∃y(y ∈ x ∧ y ∈ / Ai )).

2. (a) ∃A ∈ F(x ∈ A) ∧ ∀B ∈ G(x ∈ / B). (b) ∀x((x ∈ B ∧ x ∈ / C) → x ∈ A). (c) ∀i ∈ I(x ∈ Ai ∨ x ∈ Bi ). (d) ∀i ∈ I(x ∈ Ai ) ∨ ∀i ∈ I(x ∈ Bi ).

3. P({∅}) = {∅, {∅}}.

CHAPTER 2

19

S F = {red, green, blue} ∩ {orange, red, blue} ∩ {purple, red, green, blue} = {red, blue}. F = {red, green, blue} ∪ {orange, red, blue} ∪ {purple, red, green, blue} = {red, green, blue, orange, purple}.

T 5. S F = {3, 7, 12} ∩ {5, 7, 16} ∩ {5, 12, 23} = ∅.

F = {3, 7, 12} ∪ {5, 7, 16} ∪ {5, 12, 23} = {3, 5, 7, 12, 16, 23}.

4.

T

6. (a) A2 = {2, 3, 1, 4}, A3 = {3, 4, 2, 6}, A4 = {4, 5, 3, 8}, A5 = {5, 6, 4, 10}.

T S (b) i∈I Ai = A2 ∩ A3 ∩ A4 ∩ A5 = {2, 3, 1, 4} ∩ {3, 4, 2, 6} ∩ {4, 5, 3, 8} ∩ {5, 6, 4, 10} = {4}. i∈I Ai = A2 ∪ A3 ∪ A4 ∪ A5 = {2, 3, 1, 4} ∪ {3, 4, 2, 6} ∪ {4, 5, 3, 8} ∪ {5, 6, 4, 10} = {1, 2, 3, 4, 5, 6, 8, 10}.

7. A1750 = {Johann Sebastian Bach, Johann Wolfgang von Goethe, David Hume, George Washington}.

A1751 = A1752 = A1753 = A1754 = A1755 = {Johann Wolfgang von Goethe, David Hume, George Washington}. A1756 = A1757 = A1758 = A1759 = {Johann S Wolfgang von Goethe, David Hume, Wolfgang Amadeus Mozart, George Washington}. Therefore y∈Y Ay = {Johann Sebastian Bach, Johann Wolfgang von Goethe, David Hume, Wolfgang Amadeus Mozart, George Washington} and T y∈Y Ay = {Johann Wolfgang von Goethe, David Hume, George Washington}.

8. (a) A2 = {2, 4}, A3 = {3, 6}, B2 = {2, 3}, B3 = {3, 4}.

T (b) i∈I (Ai ∪ Bi ) = (A2 ∪ B2 ) ∩ (A3 ∪ B3 ) = {2, 3, 4} ∩ {3, 4, 6} = {3, 4}, T T ( i∈I Ai ) ∪ ( i∈I Bi ) = (A2 ∩ A3 ) ∪ (B2 ∩ B3 ) = ∅ ∪ {3} = {3}. (c) They are not equivalent.

S 9. (a) x ∈ i∈I (Ai \ Bi ) means ∃i ∈ I(x ∈ Ai ∧ x ∈ / Bi ), S S x ∈ ( i∈I Ai ) \ ( i∈I Bi ) means ∃i ∈ I(x ∈ Ai ) ∧ ¬∃i ∈ I(x ∈ Bi ), S T x ∈ ( i∈I Ai ) \ ( i∈I Bi ) means ∃i ∈ I(x ∈ Ai ) ∧ ¬∀i ∈ I(x ∈ Bi ).

S (b) i∈I (Ai \ Bi ) = (A2 \ B2 ) ∪ (A3 \ B3 ) = {4} ∪ {6} = {4, 6}, S S ( i∈I Ai ) \ ( i∈I Bi ) = (A2 ∪ A3 ) \ (B2 ∪ B3 ) = {2, 3, 4, 6} \ {2, 3, 4} = {6}, S T ( i∈I Ai ) \ ( i∈I Bi ) = (A2 ∪ A3 ) \ (B2 ∩ B3 ) = {2, 3, 4, 6} \ {3} = {2, 4, 6}.

None of the statements are equivalent.

10. One example is I = {1, 2}, A1 = {a, b}, A2 = {c, d}, B1 = {a, e}, B2 = {b, f }.

S

i∈I (Ai ∩ Bi ) = (A1 ∩ B1 ) ∪ (A2 ∩ B2 ) = {a} ∪ ∅ = {a}, S S ( i∈I Ai ) ∩ ( i∈I Bi ) = (A1 ∪ A2 ) ∩ (B1 ∪ B2 ) = {a, b, c, d} ∩ {a, b, e, f } = {a, b}.

11.

x ∈ P(A ∩ B) is equivalent to ∀y(y ∈ x → (y ∈ A ∧ y ∈ B))

(Example 2.3.3, 4),

which is equivalent to ∀y(y ∈ / x ∨ (y ∈ A ∧ y ∈ B))

(conditional law),

which is equivalent to ∀y((y ∈ / x ∨ y ∈ A) ∧ (y ∈ / x ∨ y ∈ B))

(distributive law),

which is equivalent to ∀y((y ∈ x → y ∈ A) ∧ (y ∈ x → y ∈ B))

(conditional law),

which is equivalent to ∀y(y ∈ x → y ∈ A) ∧ ∀y(y ∈ x → y ∈ B)

(distributive law),

which is equivalent to x ∈ P(A) ∩ P(B)

(Example 2.3.3, 5).

12. One example is A = {1, 2} and B = {2, 3}.

P(A ∪ B) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}, P(A) ∪ P(B) = {∅, {1}, {2}, {3}, {1, 2}, {2.3}}.

13. (a) x ∈

S

i∈I (Ai ∪ Bi ) means ∃i ∈ I(x ∈ Ai ∨ x ∈ Bi )

which is equivalent to ∃i ∈ I(x ∈ Ai ) ∨ ∃i ∈ I(x ∈ Bi ) S S which is equivalent to x ∈ ( i∈I Ai ) ∪ ( i∈I Bi )

(distributive law), S , ∪).

(definitions of

20

SOLUTIONS T T (b) x ∈ ( F) ∩ ( G) means ∀A ∈ F(x ∈ A) ∧ ∀A ∈ G(x ∈ A) which is equivalent to ∀A ∈ F ∪ G(x ∈ A) T which is equivalent to x ∈ (F ∪ G) (c) x ∈

T

(Sec. 2.2, exercise 8), T (Definition of ).

/ Bi ) i∈I (Ai \ Bi ) means ∀i ∈ I(x ∈ Ai ∧ x ∈

which is equivalent to ∀i ∈ I(x ∈ Ai ) ∧ ∀i ∈ I(x ∈ / Bi )

(distributive law),

which is equivalent to ∀i ∈ I(x ∈ Ai ) ∧ ¬∃i ∈ I(x ∈ Bi ) T S which is equivalent to x ∈ ( i∈I Ai ) \ ( i∈I Bi )

(quantifier negation law), T S (Definitions of , , \).

14. (a) B3 = A1,3 ∪ A2,3 = {1, 3, 4} ∪ {2, 3, 5} = {1, 2, 3, 4, 5}, B4 = A1,4 ∪ A2,4 = {1, 4, 5} ∪ {2, 4, 6} = {1, 2, 4, 5, 6}.

T (b) j∈J Bj = B3 ∩ B4 = {1, 2, 4, 5}.

T (c) For each i ∈ I, let Ci = j∈J Ai,j . Then C1 = A1,3 ∩ A1,4 = {1, 3, 4} ∩ {1, 4, 5} = {1, 4}, S

C2 = A2,3 ∩ A2,4 = {2, 3, 5} ∩ {2, 4, 6} = {2}, T S i∈I ( j∈J Ai,j ) = i∈I Ci = C1 ∪ C2 = {1, 2, 4}.

This is not equal to the set in part (b).

T S S T (d) x ∈ j∈J ( i∈I Ai,j ) means ∀j ∈ J∃i ∈ I(x ∈ Ai,j ) and x ∈ i∈I ( j∈J Ai,j ) means ∃i ∈ I∀j ∈ J(x ∈ Ai,j ). They are not equivalent.

S 15. (a) x ∈ F means ∃A ∈ F(x ∈ A). If F = ∅ then this statement will be false, no matter what x is, because there is no set A ∈ F. T (b) x ∈ F means ∀A ∈ F(x ∈ A). If F = ∅ then this statement will be vacuously true, no matter what x is.

16. (a) Since U is the set of all sets and R is a set, R ∈ U . Since ∀A ∈ U (A ∈ R ↔ A ∈ / A), we conclude that R ∈ R ↔ R ∈ / R. But this is impossible, since the formula P ↔ ¬P is a contradiction (make a truth table). (b) We could rephrase the reasoning in part (a) as follows: Suppose U is any set whose elements are sets. Let R = {A ∈ U | A ∈ / A}. Then ∀A ∈ U (A ∈ R ↔ A ∈ / A). If R ∈ U then, as in part (a), we conclude that R ∈ R ↔ R ∈ / R, which is impossible. Therefore R ∈ / U . What this rephrasing of the proof shows is that if U is any set whose elements are sets, then we can come up with a set R such that R ∈ / U . Thus, U could not have been the set of all sets. Many mathematicians interpret this as meaning that there cannot be a set of all sets.

Chapter 3 Section 3.1 1. (a) Hypotheses: n is an integer larger than 1 and n is not prime. Conclusion: 2n − 1 is not prime.

The hypotheses are true when n = 6, so the theorem tells us that 26 − 1 is not prime. This is correct, since 26 − 1 = 63 = 9 · 7. (b) We can conclude that 32767 is not prime. This is correct, since 32767 = 151 · 217. (c) The theorem tells us nothing; 11 is prime, so the hypotheses are not satisfied.

2. (a) Hypothesis: b2 > 4ac. Conclusion: The quadratic equation ax2 + bx + c = 0 has exactly two real solutions.

CHAPTER 3

21

(b) Because a, b, and c are free variables in the hypothesis and conclusion, but x is not. (c) b2 = 25 > 24 = 4ac, so we can conclude that the equation 2x2 − 5x + 3 has exactly two real solutions. This is correct; the solutions are x = 3/2 and x = 1. (d) b2 = 16 < 24 = 4ac, so the hypothesis is not true and we cannot conclude anything.

3. Hypotheses: n is a natural number, n > 2, and n is not prime. Conclusion: 2n + 13 is not prime. One counterexample is n = 8.

4. Suppose 0 < a < b. Then b − a > 0. Multiplying both sides by the positive number b + a, we get (b + a) · (b − a) > (b + a) · 0, or in other words b2 − a2 > 0. Since b2 − a2 > 0, it follows that a2 < b2 .

Therefore if 0 < a < b then a2 < b2 .

5. Suppose a < b < 0. Multiplying the inequality a < b by the negative number a (and reversing the direction of the inequality) we can conclude that a2 > ab, and similarly multiplying by b we get ab > b2 .

Therefore a2 > ab > b2 , so a2 > b2 , as required. Thus, if a < b < 0 then a2 > b2 .

6. Suppose 0 < a < b. Dividing the inequality a < b by the positive number ab, we get a b 1 1 = < = . b ab ab a Thus, if 0 < a < b then 1/b < 1/a.

7. Suppose a3 > a. Then a3 − a > 0. Multiplying this inequality by the positive number a2 + 1, we get a5 − a = (a3 − a)(a2 + 1) > 0 · (a2 + 1) = 0, so a5 > a. Thus, if a3 > a then a5 > a.

8. We will prove the contrapositive. Suppose x ∈ / B. Then since x ∈ A, it follows that x ∈ A \ B. But we also know that A \ B ⊆ C ∩ D, so we can conclude that x ∈ C ∩ D, and therefore x ∈ D. Thus, if x∈ / D then x ∈ B. 9. Suppose x ∈ A. To prove that if x ∈ D then x ∈ / B we will prove the contrapositive, so suppose x ∈ B. Then since x ∈ A and x ∈ B, x ∈ A ∩ B, and since A ∩ B ⊆ C \ D, we can conclude that x ∈ C \ D, and therefore x ∈ / D. Thus, if x ∈ D then x ∈ / B. Therefore if x ∈ A, then if x ∈ D then x ∈ / B. 10. Suppose a < b. Adding b to both sides of this inequality we get a + b < 2b, and then dividing by 2 we conclude that (a + b)/2 < b. Therefore if a < b then (a + b)/2 < b.

11. We prove the contrapositive. Suppose x = 8. Then √ 3 x+5 7 1 1 1 = = 6= = . x2 + 6 70 10 8 x √ Thus, if ( 3 x + 5)/(x2 + 6) = 1/x then x 6= 8.

12. We will prove the contrapositive. Suppose c ≤ d. Multiplying both sides of this inequality by the positive number a, we get ac ≤ ad. Also, multiplying both sides of the given inequality a < b by the positive number d gives us ad < bd. Combining ac ≤ ad and ad < bd, we can conclude that ac < bd.

Thus, if ac ≥ bd then c > d.

13. Suppose x > 1. Then 3x > 3, so 3x + 2y > 3 + 2y. Combining this with the fact that 3x + 2y ≤ 5, we conclude that 3 + 2y < 5. Therefore 2y < 2, so y < 1. Thus, if x > 1 then y < 1.

14. Suppose x2 + y = −3 and 2x − y = 2. Adding these equations, we get x2 + 2x = −1, so x2 + 2x + 1 = 0.

We can rewrite this as (x + 1)2 = 0, so x + 1 = 0, and therefore x = −1. Thus, if x2 + y = −3 and 2x − y = 2 then x = −1.

22

SOLUTIONS

15. Since x > 3 > 0, by the theorem in Example 3.2.1, x2 > 9. Also, multiplying both sides of the given inequality y < 2 by −2 (and reversing the direction of the inequality, since −2 is negative) we get −2y > −4. Finally, adding the inequalities x2 > 9 and −2y > −4 gives us x2 − 2y > 5.

16. (a) The reasoning is backwards. The proof proves that if x = 7 then (2x − 5)/(x − 4) = 3, not that if (2x − 5)/(x − 4) = 3 then x = 7. (b) Suppose (2x − 5)/(x − 4) = 3. Multiplying both sides of this equation by x − 4, we conclude that 2x − 5 = 3(x − 4) = 3x − 12. Subtracting 2x from both sides we get −5 = x − 12, and finally adding 12 to both sides gives us 7 = x. Thus, if (2x − 5)/(x − 4) = 3 then x = 7.

17. (a) The mistake is the step “Since x 6= 3, x2 6= 9.” If x = −3 then x 6= 3, but x2 = 9, so this step is incorrect. (b) One counterexample is x = −3, y = 1.

Section 3.2 1. (a) Suppose P . Since P → Q, it follows that Q. But then, since Q → R, we can conclude R. Thus, P → R. (b) Suppose P . To prove that Q → R, we will prove the contrapositive, so suppose ¬R. Since ¬R → (P → ¬Q), it follows that P → ¬Q, and since we know P , we can conclude ¬Q. Thus, Q → R, so P → (Q → R).

2. (a) Suppose P . Since P → Q, it follows that Q. But then since R → ¬Q, we can conclude that ¬R. Thus, P → ¬R. (b) Suppose Q. If we assume Q → ¬P , then we can conclude ¬P , which contradicts the given information that P . Therefore ¬(Q → ¬P ). Thus, Q → ¬(Q → ¬P ).

3. Suppose x ∈ A. Since A ⊆ C, x ∈ C. But then since B and C are disjoint, it follows that x ∈ / B. Thus, if x ∈ A then x ∈ / B. 4. Suppose x ∈ C. Suppose x ∈ / B. Since x ∈ A and x ∈ / B, x ∈ A \ B. But then since x ∈ C, this contradicts the fact that A \ B and C are disjoint. Therefore x ∈ B. Thus, if x ∈ C then x ∈ B. 5. Suppose x ∈ A \ B and x ∈ B \ C. Since x ∈ A \ B, x ∈ A and x ∈ / B, and since x ∈ B \ C, x ∈ B and x∈ / C. But now we have x ∈ B and x ∈ / B, which is a contradiction. Therefore it cannot be the case that x ∈ A \ B and x ∈ B \ C. 6. Suppose a ∈ A \ B. This means that a ∈ A and a ∈ / B. Since a ∈ A and a ∈ C, a ∈ A ∩ C. But then since A ∩ C ⊆ B, it follows that a ∈ B, and this contradicts the fact that a ∈ / B. Thus, a ∈ / A \ B. 7. Suppose a ∈ / C. Since a ∈ A and A ⊆ B, a ∈ B. Since a ∈ B and a ∈ / C, a ∈ B \ C, which contradicts the given information that a ∈ / B \ C. Therefore a ∈ C. 8. Suppose y = 0. Substituting into the equation y + x = 2y − x, we get x = −x. Therefore 2x = 0, so x = 0. But this contradicts the given information that x and y are not both zero. Therefore y 6= 0.

9. Suppose a < 1/a < b < 1/b. Suppose a ≥ 1. Dividing this inequality by the positive number a, we get 1 ≥ 1/a, and combining the inequalities a ≥ 1 and 1 ≥ 1/a we can conclude that a ≥ 1/a, contradicting the fact that a < 1/a. Thus, a < 1. Next, suppose that a > 0. Then dividing a < 1 by a we get 1 < 1/a, and since 1/a < b it follows that b > 1. Dividing by b gives us 1 > 1/b, and together with b > 1 this implies that b > 1/b, contradicting the fact that b < 1/b. We conclude that a ≤ 0. But we know that a is nonzero, so in fact a < 0.

We are now finally ready to prove that a < −1. Suppose that a ≥ −1. Dividing by the positive number −a, we conclude that −1 ≥ 1/a. Combining a ≥ −1 and −1 ≥ 1/a we get a ≥ 1/a, contradicting the fact that a < 1/a. Therefore a < −1.

CHAPTER 3

23

10. Suppose x2 y = 2x + y. To prove that if y 6= 0 then x 6= 0, we will prove the contrapositive. Suppose x = 0. Substituting into the equation x2 y = 2x + y we get 0 = 0 + y = y. Thus, if y 6= 0 then x 6= 0.

This proves that if x2 y = 2x + y, then if y 6= 0 then x 6= 0.

11. Suppose x 6= 0. Suppose y = (3x2 + 2y)/(x2 + 2). Multiplying by x2 + 2, we get x2 y + 2y = 3x2 + 2y, and therefore x2 y = 3x2 . Since x 6= 0, x2 6= 0, so we can divide both sides by x2 to conclude that y = 3. Thus, if x 6= 0, then if y = (3x2 + 2y)/(x2 + 2) then y = 3.

12. (a) The sentence “Then x = 3 and y = 8” is incorrect. The assumption that the conclusion is false means ¬((x 6= 3)∧(y 6= 8)), and by one of De Morgan’s laws this is equivalent to (x = 3)∨(y = 8).

So the sentence should say “Then x = 3 or y = 8,” not “Then x = 3 and y = 8.” (b) One counterexample is x = 3, y = 7.

13. (a) The sentence “Since x ∈ / B and B ⊆ C, x ∈ / C” is incorrect. It is possible to have x ∈ / B and B ⊆ C, but x ∈ C. (b) One counterexample is A = {1}, B = {2}, C = {1, 2}, and x = 1.

14.

P F F T T

Q F T F T

Premises P → Q ¬Q T T T F F T T F

P F F F F T T T T

Q F F T T F F T T

R F T F T F T F T

Premise P → (Q → R) T T T T T T F T

P F F F F T T T T

Q F F T T F F T T

R F T F T F T F T

15.

16. (a)

Conclusion ¬P T T F F Conclusion ¬R → (P → ¬Q) T T T T T T F T

Premises P →Q Q→R T T T T T F T T F T F T T F T T

Conclusion P →R T T T T F T F T

(b) The truth table is the same as the one for exercise 15, but with the premise and conclusion reversed.

24

SOLUTIONS

17. (a) P F F F F T T T T

Q F F T T F F T T

P F F T T

Q F T F T

(b)

Premises Conclusion R P → Q R → ¬Q P → ¬R F T T T T T T T F T T T T T F T F F T T T F T F F T T T T T F F Premise Conclusion P Q → ¬(Q → ¬P ) F T F F T T T T

18. No. Here is a counterexample to the proposed theorem: x = −3, y = 4.

Section 3.3 1. Suppose ∃x(P (x) → Q(x)). Then we can choose some x0 such that P (x0 ) → Q(x0 ). Now suppose that ∀xP (x). Then in particular, P (x0 ), and since P (x0 ) → Q(x0 ), it follows that Q(x0 ). Since we have found a particular value of x for which Q(x) holds, we can conclude that ∃xQ(x). Thus ∀xP (x) → ∃xQ(x).

2. Suppose A and B \ C are disjoint. Let x be an arbitrary element of A ∩ B. Then x ∈ A and x ∈ B. Suppose x ∈ / C. Then since x ∈ B and x ∈ / C, x ∈ B \ C. But we also have x ∈ A, so this contradicts the fact that A and B \ C are disjoint. Therefore x ∈ C. Since x was arbitrary, we can conclude that A ∩ B ⊆ C. 3. Suppose that A ⊆ B \ C, but A and C are not disjoint. Then we can choose some x such that x ∈ A and x ∈ C. Since x ∈ A and A ⊆ B \ C, it follows that x ∈ B \ C, which means that x ∈ B and x ∈ / C. But now we have both x ∈ C and x ∈ / C, which is a contradiction. Thus, if A ⊆ B \ C then A and C are disjoint.

4. Let X be an arbitrary element of P(A). Then X ⊆ A. Suppose x ∈ X. Then since X ⊆ A, x ∈ A, and therefore since A ⊆ P(A), x ∈ P(A). Since x was arbitrary, it follows that X ⊆ P(A), so X ∈ P(P(A)). Since X was arbitrary, we can conclude that P(A) ⊆ P(P(A)).

5. (a) The easiest example is A = ∅. (b) By exercise 4, another example is A = P(∅) = {∅}.

6. (a) Suppose x 6= 1. Let y = (2x + 1)/(x − 1), which is defined since x 6= 1. Then 2x+1 3x y+1 x−1 + 1 = 2x+1 = x−1 = x.

3 y−2 x−1 − 2 x−1

(b) Suppose y is a real number such that (y +1)/(y −2) = x. Suppose x = 1. Then (y +1)/(y −2) = 1, so y + 1 = y − 2, and therefore 1 = −2, which is a contradiction. Therefore x 6= 1. √ 7. Suppose x > 2. Let y = (x + x2 − 4)/2, which is defined since x2 − 4 > 0. Then √ √ √ 1 x + x2 − 4 2 2x2 + 2x x2 − 4 2x(x + x2 − 4) √ √ √ y+ = + = = = x. y 2 x + x2 − 4 2(x + x2 − 4) 2(x + x2 − 4)

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S 8. Suppose F is a family of sets S and A ∈ F. Let x be an arbitrary element of A. By the definitionSof F, since x ∈ A and A ∈ F, x ∈ F. Since x was an arbitrary element of A, it follows that A ⊆ F. T T 9. Suppose T F is a family of sets and A ∈ F. Suppose x ∈ F. TThen by the definitionTof F, since x ∈ F and A ∈ F, x ∈ A. But x was an arbitrary element of F, so it follows that F ⊆ A. 10. Let x be an arbitrary element of B. Suppose C ∈ F. By assumption, ∀A ∈ F(B ⊆ A), so B ⊆ C. Since x ∈ B, it follows that T x ∈ C. But C was an arbitrary element of F, so it follows that ∀C ∈TF(x ∈ C), and therefore x ∈ F. Since x was an arbitrary element of B, we can conclude that B ⊆ F. T T 11. Suppose ∅ ∈ F. Suppose F 6= ∅. Then we can choose some x such that x ∈ F. By the definition T T of F, since T x ∈ F and ∅ ∈ F, x ∈ ∅. But this is a contradiction, since ∅ has no elements.

Therefore F = ∅.

S S 12. Suppose F ⊆ G, and let x be an arbitrary element of F. By the definition of F, this means that we can Since A ∈ F and F ⊆ G, A ∈ S G. Since S choose a set A ∈ F such that x ∈ A. S S x ∈ A and A ∈ G, x ∈ G. But x was an arbitrary element of F, so we can conclude that F ⊆ G. T 13. Suppose F ⊆ G, and let x be an arbitrary elementTof G. Suppose A ∈ F. Since F ⊆ G, it follows T that A ∈ G. By the definition of G, since x ∈ G and A ∈ G, x ∈ T A. Since A was an arbitrary element of T F, we conclude thatT∀A ∈ T F(x ∈ A), which means x ∈ F. Since x was an arbitrary element of G, this shows that G ⊆ F. S 14. Suppose x ∈ i∈I P(Ai ). Then we can choose some i ∈ I such that x ∈ P(Ai ), or S in other words x ⊆ Ai . Now let a be an arbitrary element of x. S Then a ∈ Ai , and therefore a ∈ S i∈I Ai . Since a was an arbitrary element of x, it follows that x ⊆ A , which means that x ∈ P( i i∈I i∈I Ai ). Thus S S P(A ) ⊆ P( A ). i i i∈I i∈I T 15. Suppose i T ∈ I. Let x be an arbitrary element of i∈I Ai . Then T T x ∈ Ai . Since x was an arbitrary element of i∈I AiT , it followsTthat i∈I Ai ⊆ Ai , and therefore i∈I Ai ∈ P(Ai ). Since i was arbitrary, we conclude that i∈I Ai ∈ i∈A P(Ai ).

S 16. Suppose F ⊆ P(B). Let y be an arbitrary element of F. Then we can choose some x ∈ F such that y ∈ x. Since x ∈ F and F ⊆ P(B), x ∈ S P(B), which means that x ⊆SB. Since y ∈ x and x ⊆ B, y ∈ B. But y was an arbitrary element of F, so we can conclude that F ⊆ B. S 17. Suppose x ∈ F. Then we can choose some A ∈ F such that x ∈ A. Let B be an arbitrary element of G. Since every element of F is a subset of every element of G, A ⊆ B. Since x ∈ A, it follows that x ∈ B. But T B was an S arbitrary T element of G, so we can conclude that ∀B ∈ G(x ∈ B), which means that x ∈ G. Thus, F ⊆ G. 18. (a) Suppose a | b and a | c. Then we can choose integers j and k such that ja = b and ka = c.

Therefore b + c = ja + ka = (j + k)a. Since j + k is an integer, this shows that a | (b + c). (b) Suppose ac | bc and c 6= 0. Since ac | bc, we can choose some integer k such that kac = bc. Since c 6= 0, we can divide both sides of this equation by c to conclude that ka = b, and therefore a | b.

19. (a) Let x and y be arbitrary real numbers. Let z = (y − x)/2. Then x+z =x+

y+x y−x y−x = =y− = y − z.

2 2 2

(b) No. For example, if x = 1 and y = 2, then the only value of z that would make the equation x + z = y − z true is z = 1/2, which is not an integer.

20. The sentence “Then for every real number x, x2 < 0” is incorrect. The assumption that the conclusion is false means ¬∀x(x2 ≥ 0), which is equivalent to ∃x(x2 < 0). So the sentence should say “Then for some real number x, x2 < 0,” not “Then for every real number x, x2 < 0.”

26

SOLUTIONS

21. (a) In the second sentence of the proof, x was introduced as an arbitrary element of A, not an arbitrary element of B, so it is incorrect to conclude at the end of the proof that ∀x ∈ B(x 6= 0). To prove this goal, x would have to be introduced with the sentence “Let x be an arbitrary element of B.” (b) One counterexample is A = {1}, B = {0, 1}.

22. Based on the logical form of the statement to be proven, the proof should have this outline: Let x = . . ..

Let y be an arbitrary real number.

[Proof of xy 2 = y − x goes here.]

Since y was arbitrary, ∀y ∈ R(xy 2 = y − x).

Thus, ∃x ∈ R∀y ∈ R(xy 2 = y − x).

This outline makes it clear that y should be introduced into the proof after x. Therefore, x cannot be defined in terms of y, because y will not yet have been introduced into the proof when x is being defined. But in the given proof, x is defined in terms of y in the first sentence. (The mistake has been disguised by the fact that the sentence “Let y be an arbitrary real number” has been left out of the proof. If you try to add this sentence to the proof, you will find that there is nowhere it could be added that would lead to a correct proof of the incorrect theorem.)

S S 23. (a) The mistake is inSthe sentence S “But then every element of A is in both F and G, and this is impossible since F and G are disjoint.” set then it will be vacuously true S SIf A is the S empty S that every element of A is in both F and G, but F and G could still be disjoint. (b) One counterexample is F = {∅, {1}}, G = {∅, {2}}.

24. (a) The proof assumes that the values of x and y are the same, but they might not be. (b) The theorem is incorrect. One counterexample is x = 1, y = 0.

25. Let x be an arbitrary real number. Let y = 2x. Now let z be an arbitrary real number. Then (x + z)2 − (x2 + z 2 ) = (x2 + 2xz + z 2 ) − (x2 + z 2 ) = 2xz = yz.

26. (a) Goal of the form ∀xP (x) and given of the form ∃xP (x): In both cases you introduce a new variable x into the proof, but you don’t specify the value of x. Goal of the form ∃xP (x) and given of the form ∀xP (x): In both cases you specify a value to be assigned to x in the statement P (x). (b) If you are proving the goal ∀xP (x) and you use proof by contradiction, then you will assume ¬∀xP (x), which is equivalent to ∃x¬P (x). If you are proving ∃xP (x) and you use proof by contradiction, then you will assume ¬∃xP (x), which is equivalent to ∀x¬P (x). So a goal with one kind of quantifier is converted to a given with the other kind when you use proof by contradiction.

Section 3.4 1. (→) Suppose ∀x(P (x) ∧ Q(x)). Let y be arbitrary. Then since ∀x(P (x) ∧ Q(x)), P (y) ∧ Q(y), and so in particular P (y). Since y was arbitrary, this shows that ∀xP (x). A similar argument proves ∀xQ(x): for arbitrary y, P (y) ∧ Q(y), and therefore Q(y). Thus, ∀xP (x) ∧ ∀xQ(x). (←) Suppose ∀xP (x) ∧ ∀xQ(x). Let y be arbitrary. Then since ∀xP (x), P (y), and similarly since ∀xQ(x), Q(y). Thus, P (y) ∧ Q(y), and since y was arbitrary, it follows that ∀x(P (x) ∧ Q(x)).

2. Suppose A ⊆ B and A ⊆ C. Let x be an arbitrary element of A. Then since x ∈ A and A ⊆ B, x ∈ B, and since x ∈ A and A ⊆ C, x ∈ C. Thus x ∈ B ∩ C, and since x was arbitrary it follows that A ⊆ B ∩ C. 3. Let C be an arbitrary set. Suppose x ∈ C \ B. Then x ∈ C and x ∈ / B. If x ∈ A then, since A ⊆ B, x ∈ B, which is a contradiction. Therefore x ∈ / A. Since x ∈ C and x ∈ / A, x ∈ C \ A. But x was an arbitrary element of C \ B, so we can conclude that C \ B ⊆ C \ A.

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4. Suppose that A ⊆ B and A * C. Since A * C, we can choose some a ∈ A such that a ∈ / C. Since a ∈ A and A ⊆ B, a ∈ B. Since a ∈ B and a ∈ / C, B * C. 5. Suppose A ⊆ B \ C and A 6= ∅. Since A 6= ∅, we can choose some a ∈ A, and since A ⊆ B \ C, a ∈ B \ C. Therefore a ∈ B and a ∈ / C, so B * C. 6. Let x be arbitrary. Then x ∈ A \ (B ∩ C) iff x ∈ A ∧ ¬(x ∈ B ∧ x ∈ C)

(definitions of \, ∩),

iff x ∈ A ∧ (x ∈ / B∨x∈ / C)

(De Morgan’s law),

iff (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C)

(distributive law),

iff x ∈ (A \ B) ∪ (A \ C)

(definitions of \, ∪).

7. Let A and B be arbitrary sets. Let x be arbitrary, and suppose that x ∈ P(A ∩ B). Then x ⊆ A ∩ B. Now let y be an arbitrary element of x. Then since x ⊆ A ∩ B, y ∈ A ∩ B, and therefore y ∈ A. Since y was arbitrary, this shows that x ⊆ A, so x ∈ P(A). A similar argument shows that x ⊆ B, and therefore x ∈ P(B). Thus, x ∈ P(A) ∩ P(B).

Now suppose that x ∈ P(A) ∩ P(B). Then x ∈ P(A) and x ∈ P(B), so x ⊆ A and x ⊆ B. Suppose that y ∈ x. Then since x ⊆ A and x ⊆ B, y ∈ A and y ∈ B, so y ∈ A ∩ B. Thus, x ⊆ A ∩ B, so x ∈ P(A ∩ B).

8. (→) Suppose A ⊆ B. Let x be an arbitrary element of P(A). Then x ⊆ A. Suppose y ∈ x. Then since x ⊆ A, y ∈ A, and since A ⊆ B, y ∈ B. Since y was arbitrary, this shows that x ⊆ B, so x ∈ P(B).

And since x was arbitrary, this shows that P(A) ⊆ P(B). (←) Suppose P(A) ⊆ P(B). Clearly A ⊆ A, so A ∈ P(A). Since P(A) ⊆ P(B), it follows that A ∈ P(B), and therefore A ⊆ B. 9. Suppose that x and y are odd. Then we can choose integers j and k such that x = 2j +1 and y = 2k +1.

Therefore xy = (2j + 1)(2k + 1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1. Since 2jk + j + k is an integer, it follows that xy is odd.

10. Suppose that x and y are odd. Then we can choose integers j and k such that x = 2j +1 and y = 2k +1.

Therefore x − y = (2j + 1) − (2k + 1) = 2(j − k). Since j − k is an integer, x − y is even.

11. Let n be an arbitrary integer. (→) We will prove the contrapositive. Suppose n is odd. Then we can choose some integer k such that n = 2k + 1. Therefore n3 = (2k + 1)3 = 8k 3 + 12k 2 + 6k + 1 = 2(4k 3 + 6k 2 + 3k) + 1, which is odd, since 4k 3 + 6k 2 + 3k is an integer. Therefore, if n3 is even then n is even. (←) Suppose n is even. Then we can choose some integer k such that n = 2k. Therefore n3 = 8k 3 = 2(4k 3 ), so n3 is even.

12. (a) The mistake is in the sentence “Similarly, since n is odd we have n = 2k + 1.” The letter k already stands for something (it was introduced in the previous sentence), so we need to use a different letter. The sentence should be “Similarly, since n is odd, we can choose some integer j such that n = 2j + 1.” (b) The theorem is incorrect. One counterexample is m = 2, n = 5.

13. Let x ∈ R be arbitrary. (→) Suppose ∃y ∈ R(x + y = xy). Then we can let y0 be some real number such that x + y0 = xy0 .

Suppose x = 1. Then 1 + y0 = 1 · y0 = y0 , so 1 = 0, which is a contradiction. Therefore x 6= 1. (←) Suppose x 6= 1. Let y = x/(x − 1), which is defined since x 6= 1. Then x+y =x+

x x2 − x x x2 x = + = =x· = xy. x−1 x−1 x−1 x−1 x−1

28

SOLUTIONS

14. Let z = 1. Let x ∈ R+ be arbitrary. (→) Suppose ∃y ∈ R(y−x = y/x). Then we can let y0 be some real number such that y0 −x = y0 /x.

Suppose x = z = 1. Then y0 − 1 = y0 /1 = y0 , so −1 = 0, which is a contradiction. Therefore x 6= z. (←) Suppose x 6= z = 1. Let y = x2 /(x − 1), which is defined since x 6= 1. Then x2

x2 x2 − x x y x2 −x= − = = x−1 = . y−x= x−1 x−1 x−1 x−1 x x S 15. Suppose x ∈ {A \ B | A ∈ F}. Then we can choose some A ∈ F such that x ∈ A \ B, so x ∈ A and x ∈ / B. Since x ∈ A and x ∈ / B, A * B, so / P(B). Since A ∈ F and A ∈ / P(B), SA ∈ A∈F \ P(B).

Since x ∈ A and A ∈ F \ P(B), x ∈ (F \ P(B)).

Since x was arbitrary, we conclude S S that {A \ B | A ∈ F} ⊆ (F \ P(B)).

S T S 16. Suppose G are not disjoint. Then we can choose some x such that x ∈ F and T that F and S x ∈ G. Since x ∈ F, we can choose some A ∈ F such that x ∈ A. Since we are given that every element of F is disjoint from some element of G, there must T be some B ∈ G such that A ∩ B = ∅.

Since x ∈ A, it follows that x ∈ / B. ButSwe also T have x ∈ G and B ∈ G, from which it follows that x ∈ B, which is a contradiction. Thus, F and G must be disjoint.

17. Let A ∈ P(A). Since x ∈ A and A ∈ P(A), S be an arbitrary set. Suppose x ∈ A. Clearly A ⊆ A, so AS x ∈ P(A). Since xSwas arbitrary, we can conclude that A ⊆ P(A).

Now suppose x ∈ P(A). Then we can choose some B ∈ P(A) such that x ∈ B. Since S B ∈ P(A), B ⊆ A. Since x ∈ B and B ⊆ S A, x ∈ A. Since x was arbitrary, we can conclude that P(A) ⊆ A. S S We have proven that A ⊆ P(A) and P(A) ⊆ A. It follows that A = P(A).

S 18. (a) Suppose x ∈ (F ∩ G). Then we can choose some AS∈ F ∩ G such that x ∈ A. Since A ∈ F ∩ G, A∈S F and A ∈ G. Since xS∈ A and S A ∈ F, x ∈ F, and similarly since x ∈ ASand A ∈ G, xS∈ G. STherefore, x ∈ ( F) ∩ ( G). Since x was arbitrary, this shows that (F ∩ G) ⊆ ( F) ∩ ( G). (b) The sentence “Thus, we can choose a set A such that A ∈ F, A ∈ G, and x ∈ A” is incorrect. It should say “Thus, we can choose a set A1 ∈ F such that x ∈ A1 , and we can choose a set A2 ∈ G such that x ∈ A2 .” We cannot assume the sets A1 and A2 are the same, so we have to give them different names. (c) One example is F = {{1}, {2}}, G = {{1}, {1, 2}}.

S S S 19. (→) Suppose ( F) ∩ ( G) ⊆ (F ∩ G). Let A be an arbitrary element of F and let B be an S arbitrary element of G. Suppose x ∈ A ∩ B. Then x ∈ A and x ∈ B. Since x ∈ A and A ∈ F, x ∈ S S S S S S F, and since x ∈ B and B S ∈ G, x ∈ G. Therefore x ∈ ( F) ∩ ( G), and since ( F) ∩ ( S G) ⊆ (F ∩ G), it follows that x ∈ (F ∩ G). Since x was S arbitrary, we can conclude that A ∩ B ⊆ (FS∩ G). S (←) Suppose ∀A ∈ F∀B ∈ G(A ∩ B ⊆ (F ∩ G)). Let x be an arbitrary element of ( F) ∩ ( G).

S S Then x ∈ F and x S ∈ G, so we can choose sets A ∈ F and B ∈ G such that x ∈ ASand x ∈ B. By assumption, A ∩ B ⊆ (F ∩ G). S Since x ∈SA andS x ∈ B, x ∈ A ∩ B, and therefore x ∈ (F ∩ G). Since x was arbitrary, it follows that ( F) ∩ ( G) ⊆ (F ∩ G).

S S 20. (→) Suppose F and G are disjoint. x ∈ A ∩ B. Then S Let A ∈ F and B ∈ G be arbitrary. Suppose S x ∈ A and A ∈ F, and therefore x ∈ F, and also x ∈ B and B ∈ G, so x ∈ G. But this contradicts S S the fact that F and G are disjoint. Therefore there can be no elements in A ∩ B, so A and B are disjoint.

S S (←) Assume that for S S all A ∈ F and B ∈ G, A and B are disjoint. Suppose x ∈ ( F) ∩ ( G).

Then x ∈ F and x ∈ G, so we can choose A ∈ F and B ∈ G such that S x ∈ A and S x ∈ B. But then A and B are not disjoint, which contradicts our assumption. Therefore F and G are disjoint.

S S S S S 21. (a) Suppose x ∈ ( F) \ ( G). Then x ∈ F and x ∈ / S G. Since x ∈ F, we can choose some A ∈ F such that x ∈ A. If A ∈ G then since x ∈ A, x ∈ G, which is a contradiction. Therefore

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S AS∈ / G, so F \ G. Since x ∈ A and A ∈ F \ G, x ∈ (F \ G). Sionce x was arbitrary, SA ∈ S ( F) \ ( G) ⊆ (F \ G).

S S (b) The mistake is in the sentence “Since x ∈ A and A ∈ / G, x ∈ / G.” The statement x ∈ / G means ¬∃A(A ∈ G ∧ x ∈ A), but what has been proven is ∃A(A ∈ / G ∧ x ∈ A), which does not mean the same thing.

S S S (c) (→) Suppose (F \ G) ⊆ ( F) \ ( G). Let A ∈ F \ G and B ∈ G be arbitrary. SSuppose A ∩ BS6= ∅. Then we some S x ∈ A ∩SB. Since x ∈SA and A ∈ F S can choose S S \ G, x ∈ (F \ G).

Since (F \ G)S⊆ ( F) \ ( G), x ∈ ( F) \ ( G), so x ∈ F and x ∈ / G. But x ∈ B and B ∈ G, so x ∈ G, which is a contradiction. Therefore A and B must be disjoint, as required.

S (←) Suppose ∀A ∈ (F \ G)∀B ∈ G(A ∩ B = ∅). Let x be an arbitrary element of (F \ G).

Then we can chooseSsome A ∈ F \ G such that S x ∈ A. Since A ∈ F \ G, A ∈ F, and since x ∈ A, it follows that x ∈ F. Now suppose x ∈ G. Then we can choose some set B ∈ G such that x ∈ B. Since A ∈ F \ G and B ∈ G, by so we have S our assumption S A∩B = ∅ S But x ∈SA ∩ B, S reached a contradiction Therefore x ∈ / S G. Since x ∈ F and x ∈ / G, x ∈ ( F) \ ( G). Since S S x was arbitrary, we can conclude that (F \ G) ⊆ ( F) \ ( G). (d) One example is F = {{1, 2}, {3}}, G = {{1}, {2, 3}}.

S S S S S 22. Suppose that F * G. Then there is some x ∈ F such that x ∈ / G. Since x ∈ F, we can choose some A ∈ F such that x ∈ A. Now S let B ∈ G be arbitrary. If A ⊆ B, then since x ∈ A, x ∈ B. But then since x ∈ B and B ∈ G, x ∈ G, which we already know is false. Therefore A * B. Since B was arbitrary, this shows that for all B ∈ G, A * B. Thus, we have shown that there is some A ∈ F such that for all B ∈ G, A * B. S S S 23. (a) The S proof proves that B ∩ ( i∈I Ai ) = i∈I (B ∩ Ai ) by proving that ∀x(x ∈ B ∩ ( i∈I Ai ) ↔ x ∈ i∈I (B ∩ Ai )), which is proven by letting x be arbitrary and then proving both directions of the biconditional. For both directions, the crucial step is the use of existential instantiation to choose i0 .

S S (b) Let x be arbitrary. Suppose x ∈ S B \ ( i∈I Ai ). Then x ∈ B and x ∈ / i∈I Ai . Now let i ∈ I be arbitrary. If x ∈ Ai then x ∈ i∈I Ai , which is a contradiction. Therefore x ∈ / Ai . Since we also know that x ∈ B, we T can conclude that x ∈ B \ Ai . Since i was arbitrary, this proves that ∀i ∈ I(x ∈ B \ Ai ), soTx ∈ i∈I (B \ Ai ).

Now T suppose x ∈ i∈I (B \ Ai ). Since I 6= ∅, we can S choose some i0 such that i0 ∈ I. Since x ∈ i∈I (B \ Ai ), x ∈ B \ Ai0 , so T x ∈ B. Suppose x ∈ i∈I Ai . Then we can choose some i1 ∈ I such that x ∈ Ai1 . But since x ∈ S i∈I (B \ Ai ) we also know that / Ai1 . This S x ∈ B \ Ai1 , soSx ∈ is a contradiction. Therefore x ∈ / i∈I A .

Since x ∈ B and x ∈ / A , x ∈ B \ ( i i i∈ i∈I S T S T Ai ).

We have proven that ∀x(x ∈ B\( i∈I Ai ) ↔ x ∈ i∈I (B\Ai )), so B\( i∈I Ai ) = i∈I (B\Ai ).

T S T (c) B \ ( i∈ITAi ) = i∈I (B \ Ai ).T Proof: Let x be arbitrary. Suppose x ∈ B \ ( i∈I Ai ). Then x ∈ B and x ∈ / i∈I Ai . Since x ∈ / i∈I Ai , we S can choose some i0 ∈ I such that x ∈ / Ai0 . Since x ∈ B and x ∈ / Ai 0 , x ∈ B \ S Ai0 . Therefore x ∈ i∈I (B \ Ai ).

Now suppose x ∈ i∈I (B \ Ai ). Then we can choose T some i0 ∈ I such that x ∈ B \ AiT . Since 0 x ∈ B\A , x ∈ B and x ∈ / A .

Since x ∈ / A , x ∈ / A .

Since x ∈ B and x ∈ / i i i i 0 0 0 i∈I i∈I Ai , T x ∈ B \ ( i∈I Ai ).

T S T S We have shown that ∀x(x ∈ B \ ( i∈I Ai ) ↔ x ∈ i∈I (B \ Ai ). Therefore B \ ( i∈I Ai ) = i∈I (B \ Ai ).

S 24. (a) Suppose x ∈ i∈l (Ai \ Bi ). Then we can choose Ai \ Bi , which S some i ∈ I such that x ∈ T means x ∈ A and x ∈ / B .

Since x ∈ A , x ∈ A , and since x ∈ / B , x ∈ / i i i i i i∈l S T S S T i∈l Bi . Thus, x ∈ ( i∈l Ai ) \ ( i∈l Bi ). Since x was arbitrary, i∈I (Ai \ Bi ) ⊆ ( i∈I Ai ) \ ( i∈I Bi ). (b) One example is I = {1, 2}, A1 = B1 = {1}, A2 = B2 = {2}.

S 25. (a) Suppose x ∈ i∈I (Ai ∩ Bi ). Then we can choose some i ∈ I such that x ∈ Ai ∩ Bi , which means

30

SOLUTIONS S S that S x ∈ Ai and S x ∈ Bi . Since x ∈ Ai , x ∈ i∈I S Ai , and since x ∈SBi , x ∈ i∈I S Bi . Therefore x ∈ ( i∈I Ai ) ∩ ( i∈I Bi ). Since x was arbitrary, i∈I (Ai ∩ Bi ) ⊆ ( i∈I Ai ) ∩ ( i∈I Bi ). (b) One example is I = {1, 2}, A1 = {1}, A2 = {2}, B1 = {2}, B2 = {1}.

26. Let a and b be arbitrary integers. Let c = ab. Then c is an integer, a | c, and b | c.

27. (a) Let n be an arbitrary integer. (→) Suppose 15 | n. Then we can choose some integer k such that n = 15k. Therefore n = 3(5k), so 3 | n, and also n = 5(3k), so 5 | k. (←) Suppose 3 | n and 5 | n. Then we can choose integers j and k such that n = 3j = 5k.

Therefore 15(2k − j) = 30k − 15j = 6(5k) − 5(3j) = 6n − 5n = n, so 15 | n. (b) Let n = 30. Then 6 | n and 10 | n, but 60 - n.

Section 3.5 1. Suppose x ∈ A ∩ (B ∪ C). Then x ∈ A, and either x ∈ B or x ∈ C. Case 1. x ∈ B. Then since x ∈ A, x ∈ A ∩ B, so x ∈ (A ∩ B) ∪ C. Case 2. x ∈ C. Then clearly x ∈ (A ∩ B) ∪ C. Since x was arbitrary, we can conclude that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C. 2. Suppose x ∈ (A ∪ B) \ C. Then either x ∈ A or x ∈ B, and also x ∈ / C. Case 1. x ∈ A. Then x ∈ A ∪ (B \ C).

Case 2. x ∈ B. Then since x ∈ / C, x ∈ B \ C, so x ∈ A ∪ (B \ C).

Since x was arbitrary, we can conclude that (A ∪ B) \ C ⊆ A ∪ (B \ C).

3. Suppose x ∈ A \ (A \ B). Then x ∈ A and x ∈ / A \ B. Since x ∈ / A \ B, either x ∈ / A or x ∈ B. But it cannot be the case that x ∈ / A, since we know x ∈ A, so x ∈ B. Since x ∈ A and x ∈ B, x ∈ A ∩ B. This shows that A \ (A \ B) ⊆ A ∩ B. Now suppose x ∈ A ∩ B. Then x ∈ A and x ∈ B. Since x ∈ B, x ∈ / A \ B. Since x ∈ A and x∈ / A \ B, x ∈ A \ (A \ B). Thus A ∩ B ⊆ A \ (A \ B).

We have shown that A \ (A \ B) ⊆ A ∩ B and A ∩ B ⊆ A \ (A \ B), so A \ (A \ B) = A ∩ B. 4. Let x be arbitrary. Then x ∈ A \ (B \ C) iff x ∈ A ∧ ¬(x ∈ B ∧ x ∈ / C)

(definition of \)

iff x ∈ A ∧ (x ∈ / B ∨ x ∈ C)

(De Morgan’s law)

iff (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ C)

(distributive law)

iff x ∈ (A \ B) ∪ (A ∩ C)

(definitions of \, ∪, and ∩)..

Therefore A \ (B \ C) = (A \ B) ∪ (A ∩ C).

5. Suppose x ∈ A. We now consider two cases: Case 1. x ∈ C. Then x ∈ A ∩ C, so since A ∩ C ⊆ B ∩ C, x ∈ B ∩ C, and therefore x ∈ B. Case 2. x ∈ / C. Since x ∈ A, x ∈ A ∪ C, so since A ∪ C ⊆ B ∪ C, x ∈ B ∪ C. But x ∈ / C, so we must have x ∈ B. Thus, x ∈ B, and since x was arbitrary, A ⊆ B. 6. Suppose A 4 B ⊆ A. Suppose B * A. Then we can choose some x ∈ B such that x ∈ / A. Since x ∈ B and x ∈ / A, x ∈ B \ A, so x ∈ (A \ B) ∪ (B \ A) = A 4 B. But now we have x ∈ A 4 B and x ∈ / A, which contradicts the fact that A 4 B ⊆ A. Therefore B ⊆ A. 7. (→) Suppose A ∪ C ⊆ B ∪ C. Suppose x ∈ A \ C, which means x ∈ A and x ∈ / C. Since x ∈ A, x ∈ A ∪ C, and since A ∪ C ⊆ B ∪ C, x ∈ B ∪ C. This means either x ∈ B or x ∈ C. But we also know x ∈ / C, so x ∈ B. Since x ∈ B and x ∈ / C, x ∈ B \ C. Since x was arbitrary, we conclude that A \ C ⊆ B \ C.

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(←) Suppose A \ C ⊆ B \ C. Suppose x ∈ A ∪ C. Case 1. x ∈ C. Then x ∈ B ∪ C. Case 2. x ∈ / C. Since x ∈ A ∪ C and x ∈ / C, x ∈ A. Since x ∈ A and x ∈ / C, x ∈ A \ C. Since A \ C ⊆ B \ C, it follows that x ∈ B \ C. Therefore x ∈ B, so x ∈ B ∪ C. Since these cases are exhaustive, we have proven that x ∈ B ∪ C. Since x was arbitrary, we can conclude that A ∪ C ⊆ B ∪ C. 8. Suppose x ∈ P(A) ∪ P(B). Then either x ∈ P(A) or x ∈ P(B).

Case 1. x ∈ P(A). Then x ⊆ A. Let y be an arbitrary element of x. Since x ⊆ A, y ∈ A, so y ∈ A ∪ B. Since y was arbitrary, x ⊆ A ∪ B, so x ∈ P(A ∪ B).

Case 2. x ∈ P(B). A similar argument shows x ∈ P(A ∪ B).

Since x was arbitrary, we have proven that P(A) ∪ P(B) ⊆ P(A ∪ B).

9. Suppose P(A) ∪ P(B) = P(A ∪ B). Clearly A ∪ B ⊆ A ∪ B, so A ∪ B ∈ P(A ∪ B). Since P(A) ∪ P(B) = P(A ∪ B), it follows that A ∪ B ∈ P(A) ∪ P(B), so either A ∪ B ∈ P(A) or A ∪ B ∈ P(B).

Case 1. A ∪ B ∈ P(A). Then A ∪ B ⊆ A. Suppose x ∈ B. Then x ∈ A ∪ B, so since A ∪ B ⊆ A, x ∈ A. Since x was arbitrary, B ⊆ A. Case 2. A ∪ B ∈ P(B). A similar argument shows that A ⊆ B. Since these cases are exhaustive, we conclude that either A ⊆ B or B ⊆ A. 10. (→) Suppose y + 1/x = 1 + y/x. Then y − 1 = y/x − 1/x = (y − 1)/x. If y 6= 1 then y − 1 6= 0, so we can divide both sides of this equation by y − 1 and conclude that 1 = 1/x, so x = 1. Thus, either x = 1 or y = 1. (←) Suppose that either x = 1 or y = 1.

Case 1. x = 1. Then y + 1/x = y + 1 = 1 + y/x.

Case 2. y = 1. Then y + 1/x = 1 + 1/x = 1 + y/x.

11. Let x be an arbitrary real number and assume that |x − 3| > 3.

Case 1. x − 3 ≥ 0. Then |x − 3| = x − 3, so since |x − 3| > 3, x − 3 > 3, and therefore x > 6.

Multiplying by x, which is positive, we conclude that x2 > 6x.

Case 2. x − 3 < 0. Then |x − 3| = −(x − 3) = 3 − x, so since |x − 3| > 3, 3 − x > 3, so 3 > x + 3 and therefore 0 > x. Thus x2 ≥ 0 > 6x.

12. Let x be an arbitrary real number. (←) Suppose |x − 4| > 2.

Case 1. x − 4 ≥ 0. Then |x − 4| = x − 4, so we have x − 4 > 2, and therefore x > 6. Adding x to both sides gives us 2x > 6 + x, so 2x − 6 > x. Since x > 6, this implies that 2x − 6 is positive, so |2x − 6| = 2x − 6 > x.

Case 2. x − 4 < 0. Then |x − 4| = 4 − x, so we have 4 − x > 2, and therefore x < 2. Therefore 3x < 6, and subtracting 2x from both sides we get x < 6 − 2x. Also, from x < 2 we get 2x < 4, so 2x − 6 < −2. Therefore 2x − 6 is negative, so |2x − 6| = 6 − 2x > x. (→) Suppose |2x − 6| > x.

Case 1. 2x − 6 ≥ 0. Then |2x − 6| = 2x − 6, so 2x − 6 > x, and therefore x > 6. Since x > 6, x − 4 > 2 > 0, so |x − 4| = x − 4 > 2.

Case 2. 2x − 6 < 0. Then |2x − 6| = 6 − 2x, so 6 − 2x > x, and therefore 6 > 3x and x < 2. Thus x − 4 < −2 < 0, so |x − 4| = 4 − x > 2.

13. (a) Let a and b be arbitrary real numbers. (→) Suppose |a| ≤ b.

Case 1. a ≥ 0. Then |a| = a, so a ≤ b. Also, since b ≥ a and a ≥ 0, b ≥ 0, so −b ≤ 0.

Therefore −b ≤ 0 ≤ a ≤ b, so −b ≤ a ≤ b.

Case 2. a < 0. Then |a| = −a, so −a ≤ b, and therefore a ≥ −b. Also, since −b ≤ a and a < 0, −b < 0, so b > 0. Therefore −b ≤ a < 0 < b, so −b ≤ a ≤ b.

32

SOLUTIONS (←) Suppose −b ≤ a ≤ b. Then a ≤ b and −b ≤ a, so b ≥ −a.

Case 1. a ≥ 0. Then |a| = a ≤ b.

Case 2. a < 0. Then |a| = −a ≤ b. (b) Let x be an arbitrary real number. Clearly |x| ≤ |x|, and by part (a) it follows that −|x| ≤ x ≤ |x|. (c) Let x and y be arbitrary real numbers. By part (b), we have −|x| ≤ x ≤ |x| and −|y| ≤ y ≤ |y|.

Adding these two inequalities we get −(|x| + |y|) ≤ x + y ≤ |x| + |y|. By part (a), it follows that |x + y| ≤ |x| + |y|. (d) Let x and y be arbitrary real numbers. By part (c), |x| = |(x + y) + (−y)| ≤ |x + y| + |−y| = |x + y| + |y|. Therefore |x + y| ≥ |x| − |y|.

14. Let x be an arbitrary integer.

Case 1. x is even. Then we can choose some integer k such that x = 2k. Therefore x2 + x = (2k)2 + 2k = 4k 2 + 2k = 2(2k 2 + k), so x2 + x is even.

Case 2. x is odd. Then we can choose some integer k such that x = 2k + 1. Therefore x2 + x = (2k + 1)2 + (2k + 1) = 4k 2 + 4k + 1 + 2k + 1 = 4k 2 + 6k + 2 = 2(2k 2 + 3k + 1), so x2 + x is even.

Since there cases are exhaustive, we conclude that x2 + x is even.

15. Let x be an arbitrary integer.

Case 1. x is even. Then we can choose some integer k such that x = 2k. Therefore x4 = (2k)4 = 16k 4 = 8(2k 4 ), so the quotient when x4 is divided by 8 is 2k 4 , and the remainder is 0.

Case 2. x is odd. Then we can choose some integer k such that x = 2k + 1. Therefore x4 = (2k + 1)4 = 16k 4 + 32k 3 + 24k 2 + 8k + 1 = 8(2k 4 + 4k 3 + 3k 2 + k) + 1, so the quotient when x4 is divided by 8 is 2k 4 + 4k 3 + 3k 2 + k and the remainder is 1.

Thus, the remainder when x4 is divided by 8 is either 0 or 1.

S 16. (a) Suppose x ∈ (F ∪ G). Then we can choose some A ∈ F ∪ G such that x ∈ A. Since A ∈ F ∪ G, either A ∈ F or A ∈ G. S S S Case 1. A ∈ F. Since x ∈ A and A ∈ F, x ∈S F, so x ∈ S ( F) ∪ S ( G).

Case 2. A ∈SG. Since S x ∈ A and A ∈ G, x ∈ G, so x ∈ ( F) ∪ ( G).

Thus, x ∈ ( F) ∪ ( G).

S S S S Now supposeSthat x ∈ ( F) ∪ ( G). Then either x ∈ F or x ∈ G. Case 1. x ∈ F. Then we can S choose some A ∈ F such that x ∈ A. Since A ∈ F, A ∈ F ∪ G, so since x ∈ A, it follows that x ∈ (F ∪ G).

S S Case 2. x ∈ G. A similar argument shows that x ∈ (F ∪ G).

S Thus, x ∈ (F ∪ G).

T T T T (b) (F ∪ G) = ( F) ∩ ( G). Proof: Suppose x ∈ (F ∪ G). Let A be an arbitrary element of F. T Since A ∈ F, A ∈ FT∪ G, so since x ∈ (F ∪ G), x ∈ A. SinceTA was an arbitrary element of F, T T this shows that x ∈ T F. A similar argument shows that x ∈ G, so x ∈ ( F) ∩ ( G).

T T T Now suppose x ∈ ( F) ∩ ( G). Then x ∈ F and x ∈ G. Let A be an arbitrary element of F ∪ G. Then either A ∈ F or T A ∈ G. Case 1. A ∈ F. Since x ∈ T F and A ∈ F, x ∈ A. Case 2. A ∈ G. Since x ∈ G and A ∈ G, x ∈ A. T Thus x ∈ A, and since A was an arbitrary element of F ∪ G, this prove that x ∈ (F ∪ G).

S S 17. (a) Suppose x ∈ B ∪ ( F). Then either x ∈ B or x ∈ F.S Case 1. x ∈ B. S Since x ∈ B and B ∈ F ∪ {B}, x ∈ (F ∪ {B}).

Case 2. x ∈ S F. Then we can choose is some A ∈ F such that x ∈ A. Since x ∈ A and A ∈ F ∪ {B}, S x ∈ (F ∪ {B}).

Thus, x ∈ (F ∪ S {B}).

Now suppose x ∈ (F ∪ {B}). Then we can choose some A ∈ F ∪ {B} such that x ∈ A. Since A ∈ F ∪ {B}, either A ∈ F or A = B. S Case 1. A ∈ F. Then since x ∈ A and A ∈ F, x ∈ F.

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Case 2. A = B. Then since x ∈ A, x ∈ B. S S Thus, either x ∈ F or x ∈ B, so x ∈ B ∪ ( F).

T T (b) Suppose x ∈ B ∪ ( F). Then either x ∈ B or x ∈ F. Case 1. x ∈ B. Let A ∈ F be arbitrary. Since xT∈ B, x ∈ B ∪ A. Since A was arbitrary, we can conclude that (B ∪ A).

T for all A ∈ F, x ∈ B ∪ A, so x ∈ A∈FT Case 2. x ∈ F. Let A ∈ F be arbitrary. Since x ∈ F, x ∈TA, so x ∈ B ∪ A. Since A was arbitrary, we can T conclude that for all A ∈ F, x ∈ B ∪ A, so x ∈ A∈F (B ∪ A).

Thus, x ∈ A∈F (B T ∪ A).

T Now suppose x ∈ A∈F (B ∪ A). If / B. T x ∈ B, then of course x ∈ B ∪ ( F). Now suppose x ∈ Let A ∈ F be arbitrary. Since x ∈ A∈F (B ∪ A), x ∈ B ∪ A, soTeither x ∈ B or T x ∈ A. But x∈ / B, so x ∈ A. Since A was arbitrary, we can conclude that x ∈ F, so x ∈ B ∪ ( F).

S S T T (c) B ∩ ( F) = A∈F S(B ∩ A) and B ∩ ( F) = S(F ∪ {B}). Proof: S To prove the first equation, suppose x ∈ B ∩ ( F). Then x ∈ B and x ∈ F. Since x ∈ F, we can choose some A ∈ F suchSthat x ∈ A. Since x ∈ B and x ∈ A, x ∈ B ∩ A. And since A ∈ F, it follows that x ∈ A∈F (B ∩ A). S Now suppose x ∈ A∈F (B ∩ A). Then we can A, so S choose some A ∈ F suchSthat x ∈ B ∩ S x ∈ B and x ∈ A. Since x ∈ A and S A ∈ F, S x ∈ F. Since x ∈ B and x ∈ F, x ∈ B ∩ ( F).

This completes the proof that B ∩ ( F) = A∈F (B T ∩ A).

T To prove the second equation, suppose x ∈ B ∩ ( F). Then x ∈ B and x ∈ F. Let A be an arbitrary element of F ∪ {B}. Then T either A ∈ F or A = B. Case 1. A ∈ F. Then since x ∈ F, x ∈ A. Case 2. A = B. Then since x ∈ B, x ∈ A. Since T these cases are exhaustive, we conclude that x ∈ A. And since A was arbitrary, it follows that x ∈ (F ∪ {B]}).T Now suppose x ∈ (F ∪ {B}). Since B ∈ F ∪ {B}, it follows that x ∈ B. Now let A be T an arbitrary element of F. Then A ∈ F ∪ {B}, so since x ∈ (F ∪ {B}), T T T x ∈ A. Since A was arbitrary, this proves that x ∈ F. Since x ∈ B and x ∈ F, x ∈ B ∩ ( F).

T T T T T 18. Suppose x ∈ H. If x ∈ F, then of course x ∈ ( F) ∪ ( G). Now suppose x ∈ / F. Then we can choose some A ∈ F such that T x∈ / A. Now let B ∈ G be arbitrary. Since A ∈ F and B ∈ G, by assumption A ∪ B ∈ H. Since x ∈ H, it follows that x ∈ A ∪ B, so either ∈ B. But T x ∈ A or x T T we know x ∈ / A, so x ∈ B. Since B was arbitrary, we can conclude that x ∈ G. Thus x ∈ ( F) ∪ ( G).

T T T Since x was arbitrary, we have proven that H ⊆ ( F) ∪ ( G).

19. Let x be arbitrary. (→) Suppose x ∈ A 4 B. Then x ∈ (A \ B) ∪ (B \ A), so either x ∈ A \ B or x ∈ B \ A. We now must prove x ∈ A ↔ x ∈ / B, so we will prove both directions of this biconditional. (→) Suppose x ∈ A. Then x ∈ / B \ A, so it must be the case that x ∈ A \ B, and therefore x ∈ / B. (←) Suppose x ∈ / B. Then x ∈ / B \ A, so it must be the case that x ∈ A \ B, and therefore x ∈ A. This completes the → direction of the proof. (←) Suppose x ∈ A ↔ x ∈ / B. We now consider two cases.

Case 1. x ∈ A. Then since x ∈ A ↔ x ∈ / B, x ∈ / B. Therefore x ∈ A \ B, so x ∈ A 4 B. Case 2. x ∈ / A. Then since x ∈ A ↔ x ∈ / B, x ∈ B. Therefore x ∈ B \ A, so x ∈ A 4 B. Since these cases are exhaustive, x ∈ A 4 B. 20. (→) Suppose that A 4 B and C are disjoint. Let x be an arbitrary element of A ∩ C. Then x ∈ A and x ∈ C. If x ∈ / B, then since x ∈ A, x ∈ A \ B, and therefore x ∈ A 4 B. But also x ∈ C, so this contradicts our assumption that A 4 B and C are disjoint. Therefore x ∈ B. Since we also know x ∈ C, we have x ∈ B ∩ C. Since x was an arbitrary element of A ∩ C, this shows that A ∩ C ⊆ B ∩ C. A similar argument shows that B ∩ C ⊆ A ∩ C. (←) Suppose that A ∩ C = B ∩ C. Suppose that A 4 B and C are not disjoint. Then we can choose

34

SOLUTIONS some x such that x ∈ A 4 B and x ∈ C. Since x ∈ A 4 B, either x ∈ A \ B or x ∈ B \ A. Case 1. x ∈ A \ B. Then x ∈ A and x ∈ / B. Since we also know x ∈ C, we can conclude that x ∈ A ∩ C but x ∈ / B ∩ C. This contradicts the fact that A ∩ C = B ∩ C. Case 2. x ∈ B \ A. Similarly, this leads to a contradiction.

Thus we can conclude that A 4 B and C are disjoint.

21. (→) Suppose A 4 B ⊆ C. Let x be an arbitrary element of A ∪ C. Suppose x ∈ / B ∪ C. Then x ∈ /B and x ∈ / C. Since x ∈ A ∪ C and x ∈ / C, x ∈ A. But then since x ∈ A and x ∈ / B, x ∈ A \ B, so x ∈ A 4 B. Since A 4 B ⊆ C, x ∈ C, which contradicts the fact that x ∈ / C. Therefore x ∈ B ∪ C. Since x was an arbitrary element of A ∪ C, this shows that A ∪ C ⊆ B ∪ C. A similar proof shows that B ∪ C ⊆ A ∪ C, so A ∪ C = B ∪ C. (←) Suppose A ∪ C = B ∪ C. Let x be an arbitrary element of A 4 B. Then x ∈ (A \ B) ∪ (B \ A), so either x ∈ A \ B or x ∈ B \ A. Case 1. x ∈ A \ B. Then x ∈ A and x ∈ / B. Since x ∈ A, x ∈ A ∪ C, so since A ∪ C = B ∪ C, x ∈ B ∪ C. But x ∈ / B, so x ∈ C. Case 2. x ∈ B \ A. A similar argument shows that x ∈ C. Since x was an arbitrary element of A 4 B, we can conclude that A 4 B ⊆ C. 22. (→) Suppose C ⊆ A 4 B. Let x be an arbitrary element of C. Then since C ⊆ A 4 B, x ∈ A 4 B = (A \ B) ∪ (B \ A), so either x ∈ A and x ∈ / B or x ∈ B and x ∈ / A. Thus either x ∈ A or x ∈ B, so x ∈ A ∪ B. Since x was arbitrary, this shows that C ⊆ A ∪ B. Now suppose A ∩ B ∩ C 6= ∅. Then we can choose some x such that x ∈ A, x ∈ B, and x ∈ C. Since x ∈ C and C ⊆ A 4 B, x ∈ A 4 B, so as before either x ∈ A and x ∈ / B or x ∈ B and x ∈ / A. So either x ∈ / A or x ∈ / B, which contradicts the fact that x ∈ A and x ∈ B. Therefore A ∩ B ∩ C = ∅. (←) Suppose C ⊆ A ∪ B and A ∩ B ∩ C = ∅. Let x ∈ C be arbitrary. Then since C ⊆ A ∪ B, x ∈ A ∪ B, so either x ∈ A or x ∈ B. Case 1. x ∈ A. If x ∈ B then x ∈ A ∩ B ∩ C, which contradicts the fact that A ∩ B ∩ C = ∅.

Therefore x ∈ / B, so x ∈ A \ B and therefore x ∈ A 4 B. Case 2. x ∈ B. Then a similar argument shows that x ∈ B \ A, so x ∈ A 4 B. Since x was arbitrary, this shows that C ⊆ A 4 B. 23. (a) Suppose x ∈ A \ C. Then x ∈ A and x ∈ / C. Case 1. x ∈ B. Then since x ∈ / C, x ∈ B \ C, so x ∈ (A \ B) ∪ (B \ C).

Case 2. x ∈ / B. Then since x ∈ A, x ∈ A \ B, so x ∈ (A \ B) ∪ (B \ C).

Since x was arbitrary, we conclude that A \ C ⊆ (A \ B) ∪ (B \ C). (b) Suppose x ∈ A 4 C. Then either x ∈ A \ C or x ∈ C \ A. Case 1. x ∈ A \ C. Then by part (a), x ∈ (A \ B) ∪ (B \ C), so either x ∈ A \ B or x ∈ B \ C. But if x ∈ A \ B then x ∈ A 4 B and if x ∈ B \ C then x ∈ B 4 C, so either x ∈ A 4 B or x ∈ B 4 C. Therefore x ∈ (A 4 B) ∪ (B 4 C).

Case 2. x ∈ C \ A. A similar argument shows that either x ∈ B \ A or x ∈ C \ B, so x ∈ (A 4 B) ∪ (B 4 C).

Since x was arbitrary, A 4 C ⊆ (A 4 B) ∪ (B 4 C).

24. (a) Suppose x ∈ (A ∪ B) 4 C. Then either x ∈ (A ∪ B) \ C or x ∈ C \ (A ∪ B).

Case 1. x ∈ (A ∪ B) \ C. Then either x ∈ A or x ∈ B, and x ∈ / C. We now break case 1 into two subcases, depending on whether x ∈ A or x ∈ B: Case 1a. x ∈ A. Then x ∈ A \ C, so x ∈ A 4 C, so x ∈ (A 4 C) ∪ (B 4 C).

Case 1b. x ∈ B. Similarly, x ∈ B 4 C, so x ∈ (A 4 C) ∪ (B 4 C).

Case 2. x ∈ C \ (A ∪ B). Then x ∈ C, x ∈ / A, and x ∈ / B. It follows that x ∈ A 4 C and x ∈ B 4 C, so certainly x ∈ (A 4 C) ∪ (B 4 C). (b) Here is one example: A = {1}, B = {2}, C = {1, 2}.

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25. (a) Suppose x ∈ (A 4 C) ∩ (B 4 C). Then either x ∈ A \ C or x ∈ C \ A, and also either x ∈ B \ C or x ∈ C \ B. Case 1. x ∈ C. Then x ∈ / A \ C, so x ∈ C \ A, which implies that x ∈ / A. Since x ∈ / A, x∈ / A ∩ B. Therefore x ∈ C \ (A ∩ B), so x ∈ (A ∩ B) 4 C. Case 2. x ∈ / C. Then x ∈ / C \ A, so x ∈ A \ C, which implies that x ∈ A. Similarly, x ∈ / C \ B, so x ∈ B \ C and therefore x ∈ B. Since x ∈ A, x ∈ B, and x ∈ / C, x ∈ (A ∩ B) \ C, so x ∈ (A ∩ B) 4 C. Since x was arbitrary, this shows that (A 4 C) ∩ (B 4 C) ⊆ (A ∩ B) 4 C. (b) No. One counterexample is A = {1}, B = {2}, C = {1, 2}.

26. (A 4 C) \ (B 4 C) ⊆ (A \ B) 4 C, but the two sets are not always equal. To prove the subset statement, suppose x ∈ (A 4 C) \ (B 4 C). Then x ∈ A 4 C and x ∈ / B 4 C. Case 1. x ∈ C. Then since x ∈ A 4 C, x ∈ / A. Therefore x ∈ / A \ B, so x ∈ (A \ B) 4 C. Case 2. x ∈ / C. Then since x ∈ A 4 C, x ∈ A, and since x ∈ / B 4 C, x ∈ / B. Therefore x ∈ A \ B, so x ∈ (A \ B) 4 C. Since x was an arbitrary element of (A 4 C) \ (B 4 C), (A 4 C) \ (B 4 C) ⊆ (A \ B) 4 C. To show that the sets need not be equal, we just need a counterexample. One counterexample is A = {1}, B = {1}, C = {1}.

27. The proof is incorrect, because it only establishes that either 0 < x or x < 6, but what must be proven is that 0 < x and x < 6. However, it can be fixed. In case 1 we have x − 3 ≥ 0, so x ≥ 3 > 0. Since the proof for this case already establishes x < 6, we have 0 < x < 6. In case 2 we have x − 3 < 0, so x < 3 < 6, and the proof already shows 0 < x, so 0 < x < 6. Thus, the theorem is correct.

28. The proof is correct. When the statement A * C is reexpressed as a positive statement it is an existential statement, so the proof uses existential instantiation to introduce x. Since A \ B ⊆ C, x ∈ A \ B → x ∈ C. The proof uses modus tollens to conclude x ∈ / A \ B, and when this statement is reexpressed as a positive statement it becomes a disjunction. The proof then uses our last strategy for givens of the form P ∨ Q to conclude x ∈ B. Finally, when the goal A ∩ B 6= ∅ is expressed as a positive statement it is an existential statement, and the proof establishes this by showing that x is an instance of the existential statement.

29. The proof is correct. Since the goal is a universal statement, the proof starts by introducing x as an arbitrary object. Then the proof proceeds by cases. In each case, the existential statement ∃y ∈ R(xy 2 6= y − x) is proven by giving an example of the required real number y.

30. Suppose ∀xP (x) → ∃xQ(x).

Case 1. ∀xP (x). Then by our assumption, ∃xQ(x), so we can choose some object a such that Q(a).

Since Q(a) is true, P (a) → Q(a). Therefore ∃x(P (x) → Q(x)).

Case 2. ¬∀xP (x). Then ∃x¬P (x), so we can choose some a such that ¬P (a). Since P (a) is false, P (a) → Q(a). Therefore ∃x(P (x) → Q(x)).

31. The proof is incorrect. In case 1, we cannot conclude that for all x ∈ A, x ∈ B. All we can say is that for all x ∈ A that fall under case 1, x ∈ B. Similarly, in case 2 we cannot conclude that ∀x ∈ A(x ∈ C).

One way to understand the problem is to think about the structure of the proof. The proof begins with “Let x be an arbitrary element of A,” but this beginning is inappropriate because the goal is not a universal statement. The goal is the disjunction A ⊆ B ∨ A ⊆ C, so it would be appropriate to use proof by cases, proving A ⊆ B in one case and A ⊆ C in the other. But to use that strategy the proof should be structured like this: Case 1.

Let x be an arbitrary element of A. [Proof of x ∈ B goes here.]

Since x was arbitrary, we can conclude that A ⊆ B.

36

SOLUTIONS Case 2.

Let x be an arbitrary element of A. [Proof of x ∈ C goes here.]

Since x was arbitrary, we can conclude that A ⊆ C. Since these cases are exhaustive, either A ⊆ B or A ⊆ C. Notice that the arbitrary objects called x are introduced within each case. Therefore the division into cases cannot be based on a property of x, since x has not yet been introduced when the proof is broken into cases.

The proof cannot be fixed, because the theorem is false. Here is a counterexample to the theorem: A = {1, 2}, B = {1}, C = {2}.

32. Suppose A * B. Then we can choose some a ∈ A such that a ∈ / B. Since a ∈ A and A ⊆ B ∪ C, a ∈ B ∪ C, so either a ∈ B or x ∈ C. But we also know a ∈ / B, so a ∈ C. Since a ∈ A and a ∈ C, A ∩ C 6= ∅. Thus, either A ⊆ B or A ∩ C 6= ∅.

33. Case 1. ∀yP (y). Since we have assumed the universe of discourse is not the empty set, we can let a be some element of the universe of discourse. Then since ∀yP (y) is true, P (a) → ∀yP (y), so ∃x(P (x) → ∀yP (y)).

Case 2. ¬∀yP (y). Then ∃x¬P (x), so we can choose some object a such that P (a) is false. Therefore P (a) → ∀yP (y) is true, so ∃x(P (x) → ∀yP (y)).

Section 3.6 1. Let x be an arbitrary real number. Let y = x/(x2 + 1). Then x−y =x−

x x2 + 1

=

x3 + x x x3 x − 2 = 2 = x2 · 2 = x2 y.

2 x +1 x +1 x +1 x +1

To see that y is unique, suppose that x2 z = x − z. Then z(x2 + 1) = x, and since x2 + 1 6= 0, we can divide both sides by x2 + 1 to conclude that z = x/(x2 + 1) = y.

2. Let x = 4. Let y be an arbitrary real number. Then xy + x − 4 = 4y + 4 − 4 = 4y.

To prove that x is unique, suppose z is a real number such that for every real number y, zy +z −4 = 4y. Then in particular, setting y = 0 we get z − 4 = 0, so z = 4 = x.

3. Let x be an arbitrary real number, and assume that x 6= 0 and x 6= 1. Let y = x2 /(x − 1), which is defined since x 6= 1. Then x2

x2 x2 x2 − x x y y−x= −x= − = = x−1 = . x−1 x−1 x−1 x−1 x x To see that y is unique, suppose that z is a real number such that z/x = z − x. Multiplying both sides by x, we conclude that z = zx − x2 , so x2 = z(x − 1) and therefore z = x2 /(x − 1) = y.

4. Suppose x 6= 0. Let y = 1/x. Now let z be an arbitrary real number. Then zy = z(1/x) = z/x, as required.

To see that y is unique, suppose that y 0 is a number with the property that ∀z ∈ R(zy 0 = z/x).

Then in particular, taking z = 1, we have y 0 = 1/x, so y 0 = y.

S 5. (a) Suppose x ∈ !F. Then ∃!A(A ∈ F ∧ x ∈SA), so we can let A be the S unique S set such that A ∈ F and x ∈ A. Since x ∈ A and A ∈ F, x ∈ F. Since x was arbitrary, !F ⊆ F. (b) Let F be an arbitrary family of sets.

S S (→) Suppose !F = F. Let A and B be arbitrary elements of F and assume that A∩B 6= ∅.

Then we can choose some x such that x ∈ A ∩ B, so x ∈ A and x ∈ B. Since x ∈ A and A ∈ F,

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S S S S x ∈ F. But !F = F, so we conclude that x ∈ !F, and therefore there is only one set in F that contains x as an element. Since x ∈ A and x ∈ B, this implies that A = B. Since A and B were arbitrary, we conclude that F is pairwise disjoint.

S S (←) Suppose S F isSpairwise disjoint. We know from partS(a) that !F ⊆ F, so we only need to show that F ⊆ !F. To prove this, suppose that x ∈ F. Then we can choose some A ∈ F such that x ∈ A. To see that A is unique, suppose B ∈ F and x ∈ B. Then since xS ∈ A and x ∈ B, A ∩ B 6= ∅. Since F is pairwise disjoint, it follows that B = A. Therefore x ∈ !F. 6. (a) Let A = ∅ ∈ P(U ). Then clearly for any B ∈ P(U ), A ∪ B = ∅ ∪ B = B. To see that A is unique, suppose that A0 ∈ P(U ) and for all B ∈ P(U ), A0 ∪ B = B. Then in particular, taking B = ∅, we can conclude that A0 ∪ ∅ = ∅. But clearly A0 ∪ ∅ = A0 , so we have A0 = ∅ = A. (b) Let A = U . Let B ∈ P(U ) be arbitrary. Then B ⊆ U = A, so A ∪ B = A. To see that A is unique, suppose A0 ∈ P(U ) and ∀B ∈ P(U )(A0 ∪ B = A0 ). In particular, since U ∈ P(U ), A0 ∪ U = A0 . But A0 ∈ P(U ), so A0 ⊆ U and therefore A0 ∪ U = U . Thus A0 = U = A. 7. (a) Let A = U . Let B ∈ P(U ) be arbitrary. Then B ⊆ U = A, so A ∩ B = B. To see that A is unique, suppose A0 ∈ P(U ) and ∀B ∈ P(U )(A0 ∩ B = B). In particular, since U ∈ P(U ), A0 ∩ U = U . But A0 ∈ P(U ), so A0 ⊆ U and therefore A0 ∩ U = A0 . Thus A0 = U = A. (b) Let A = ∅. Let B ∈ P(U ) be arbitrary. Then A ∩ B = ∅ ∩ B = ∅ = A. To see that A is unique, suppose A0 ∈ P(U ) and ∀B ∈ P(U )(A0 ∩ B = A0 ). Then in particular, since ∅ ∈ P(U ), A0 ∩ ∅ = A0 . But A0 ∩ ∅ = ∅, so A0 = ∅ = A. 8. (a) Suppose A ∈ P(U ). Let B = U \ A. Let C ∈ P(U ) be arbitrary. Then C ⊆ U . We claim now that C \ A = C ∩ B. To prove this, suppose that x ∈ C \ A. Then x ∈ C and x ∈ / A. Since x ∈ C and C ⊆ U , x ∈ U . Since x ∈ U and x ∈ / A, x ∈ U \ A = B. Therefore x ∈ C ∩ B. Now suppose x ∈ C ∩ B. Then x ∈ C and x ∈ B = U \ A, so x ∈ / A. Therefore x ∈ C \ A. This completes the proof that C \ A = C ∩ B. Finally, we must prove that B is unique. So suppose that B 0 ∈ P(U ) and ∀C ∈ P(U )(C \A = C ∩ B 0 ). In particular, U \ A = U ∩ B 0 . But B 0 ⊆ U , so U ∩ B 0 = B 0 . Therefore B 0 = U \ A = B. (b) Suppose A ∈ P(U ). Let B = U \ A. Let C ∈ P(U ) be arbitrary. Then C ⊆ U . We claim now that C ∩ A = C \ B. To prove this, suppose that x ∈ C ∩ A. Then x ∈ C and x ∈ A. Since x ∈ A, x∈ / U \ A = B, so x ∈ C \ B. Now suppose x ∈ C \ B. Then x ∈ C and x ∈ / B = U \ A, so either x ∈ / U or x ∈ A. But since x ∈ C and C ⊆ U , x ∈ U . Therefore x ∈ A, so x ∈ C ∩ A. Finally, we must prove that B is unique. So suppose that B 0 ∈ P(U ) and ∀C ∈ P(U )(C ∩A = C \ B 0 ). Then in particular, U ∩ A = U \ B 0 . Since A ⊆ U , U ∩ A = A, so U \ B 0 = A. We claim now that B 0 = U \ A. To prove this, suppose x ∈ B 0 . Then x ∈ / U \ B 0 = A. Also, since x ∈ B 0 0 and B ⊆ U , x ∈ U . Thus x ∈ U \ A. Next, suppose x ∈ U \ A. Then x ∈ U and x ∈ / A = U \ B0, 0 0 0 so either x ∈ / U or x ∈ B . Since x ∈ U , x ∈ B . This proves that B = U \ A = B. 9. (a) Existence: Let X = ∅. Then for every set A, A 4 X = (A \ ∅) ∪ (∅ \ A) = A ∪ ∅ = A. Uniqueness: Suppose X1 and X2 are both identity elements for symmetric difference. Then ∀A(A 4 X1 = A) and ∀A(A 4 X2 = A). Applying the first statement to X2 and the second to X1 we conclude that X2 4 X1 = X2 and X1 4 X2 = X1 . But X1 4 X2 = X2 4 X1 , so X1 = X2 . (b) Let A be an arbitrary set.

Existence: Let B = A. Then A 4 B = (A \ A) ∪ (A \ A) = ∅ ∪ ∅ = ∅.

Uniqueness: Suppose B1 and B2 are both inverses of A. Then A 4 B1 = A 4 B2 = ∅, so B1 = B1 4 ∅ = B1 4 (A 4 B2 ) = (B1 4 A) 4 B2 = (A 4 B1 ) 4 B2 = ∅ 4 B2 = B2 4 ∅ = B2 .

38

SOLUTIONS (c) Let A and B be arbitrary sets.

Existence: Let C = A 4 B. Then A 4 C = A 4 (A 4 B) = (A 4 A) 4 B = ∅ 4 B = B. Uniqueness: Suppose C1 and C2 are sets such that A 4 C1 = A 4 C2 = B. Then C1 = ∅ 4 C1 = (A 4 A) 4 C1 = A 4 (A 4 C1 ) = A 4 (A 4 C2 ) = (A 4 A) 4 C2 = ∅ 4 C2 = C2 . (d) Let A be an arbitrary set. Let B = A. Suppose C ⊆ A. Then B 4 C = (A \ C) ∪ (C \ A) = (A \ C) ∪ ∅ = A \ C.

To see that B is unique, suppose B 0 ⊆ A and ∀C ⊆ A(B 0 4 C = A \ C). Applying this statement with C = ∅ we get B 0 4 ∅ = A \ ∅ = A. But also B 0 4 ∅ = B 0 , so B 0 = A = B. S 10. Suppose first that A = ∅. Let F = ∅. Then F = ∅ = A, but A ∈ / F, which is a contradiction.

Therefore A 6= ∅, so A has at least one element.

Let x be an element of A. S Now let F = {{x}, A \ {x}}. Then F = {x} ∪ (A \ {x}) = A, so by assumption A ∈ F, and therefore either A = {x} or A = A \ {x}. But clearly A 6= A \ {x}, because x ∈ A but x ∈ / A \ {x}.

Therefore A = {x}, so A has exactly one element.

S S 11. Existence: We are given that for every G ⊆ F, G ∈ F, so in particular, S S since F ⊆ F, F ∈ F. Let A = F. Now suppose B ∈ F. Then by exercise 8 of Section 3.3, B ⊆ F = A, as required.

Uniqueness: Suppose that A1 ∈ F, A2 ∈ F, ∀B ∈ F(B ⊆ A1 ), and ∀B ∈ F(B ⊆ A2 ). Applying this last fact with B = A1 we can conclude that A1 ⊆ A2 , and similarly the previous fact implies that A2 ⊆ A1 . Thus A1 = A2 .

12. (a) ∃x∃y[P (x) ∧ P (y) ∧ x 6= y ∧ ∀z(P (z) → (z = x ∨ z = y))]. (b) To prove that there are exactly two values of x for which P (x) is true, find two objects a and b such that you can prove P (a), P (b), a 6= b, and ∀z(P (z) → (z = a ∨ z = b)). (c) First, note that 0 and 1 are clearly solutions to the equation, and of course 0 6= 1. Now suppose z is also a solution, so z 3 = z 2 . If z 6= 0 then we can divide both sides of this equation by z 2 to conclude that z = 1. Therefore either z = 0 or z = 1.

13. (a) Let c = 9/4. Then the equation is x2 + 3x + 9/4 = 0, or equivalently (x + 3/2)2 = 0. The unique solution to this equation is x = −3/2.

To see that c is unique, suppose c0 6= 9/4, and consider the equation x2 + 3x + c0 = 0.

Case 1. c0 > 9/4. Then the equation can be rewritten (x + 3/2)2 + (c0 − 9/4) = 0, and this equation has no solutions because for all values of x, (x + 3/2)2 ≥ 0 and c0 − 9/4 > 0.

0 Case has the two solutions x = √ 2. c < 9/4. Then by the quadratic formula the equation (−3 ± 9 − 4c0 )/2, and these solutions are distinct because 9 − 4c0 > 0.

Thus, c = 9/4 is the only value of c for which the equation x2 + 3x + c = 0 has exactly one solution. (b) For every real number x, there is a unique value of c for which x2 +3x+c = 0, namely c = −x2 −3x.

Thus, there is more than one real number x such that there is a unique real number c such that x2 + 3x + c = 0, so it is not the case that there is a unique such real number x.

Section 3.7 S 1. Existence: Let A = F. To prove (a), let X be an arbitrary element of F. By exercise 8 in Section 3.3, S X ⊆ F = A, so X ∈ P(A). Since X was arbitrary, this shows that F ⊆ P(A). ToSprove (b), let B be an arbitrary set and assume that F ⊆ P(B). By exercise 16 in Section 3.3, A = F ⊆ B.

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Uniqueness: Suppose A1 and A2 are sets satisfying (a) and (b). In other words, (a1) F ⊆ P(A1 ), (b1) ∀B(F ⊆ P(B) → A1 ⊆ B), (a2) F ⊆ P(A2 ), and (b2) ∀B(F ⊆ P(B) → A2 ⊆ B). By (a1), we can plug in A1 for B in (b2) to conclude that A2 ⊆ A1 . Similarly, using (a2) and (b1) we get A1 ⊆ A2 , so A1 = A2 .

2. Existence: Let m = 1. To prove (a), let x be an arbitrary positive real number. Then x < x + 1, so x/(x + 1) < 1 = m. To prove (b), suppose y is a positive real number and for every positive real number x, x/(x + 1) < y. Suppose y < 1. Let x = y/(1 − y). Then x is a positive real number, so x/(x + 1) < y, but also y y y x 1−y 1−y = y = 1 = = y. x+1 1 1−y + 1 1−y This is a contradiction. Therefore m = 1 ≤ y.

Uniqueness: Suppose m1 and m2 are positive real numbers with properties (a) and (b). In other words, (a1) for every positive real number x, x/(x + 1) < m1 , (b1) if y is any positive real number with the property that for every positive real number x, x/(x + 1) < y, then m1 ≤ y, (a2) for every positive real number x, x/(x + 1) < m2 , and (b2) if y is any positive real number with the property that for every positive real number x, x/(x + 1) < y, then m2 ≤ y. By (a1) and (b2), m2 ≤ m1 , and by (a2) and (b1), m1 ≤ m2 . Therefore m1 = m2 .

3. P(A \ B) \ (P(A) \ P(B)) = {∅}. Proof: Since ∅ is a subset of every set, ∅ ∈ P(B), so ∅ ∈ / P(A) \ P(B), and also ∅ ∈ P(A \ B). Therefore ∅ ∈ P(A \ B) \ (P(A) \ P(B)), so {∅} ⊆ P(A \ B) \ (P(A) \ P(B)). For the other direction, suppose X ∈ P(A \ B) \ (P(A) \ P(B)). Then X ∈ P(A \ B), so X ⊆ A \ B, and X ∈ / P(A) \ P(B). Clearly A \ B ⊆ A, so since X ⊆ A \ B, X ⊆ A, and therefore X ∈ P(A). Since X ∈ / P(A) \ P(B), it follows that X ∈ P(B), so X ⊆ B. We claim now that X = ∅. To see why, suppose x ∈ X. Then since X ⊆ A \ B, x ∈ / B, and since X ⊆ B, x ∈ B, which is a contradiction. Therefore X cannot have any elements, so X = ∅. This proves that P(A \ B) \ (P(A) \ P(B)) ⊆ {∅}, so the two sets are equal.

4. (a) → (b). Suppose (a) is true. Suppose x ∈ A ∩ B, which means x ∈ A and x ∈ B. Suppose x ∈ / C. Then since x ∈ A and x ∈ / C, x ∈ A \ C, so x ∈ A 4 C. Similarly, x ∈ B 4 C, so this contradicts (a).

Therefore x ∈ C. Since x was an arbitrary element of A ∩ B, this shows that A ∩ B ⊆ C. Now suppose x ∈ C. Suppose x ∈ / A ∪ B. Then x ∈ / A and x ∈ / B. Since x ∈ C and x ∈ / A, x ∈ A 4 C. Similarly x ∈ B 4 C, so this contradicts (a). Therefore x ∈ A ∪ B. Since x was an arbitrary element of C, we conclude that C ⊆ A ∪ B. (b) → (c). Suppose (b) is true. Suppose x ∈ A 4 C. Then either x ∈ A \ C or x ∈ C \ A. Case 1. x ∈ A \ C. Then x ∈ A and x ∈ / C. If x ∈ B then x ∈ A ∩ B, so by (b), x ∈ C, which is a contradiction. Therefore x ∈ / B. Since x ∈ A and x ∈ / B, x ∈ A 4 B. Case 2. x ∈ C \ A. Then x ∈ C and x ∈ / A. By (b), since x ∈ C, x ∈ A ∪ B. But then since x ∈ / A, x ∈ B. Since x ∈ B and x ∈ / A, x ∈ A 4 B. Thus, x ∈ A 4 B. Since x was an arbitrary element of A 4 C, this shows that A 4 C ⊆ A 4 B. (c) → (a). Suppose (c) is true. Suppose (A 4 C) ∩ (B 4 C) 6= ∅. Then we can choose some x ∈ (A 4 C) ∩ (B 4 C), so x ∈ A 4 C and x ∈ B 4 C. By (c), since x ∈ A 4 C, x ∈ A 4 B. Case 1. x ∈ C. Since x ∈ A 4 C and x ∈ B 4 C, it follows that x ∈ / A and x ∈ / B, which contradicts x ∈ A 4 B. Case 2. x ∈ / C. Since x ∈ A 4 C and x ∈ B 4 C, it follows that x ∈ A and x ∈ B, which contradicts x ∈ A 4 B. Thus we have reached a contradiction, so (A 4 C) ∩ (B 4 C) = ∅.

S S S S S S 5. Suppose P( S i∈l Ai )S⊆ i∈l P(Ai ). It is clear that i∈l Ai ⊆ i∈l Ai , so i∈l Ai ∈ P( i∈l Ai ) and therefore i∈l Ai ∈ i∈l P(A S i ). By the definition of the union of a family, this means that we can chooseSsome i ∈ I such that i∈l Ai ⊆ Ai . Now let j ∈ I be arbitrary. Then by exercise 8 in Section 3.3, Aj ⊆ i∈l Ai , so Aj ⊆ Ai .

40

SOLUTIONS S 6. (a) Suppose S x ∈ i∈I Ai . Then we can choose some i0 ∈ I such that x ∈ Ai0 . Since i0S∈ I and I = F, we can choose some X0 ∈ F S such that S i0 ∈ X0 . Since x ∈ Ai0 and i0 ∈ X0 , x ∈ i∈X0 Ai , and since X0 ∈ F it follows that x ∈ ( X∈F i∈X Ai ).

S S S Next, suppose x ∈ X∈F ( i∈I Ai ). Then we can choose some X0 ∈ F such that x ∈ i∈X0 Ai , which S implies that we can choose i0 ∈ X0 such S that x ∈ Ai0 . Since i0 ∈ X0 and X0 ∈ F, i0 ∈ F = I. Since x ∈ Ai0 , it follows that x ∈ i∈I Ai .

T (b) Suppose x ∈S i∈I Ai . Let X ∈ FTbe arbitrary. Let i ∈ X be arbitrary. Then since i ∈ X and X ∈ F, i ∈ F =TI. Since x ∈ i∈I Ai , it follows that x ∈ Ai . Since T i wasTarbitrary, we can conclude that x ∈ i∈X A .

Since X was arbitrary, it follows that x ∈ i X∈F T T S( i∈X Ai ).

Now suppose x ∈ X∈F ( i∈X Ai ). Let i ∈ I beTarbitrary.

Since I = F, this means that T T we can choose X0 ∈ F such that i ∈ X0 . Since x ∈ X∈F ( i∈X Ai ) and X0 ∈ F, x ∈ i∈X0 Ai .

T But then since i ∈ X0 , x ∈ Ai . Since i was arbitrary, we can conclude that x ∈ i∈I Ai .

S (c) Suppose x ∈ i∈J T Ai . Then we can choose some j ∈ J such that x ∈ Aj . Let S X ∈ F be arbitrary.

Since j ∈ J = F and X ∈ T F, j ∈ S X. Since j ∈ X and x ∈ Aj , x ∈ i∈X Ai . Since X was arbitrary, this shows that x ∈ X∈F ( i∈X Ai ).

The two sets are not always equal. Here’s an example in which they are not equal: F = {{1, 2}, {2, 3}}, A1 = {a}, A2 = {b}, A3 = {a}.

S T T S T (d) X∈F ( i∈X Ai ) ⊆ i∈JTAi . Proof: Suppose x ∈ X∈F ( i∈X Ai ). Then weT can choose some X0 ∈ F such that x ∈ i∈X0 Ai . Now T let j ∈ J be arbitrary. Since J = F and X0 ∈ F, this implies that j T∈ X0 . Since x ∈ i∈X0 Ai , it follows that x ∈ Aj . Since j was arbitrary, we conclude that x ∈ i∈J Ai .

The two sets need not be equal. The example from part (c) shows this.

7. Suppose > 0. Let δ = /3, which is also clearly positive. Let x be an arbitrary real number, and suppose that 0 < |x − 2| < δ. Since |x − 2| > 0, x 6= 2, so (3x2 − 12)/(x − 2) is defined, and

3(x + 2)(x − 2) 3x2 − 12 − 12 = − 12 = |3(x + 2) − 12| = |3x − 6| = 3|x − 2| < 3δ = 3 = . x−2 x−2 3 8. Suppose that limx→c f (x) = L > 0. Let = L. Then by the definition of limit, we can choose some δ > 0 such that for all x, if 0 < |x − c| < δ then |f (x) − L| < = L. Now let x be an arbitrary real number and suppose 0 < |x − c| < δ. Then |f (x) − L| < L, so −L < f (x) − L < L and therefore 0 < f (x) < 2L. Therefore, for every real number x, if 0 < |x − c| < δ then f (x) > 0.

9. Suppose > 0. Then /7 > 0, so since limx→c f (x) = L, we can choose some δ > 0 such that for all x, if 0 < |x − c| < δ then |f (x) − L| < /7. Now let x be an arbitrary real number and suppose 0 < |x − c| < δ. Then |f (x) − L| < /7, so |7f (x) − 7L| = 7|f (x) − L| < 7( /7) = .

10. The proof is correct. The proof uses proof by cases. Within each case, the proof proves an existential statement by giving a specific instance of the statement that is true.

Chapter 4 Section 4.1 1. (a) {(x, y) ∈ P × P | x is a parent of y} = {(Bill Clinton, Chelsea Clinton), (Goldie Hawn, Kate Hudson), . . . }. (b) {(x, y) ∈ C × U | there is someone who lives in x and attends y}. If you are a university student, then let x be the city you live in, and let y be the university you attend; (x, y) will then be an element of this truth set.

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Figure 1: Truth sets for exercise 3.

2. (a) {(x, y) ∈ P × C | x lives in y}. For example, if you live in some city c, then (you, c) is an element of this truth set. (b) {(x, y) ∈ C × N | the population of x is y} = {(Tokyo, 37,468,302), (Paris, 10,900,952), . . . }.

(Population data from worldpopulationreview.com.)

3. For pictures, see Figure 1. Here are a few elements of each truth set: (a) (−1, 0), (0, −2), (1, −2), (2, 0). (b) (1, 0), (3, 1), (0, −37). (c) (1, −2), (1, 1), (0, −2). (d) (1, −2), (0, −2), (−1, −5).

4. A × (B ∩ C) = (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}, A × (B ∪ C) = (A × B) ∪ (A × C) = {(1, 1), (2, 1), (3, 1), (1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4)}, (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D) = ∅, (A × B) ∪ (C × D) = {(1, 1), (2, 1), (3, 1), (1, 4), (2, 4), (3, 4), (3, 5), (4, 5)}, (A ∪ C) × (B ∪ D) = {(1, 1), (2, 1), (3, 1), (4, 1), (1, 4), (2, 4), (3, 4), (4, 4), (1, 5), (2, 5), (3, 5), (4, 5)}.

5. Proof of 2: For any ordered pair (x, y), (x, y) ∈ A × (B ∪ C) iff x ∈ A ∧ (y ∈ B ∨ y ∈ C)

(definitions of ×, ∪)

iff (x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C)

(distributive law)

iff (x, y) ∈ (A × B) ∪ (A × C)

(definitions of ×, ∪).

Proof of 3: For any ordered pair (x, y), (x, y) ∈ (A × B) ∩ (C × D) iff (x ∈ A ∧ y ∈ B) ∧ (x ∈ C ∧ y ∈ D)

(definitions of ×, ∩)

iff (x ∈ A ∧ x ∈ C) ∧ (y ∈ B ∧ y ∈ D)

(commutative, associative laws)

iff (x, y) ∈ (A ∩ C) × (B ∩ D)

(definitions of ∩, ×).

6. The cases are not exhaustive. For example, it is possible that x ∈ A and y ∈ D. 7. mn. To see why, imagine writing the elements of A × B in a table with m rows and n columns.

8. Yes, it is true. Proof: Suppose (x, y) ∈ A × (B \ C). Then x ∈ A and y ∈ B \ C, so y ∈ B and y ∈ / C. Since x ∈ A and y ∈ B, (x, y) ∈ A×B. Since y ∈ / C, (x, y) ∈ / A×C. Therefore (x, y) ∈ (A×B)\(A×C).

Now suppose (x, y) ∈ (A × B) \ (A × C). Then (x, y) ∈ A × B, so x ∈ A and y ∈ B, and also (x, y) ∈ / A × C, so either x ∈ / A or y ∈ / C. But we already know x ∈ A, so y ∈ / C. Since y ∈ B and y∈ / C, y ∈ B \ C, so (x, y) ∈ A × (B \ C).

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SOLUTIONS 9.

A × (B 4 C) = A × ((B \ C) ∪ (C \ B))

(definition of 4)

= (A × (B \ C)) ∪ (A × (C \ B))

(Theorem 4.1.3, part 2)

= ((A × B) \ (A × C)) ∪ ((A × C) \ (A × B))

(exercise 8)

= (A × B) 4 (A × C)

(definition of 4).

10. Suppose (x, y) ∈ (A \ C) × (B \ D). Then x ∈ A \ C and y ∈ B \ D, which means x ∈ A, x ∈ / C, y ∈ B, and y ∈ / D. Since x ∈ A and y ∈ B, (x, y) ∈ A × B. And since x ∈ / C, (x, y) ∈ / C × D. Therefore (x, y) ∈ (A × B) \ (C × D).

11. Suppose (x, y) ∈ (A × B) \ (C × D). Then (x, y) ∈ A × B, so x ∈ A and y ∈ B, and also (x, y) ∈ / C × D, so either x ∈ / C or y ∈ / D. Case 1. x ∈ / C. Then since x ∈ A and x ∈ / C, x ∈ A \ C. Since we also have y ∈ B, (x, y) ∈ (A \ C) × B, so (x, y) ∈ [A × (B \ D)] ∪ [(A \ C) × B].

Case 2. y ∈ / D. Then since y ∈ B and y ∈ / D, y ∈ B \ D. Since we also have x ∈ A, (x, y) ∈ A × (B \ D), so (x, y) ∈ [A × (B \ D)] ∪ [(A \ C) × B].

Since these cases are exhaustive, (x, y) ∈ [A × (B \ D)] ∪ [(A \ C) × B].

Now suppose (x, y) ∈ [A × (B \ D)] ∪ [(A \ C) × B]. Then either (x, y) ∈ A × (B \ D) or (x, y) ∈ (A \ C) × B. Case 1. (x, y) ∈ A × (B \ D). Then x ∈ A and y ∈ B \ D, so y ∈ B and y ∈ / D. Since x ∈ A and y ∈ B, (x, y) ∈ A × B, and since y ∈ / D, (x, y) ∈ / C × D. Therefore (x, y) ∈ (A × B) \ (C × D).

Case 2. (x, y) ∈ (A \ C) × B. Then x ∈ A \ C, so x ∈ A and x ∈ / C, and also y ∈ B. Since x ∈ A and y ∈ B, (x, y) ∈ A × B, and since x ∈ / C, (x, y) ∈ / C × D. Therefore (x, y) ∈ (A × B) \ (C × D).

Thus (x, y) ∈ (A × B) \ (C × D).

12. Suppose A × B and C × D are disjoint. Suppose A and C are not disjoint, and also B and D are not disjoint. Then we can choose some x ∈ A ∩ C and y ∈ B ∩ D. But then (x, y) ∈ A × B and also (x, y) ∈ C × D, which contradicts the fact that A × B and C × D are disjoint. Therefore either A and C are disjoint or B and D are disjoint.

T 13. Let the (x, y) ∈ ( i∈I Ai ) × B. Then T indexed family {Ai | i ∈ I} and the set B be arbitrary. Suppose T x ∈ i∈I Ai and y ∈ B. Now let i ∈ I be arbitrary.

Since x ∈ i∈I Ai and i ∈ I, x ∈ Ai . Therefore T (x, y) ∈ Ai × B. Since i was arbitrary, (x, y) ∈ (A i × B). i∈I T Now suppose (x, y) ∈ (A × B).

Since I = 6 ∅, we can choose some i0 ∈ I. Since (x, y) ∈ i i∈I T T (A × B), (x, y) ∈ A × B, so y ∈ B. Now let i ∈ I be arbitrary.

Then since (x, y) ∈ (A i i i × B), 0 i∈I i∈I T (x, y) ∈ A × B, so x ∈ A .

Since i was arbitrary, x ∈ A .

Since we also have y ∈ B, (x, y) ∈ i i i i∈I T ( i∈I Ai ) × B. S 14. (a) Suppose (x, y) ∈ i∈I (Ai × Bi ). Then we can choose someSi0 ∈ I such that (x, y) ∈ Ai0 × Bi0 , so xS∈ Ai0 and y ∈ Bi0 . Since S x ∈ Ai0 andSi0 ∈ I, x ∈ i∈I Ai , and similarly since y ∈ Bi0 , y ∈ i∈I Bi . Therefore (x, y) ∈ ( i∈I Ai ) × ( i∈I Bi ).

S (b) Suppose (x, y) ∈ p∈P Cp . Then we can choose some (i0 , i1S ) ∈ P = I × I such S that (x, y) ∈ C(i0 ,i1 ) =SAi0 × Bi1 . S Therefore x ∈ Ai0 and y ∈ Bi1 , so x ∈ i∈I Ai and y ∈ i∈I Bi , and thus (x, y) ∈ ( i∈I Ai ) × ( i∈I B Si ).

S S S Now suppose (x, y) ∈ ( i∈I Ai ) × ( i∈I Bi ). Then x ∈ i∈I Ai and y ∈ i∈I Bi , so we can choose i0 and i1 in I such that S x ∈ Ai0 and y ∈ Bi1 . Therefore (x, y) ∈ Ai0 × Bi1 = C(i0 ,i1 ) and (i0 , i1 ) ∈ I × I = P , so (x, y) ∈ p∈P Cp .

15. The theorem is incorrect. Counterexample: A = {1}, B = C = D = ∅. To understand the mistake in the proof, it is helpful to discuss how the proof should have been structured. After we suppose that A × B ⊆ C × D, the goal is (A ⊆ C) ∧ (B ⊆ D), so the rest of the proof should be structured as follows: Let a be an arbitrary element of A. [Proof that a ∈ C goes here.]

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43

Since a was arbitrary, A ⊆ C. Let b be an arbitrary element of B. [Proof that b ∈ D goes here.]

Since b was arbitrary, B ⊆ D. Therefore A ⊆ C and B ⊆ D. Notice that the proofs that A ⊆ C and B ⊆ D should be separate; each must stand on its own.

The proof violates this by using the arbitrary object b in the proof that A ⊆ C and using the arbitrary object a in the proof that B ⊆ D.

Section 4.2 1. (a) Domain = {p ∈ P | p has a living child}; Range = {p ∈ P | p has a living parent}. (b) Domain = R; Range = R+ .

2. (a) Domain = {p ∈ P | p is male and has a living sibling}; Range = {p ∈ P | p has a male living sibling}. (b) Domain = {x ∈ R | x ≥ 1 or x ≤ −1}; Range = {x ∈ R | −1 < x < 1}.

3. (a) L−1 ◦ L = {(s, t) ∈ S × S | ∃r ∈ R((s, r) ∈ L ∧ (r, t) ∈ L−1 )} = {(s, t) ∈ S × S | ∃r ∈ R((s, r) ∈ L ∧ (t, r) ∈ L)} = {(s, t) ∈ S × S | the students s and t live in the same dorm room}. (b) E ◦ (L−1 ◦ L) = {(s, c) ∈ S × C | ∃t ∈ S((s, t) ∈ L−1 ◦ L ∧ (t, c) ∈ E)} = {(s, c) ∈ S × C | someone who lives in the same dorm room as the student s is taking the course c}.

4. (a) M ◦ E = {(s, d) ∈ S × D | ∃c ∈ C((s, c) ∈ E ∧ (c, d) ∈ M )} = {(s, d) ∈ S × D | the student s is enrolled in some course that meets on the day d}. (b) M ◦ T −1 = {(p, d) ∈ P × D | ∃c ∈ C((p, c) ∈ T −1 ∧ (c, d) ∈ M )} = {(p, d) ∈ P × D | the professor p is teaching some course that meets on the day d}.

5. (a) S ◦ R = {(1, 4), (1, 5), (1, 6), (2, 4), (3, 6)}. (b) S ◦ S −1 = {(4, 4), (5, 5), (5, 6), (6, 5), (6, 6)}.

6. (a) S −1 ◦ R = {(1, 4), (3, 4), (3, 5)}. (b) R−1 ◦ S = {(4, 1), (4, 3), (5, 3)}.

7. (a) First note that Ran(R−1 ) and Dom(R) are both subsets of A. Now let a be an arbitrary element of A. Then a ∈ Ran(R−1 ) iff ∃b ∈ B((b, a) ∈ R−1 ) iff ∃b ∈ B((a, b) ∈ R) iff a ∈ Dom R. (b)

Ran(R−1 ) = Dom((R−1 )−1 ) = Dom(R)

(part 2) (part 1).

(c) Suppose (a, d) ∈ (T ◦S)◦R. Then we can choose some b ∈ B such that (a, b) ∈ R and (b, d) ∈ T ◦S. Since (b, d) ∈ T ◦ S, we can choose some c ∈ C such that (b, c) ∈ S and (c, d) ∈ T . Since (a, b) ∈ R and (b, c) ∈ S, (a, c) ∈ S ◦ R. And since (a, c) ∈ S ◦ R and (c, d) ∈ T , (a, d) ∈ T ◦ (S ◦ R). (d) First note that (S ◦R)−1 and R−1 ◦S −1 are both relations from C to A. Now consider an arbitrary (c, a) ∈ C × A. Then (c, a) ∈ (S ◦ R)−1 iff (a, c) ∈ S ◦ R iff ∃b ∈ B((a, b) ∈ R ∧ (b, c) ∈ S) iff ∃b ∈ B((c, b) ∈ S −1 ∧ (b, a) ∈ R−1 ) iff (c, a) ∈ R−1 ◦ S −1 .

44

SOLUTIONS 8. E ◦ E ⊆ F .

9. (a) Suppose a ∈ Dom(S ◦ R). Then we can choose some c ∈ C such that (a, c) ∈ S ◦ R, which means that we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Since (a, b) ∈ R, a ∈ Dom(R). (b) Suppose Ran(R) ⊆ Dom(S). We already know that Dom(S ◦ R) ⊆ Dom(R), so all we have to prove is that Dom(R) ⊆ Dom(S ◦ R). To prove this, suppose a ∈ Dom(R). Then we can choose some b ∈ B such that (a, b) ∈ R. Since (a, b) ∈ R, b ∈ Ran(R), and since Ran(R) ⊆ Dom(S), it follows that b ∈ Dom(S). Thus, we can choose some c ∈ C such that (b, c) ∈ S. Since (a, b) ∈ R and (b, c) ∈ S, (a, c) ∈ S ◦ R, and therefore a ∈ Dom(S ◦ R). (c) Ran(S ◦ R) ⊆ Ran(S), and if Dom(S) ⊆ Ran(R) then Ran(S ◦ R) = Ran(S). Proof: Suppose c ∈ Ran(S ◦ R). Then we can choose some a ∈ A such that (a, c) ∈ S ◦ R. Therefore we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Since (b, c) ∈ S, c ∈ Ran(S).

Now suppose Dom(S) ⊆ Ran(R), and suppose that c ∈ Ran(S). Then we can choose some b ∈ B such that (b, c) ∈ S. Therefore b ∈ Dom(S), so since Dom(S) ⊆ Ran(R), b ∈ Ran(R), and therefore we can choose some a ∈ A such that (a, b) ∈ R. Since (a, b) ∈ R and (b, c) ∈ S, (a, c) ∈ S ◦ R, so c ∈ Ran(S ◦ R).

10. (a) True. Proof: Suppose (a, b) ∈ R. Then a ∈ Dom(R) and b ∈ Ran(R), so (a, b) ∈ Dom(R) × Ran(R). (b) True. Proof: Suppose R ⊆ S, and suppose (b, a) ∈ R−1 . Then (a, b) ∈ R, so since R ⊆ S, (a, b) ∈ S, and therefore (b, a) ∈ S −1 . (c) True. Proof: Let (b, a) be an arbitrary element of B × A. Then (b, a) ∈ (R ∪ S)−1 iff (a, b) ∈ R ∪ S

(definition of inverse)

iff (a, b) ∈ R ∨ (a, b) ∈ S iff (b, a) ∈ R

−1

∨ (b, a) ∈ S

iff (b, a) ∈ R

−1

−1

∪S

(definition of ∪) −1

(definition of inverse) (definition of ∪).

11. We prove the contrapositives of both directions. (→) Suppose Ran(R) and Dom(S) are not disjoint. Then we can choose some b ∈ Ran(R)∩Dom(S).

Since b ∈ Ran(R), we can choose some a ∈ A such that (a, b) ∈ R. Similarly, since b ∈ Dom(S), we can choose some c ∈ C such that (b, c) ∈ S. But then (a, c) ∈ S ◦ R, so S ◦ R 6= ∅. (←) Suppose S ◦ R 6= ∅. Then we can choose some (a, c) ∈ S ◦ R. By definition of S ◦ R, this means that we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. But then b ∈ Ran(R) and b ∈ Dom(S), so Ran(R) and Dom(S) are not disjoint.

12. (a) Suppose (a, c) ∈ (S ◦ R) \ (T ◦ R). Then (a, c) ∈ S ◦ R and (a, c) ∈ / T ◦ R. Since (a, c) ∈ S ◦ R, we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. If (b, c) ∈ T then (a, c) ∈ T ◦ R, which is a contradiction. Therefore (b, c) ∈ / T , so (b, c) ∈ S \ T . Since (a, b) ∈ R and (b, c) ∈ S \ T , (a, c) ∈ (S \ T ) ◦ R. (b) The mistake is in the sentence “Similarly, since (a, b) ∈ R and (b, c) ∈ / T , (a, c) ∈ / T ◦ R.” We can’t be sure that (a, c) ∈ / T ◦ R, because there could be some b0 ∈ B such that (a, b0 ) ∈ R and (b0 , c) ∈ T . (c) No, it need not be true. Counterexample: A = {1}, B = {2, 3}, C = {4}, R = {(1, 2), (1, 3)}, S = {(2, 4)}, T = {(3, 4)}.

13. (a) True. Proof: Suppose R and S and disjoint, but R−1 and S −1 are not. Then we can choose some (b, a) ∈ B × A such that (b, a) ∈ R−1 and (b, a) ∈ S −1 . But then (a, b) ∈ R and (a, b) ∈ S, which contradicts the fact that R and S are disjoint. Therefore if R and S are disjoint then so are R−1 and S −1 .

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(b) Not necessarily true. Counterexample: A = {1}, B = {2, 3}, C = {4}, R = {(1, 2)}, S = {(1, 3)}, T = {(2, 4), (3, 4)}. (c) Not necessarily true. Counterexample: A = {1}, B = {2, 3}, C = {4}, R = S = {(1, 2)}, T = {(3, 4)}.

14. (a) True. Proof: Suppose S ⊆ T , and suppose (a, c) ∈ S ◦ R. Then we can choose b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Since (b, c) ∈ S and S ⊆ T , (b, c) ∈ T . Since (a, b) ∈ R and (b, c) ∈ T , (a, c) ∈ T ◦ R. (b) True. Proof: Suppose (a, c) ∈ (S ∩ T ) ◦ R. Then we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S ∩ T . Since (b, c) ∈ S ∩ T , (b, c) ∈ S and (b, c) ∈ T . Since (a, b) ∈ R and (b, c) ∈ S, (a, c) ∈ S ◦ R. Similarly, since (a, b) ∈ R and (b, c) ∈ T , (a, c) ∈ T ◦ R. Therefore (a, c) ∈ (S ◦ R) ∩ (T ◦ R). (c) Not necessarily true. Counterexample: A = {1}, B = {2, 3}, C = {4}, R = {(1, 2), (1, 3)}, S = {(2, 4)}, T = {(3, 4)}. (d) True. Proof: Suppose (a, c) ∈ (S ∪ T ) ◦ R. Then we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S ∪ T , which means either (b, c) ∈ S or (b, c) ∈ T .

Case 1. (b, c) ∈ S. Since (a, b) ∈ R and (b, c) ∈ S, (a, c) ∈ S ◦ R, so (a, c) ∈ (S ◦ R) ∪ (T ◦ R).

Case 2. (b, c) ∈ T . Since (a, b) ∈ R and (b, c) ∈ T , (a, c) ∈ T ◦ R, so (a, c) ∈ (S ◦ R) ∪ (T ◦ R).

Therefore (a, c) ∈ (S ◦ R) ∪ (T ◦ R).

Now suppose (a, c) ∈ (S ◦ R) ∪ (T ◦ R). Then either (a, c) ∈ S ◦ R or (a, c) ∈ T ◦ R. Case 1. (a, c) ∈ S ◦ R. Then we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Since (b, c) ∈ S, (b, c) ∈ S ∪ T . Since (a, b) ∈ R and (b, c) ∈ S ∪ T , (a, c) ∈ (S ∪ T ) ◦ R. Case 2. (a, c) ∈ T ◦ R. Then we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ T .

Since (b, c) ∈ T , (b, c) ∈ S ∪ T . Since (a, b) ∈ R and (b, c) ∈ S ∪ T , (a, c) ∈ (S ∪ T ) ◦ R. Therefore (a, c) ∈ (S ∪ T ) ◦ R. 15. Let E = B ∪ C. Then B ⊆ E and C ⊆ E, so R ⊆ A × B ⊆ A × E and S ⊆ C × D ⊆ E × D. Now suppose E1 and E2 are two sets such that R ⊆ A × E1 , S ⊆ E1 × D, R ⊆ A × E2 , and S ⊆ E2 × D. Let T1 = {(a, d) ∈ A × D | ∃e ∈ E1 ((a, e) ∈ R ∧ (e, d) ∈ S)}, T2 = {(a, d) ∈ A × D | ∃e ∈ E2 ((a, e) ∈ R ∧ (e, d) ∈ S)}.

We must prove that T1 = T2 .

Suppose (a, d) ∈ T1 . Then we can choose some e ∈ E1 such that (a, e) ∈ R and (e, d) ∈ S. Since (a, e) ∈ R and R ⊆ A × E2 , (a, e) ∈ A × E2 , and therefore e ∈ E2 . Since e ∈ E2 , (a, e) ∈ R, and (e, d) ∈ S, (a, d) ∈ T2 . A similar argument shows that if (a, d) ∈ T2 then (a, d) ∈ T1 , so T1 = T2 .

Section 4.3 1.

L a b c d e

W bad bed cab

46

SOLUTIONS 2.

bird cat dog rat R is reflexive and symmetric, but not transitive.

3. 1 2 3 4 4. (a) {(a, c), (b, d), (c, c), (d, a), (d, b)}. Not reflexive, symmetric, or transitive. (b) {(a, b), (a, d), (b, a), (b, d)}. Not reflexive, symmetric, or transitive. (c) {(a, a), (b, b), (b, d), (c, c), (d, b), (d, d)}. Reflexive, symmetric, and transitive. (d) {(a, b), (a, c), (a, d), (b, d), (c, d)}. Not reflexive or symmetric, but transitive.

5. S ◦ R = {(a, y), (a, z), (b, x), (c, y), (c, z)}.

6. Dr ◦ Ds = Dr+s . Proof: Suppose (x, y) ∈ Dr ◦ Ds . Then we can choose some real number z such that (x, z) ∈ Ds and (z, y) ∈ Dr . This means that |x − z| < s and |z − y| < r. Therefore by the triangle inequality, |x − y| = |(x − z) + (z − y)| ≤ |x − z| + |z − y| < r + s, so (x, y) ∈ Dr+s .

Now suppose (x, y) ∈ Dr+s . Then |x − y| < r + s. Let z = (rx + sy)/(r + s). Then |x − z| = x −

rx + sy sx − sy s s = = |x − y| < · (r + s) = s, r+s r+s r+s r+s

so (x, z) ∈ Ds , and also |z − y| =

rx + sy rx − ry r r −y = = |x − y| < · (r + s) = r, r+s r+s r+s r+s

so (z, y) ∈ Dr . Therefore (x, y) ∈ Dr ◦ Ds .

7. (→) Suppose R is reflexive. Let (x, y) be an arbitrary element of iA . Then by the definition of iA , x = y ∈ A. Since R is reflexive, (x, y) = (x, x) ∈ R. Since (x, y) was arbitrary, this shows that iA ⊆ R. (←) Suppose iA ⊆ R. Let x ∈ A be arbitrary. Then (x, x) ∈ iA , so since iA ⊆ R, (x, x) ∈ R. Since x was arbitrary, this shows that R is reflexive.

8. (→) Suppose R is transitive. Let (x, y) be an arbitrary element of R ◦ R. Then we can choose some z ∈ A such that (x, z) ∈ R and (z, y) ∈ R. Since R is transitive, it follows that (x, y) ∈ R. Since (x, y) was arbitrary, this shows that R ◦ R ⊆ R. (←) Suppose R ◦ R ⊆ R. Suppose (x, y) ∈ R and (y, z) ∈ R. Then (x, z) ∈ R ◦ R, and since R ◦ R ⊆ R it follows that (x, z) ∈ R. Since x, y, and z were arbitrary, this shows that R is transitive.

9. (a) Suppose R is a relation from A to B. Let (a, b) be an arbitrary element of A × B. Suppose (a, b) ∈ R ◦ iA . Then we can choose some x ∈ A such that (a, x) ∈ iA and (x, b) ∈ R. Since (a, x) ∈ iA , a = x, and therefore (a, b) = (x, b) ∈ R. Now suppose (a, b) ∈ R. Since (a, a) ∈ iA , (a, b) ∈ R ◦ iA . Thus, R ◦ ia = R.

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(b) Suppose R is a relation from A to B. Let (a, b) be an arbitrary element of A × B. Suppose (a, b) ∈ iB ◦ R. Then we can choose some x ∈ B such that (a, x) ∈ R and (x, b) ∈ iB . Since (x, b) ∈ iB , b = x, and therefore (a, b) = (a, x) ∈ R. Now suppose (a, b) ∈ R. Since (b, b) ∈ iB , (a, b) ∈ iB ◦ R. Thus, iB ◦ R = R. 10. Suppose (x, y) ∈ iD . Then x = y ∈ D = Dom(S), so there is some z ∈ A such that (x, z) ∈ S. Therefore (z, x) ∈ S −1 , so (x, y) = (x, x) ∈ S −1 ◦ S. Thus, iD ⊆ S −1 ◦ S. The proof of the other statement is similar.

11. Suppose R is a relation on A and R is reflexive. Suppose (x, y) ∈ R. Since R is reflexive, (x, x) ∈ R. Since (x, x) ∈ R and (x, y) ∈ R, (x, y) ∈ R ◦ R. Thus, R ⊆ R ◦ R. 12. (a) Suppose R is reflexive. Let x ∈ A be arbitrary. Since R is reflexive, (x, x) ∈ R, so (x, x) ∈ R−1 .

Thus, R−1 is reflexive. (b) Suppose R is symmetric. Let x, y ∈ A be arbitrary, and suppose (x, y) ∈ R−1 . Then (y, x) ∈ R. Since R is symmetric, (x, y) ∈ R, so (y, x) ∈ R−1 . Thus, R−1 is symmetric. (c) Suppose R is transitive. Let x, y, z ∈ A be arbitrary, and suppose (x, y) ∈ R−1 and (y, z) ∈ R−1 .

Then (z, y) ∈ R and (y, x) ∈ R. Since R is transitive, it follows that (z, x) ∈ R, so (x, z) ∈ R−1 .

Thus, R−1 is transitive.

13. (a) Yes. Proof: Suppose R1 and R2 are reflexive, and suppose a ∈ A. Since R1 is reflexive, (a, a) ∈ R1 , so (a, a) ∈ R1 ∪ R2 . (b) Yes. Proof: Suppose R1 and R2 are symmetric, and suppose (x, y) ∈ R1 ∪R2 . Then either (x, y) ∈ R1 or (x, y) ∈ R2 . If (x, y) ∈ R1 then since R1 is symmetric, (y, x) ∈ R1 , so (y, x) ∈ R1 ∪ R2 .

Similar reasoning shows that if (x, y) ∈ R2 then (y, x) ∈ R1 ∪ R2 . (c) No. Counterexample: A = {1, 2, 3}, R1 = {(1, 2)}, R2 = {(2, 3)}.

14. (a) Yes. Proof: Suppose R1 and R2 are reflexive, and suppose a ∈ A. Since R1 and R2 are reflexive, (a, a) ∈ R1 and (a, a) ∈ R2 , so (a, a) ∈ R1 ∩ R2 . (b) Yes. Proof: Suppose R1 and R2 are symmetric, and suppose (x, y) ∈ R1 ∩ R2 . Then (x, y) ∈ R1 and (x, y) ∈ R2 . Since R1 and R2 are symmetric, it follows that (y, x) ∈ R1 and (y, x) ∈ R2 , so (y, x) ∈ R1 ∩ R2 . (c) Yes. Proof: Suppose R1 and R2 are transitive, and suppose (x, y) ∈ R1 ∩ R2 and (y, z) ∈ R1 ∩ R2 .

Then (x, y) ∈ R1 and (y, z) ∈ R1 , so since R1 is transitive, (x, z) ∈ R1 . Similarly, (x, y) ∈ R2 and (y, z) ∈ R2 , so since R2 is transitive, (x, z) ∈ R2 . Therefore (x, z) ∈ R1 ∩ R2 .

15. (a) No. Counterexample: A = {1}, R1 = R2 = {(1, 1)}. (b) Yes. Proof: Suppose R1 and R2 are symmetric, and suppose (x, y) ∈ R1 \ R2 . Then (x, y) ∈ R1 and (x, y) ∈ / R2 . Since (x, y) ∈ R1 and R1 is symmetric, (y, x) ∈ R1 . Now suppose (y, x) ∈ R2 .

Then since R2 is symmetric, (x, y) ∈ R2 , which is a contradiction. Therefore (y, x) ∈ / R2 . Since (y, x) ∈ R1 and (y, x) ∈ / R2 , (y, x) ∈ R1 \ R2 . (c) No. Counterexample: A = {1, 2, 3}, R1 = {(1, 2), (2, 3), (1, 3)}, R2 = {(1, 3)}.

16. Suppose a ∈ A. Since R and S are reflexive, (a, a) ∈ R and (a, a) ∈ S, so (a, a) ∈ R ◦ S. 17. First note that by part 2 of Theorem 4.3.4, since R and S are symmetric, R = R−1 and S = S −1 .

Therefore R ◦ S is symmetric iff R ◦ S = (R ◦ S)−1 iff R ◦ S = S

−1

◦R

−1

(Theorem 4.3.4, part 2) (Theorem 4.2.5, part 5)

iff R ◦ S = S ◦ R. 18. Suppose S ◦ R ⊆ R ◦ S. Suppose (x, y) ∈ R ◦ S and (y, z) ∈ R ◦ S. Then we can choose u, v ∈ A such that (x, u) ∈ S, (u, y) ∈ R, (y, v) ∈ S, and (v, z) ∈ R. Since (u, y) ∈ R and (y, v) ∈ S, (u, v) ∈ S ◦ R,

48

SOLUTIONS so since S ◦ R ⊆ R ◦ S, (u, v) ∈ R ◦ S. This means that we can choose some w ∈ A such that (u, w) ∈ S and (w, v) ∈ R. Since (x, u) ∈ S, (u, w) ∈ S, and S is transitive, (x, w) ∈ S. And since (w, v) ∈ R, (v, z) ∈ R, and R is transitive, (w, z) ∈ R. Since (x, w) ∈ S and (w, z) ∈ R, (x, z) ∈ R ◦ S.

19. (a) The mistake is in the sentence “Then by the definition of S, xRy and yRz, where x ∈ X, y ∈ Y , and z ∈ Z.” The correct sentence would be “Then by the definition of S, xRy1 and y2 Rz for some x ∈ X, y1 , y2 ∈ Y , and z ∈ Z.” (b) The theorem is incorrect. Counterexample: A = {1, 2, 3, 4}, R = {(1, 2), (3, 4)}. Let X = {1}, Y = {2, 3}, and Z = {4}. Then (X, Y ) ∈ S and (Y, Z) ∈ S, but (X, Z) ∈ / S. 20. Suppose R is transitive, and suppose (X, Y ) ∈ S and (Y, Z) ∈ S. To prove that (X, Z) ∈ S we must show that ∀x ∈ X∀z ∈ Z(xRz), so let x ∈ X and z ∈ Z be arbitrary. Since Y ∈ B, Y 6= ∅, so we can choose y ∈ Y . Since (X, Y ) ∈ S and (Y, Z) ∈ S, by the definition of S we have xRy and yRz. But then since R is transitive, xRz, as required. The empty set had to be excluded from B so that we could come up with y ∈ Y in this proof.

21. (a) Yes. Proof: Suppose R is reflexive. Let X ∈ P(A) be arbitrary. Let x ∈ X be arbitrary. Since R is reflexive, xRx, so ∃y ∈ X(xRy). Since x was arbitrary, ∀x ∈ X∃y ∈ X(xRy), so (X, X) ∈ S. (b) No. Counterexample: A = {1, 2, 3}, R = {(1, 2), (2, 1)}, which is symmetric. Let X = {1} and Y = {2, 3}. Then (X, Y ) ∈ S but (Y, X) ∈ / S, so S is not symmetric. (c) Yes. Proof: Suppose R is transitive, and suppose (X, Y ) ∈ S and (Y, Z) ∈ S. Let x ∈ X be arbitrary. Then since (X, Y ) ∈ S, we can choose some y ∈ Y such that xRy. Since (Y, Z) ∈ S, we can choose some z ∈ Z such that yRz. By transitivity of R, since xRy and yRz, xRz. Since x was arbitrary, we conclude that ∀x ∈ X∃z ∈ Z(xRz)}, so (X, Z) ∈ S. 22. The proof is incorrect. The mistake is in the sentence “Let y be any element of A such that xRy.” We don’t know that such a y exists. The theorem is incorrect. Counterexample: A = {1}, R = ∅.

23. Suppose aRb and bRc. To prove aRc, suppose that X ⊆ A \ {a, c} and X ∪ {a} ∈ F. We must prove that X ∪ {c} ∈ F. If c = a then X ∪ {c} = X ∪ {a} ∈ F, as required. Now suppose c 6= a. We consider two cases.

Case 1. b ∈ / X. Then X ⊆ A \ {a, b}. Since aRb and X ∪ {a} ∈ F, X ∪ {b} ∈ F. But also bRc and X ⊆ A \ {b, c}, so since X ∪ {b} ∈ F, X ∪ {c} ∈ F. Case 2. b ∈ X. Since a ∈ / X and c ∈ / X, b 6= a and b 6= c. Let X1 = X∪{a}\{b}. Then X1 ⊆ A\{b, c} and X1 ∪ {b} = X ∪ {a} ∈ F, so since bRc, X1 ∪ {c} ∈ F. Note that X1 ∪ {c} = X ∪ {a, c} \ {b}. Let X2 = X ∪ {c} \ {b}. Then X2 ⊆ A \ {a, b} and X2 ∪ {a} = X ∪ {a, c} \ {b} = X1 ∪ {c} ∈ F, so since aRb, X2 ∪ {b} ∈ F. But X2 ∪ {b} = X ∪ {c}, so X ∪ {c} ∈ F. Therefore X ∪ {c} ∈ F. Since X was arbitrary, this proves that aRc. 24. (a) Yes, because for every n ∈ N, |n − n| = 0 ≤ 1, so (n, n) ∈ R. (b) No, because, for example, (−1, −1) ∈ / R.

Section 4.4 1. (a) Partial order, but not total order since (a, c) ∈ / R and (c, a) ∈ / R. (b) Not a partial order, since (−1, 1) ∈ R and (1, −1) ∈ R, but 1 6= −1. (c) Partial order, but not total order since (−1, 1) ∈ / R and (1, −1) ∈ / R. 2. (a) Total order. (This is the order in which words appear in a dictionary.) (b) Not a partial order, because (bat, bed) ∈ R and (bed, bat) ∈ R, but of course bed 6= bat. (c) Probably a total order. If there were two different countries C1 and C2 with exactly the same population, then we would have (C1 , C2 ) ∈ R and (C2 , C1 ) ∈ R but C1 6= C2 , so R would fail to be antisymmetric. But this seems unlikely.

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3. (a) 2 is the smallest element of B and also a minimal element. 3 and 4 are maximal elements, but there is no largest element. 2 is the greatest lower bound. There are no upper bounds, so no least upper bound. (b) 1 is the smallest element of B, and also a minimal element. There are no maximal or largest elements. 1 is the greatest lower bound, and 2 is the least upper bound. (c) ∅ is the smallest element and also a minimal element. Every 5-element set is a maximal element, but there is no largest element. ∅ is the greatest lower bound and N is the least upper bound.

4. (→) Suppose that R is both antisymmetric and symmetric. Suppose that (x, y) ∈ R. Then since R is symmetric, (y, x) ∈ R, and since R is antisymmetric, it follows that x = y. Therefore (x, y) ∈ iA .

Since (x, y) was arbitrary, this shows that R ⊆ iA . (←) Suppose that R ⊆ iA . Suppose (x, y) ∈ R. Then (x, y) ∈ iA , so x = y, and therefore (y, x) = (x, y) ∈ R. This shows that R is symmetric. To see that R is antisymmetric, suppose that (x, y) ∈ R and (y, x) ∈ R Then (x, y) ∈ iA , so x = y.

5. Suppose x ∈ B. Then since B ⊆ A, x ∈ A. Since R is reflexive on A, (x, x) ∈ R. Also, (x, x) ∈ B × B, so (x, x) ∈ R ∩ (B × B). Since x was an arbitrary element of B, this shows that R ∩ (B × B) is reflexive on B. Suppose (x, y) ∈ R ∩ (B × B) and (y, z) ∈ R ∩ (B × B). Then (x, y) ∈ R and (y, z) ∈ R, so since R is transitive, (x, z) ∈ R. Also, (x, y) ∈ B × B and (y, z) ∈ B × B, so x ∈ B and z ∈ B, and therefore (x, z) ∈ B × B. Thus (x, z) ∈ R ∩ (B × B), so R ∩ (B × B) is transitive.

Finally, suppose (x, y) ∈ R ∩ (B × B) and (y, x) ∈ R ∩ (B × B). Then (x, y) ∈ R and (y, x) ∈ R, so since R is antisymmetric, x = y. Therefore R ∩ (B × B) is antisymmetric.

6. (a) Yes. By parts (a) and (c) of exercise 14 in Section 4.3, R1 ∩ R2 is reflexive and transitive. To prove that it is antisymmetric, suppose (x, y) ∈ R1 ∩ R2 and (y, x) ∈ R1 ∩ R2 . Then (x, y) ∈ R1 and (y, x) ∈ R1 , so since R1 is antisymmetric, x = y. (b) No. Counterexample: A = R, R1 = {(x, y) ∈ R × R | x ≤ y}, R2 = {(x, y) ∈ R × R | x ≥ y}.

7. (a) Suppose x ∈ A1 ∪ A2 . Then either x ∈ A1 or x ∈ A2 .

Case 1. x ∈ A1 . Then since R1 is reflexive on A1 , (x, x) ∈ R1 , so (x, x) ∈ R1 ∪ R2 .

Case 2. x ∈ A2 . Then since R2 is reflexive on A2 , (x, x) ∈ R2 , so (x, x) ∈ R1 ∪ R2 .

Thus (x, x) ∈ R1 ∪ R2 , so R1 ∪ R2 is reflexive on A1 ∪ A2 .

Now suppose (x, y) ∈ R1 ∪ R2 and (y, z) ∈ R1 ∪ R2 . Then either (x, y) ∈ R1 or (x, y) ∈ R2 .

Case 1. (x, y) ∈ R1 . Then since R1 ⊆ A1 × A1 , y ∈ A1 . Since A1 and A2 are disjoint, y ∈ / A2 .

Since R2 ⊆ A2 ×A2 , it follows that (y, z) ∈ / R2 , so since (y, z) ∈ R1 ∪R2 , we must have (y, z) ∈ R1 .

Since (x, y) ∈ R1 , (y, z) ∈ R1 , and R1 is transitive, (x, z) ∈ R1 , so (x, z) ∈ R1 ∪ R2 .

Case 2. (x, y) ∈ R2 . Then a similar argument shows that (y, z) ∈ R2 , so by transitivity of R2 , (x, z) ∈ R2 , so (x, z) ∈ R1 ∪ R2 .

Thus (x, z) ∈ R1 ∪ R2 , so R1 ∪ R2 is transitive.

Finally, suppose (x, y) ∈ R1 ∪ R2 and (y, x) ∈ R1 ∪ R2 . Then either (x, y) ∈ R1 or (x, y) ∈ R2 .

Case 1. (x, y) ∈ R1 . As in the proof of transitivity, it follows that (y, x) ∈ / R2 , so (y, x) ∈ R1 .

Therefore, since R1 is antisymmetric, x = y.

Case 2. (x, y) ∈ R2 . Similar to case 1.

Thus, x = y, so R1 ∪ R2 is antisymmetric. (b) The proof that R1 ∪ R2 ∪ (A1 × A2 ) is reflexive is the same as in part (a). For transitivity, suppose (x, y) ∈ R1 ∪ R2 ∪ (A1 × A2 ) and (y, z) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

Case 1. (x, y) ∈ R1 . Then y ∈ A1 , so since A1 and A2 are disjoint, y ∈ / A2 , and therefore (y, z) ∈ / R2 . Thus either (y, z) ∈ R1 or (y, z) ∈ A1 × A2 .

Case 1a. (y, z) ∈ R1 . Then since R1 is transitive, (x, z) ∈ R1 , so (x, z) ∈ R1 ∪R2 ∪(A1 ×A2 ).

Case 1b. (y, z) ∈ A1 × A2 . Then z ∈ A2 . Since (x, y) ∈ R1 , x ∈ A1 , so (x, z) ∈ A1 × A2 , and therefore (x, z) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

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SOLUTIONS Case 2. (x, y) ∈ R2 . Then y ∈ A2 , so since A1 and A2 are disjoint, y ∈ / A1 , and therefore (y, z) ∈ / R1 and (y, z) ∈ / A1 × A2 . Thus (y, z) ∈ R2 . By transitivity of R2 , (x, z) ∈ R2 , so (x, z) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

Case 3. (x, y) ∈ A1 × A2 . Then x ∈ A1 and y ∈ A2 . As in case 2, (y, z) ∈ R2 , and therefore z ∈ A2 . Thus (x, z) ∈ A1 × A2 , so (x, z) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

Thus (x, z) ∈ R1 ∪ R2 ∪ (A1 × A2 ), so R1 ∪ R2 ∪ (A1 × A2 ) is transitive.

Finally, suppose (x, y) ∈ R1 ∪R2 ∪(A1 ×A2 ) and (y, x) ∈ R1 ∪R2 ∪(A1 ×A2 ). If (x, y) ∈ A1 ×A2 then x ∈ A1 and y ∈ A2 , so (y, x) ∈ / R1 ∪ R2 ∪ (A1 × A2 ), which is a contradiction. Therefore either (x, y) ∈ R1 or (x, y) ∈ R2 .

Case 1. (x, y) ∈ R1 . Then x ∈ A1 , so (y, x) ∈ / A1 × A2 and (y, x) ∈ / R2 . Therefore (y, x) ∈ R1 , so by antisymmetry of R1 , x = y.

Case 2. (x, y) ∈ R2 . Then y ∈ A2 , so (y, x) ∈ / A1 × A2 and (y, x) ∈ / R1 . Therefore (y, x) ∈ R2 , so by antisymmetry of R2 , x = y.

Thus x = y, so R1 ∪ R2 ∪ (A1 × A2 ) is antisymmetric. (c) If a1 ∈ A1 and a2 ∈ A2 then (a1 , a2 ) ∈ / R1 ∪ R2 and (a2 , a1 ) ∈ / R1 ∪ R2 . Therefore as long as A1 6= ∅ and A2 6= ∅, R1 ∪ R2 is not a total order. However, R1 ∪ R2 ∪ (A1 × A2 ) is a total order.

To prove this, suppose x, y ∈ A1 ∪ A2 .

Case 1. x ∈ A1 and y ∈ A1 . Then since R1 is a total order on A1 , either (x, y) ∈ R1 or (y, x) ∈ R1 , so either (x, y) ∈ R1 ∪ R2 ∪ (A1 × A2 ) or (y, x) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

Case 2. x ∈ A2 and y ∈ A2 . Similar to case 1, using R2 instead of R1 .

Case 3. x ∈ A1 and y ∈ A2 . Then (x, y) ∈ A1 × A2 , so (x, y) ∈ R1 ∪ R2 ∪ (A1 × A2 ).

Case 4. x ∈ A2 and y ∈ A1 . Similar to case 3, using (y, x) instead of (x, y).

8. To see that T is reflexive, consider an arbitrary (a, b) ∈ A × B. Since R and S are both reflexive, we have aRa and bSb. By the definition of T , it follows that (a, b)T (a, b). To see that T is antisymmetric, suppose that (a, b)T (a0 , b0 ) and (a0 , b0 )T (a, b). Then aRa0 and a0 Ra, so since R is antisymmetric, a = a0 .

Similarly, bSb0 and b0 Sb, so since S is antisymmetric, we also have b = b0 . Thus (a, b) = (a0 , b0 ), as required. Finally, to see that T is transitive, suppose that (a, b)T (a0 , b0 ) and (a0 , b0 )T (a00 , b00 ). Then aRa0 and a0 Ra00 , so since R is transitive, aRa00 . Similarly, bSb0 and b0 Sb00 , so bSb00 , and therefore (a, b)T (a00 , b00 ).

Even if both R and S are total orders, T need not be a total order. For example, let A = B = R and R = S = {(x, y) ∈ R × R | x ≤ y}. Then ((1, 2), (2, 1)) ∈ / T and ((2, 1), (1, 2)) ∈ / T. 9. Suppose (a, b) ∈ A × B. Since R is reflexive on A and S is reflexive on B, aRa and bSb. Therefore ((a, b), (a, b)) ∈ L. Now suppose ((a, b), (a0 , b0 )) ∈ L and ((a0 , b0 ), (a00 , b00 )) ∈ L. Then aRa0 and a0 Ra00 , so by transitivity of R, aRa00 . Now suppose a = a00 . Then since a0 Ra00 , a0 Ra. Since aRa0 , a0 Ra, and R is antisymmetric, a0 = a = a00 . So since ((a, b), (a0 , b0 )) ∈ L and ((a0 , b0 ), (a00 , b00 )) ∈ L, bSb0 and b0 Sb00 . By transitivity of S, bSb00 . Thus ((a, b), (a00 , b00 )) ∈ L, which proves that L is transitive.

Next, suppose that ((a, b), (a0 , b0 )) ∈ L and ((a0 , b0 ), (a, b)) ∈ L. Then aRa0 and a0 Ra, so by the antisymmetry of R, a = a0 . But then we also have bSb0 and b0 Sb, so by antisymmetry of S, b = b0 , and therefore (a, b) = (a0 , b0 ). Thus L is antisymmetric.

If R and S are total orders, then L is also total. To see why, suppose (a, b), (a0 , b0 ) ∈ A × B. Case 1. a 6= a0 . Since R is a total order, either aRa0 or a0 Ra. Since a 6= a0 , it follows that either ((a, b), (a0 , b0 )) ∈ L or ((a0 , b0 ), (a, b)) ∈ L. Case 2. a = a0 . Since R is reflexive, aRa0 and a0 Ra. Since S is a total order, either bSb0 or b0 Sb. Therefore either ((a, b), (a0 , b0 )) ∈ L or ((a0 , b0 ), (a, b)) ∈ L.

10. Let x and y be arbitrary elements of A. (→) Suppose xRy. Suppose a ∈ Px . Then aRx. Since aRx, xRy, and R is transitive, aRy, so a ∈ Py . Since a was arbitrary, Px ⊆ Py . (←) Suppose Px ⊆ Py . Since R is reflexive, xRx, so x ∈ Px . Since Px ⊆ Py , x ∈ Py , so xRy.

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11. The minimal elements of B are the prime numbers. B has no smallest element.

12. Let F = {X ⊆ R | X 6= ∅ and ∀x∀y((x ∈ X ∧ x < y) → y ∈ X)}. Suppose X is a minimal element of F. Then X 6= ∅, so we can choose some x ∈ X. Also, ∀y(x < y → y ∈ X), so {y ∈ R | y ≥ x} ⊆ X. Let X 0 = {y ∈ R | y ≥ x + 1}. Then X 0 ∈ F, X 0 ⊆ X, and X 0 6= X. This contradicts the assumption that X is minimal. Therefore F has no minimal element.

13. By exercise 12 of Section 4.3, R−1 is reflexive and transitive. Now suppose (x, y) ∈ R−1 and (y, x) ∈ R−1 . Then (y, x) ∈ R and (x, y) ∈ R, so since R is antisymmetric, x = y. Therefore R−1 is antisymmetric. If R is a total order, then ∀x ∈ A∀y ∈ A((x, y) ∈ R ∨ (y, x) ∈ R), so ∀x ∈ A∀y ∈ A((y, x) ∈ R−1 ∨ (x, y) ∈ R−1 ) and therefore R−1 is also a total order.

14. (a) b is the R-largest element of B iff b ∈ B and ∀x ∈ B(xRb) iff b ∈ B and ∀x ∈ B(bR−1 x) iff b is the R−1 -smallest element of B. (b) b is an R-maximal element of B iff b ∈ B and ¬∃x ∈ B(bRx ∧ x 6= b) iff b ∈ B and ¬∃x ∈ B(xR−1 b ∧ x 6= b) iff b is an R−1 -minimal element of B. 15. (a) Suppose b is the R1 -smallest element of B. Let x ∈ B be arbitrary. Then (b, x) ∈ R1 , so since R1 ⊆ R2 , (b, x) ∈ R2 . Since x was arbitrary, ∀x ∈ B((b, x) ∈ R2 ), so b is the R2 -smallest element of B. (b) Suppose b is an R2 -minimal element of B. Suppose x ∈ B, (x, b) ∈ R1 , and x 6= b. Then since R1 ⊆ R2 , (x, b) ∈ R2 , which contradicts the fact that b is R2 -minimal. Therefore ¬∃x ∈ B((x, b) ∈ R1 ∧ x 6= b), so b is an R1 -minimal element of B. 16. Suppose b is the largest element of B. Let x be an arbitrary element of B and suppose that bRx. Since b is the largest element of B, we must have xRb, and now by antisymmetry it follows that x = b. Thus, there can be no x ∈ B such that bRx and x 6= b, so b is a maximal element.

To see that it is the only one, suppose c is also a maximal element. Since b is the largest element of B, cRb. But then since c is maximal we must have b = c. Thus b is the only maximal element of B. 17. No. Let A = R × R, and let R = {((x, y), (x0 , y 0 )) ∈ A × A | x ≤ x0 and y ≤ y 0 }. By exercise 8, R is a partial order on A. Let B = {(0, 0)} ∪ ({1} × R). (0, 0) is a minimal element of B because for every real number x, ((1, x), (0, 0)) ∈ / R. But it is not smallest since, for example, ((0, 0), (1, −1)) ∈ / R. For every real number x, (1, x − 1)R(1, x), so (1, x) is not a minimal element of B, and therefore (0, 0) is the only minimal element.

18. (a) Let x ∈ A be arbitrary. (→) Suppose x is an upper bound for B1 . Suppose b ∈ B2 . Then since ∀x ∈ B2 ∃y ∈ B1 (xRy), we can choose some y ∈ B1 such that bRy. Since y ∈ B1 and x is an upper bound for B1 , yRx. Since bRy and yRx, by transitivity of R it follows that bRx. Since b was arbitrary, this shows that x is an upper bound for B2 . (←) Similar to (→). (b) Suppose B1 and B2 are disjoint. Suppose b ∈ B1 . Then since ∀x ∈ B1 ∃y ∈ B2 (xRy), we can choose some c ∈ B2 such that bRc. And then since ∀x ∈ B2 ∃y ∈ B1 (xRy), we can choose some d ∈ B1 such that cRd. Since bRc and cRd, by transitivity of R, bRd. If b = d then bRc and cRb, so by antisymmetry of R, b = c. But b ∈ B1 and c ∈ B2 , so this would contradict the fact that B1 and B2 are disjoint. Thus b 6= d. Since bRd and b 6= d, b is not maximal. Since b was arbitrary,

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SOLUTIONS this shows that B1 does not have a maximal element. A similar argument shows that B2 does not have a maximal element.

19. (a) In case 1, we cannot conclude that for all x ∈ B, bRx; all we can say is that for all x ∈ B that fall under case 1, bRx. Similarly, the conclusion in case 2 is also not justified. (The mistake here is essentially the same as the one in exercise 31 of Section 3.5.) (b) The theorem is incorrect. Counterexample: A = R, R = {(x, y) ∈ R × R | x ≤ y}, B = R. 20. (a) Suppose b is the smallest element of B. Then ∀x ∈ B(bRx), so b is a lower bound for B. Now suppose c is a lower bound. Then since b ∈ B, cRb. Thus b is the greatest lower bound. (b) Similar to (a).

21. (a) Suppose that x ∈ U and xRy. To prove that y ∈ U , we must show that y is an upper bound for B, so suppose that b ∈ B. Since x ∈ U , x is an upper bound for B, so bRx. But we also have xRy, so by transitivity of R we can conclude that bRy. Since b was arbitrary, this shows that y is an upper bound for B. (b) Suppose b ∈ B. To prove that b is a lower bound for U , let x be an arbitrary element of U . Then by definition of U , x is an upper bound for B, so bRx. Since x was arbitrary, this shows that b is a lower bound for U . (c) Suppose x is the greatest lower bound of U . Suppose b ∈ B. Then by part (b), b is a lower bound for U . Since x is the greatest lower bound, bRx. Since b was arbitrary, this shows that x is an upper bound for B. Now suppose c is an upper bound for B. Then c ∈ U . Since x is a lower bound for U , xRc. Since c was arbitrary, this shows that x is the least upper bound of B. 22. Suppose B1 ⊆ B2 . We first show that x2 is an upper bound for B1 . To prove this, let b ∈ B1 be arbitrary. Since B1 ⊆ B2 , b ∈ B2 , and since x2 is an upper bound for B2 , bRx2 . Since b was arbitrary, this shows that x2 is an upper bound for B1 . But x1 is the least upper bound of B1 , so x1 Rx2 .

S S 23. By exercise 8 in Section 3.3, for every X ∈ F, X ⊆ F. This shows that F is an upper bound for F. Now suppose B is an upper bound. S Then for all X ∈SF, X ⊆ B, so X ∈ P(B). Therefore F ⊆ P(B), so by exercise 16 in Section 3.3, FT⊆ B. Thus, F is the least T upper bound of F. By exercise 9 in Section 3.3, for all X ∈ F, F ⊆ X. This shows that F is a lower bound for F. Now B is a lower bound. Then for all X ∈ F, B ⊆ X, so by exercise 10 in Section 3.3, T suppose T B ⊆ F. Thus, F is the greatest lower bound of F. 24. (a) Suppose (x, y) ∈ S. Then either (x, y) ∈ R or (x, y) ∈ R−1 . If (x, y) ∈ R, then (y, x) ∈ R−1 , so (y, x) ∈ S. If (x, y) ∈ R−1 , then (y, x) ∈ R, so (y, x) ∈ S. Therefore S is symmetric. Since S = R ∪ R−1 , it is clear that R ⊆ S. (b) Suppose T is a symmetric relation on A and R ⊆ T . To show that S ⊆ T , let (x, y) be an arbitrary element of S. Then either (x, y) ∈ R or (x, y) ∈ R−1 . If (x, y) ∈ R, then since R ⊆ T , (x, y) ∈ T .

If (x, y) ∈ R−1 , then (y, x) ∈ R, so since R ⊆ T , (y, x) ∈ T . But T is symmetric, so it follows that (x, y) ∈ T .

25. (a) A × A ∈ F, so F 6= ∅.

T T (b) Since A×A T ∈ F, by exercise T 9 in Section 3.3, F ⊆ A×A. To see thatT F is transitive, suppose (x, y) ∈ F and (y, z) ∈ F. Let T ∈ F be arbitrary. Since (x, y) ∈ F, (x, y) ∈ T . Similarly, T (y, z) ∈ T . Since T ∈ F, T is transitive, so (x, z) ∈ T . Since T T was arbitrary, (x, z) ∈ F. Finally, for all T ∈ F, R ⊆ T , so by exercise 10 in Section 3.3, R ⊆ F. (c) Suppose T is a T transitive relationTon A that contains R as a subset. Then T ∈ F, so by exercise 9 in Section 3.3, F ⊆ T . Thus, F is the smallest transitive relation on A that contains R. 26. (a) S2 is a symmetric relation on A and R2 ⊆ S2 . Since R1 ⊆ R2 , R1 ⊆ S2 . Since S1 is the smallest symmetric relation on A that contains R1 as a subset, S1 ⊆ S2 .

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(b) T2 is a transitive relation on A and R2 ⊆ T2 . Since R1 ⊆ R2 , R1 ⊆ T2 . Since T1 is the smallest transitive relation on A that contains R1 as a subset, T1 ⊆ T2 .

27. (a) First, note that R1 ⊆ R and R2 ⊆ R. It follows, by exercise 26, that S1 ⊆ S and S2 ⊆ S, so S1 ∪ S2 ⊆ S. For the other direction, note that R = R1 ∪ R2 ⊆ S1 ∪ S2 , and by exercise 13(b) of Section 4.3, S1 ∪ S2 is symmetric. Therefore, since S is the smallest symmetric relation on A that contains R, S ⊆ S1 ∪ S2 . (b) Imitating the first half of the proof in part (a), we can use exercise 26 to show that T1 ∪ T2 ⊆ T .

However, the answer to exercise 13(c) of Section 4.3 was no, so we can’t imitate the second half of the proof. In fact, the example given in the solution to exercise 13(c) works as an example for which T1 ∪ T2 6= T .

28. (a) Let a and b be distinct elements of A. Suppose R is the largest antisymmetric relation on A. Let S1 = {(a, b)}. Then S1 is antisymmetric, so since R is the largest antisymmetric relation, S1 ⊆ R and therefore (a, b) ∈ R. Similarly, if we let S2 = {(b, a)} then S2 is antisymmetric, so S2 ⊆ R and therefore (b, a) ∈ R. But now we have (a, b) ∈ R, (b, a) ∈ R, and a 6= b, so this contradicts the fact that R is antisymmetric. Therefore there is no largest antisymmetric relation on A. (b) Suppose S is an antisymmetric relation on A such that R ⊆ S and R 6= S. Then there must be some ordered pair (a, b) ∈ A × A such that (a, b) ∈ S and (a, b) ∈ / R. Since R is a total order on A, either (a, b) ∈ R or (b, a) ∈ R, so since (a, b) ∈ / R, (b, a) ∈ R. But R ⊆ S, so it follows that (b, a) ∈ S. Since (a, b) ∈ S, (b, a) ∈ S, and S is antisymmetric, a = b. But then since R is reflexive, (a, b) ∈ R, which is a contradiction. Therefore there is no antisymmetric relation S on A such that R ⊆ S and R 6= S. 29. (a) L is irreflexive because for all x ∈ R, x 6< x, so (x, x) ∈ / L. L is transitive because for all x, y, z ∈ R, if x < y and y < z then x < z. L is a strict total order because for all x, y ∈ R, either x < y, y < x, or y = x. (b) Suppose R is a partial order on A. Let x ∈ A be arbitrary. Then (x, x) ∈ iA , so (x, x) ∈ / R \ iA .

Therefore R \ iA is irreflexive. Now suppose (x, y) ∈ R \ iA and (y, z) ∈ R \ iA . Then (x, y) ∈ R and (y, z) ∈ R, so since R is transitive, (x, z) ∈ R. Also, (x, y) ∈ / iA and (y, z) ∈ / iA , so x 6= y and y 6= z. If x = z then (x, y) ∈ R and (y, x) = (y, z) ∈ R, which contradicts the fact that R is antisymmetric. Therefore x 6= z, so (x, z) ∈ / iA . Since (x, z) ∈ R and (x, z) ∈ / iA , (x, z) ∈ R \ iA .

This prove that R \ iA is transitive, so it is a strict partial order. Finally, suppose R is a total order. Let x, y ∈ A be arbitrary. Then since R is a total order, either (x, y) ∈ R or (y, x) ∈ R. If x 6= y then (x, y) ∈ / iA , so either (x, y) ∈ R\iA or (y, x) ∈ R\iA . Thus either either (x, y) ∈ R\iA , or (y, x) ∈ R \ iA , or x = y, so R \ iA is a strict total order. (c) Suppose R is a strict partial order on A. Since iA ⊆ R ∪ iA , by part 1 of Theorem 4.3.4, R ∪ iA is reflexive. To prove that R ∪ iA is transitive, suppose (x, y) ∈ R ∪ iA and (y, z) ∈ R ∪ iA . Since (x, y) ∈ R ∪ iA , either (x, y) ∈ R or (x, y) ∈ iA .

Case 1. (x, y) ∈ R. Since (y, z) ∈ R ∪ iA , either (y, z) ∈ R or (y, z) ∈ iA .

Case 1a. (y, z) ∈ R. By transitivity of R, (x, z) ∈ R, so (x, z) ∈ R ∪ iA .

Case 1b. (y, z) ∈ iA . Then y = z, so (x, z) = (x, y) ∈ R ∪ iA .

Case 2. (x, y) ∈ iA . Then x = y, so (x, z) = (y, z) ∈ R ∪ iA .

Thus (x, z) ∈ R ∪ iA . This proves that R ∪ iA is transitive.

To prove that R ∪ iA is antisymmetric, suppose (x, y) ∈ R ∪ iA and (y, x) ∈ R ∪ iA , but x 6= y.

Then (x, y) ∈ / iA and (y, x) ∈ / iA , so (x, y) ∈ R and (y, x) ∈ R. By transitivity of R, (x, x) ∈ R. But this contradicts the fact that R is irreflexive. Therefore R is antisymmetric, so it is a partial order.

Finally, suppose that R is a strict total order on A. Let x, y ∈ A be arbitrary. Since R is a strict total order, either (x, y) ∈ R, (y, x) ∈ R, or x = y. But if x = y then (x, y) ∈ iA , so either (x, y) ∈ R ∪ iA or (y, x) ∈ R ∪ iA . Therefore R is a total order on A.

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30. Since T is the transitive closure of R, R ⊆ T . Suppose R is symmetric. Let (x, y) ∈ R be arbitrary.

Then since R is symmetric, (y, x) ∈ R, and since R ⊆ T , it follows that (y, x) ∈ T . Thus (x, y) ∈ T −1 .

Since (x, y) was arbitrary, this shows that R ⊆ T −1 . Next, suppose (x, y) ∈ T −1 and (y, z) ∈ T −1 .

Then (z, y) ∈ T and (y, x) ∈ T , so by transitivity of T , (z, x) ∈ T , and therefore (x, z) ∈ T −1 . Thus T −1 is transitive. Since T is the smallest transitive relation on A that contains R, we conclude that T ⊆ T −1 . Finally, to prove that T is symmetric, suppose (x, y) ∈ T . Then since T ⊆ T −1 , (x, y) ∈ T −1 , so (y, x) ∈ T .

Section 4.5 1. Here is a list of all partitions: {{1, 2, 3}} {{1, 2}, {3}} {{1, 3}, {2}} {{2, 3}, {1}} {{1}, {2}, {3}} 2. Each equivalence relation corresponds to one of the partitions listed in exercise 1: {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} {(1, 1), (1, 3), (3, 1), (3, 3), (2, 2)} {(2, 2), (2, 3), (3, 2), (3, 3), (1, 1)} {(1, 1), (2, 2), (3, 3)} 3. (a) R is an equivalence relation. There are 26 equivalence classes – one for each letter of the alphabet.

The equivalence classes are: the set of all words that start with a, the set of all words that start with b, . . . , the set of all words that start with z. (b) S is not an equivalence relation, because it is not transitive. (c) T is an equivalence relation. The equivalence classes are: the set of all one-letter words, the set of all two-letter words, and so on. For every positive integer n, if there is at least one English word of length n, then the set of all words of length n is an equivalence class.

4. (a) R is not an equivalence relation: (2, 1) ∈ R but (1, 2) ∈ / R, so R is not symmetric. (b) S is an equivalence relation. For any x ∈ R, [x]S = {x + q | q ∈ Q}. (c) T is an equivalence relation. For any x ∈ R, [x]T is the set of numbers that have the same digits as x, but perhaps with the decimal point in a different location. For example, [π]T = {3.14159 . . . , 31.4159 . . . , 314.159 . . . , . . . , 0.314159 . . . , 0.0314159 . . . , . . .} 5. (a) R is an equivalence relation. The equivalence classes are families of parallel lines. (b) S is not an equivalence relation, because it is not reflexive and it is not transitive. (c) T is an equivalence relation. The set of all lines passing through the origin that are neither horizontal nor vertical is one equivalence class. Every other equivalence class contains just a single line.

6. Suppose X ∈ P/B. Then we can choose some x ∈ P such that X = [x]. Let d be x’s birthday. Then for every y ∈ P , y ∈ [x] iff yBx iff y ∈ Pd , and therefore X = [x] = Pd . Since X was arbitrary, this shows that P/B ⊆ {Pd | d ∈ D}.

Now suppose d ∈ D. At this point, we must make the very reasonable assumption that some person x was born on the day d. As before, it follows that Pd = [x] ∈ P/B. Since d was arbitrary, we conclude that {Pd | d ∈ D} ⊆ P/B.

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7. Every triangle is similar to itself, if s is similar to t then t is similar to s, and if r is similar to s and s is similar to t then r is similar to t.

S 8. To see that R is symmetric, suppose (x, y) ∈ R. Since R = X∈F (X × X), this means that there is some S X ∈ F such that (x, y) ∈ X × X, so x ∈ X and y ∈ X. Therefore (y, x) ∈ X × X, so (y, x) ∈ X∈F (X × X) = R. To prove that R is transitive, suppose (x, y) ∈ R and (y, z) ∈ R. Then we can choose sets X, Y ∈ F such that (x, y) ∈ X × X and (y, z) ∈ Y × Y . Therefore x ∈ X, y ∈ X, y ∈ Y , and z ∈ Y . Since y ∈ X and y ∈ Y , X ∩ Y 6= ∅. Since F is pairwise disjoint, it follows that X = Y . Thus x ∈ X and S z ∈ Y = X, so (x, z) ∈ X × X and therefore (x, z) ∈ X∈F (X × X) = R. 9. We first claim that ∀x ∈ A([x]R = [x]S ). To prove this, let x ∈ A be arbitrary. Then [x]R ∈ A/R = A/S, so there is some y ∈ A such that [x]R = [y]S . By part 1 of Lemma 4.5.5, x ∈ [x]R , so x ∈ [y]S , and therefore by part 2 of Lemma 4.5.5, [x]S = [y]S = [x]R . Since x was arbitrary, this proves our claim.

Now, to prove that R = S, let (x, y) ∈ A × A be arbitrary. Then (x, y) ∈ R iff x ∈ [y]R iff x ∈ [y]S iff (x, y) ∈ S. 10. Since S is the equivalence relation determined by F, the proof of Theorem 4.5.6 shows that A/S = F = A/R. The desired conclusion now follows from exercise 9.

11. (a) Let x ∈ Z be arbitrary. Since x − x = 0 = 0 · m, x ≡m x, so ≡m is reflexive. Now suppose x ≡m y. Then m | (x − y), so we can choose some integer k such that x − y = km. Therefore y − x = (−k)m, so m | (y − x) and y ≡m x. This shows that ≡m is symmetric. (b) Equivalence classes for ≡2 : [0]≡2 = {. . . , −4, −2, 0, 2, 4, . . .}, [1]≡2 = {. . . , −3, −1, 1, 3, . . .}.

Equivalence classes for ≡3 : [0]≡3 = {. . . , −6, −3, 0, 3, 6, . . .}, [1]≡3 = {. . . , −5, −2, 1, 4, 7, . . .}, [2]≡3 = {. . . , −4, −1, 2, 5, 8, . . .}.

There are m equivalence classes for ≡m . For more on this, see Section 7.3.

12. Let n be an arbitrary integer. Then n is either even or odd.

Case 1. n is even. Then we can choose an integer k such that n = 2k. Therefore n2 − 0 = 4k 2 , so 4 | (n2 − 0), which means that n2 ≡ 0 (mod 4).

Case 2. n is odd. Then we can choose some integer k such that n = 2k + 1. Therefore n2 − 1 = 2 (4k + 4k + 1) − 1 = 4(k 2 + k), so 4 | (n2 − 1), which means that n2 ≡ 1 (mod 4).

Thus, either n2 ≡ 0 (mod 4) or n2 ≡ 1 (mod 4).

13. See Lemma 7.3.4.

14. (a) Suppose x ∈ B. Then x ∈ A, so (x, x) ∈ R, since R is reflexive on A, and also (x, x) ∈ B × B. Therefore (x, x) ∈ R ∩ (B × B) = S, so S is reflexive on B. Suppose (x, y) ∈ S. Then (x, y) ∈ R, so since R is symmetric, (y, x) ∈ R. Also, (x, y) ∈ B × B, so x ∈ B and y ∈ B, and therefore (y, x) ∈ B × B. Thus (y, x) ∈ R ∩ (B × B) = S. Finally, suppose (x, y) ∈ S and (y, z) ∈ S. Then (x, y) ∈ R and (y, z) ∈ R, so by transitivity of R, (x, z) ∈ R. Also, x ∈ B and z ∈ B, so (x, z) ∈ B × B. Therefore (x, z) ∈ S. (b) Suppose x ∈ B. Then for every y, y ∈ [x]S iff (y, x) ∈ R ∩ (B × B) iff (y, x) ∈ R ∧ y ∈ B iff y ∈ [x]R ∧ y ∈ B iff y ∈ [x]R ∩ B. 15. (a) For every X ∈ P(A), X 4 X = ∅ ⊆ B, so (X, X) ∈ R. Suppose (X, Y ) ∈ R. Then X 4 Y ⊆ B. But Y 4 X = X 4 Y , so Y 4 X ⊆ B and therefore (Y, X) ∈ R. Finally, suppose (X, Y ) ∈ R and (Y, Z) ∈ R. Then X 4 Y ⊆ B and Y 4 Z ⊆ B. Therefore (X 4 Y ) 4 (Y 4 Z) ⊆ (X 4 Y ) ∪ (Y 4 Z) ⊆ B. But (X 4 Y ) 4 (Y 4 Z) = (X 4 (Y 4 Y )) 4 Z = (X 4 ∅) 4 Z = X 4 Z.

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Therefore X 4 Z ⊆ B, so (X, Z) ∈ R. (b) Suppose X ∈ P(A). Let Y = X \ B. Then Y ⊆ X, so Y \ X = ∅, and by exercise 8(c) of Section 1.4, X \ Y = X \ (X \ B) = X ∩ B. Therefore Y 4 X = (Y \ X) ∪ (X \ Y ) = ∅ ∪ (X ∩ B) = X ∩ B ⊆ B, so (Y, X) ∈ R and Y ∈ [X]R . Also, Y ∩ B = (X \ B) ∩ B = ∅. Finally, to see that Y is unique, suppose that Y 0 ∈ [X]R and Y 0 ∩ B = ∅. Then (Y 0 , X) ∈ R, so since (Y, X) ∈ R, by symmetry and transitivity of R, (Y 0 , Y ) ∈ R, so Y 0 4 Y ⊆ B. Suppose y ∈ Y . If y ∈ / Y 0 then 0 0 y ∈ Y \ Y ⊆ Y 4 Y ⊆ B. But this contradicts the fact that Y and B are disjoint. Therefore y ∈ Y 0 . Since y was arbitrary, this proves that Y ⊆ Y 0 . A similar argument shows that Y 0 ⊆ Y , so Y 0 = Y .

S S S 16. By exercise 16(a) of Section 3.5, (F ∪ G) = ( F) ∪ ( G) = A ∪ B. To see that F ∪ G is pairwise disjoint, suppose that X ∈ F ∪ G, Y ∈ F ∪ G, and X ∩ Y 6= ∅. If X ∈ F and Y ∈ G then X ⊆ A and Y ⊆ B, and since A and B are disjoint it follows that X and Y are disjoint, which is a contradiction.

Thus it cannot be the case that X ∈ F and Y ∈ G, and a similar argument can be used to rule out the possibility that X ∈ G and Y ∈ F. Thus, X and Y are either both elements of F or both elements of G. If they are both in F, then since F is pairwise disjoint, X = Y . A similar argument applies if they are both in G. Finally, we have ∀X ∈ F(X 6= ∅) and ∀X ∈ G(X 6= ∅), and it follows by exercise 8 of Section 2.2 that ∀X ∈ F ∪ G(X 6= ∅).

17. (a) Suppose x ∈ A ∪ B. Then either x ∈ A or x ∈ B. If x ∈ A then, since R is reflexive on A, (x, x) ∈ R, and if x ∈ B then, since S is reflexive on B, (x, x) ∈ S. Therefore either (x, x) ∈ R or (x, x) ∈ S, so (x, x) ∈ R ∪ S. This proves that R ∪ S is reflexive on A ∪ B. Now suppose (x, y) ∈ R ∪ S. Then either (x, y) ∈ R or (x, y) ∈ S. If (x, y) ∈ R then, by symmetry of R. (y, x) ∈ R, and if (x, y) ∈ S then, by symmetry of S, (y, x) ∈ S. Therefore (y, x) ∈ R ∪ S, so R ∪ S is symmetric. Finally, suppose (x, y) ∈ R ∪ S and (y, z) ∈ R ∪ S. Then either (x, y) ∈ R or (x, y) ∈ S. Case 1. (x, y) ∈ R. Since R ⊆ A × A, y ∈ A, and since A and B are disjoint, y ∈ / B. Since S ⊆ B × B, it follows that (y, z) ∈ / S, so since (y, z) ∈ R ∪ S, (y, z) ∈ R. Since (x, y) ∈ R, (y, z) ∈ R, and R is transitive, (x, z) ∈ R, so (x, z) ∈ R ∪ S. Case 2. (x, y) ∈ S. A similar argument shows that (y, z) ∈ S, so by transitivity of S, (x, z) ∈ S, and therefore (x, z) ∈ R ∪ S. Thus (x, z) ∈ R ∪ S, so R ∪ S is transitive. (b) Suppose x ∈ A. Suppose y ∈ [x]R∪S . Then (y, x) ∈ R ∪ S, so either (y, x) ∈ R or (y, x) ∈ S. But since x ∈ A and A and B are disjoint, x ∈ / B, so since S ⊆ B × B, (y, x) ∈ / S. Therefore (y, x) ∈ R, so y ∈ [x]R . Now suppose y ∈ [x]R . Then (y, x) ∈ R, so (y, x) ∈ R ∪ S, and therefore y ∈ [x]R∪S . Thus [x]R∪S = [x]R . A similar proof shows that if x ∈ B then [x]R∪S = [x]S . (c) Suppose X ∈ (A ∪ B)/(R ∪ S). Then for some x ∈ A ∪ B, X = [x]R∪S . Since x ∈ A ∪ B, either x ∈ A or x ∈ B. By part (b), if x ∈ A then [x]R∪S = [x]R and if x ∈ B then [x]R∪S = [x]S . Thus either X = [x]R ∈ A/R or X = [x]S ∈ B/S, so X ∈ (A/R) ∪ (B/S).

Now suppose X ∈ (A/R) ∪ (B/S). Then either X ∈ A/R or X ∈ B/S. If X ∈ A/R then for some x ∈ A, X = [x]R , and by part (b) it follows that X = [x]R∪S ∈ (A ∪ B)/(R ∪ S). Similarly if X ∈ B/S then for some x ∈ B, X = [x]R∪S ∈ (A ∪ B)/(R ∪ S). Thus (A ∪ B)/(R ∪ S) = (A/R) ∪ (B/S).

S 18. Clearly F · G ⊆ P(A), so (F · G) ⊆ A. Now suppose x ∈ A. Since F and G are both partitions of A, there S are sets X ∈ F and Y S∈ G such that x ∈ X and x ∈ Y . Therefore x ∈ X ∩ Y ∈ F · G, so x ∈ (F · G). This proves that (F · G) = A. To prove that F · G is pairwise disjoint, suppose that Z1 , Z2 ∈ F · G and Z1 ∩ Z2 6= ∅. Since Z1 ∈ F · G, we can choose sets X1 ∈ F and Y1 ∈ G such that Z1 = X1 ∩ Y1 . Similarly, since Z2 ∈ F · G we can choose X2 ∈ F and Y2 ∈ G such that Z2 = X2 ∩ Y2 . Since Z1 ∩ Z2 6= ∅, we can choose some z ∈ Z1 ∩ Z2 . Thus z ∈ Z1 = X1 ∩ Y1 and z ∈ Z2 = X2 ∩ Y2 , so z ∈ X1 , z ∈ Y1 , z ∈ X2 , and z ∈ Y2 .

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Since z ∈ X1 and z ∈ X2 , X1 and X2 are not disjoint. Since F is pairwise disjoint, it follows that X1 = X2 . Similarly, Y1 and Y2 are not disjoint, so since G is pairwise disjoint, Y1 = Y2 . Therefore Z1 = X1 ∩ Y1 = X2 ∩ Y2 = Z2 . This proves that F · G is pairwise disjoint.

Finally, it follows directly from the definition of F · G that ∀Z ∈ F · G(Z 6= ∅). Thus, F · G is a partition of A. 19. F · G = {Z− , R− \ Z− , Z+ , R+ \ Z+ , {0}}.

20. (a) Suppose x ∈ A. Since R is an equivalence relation on A, R is reflexive on A, so (x, x) ∈ R. Similarly, S is reflexive, so (x, x) ∈ S. Therefore (x, x) ∈ R ∩ S = T , so T is reflexive on A. Next, suppose (x, y) ∈ T = R ∩ S. Then (x, y) ∈ R and (x, y) ∈ S, so since R and S are symmetric, (y, x) ∈ R and (y, x) ∈ S. Therefore (y, x) ∈ R ∩ S = T , so T is symmetric.

Finally, to prove that T is transitive, suppose (x, y) ∈ T and (y, z) ∈ T . Then since T = R ∩ S, (x, y) ∈ R and (y, z) ∈ R, so since R is transitive, (x, z) ∈ R. Similarly, (x, z) ∈ S, so (x, z) ∈ R ∩ S = T. (b) Suppose x ∈ A. Then for all y ∈ A, y ∈ [x]T iff (y, x) ∈ T iff (y, x) ∈ R ∧ (y, x) ∈ S iff y ∈ [x]R ∧ y ∈ [x]S iff y ∈ [x]R ∩ [x]S . (c) Suppose X ∈ A/T . Then since A/T is a partition, X 6= ∅. Also, for some x ∈ A, X = [x]T = [x]R ∩ [x]S , so since [x]R ∈ A/R and [x]S ∈ A/S, X ∈ (A/R) · (A/S).

Now suppose X ∈ (A/R) · (A/S). Then for some y and z in A, X = [y]R ∩ [z]S . Also, X 6= ∅, so we can choose some x ∈ X. Therefore x ∈ [y]R and x ∈ [z]S , and by part 2 of Lemma 4.5.5 it follows that [x]R = [y]R and [x]S = [z]S . Therefore X = [x]R ∩ [x]S = [x]T ∈ A/T .

S 21. Since F ⊗ G ⊆ P(A × B), (F ⊗ G) ⊆ A × B. Now suppose (a, b) ∈ A × B. Then a ∈ A, so since F is a partition of A, we can choose some X ∈ F such that a ∈ X. Similarly, bS ∈ B, so we can choose someSY ∈ G such that b ∈ Y . Therefore (a, b) ∈ X × Y ∈ F ⊗ G, so (a, b) ∈ (F ⊗ G). This proves that (F ⊗ G) = A × B. To prove that F ⊗G is pairwise disjoint, suppose Z1 , Z2 ∈ F ⊗G and Z1 ∩Z2 6= ∅. Since Z1 ∈ F ⊗G, we can choose X1 ∈ F and Y1 ∈ G such that Z1 = X1 × Y1 . Similarly, since Z2 ∈ F ⊗ G, we can choose X2 ∈ F and Y2 ∈ G such that Z2 = X2 × Y2 . Since Z1 ∩ Z2 6= ∅, we can choose some (a, b) ∈ A × B such that (a, b) ∈ Z1 ∩ Z2 . Thus (a, b) ∈ Z1 = X1 × Y1 and (a, b) ∈ Z2 = X2 × Y2 , so a ∈ X1 , b ∈ Y1 , a ∈ X2 , and b ∈ Y2 . Since a ∈ X1 and a ∈ X2 , X1 and X2 are not disjoint, so since F is pairwise disjoint, X1 = X2 . Similarly, Y1 and Y2 are not disjoint, so since G is pairwise disjoint, Y1 = Y2 .

Therefore Z1 = X1 × Y1 = X2 × Y2 = Z2 , so F ⊗ G is pairwise disjoint.

Finally, suppose Z ∈ F ⊗ G. Then we can choose X ∈ F and Y ∈ G such that Z = X × Y .

Since F and G are partitions, X and Y are nonempty, so we can choose a ∈ X and b ∈ Y . Therefore (a, b) ∈ X × Y = Z, so Z 6= ∅.

22. F ⊗ F = {R+ × R+ , R− × R+ , R− × R− , R+ × R− , R+ × {0}, R− × {0}, {0} × R+ , {0} × R− , {(0, 0)}}.

In geometric terms these are the four quadrants of the plane, the positive and negative x-axes, the positive and negative y-axes, and the origin.

23. (a) Suppose (a, b) ∈ A × B. Then a ∈ A and b ∈ B, so since R is reflexive on A and S is reflexive on B, aRa and bRb. Thus ((a, b), (a, b)) ∈ T , so T is reflexive on A × B. Next, suppose that ((a, b), (a0 , b0 )) ∈ T . Then aRa0 and bSb0 . Since R and S are symmetric, 0 a Ra and b0 Sb, so ((a0 , b0 ), (a, b)) ∈ T . Therefore T is symmetric.

Finally, suppose ((a, b), (a0 , b0 )) ∈ T and ((a0 , b0 ), (a00 , b00 )) ∈ T . Then aRa0 , bSb0 , a0 Ra00 , and 0 b Sb00 . Since R and S are transitive, aRa00 and bSb00 , so ((a, b), (a00 , b00 )) ∈ T . Therefore T is transitive.

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SOLUTIONS (b) Suppose a ∈ A and b ∈ B. Then for all (x, y) ∈ A × B, (x, y) ∈ [(a, b)]T iff ((x, y), (a, b)) ∈ T iff xRa ∧ ySb iff x ∈ [a]R ∧ y ∈ [b]S iff (x, y) ∈ [a]R × [b]S . (c) Suppose Z ∈ (A × B)/T . Then there is some (a, b) ∈ A × B such that Z = [(a, b)]T = [a]R × [b]S ∈ (A/R) ⊗ (B/S). Now suppose Z ∈ (A/R) ⊗ (B/S). Then we can choose X ∈ A/R and Y ∈ B/S such that Z = X × Y . Since X ∈ A/R and Y ∈ B/S, we can choose a ∈ A and b ∈ B such that X = [a]R and Y = [b]S . Therefore Z = X × Y = [a]R × [b]S = [(a, b)]T ∈ (A × B)/T . Thus (A × B)/T = (A/R) ⊗ (B/S).

24. (a) Suppose R is compatible with S. Let T = {(X, Y ) ∈ A/S × A/S | ∃x ∈ X∃y ∈ Y (xRy)}. Let x and y be arbitrary elements of A. Suppose [x]S T [y]S . Then by the definition of T , there are x0 ∈ [x]S and y 0 ∈ [y]S such that x0 Ry 0 . Since x0 ∈ [x]S and y 0 ∈ [y]S , x0 Sx and y 0 Sy, so since R is compatible with S, xRy. Now suppose xRy. Since x ∈ [x]S and y ∈ [y]S , it follows that [x]S T [y]S .

To see that T is unique, suppose T 0 is another relation with the required properties. Let X and Y be arbitrary elements of A/S. Then we can choose x, y ∈ A such that X = [x]S and Y = [y]S . Therefore (X, Y ) ∈ T 0 iff [x]S T 0 [y]S iff xRy iff [x]S T [y]S iff (X, Y ) ∈ T, so T 0 = T . (b) Suppose x, y, x0 , y 0 ∈ A, xSx0 , and ySy 0 . Then [x]S = [x0 ]S and [y]S = [y 0 ]S , so xRy iff [x]S T [y]S iff [x0 ]S T [y 0 ]S iff x0 Ry 0 .

25. (a) By exercises 12 and 14 in Section 4.3, S is reflexive and transitive. To see that S is symmetric, suppose (x, y) ∈ S = R ∩ R−1 . Then (x, y) ∈ R and (x, y) ∈ R−1 , so (y, x) ∈ R−1 and (y, x) ∈ R, and therefore (y, x) ∈ R ∩ R−1 = S. (b) Let x, y, x0 , and y 0 be arbitrary elements of A, and suppose that xSx0 and ySy 0 . Then (x, x0 ) ∈ S = R ∩ R−1 and (y, y 0 ) ∈ S = R ∩ R−1 , so xRx0 , x0 Rx, yRy 0 , and y 0 Ry. Now suppose xRy. By transitivity of R, since x0 Rx and xRy, x0 Ry. But then since yRy 0 , by another application of transitivity, x0 Ry 0 . A similar argument shows that if x0 Ry 0 then xRy. Therefore R is compatible with S, so the required conclusion follows from exercise 24. (c) Suppose X ∈ A/S. Then we can choose some x ∈ A such that X = [x]S . Since R is reflexive, xRx, so [x]S T [x]S , which means XT X. Thus T is reflexive. Next, suppose X, Y, Z ∈ A/S, XT Y , and Y T Z. Choose x, y, z ∈ A such that X = [x]S , Y = [y]S , and Z = [z]S . Then [x]S T [y]S and [y]S T [z]S , so xRy and yRz. Since R is transitive, xRz. Therefore [x]S T [z]S , which means XT Z. This proves that T is transitive. Finally, to see that T is antisymmetric, suppose X, Y ∈ A/S, XT Y , and Y T X. Choose x, y ∈ A such that X = [x]S and Y = [y]S . Then [x]S T [y]S and [y]S T [x]S , so xRy and yRx. Therefore (x, y) ∈ R ∩ R−1 = S, so [x]S = [y]S , which means that X =Y. 26. (a) For every X ∈ A, X has at least as many elements as X, so R is reflexive. For all X, Y, Z ∈ A, if Y has at least as many elements as X and Z has at least as many elements as Y then Z has at least as many elements as X, so R is transitive. (b) For all X and Y in A, (X, Y ) ∈ S iff X and Y have the same number of elements. Let J = {0, 1, . . . , 100}, and for each j ∈ J let Pj = {X ∈ P(I) | X has exactly j elements}. Then A/S = {P0 , P1 , . . . , P100 } = {Pj | j ∈ J}, so A/S has 101 elements. T = {(Pj , Pk ) | j ∈ J, k ∈ J, and j ≤ k}. T is a total order.

27. (a) Suppose F ∈ P . Then for every X ∈ F, X ⊆ X, so ∀X ∈ F∃Y ∈ F(X ⊆ Y ) and therefore (F, F) ∈ R. Thus R is reflexive. Suppose (F, G) ∈ R and (G, H) ∈ R. Then F refines G and G

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refines H. Let X ∈ F be arbitrary. Then since F refines G, we can choose some Y ∈ G such that X ⊆ Y . Since G refines H, we can choose some Z ∈ H such that Y ⊆ Z. Since X ⊆ Y and Y ⊆ Z, X ⊆ Z. Therefore ∀X ∈ F∃Z ∈ H(X ⊆ Z), so F refines H and therefore (F, H) ∈ R. Thus, R is transitive. Finally, suppose (F, G) ∈ R and (G, F) ∈ R. Then F refines G and G refines F. Let X ∈ F be arbitrary. Then since F refines G, there is some Y ∈ G such that X ⊆ Y . Since G refines F, there is some X 0 ∈ F such that Y ⊆ X 0 . Since X ⊆ Y and Y ⊆ X 0 , X ⊆ X 0 . But then X ∩ X 0 = X 6= ∅, so since F is pairwise disjoint, X 0 = X. Therefore X ⊆ Y and Y ⊆ X 0 = X, so X = Y ∈ G. Since X was an arbitrary element of F, this shows that F ⊆ G. A similar argument shows that G ⊆ F, so F = G, and therefore R is antisymmetric. (b) (→) Suppose S ⊆ T . Suppose X ∈ F = A/S. Then we can choose some x ∈ A such that X = [x]S . Let Y = [x]T ∈ A/T = G. Let a be an arbitrary element of X. Then a ∈ [x]S , so (a, x) ∈ S. Since S ⊆ T , (a, x) ∈ T , so a ∈ [x]T = Y . Since a was arbitrary, X ⊆ Y . Since X was arbitrary, this shows that F refines G. (←) Suppose F refines G. Suppose (x, y) ∈ S. Let Y = [y]S ∈ A/S = F. Since F refines G, there is some Z ∈ G = A/T such that Y ⊆ Z. Since Z ∈ A/T , we can choose some z ∈ A such that Z = [z]T . By Lemma 4.5.5, y ∈ [y]S = Y ⊆ Z = [z]T , so [y]T = [z]T . Since (x, y) ∈ S, x ∈ [y]S ⊆ [z]T = [y]T , so (x, y) ∈ T . Since (x, y) was arbitrary, S ⊆ T . (c) Suppose Z ∈ F · G. Then we can choose X ∈ F and Y ∈ G such that Z = X ∩ Y . Therefore Z ⊆ X and Z ⊆ Y . Since Z was arbitrary, this shows that F · G refines both F and G, so F · G is a lower bound for {F, G} in the partial order R. To see that it is the greatest lower bound, suppose H is a lower bound. Then H refines both F and G. Let Z ∈ H be arbitrary. Since H refines both F and G, we can choose X ∈ F and Y ∈ G such that Z ⊆ X and Z ⊆ Y . Therefore Z ⊆ X ∩ Y . Since H is a partition and Z ∈ H, Z 6= ∅, so X ∩ Y 6= ∅ and therefore X ∩ Y ∈ F · G. Since Z was arbitrary, this shows that H refines F · G, so (H, F · G) ∈ R. Since H was arbitrary, this shows that F · G is the greatest lower bound of {F, G} in the partial order R.

Chapter 5 Section 5.1 1. (a) Yes. (b) No, because 1 appears as the first coordinate of multiple ordered pairs in f . (c) Yes. 2. (a) No, because d is not paired with anything. (b) f is not a function, because many words are paired with multiple letters, but g is a function. (c) Yes. 3. (a) f (a) = b, f (b) = b, f (c) = a. (b) f (2) = 0. (c) f (π) = 3 and f (−π) = −4.

4. (a) H(Italy) = Rome. (b) F ({1, 3}) = {2}. (c) f (2) = (3, 1).

5. L ◦ H : N → N , and for every n ∈ N , (L ◦ H)(n) = n. Thus, L ◦ H = iN .

H ◦ L : C → C, and for every c ∈ C, (H ◦ L)(c) = the capital of the country in which c is located.

1 1 6. = 2 , (f ◦ g)(x) = f (g(x)) = f (2x − 1) = (2x − 1)2 + 2 4x − 4x + 3

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(g ◦ f )(x) = g(f (x)) = g

1 x2 + 2

=2·

1 x2 − 1 = − . x2 + 2 x2 + 2

7. (a) Suppose that c ∈ C. We must prove that there is a unique b ∈ B such that (c, b) ∈ f C. Existence: Let b = f (c) ∈ B. Then (c, b) ∈ f and (c, b) ∈ C × B, and therefore (c, b) ∈ f ∩ (C × B) = f C. Uniqueness: Suppose that (c, b1 ) ∈ f C and (c, b2 ) ∈ f C. Then (c, b1 ) ∈ f and (c, b2 ) ∈ f , so since f is a function, b1 = b2 .

This proves that f C is a function from C to B. Finally, to derive the formula for (f C)(c), suppose that c ∈ C, and let b = f (c). We showed in the existence half of the proof that (c, b) ∈ f C. It follows that f (c) = b = (f C)(c). (b) (→) Suppose g = f C. Then g = f ∩ (C × B), so clearly g ⊆ f . (←) Suppose g ⊆ f . Suppose c ∈ C, and let b = g(c). Then (c, b) ∈ g, so (c, b) ∈ f , and therefore f (c) = b. But then by part (a), (f C)(c) = f (c) = b = g(c). Since c was arbitrary, it follows by Theorem 5.1.4 that g = f C. (c) h Z = h ∩ (Z × R) = {(x, y) ∈ R × R | y = 2x + 3} ∩ (Z × R) = {(x, y) ∈ Z × R | y = 2x + 3} = g.

8. Let A0 = Dom(g). Let (a, b) ∈ g be arbitrary. Then (a, b) ∈ f ⊆ A × B, so b ∈ B, and also a ∈ Dom(g) = A0 . Therefore (a, b) ∈ A0 × B. Since (a, b) was arbitrary, g ⊆ A0 × B. Now let a ∈ A0 be arbitrary. Then a ∈ Dom(g), so there is some b ∈ B such that (a, b) ∈ g. Now suppose (a, b1 ) ∈ g and (a, b2 ) ∈ g. Then (a, b1 ) ∈ f and (a, b2 ) ∈ f , so since f is a function, b1 = b2 . Therefore g is a function from A0 to B. 9. Since B 6= ∅, we can choose some b0 ∈ B. Let g = f ∪ ((A0 \ A) × {b0 }). Clearly f ⊆ g. To prove that g : A0 → B, suppose a ∈ A0 .

Case 1. a ∈ A. Then (a, f (a)) ∈ f ⊆ g. Now suppose (a, b1 ) ∈ g and (a, b2 ) ∈ g. Then since a ∈ A, a ∈ / A0 \ A, so (a, b1 ) ∈ / (A0 \ A) × {b0 } and (a, b2 ) ∈ / (A0 \ A) × {b0 }. Therefore (a, b1 ) ∈ f and (a, b2 ) ∈ f , so since f is a function, b1 = b2 .

Case 2. a ∈ / A. Then a ∈ A0 \ A, so (a, b0 ) ∈ (A0 \ A) × {b0 } ⊆ g. Now suppose (a, b1 ) ∈ g and (a, b2 ) ∈ g. Then since a ∈ / A, (a, b1 ) ∈ / f and (a, b2 ) ∈ / f . Therefore (a, b1 ) ∈ (A0 \ A) × {b0 } and 0 (a, b2 ) ∈ (A \ A) × {b0 }, so b1 = b0 = b2 .

10. Since f 6= g, by Theorem 5.1.4 we can choose some a ∈ A such that f (a) 6= g(a). Therefore (a, f (a)) ∈ f and (a, f (a)) ∈ / g, so by the definition of symmetric difference, (a, f (a)) ∈ f 4 g, and similarly (a, g(a)) ∈ f 4 g. Since f (a) 6= g(a), it follows that f 4 g is not a function.

11. We already know that iA is both an equivalence relation on A and also a function from A to A. Now suppose R is an equivalence relation on A and also a function from A to A. Then R is reflexive on A, so by Theorem 4.3.4, iA ⊆ R. Now suppose (x, y) ∈ R. Since R is reflexive, (x, x) ∈ R. Since R is a function, it follows that y = x, so (x, y) = (x, x) ∈ iA . Since (x, y) was arbitrary, this shows that R ⊆ iA . Since iA ⊆ R and R ⊆ iA , R = iA .

12. We solve part (b) first. (b) (→) Suppose f ∪ g : A ∪ B → C. We already know from exercise 7 that f (A ∩ B) and g (A ∩ B) are functions from A ∩ B to C. Let x ∈ A ∩ B be arbitrary. Let c1 = (f (A ∩ B))(x) and c2 = (g (A ∩ B))(x). Then (x, c1 ) ∈ f (A ∩ B) ⊆ f , so (x, c1 ) ∈ f ∪ g, and similarly (x, c2 ) ∈ f ∪ g. Since f ∪ g is a function, c1 = c2 , so (f (A ∩ B))(x) = (g (A ∩ B))(x). Since x was arbitrary, it follows by Theorem 5.1.4 that f (A ∩ B) = g (A ∩ B). (←) Suppose f (A∩B) = g (A∩B). It is clear that f ∪g ⊆ (A×C)∪(B ×C) = (A∪B)×C. Now suppose x ∈ A ∪ B. Then either x ∈ A or x ∈ B. If x ∈ A then (x, f (x)) ∈ f ⊆ f ∪ g, and if x ∈ B then (x, g(x)) ∈ g ⊆ f ∪ g. Thus, there exists some c ∈ C such that (x, c) ∈ f ∪ g. Now suppose (x, c1 ) ∈ f ∪ g and (x, c2 ) ∈ f ∪ g.

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Case 1. (x, c1 ) ∈ f and (x, c2 ) ∈ f . Then since f is a function, c1 = c2 .

Case 2. (x, c1 ) ∈ g and (x, c2 ) ∈ g. Then since g is a function, c1 = c2 .

Case 3. (x, c1 ) ∈ f and (x, c2 ) ∈ g. Since (x, c1 ) ∈ f , x ∈ A, and since (x, c2 ) ∈ g, x ∈ B. Therefore x ∈ A ∩ B. It follows that c1 = (f (A ∩ B))(x) = (g (A ∩ B))(x) = c2 .

Case 4. (x, c1 ) ∈ g and (x, c2 ) ∈ f . Similar to case 3.

Thus, c1 = c2 , so f ∪ g : A ∪ B → C. (a) Suppose A and B are disjoint. Then A ∩ B = ∅, so f (A ∩ B) = ∅ = g (A ∩ B), and therefore by part (b), f ∪ g : A ∪ B → C. 13. (a) Suppose b ∈ B. Since Dom(S) = B, we know that there is some c ∈ C such that (b, c) ∈ S. To see that it is unique, suppose that c0 ∈ C and (b, c0 ) ∈ S. Since Ran(R) = B, we can choose some a ∈ A such that (a, b) ∈ R. But then (a, c) ∈ S ◦ R and (a, c0 ) ∈ S ◦ R, and since S ◦ R is a function, it follows that c = c0 . (b) A = {1}, B = {2, 3}, C = {4}, R = {(1, 2), (1, 3)}, S = {(2, 4), (3, 4)}.

14. (a) Suppose S is reflexive. Suppose x ∈ A. Then since S is reflexive, (f (x), f (x)) ∈ S, so (x, x) ∈ R. Therefore R is reflexive. (b) Suppose S is symmetric. Suppose (x, y) ∈ R. Then (f (x), f (y)) ∈ S, so since S is symmetric, (f (y), f (x)) ∈ S, and therefore (y, x) ∈ R. Thus R is symmetric. (c) Suppose S is transitive. Suppose (x, y) ∈ R and (y, z) ∈ R. Then (f (x), f (y)) ∈ S and (f (y), f (z)) ∈ S, so since S is transitive, (f (x), f (z)) ∈ S. Therefore (x, z) ∈ R, so R is transitive.

15. (a) No. Example: A = {1}, B = {2, 3}, f = {(1, 2)}, R = {(1, 1)}. (b) Yes. Proof: Suppose R is symmetric. Suppose (x, y) ∈ S. Then we can choose some u and v in A such that f (u) = x, f (v) = y, and (u, v) ∈ R. Since R is symmetric, (v, u) ∈ R, and therefore (y, x) ∈ S. (c) No. Example: A = {1, 2, 3, 4}, B = {5, 6, 7}, f = {(1, 5), (2, 6), (3, 6), (4, 7)}, R = {(1, 2), (3, 4)}.

16. (a) Yes. Proof: Suppose R is reflexive. Let f ∈ F be arbitrary. Let x ∈ A be arbitrary. Since R is reflexive, (f (x), f (x)) ∈ R. Since x was arbitrary, ∀x ∈ A((f (x), f (x)) ∈ R), so (f, f ) ∈ S. Since f was arbitrary, this shows that S is reflexive. (b) Yes. Proof: Suppose R is symmetric. Suppose (f, g) ∈ S. Let x ∈ A be arbitrary. Then (f (x), g(x)) ∈ R, so since R is symmetric, (g(x), f (x)) ∈ R. Since x was arbitrary, (g, f ) ∈ S. Thus, S is symmetric. (c) Yes. Proof: Suppose R is transitive. Suppose (f, g) ∈ S and (g, h) ∈ S. Let x ∈ A be arbitrary.

Then (f (x), g(x)) ∈ R and (g(x), h(x)) ∈ R, so since R is transitive, (f (x), h(x)) ∈ R. Since x was arbitrary, (f, h) ∈ S. Therefore S is transitive.

17. (a) Suppose g : A → A. Let x ∈ A be arbitrary. Then (f ◦ g)(x) = f (g(x)) = a = f (x). Since x was arbitrary, this shows that f ◦ g = f . (b) Since A is nonempty, we can choose some a0 ∈ A. Let a = f (a0 ). Define g : A → A by the formula g(x) = a0 . By assumption, f ◦ g = f , so for every x ∈ A, f (x) = (f ◦ g)(x) = f (g(x)) = f (a0 ) = a.

Thus, f is a constant function.

18. (a) Clearly for all x > 0, f (x) = |x| = x = g(x), so (f, g) ∈ R. (b) Suppose f ∈ F. Then for all x > 0, f (x) = f (x), so (f, f ) ∈ R. Therefore R is reflexive.

Next, suppose (f, g) ∈ R. Then we can choose some a ∈ R such that ∀x > a(f (x) = g(x)).

Then ∀x > a(g(x) = f (x)), so (g, f ) ∈ R. Thus R is symmetric. Finally, suppose (f, g) ∈ R and (g, h) ∈ R. Then we can choose a1 ∈ R and a2 ∈ R such that ∀x > a1 (f (x) = g(x)) and ∀x > a2 (g(x) = h(x)). Let a be the larger of a1 and a2 . Then for every x > a, x > a1 and x > a2 , so f (x) = g(x) and g(x) = h(x), and therefore f (x) = h(x). Thus (f, h) ∈ R, so R is transitive.

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19. (a) Let a = 3 and c = 8. Then for any x > a = 3, |f (x)| = |7x + 3| = 7x + 3 < 7x + x = 8x < 8x2 = c|g(x)|.

This shows that f ∈ O(g).

Now suppose that g ∈ O(f ). Then we can choose a ∈ Z+ and c ∈ R+ such that ∀x > a(|g(x)| ≤ c|f (x)|), or in other words, ∀x > a(x2 ≤ c(7x + 3)). Let x be any positive integer larger than both a and 10c. Multiplying both sides of the inequality x > 10c by x, we can conclude that x2 > 10cx. But since x > a, we also have x2 ≤ c(7x + 3) ≤ c(7x + 3x) = 10cx, so we have reached a contradiction. Therefore g ∈ / O(f ). (b) Clearly for any function f ∈ F we have ∀x ∈ Z+ (|f (x)| ≤ 1 · |f (x)|), so f ∈ O(f ), and therefore (f, f ) ∈ S. Thus, S is reflexive. To see that it is also transitive, suppose (f, g) ∈ S and (g, h) ∈ S. Then there are positive integers a1 and a2 and positive real numbers c1 and c2 such that ∀x > a1 (|f (x)| ≤ c1 |g(x)|) and ∀x > a2 (|g(x)| ≤ c2 |h(x)|). Let a be the maximum of a1 and a2 , and let c = c1 c2 . Then for all x > a, |f (x)| ≤ c1 |g(x)| ≤ c1 c2 |h(x)| = c|h(x)|.

Thus, (f, h) ∈ S, so S is transitive. Finally, to see that S is not a partial order, we show that it is not antisymmetric. Let f and g be the functions from Z+ to R defined by the formulas f (x) = x and g(x) = 2x. Then for all x ∈ Z+ , |f (x)| ≤ |g(x)| and |g(x)| ≤ 2|f (x)|, so f ∈ O(g) and also g ∈ O(f ). Therefore (f, g) ∈ S and (g, f ) ∈ S, but f 6= g. (c) Since f1 ∈ O(g), we can choose a1 ∈ Z+ and c1 ∈ R+ such that ∀x > a1 (|f1 (x)| ≤ c1 |g(x)|).

Similarly, since f2 ∈ O(g) we can choose a2 ∈ Z+ and c2 ∈ R+ such that ∀x > a2 (|f2 (x)| ≤ c2 |g(x)|). Let a be the maximum of a1 and a2 , and let c = |s|c1 + |t|c2 + 1. (We have added 1 here just to make sure that c is positive, as required in the definition of O.) Then for all x > a, |f (x)| = |sf1 (x) + tf2 (x)| ≤ |s||f1 (x)| + |t||f2 (x)| ≤ |s|c1 |g(x)| + |t|c2 |g(x)| = (|s|c1 + |t|c2 )|g(x)| ≤ c|g(x)|.

Therefore f ∈ O(g).

20. (a) Suppose x ∈ A. Since g(x) = g(x), (x, x) ∈ R, so R is reflexive. Suppose (x, y) ∈ R. Then g(x) = g(y). Therefore g(y) = g(x), so (y, x) ∈ R. This proves that R is symmetric. Finally, suppose (x, y) ∈ R and (y, z) ∈ R. Then g(x) = g(y) and g(y) = g(z), and therefore g(x) = g(z).

Thus (x, z) ∈ R, so R is transitive. (b) Let (x, y) ∈ A × A be arbitrary. Then by part 2 of Lemma 4.5.5, (x, y) ∈ R iff x ∈ [y]R iff [x]R = [y]R iff g(x) = g(y).

21. (a) Let h = {(X, y) ∈ A/R × B | ∃x ∈ X(f (x) = y)}. We first show that h : A/R → B. Suppose X ∈ A/R. Then we can choose some x ∈ A such that X = [x]R . Let y = f (x). Since x ∈ [x]R = X and f (x) = y, we have (X, y) ∈ h. To see that y is unique, suppose (X, y 0 ) ∈ h. Then we can choose some x0 ∈ X such that f (x0 ) = y 0 . Since x0 ∈ X = [x]R , x0 Rx, so since f is compatible with R, y 0 = f (x0 ) = f (x) = y.

To see that h is unique, suppose that h0 : A/R → B and for all x ∈ A, h0 ([x]R ) = f (x).

Let X ∈ A/R be arbitrary. Then we can choose some x ∈ A such that X = [x]R . Therefore h0 (X) = h0 ([x]R ) = f (x) = h([x]R ) = h(X). Since X was arbitrary, h0 = h. (b) Suppose x, x0 ∈ R and xRx0 . Then x ∈ [x0 ]R , so by Lemma 4.5.5, [x]R = [x0 ]R . Therefore f (x) = h([x]R ) = h([x0 ]R ) = f (x0 ). Since x and x0 were arbitrary, this shows that f is compatible with R.

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22. (a) Define f : N → N/R by the formula f (x) = [x2 ]R . We first show that f is compatible with R (see exercise 21 for the definition). Suppose x, y ∈ N and xRy. Then x ≡ y (mod 5), so 5 | (x − y) and we can choose an integer k such that x − y = 5k. Therefore x2 − y 2 = (x − y)(x + y) = 5k(x + y), so 5 | (x2 − y 2 ), which means that x2 ≡ y 2 (mod 5). Thus f (x) = [x2 ]R = [y 2 ]R = f (y), and since x and y were arbitrary, this shows that f is compatible with R. Finally, the existence and uniqueness of the function h follow from exercise 21(a). (b) Suppose there is such a function. Define f : N → N/R by the formula f (x) = [2x ]R . By exercise 21(b), f must be compatible with R. But (0, 5) ∈ R, since 0 ≡ 5 (mod 5), and 20 − 25 = −31, which is not divisible by 5, so 20 6≡ 25 (mod 5), and therefore f (0) = [20 ]R 6= [25 ]R = f (5).

This contradicts the compatibility of f with R.

Section 5.2 1. (a) f is not one-to-one, but it is onto. (b) f is not a function. (c) f is one-to-one, but it is not onto.

2. (a) f is not a function from A to B. (b) f is not a function. g is a function that is onto, but not one-to-one. (c) R is one-to-one and onto.

3. (a) f is not one-to-one, but it is onto. (b) f is neither one-to-one nor onto. (c) f is not one-to-one, but it is onto.

4. (a) H is one-to-one, but it is not onto. (b) F is one-to-one and onto. (c) f is one-to-one, but it is not onto.

5. (a) Suppose that x1 ∈ A, x2 ∈ A, and f (x1 ) = f (x2 ). Then we can perform the following algebraic steps: x2 + 1 x1 + 1 = , x1 − 1 x2 − 1 (x1 + 1)(x2 − 1) = (x2 + 1)(x1 − 1), x1 x2 − x1 + x2 − 1 = x1 x2 − x2 + x1 − 1, 2x2 = 2x1 , x2 = x1 .

This shows that f is one-to-one.

To show that f is onto, suppose that y ∈ A. Let x=

y+1 . y−1

Notice that this is defined, since y 6= 1, and also clearly x 6= 1, so x ∈ A. Then f (x) =

y+1 2y x+1 y−1 + 1 y−1 = y+1 = 2 = y. x−1 y−1 y−1 − 1

(b) For any x ∈ A, x+1

(f ◦ f )(x) = x−1 x+1

+1

x−1 − 1

2x

= x−1 = x = iA (x).

2 x−1

64

SOLUTIONS 6. Suppose x1 , x2 ∈ R and f (x1 ) = f (x2 ). This means ax1 + b = ax2 + b, so ax1 = ax2 . Since a 6= 0, we can divide by a to conclude that x1 = x2 . This proves that f is one-to-one. To prove that it is onto, suppose y ∈ R. Let x = (y − b)/a. Then f (x) = a((y − b)/a) + b = y − b + b = y. Since y was arbitrary, this shows that f is onto.

7. (a) Following the hint, we first prove that if 0 < a < b then f (a) > f (b). Suppose 0 < a < b. Then by exercise 6 in Section 3.1, 1/a > 1/b, and also −a > −b. Adding these inequalities, we conclude that f (a) = 1/a − a > 1/b − b = f (b).

Now suppose f is not one-to-one. Then there are positive real numbers x1 and x2 such that x1 6= x2 and f (x1 ) = f (x2 ). Since x1 6= x2 , either x1 < x2 or x2 < x1 . But then by the previous paragraph, either f (x1 ) > f (x2 ) or f (x1 ) < f (x2 ), which is a contradiction. p p p (b) Suppose y ∈ R. Let x = ( y 2 + 4 − y)/2. Note that y 2 + 4 > y 2 = |y| ≥ y, so x ∈ R+ . Also, p p p 2 y2 + 4 − y y2 + 4 + y y2 + 4 − y 2y p f (x) = − = − = = y.

2 2 2 2 2 y +4−y (c) g(2) = 5/2 = g(1/2), so g is not one-to-one. For every x ∈ R+ , g(x) = 1/x + x > 0, so −1 ∈ / Ran(g). Therefore g is not onto. √ √ 8. (a) f (2) = {y ∈ R | y 2 < 2} = {y ∈ R | − 2 < y < 2}. (b) f (0) = f (−1) = ∅, so f is not one-to-one. And {y ∈ R | −1 < y < 2} is not in Ran(f ), so f is not onto.

S 9. (a) f ({{1, 2}, {3, 4}}) = {{1, 2}, {3, 4}} = {1, 2} ∪ {3, 4} = {1, 2, 3, 4}. (b) f ({{1}, {2, 3, 4}}) = {1} ∪ {2, 3, 4} S = {1, 2, 3, 4} = f ({{1, 2}, {3, 4}}), so f is not one-to-one. For every set X ∈ P(R), f ({X}) = {X} = X, so f is onto.

10. (a) Suppose g ◦ f is onto. Let c ∈ C be arbitrary. Since g ◦ f is onto, we can choose some a ∈ A such that (g ◦ f )(a) = c. Let b = f (a) ∈ B. Then g(b) = g(f (a)) = (g ◦ f )(a) = c. Therefore g is onto. (b) Suppose g ◦ f is one-to-one. Suppose a1 , a2 ∈ A and f (a1 ) = f (a2 ). Then (g ◦ f )(a1 ) = g(f (a1 )) = g(f (a2 ) = (g ◦ f )(a2 ), so since g ◦ f is one-to-one, a1 = a2 . Therefore f is one-to-one.

11. (a) Suppose f is onto and g is not one-to-one. Since g is not one-to-one, we can choose b1 , b2 ∈ B such that b1 6= b2 and g(b1 ) = g(b2 ). Since f is onto, we can choose a1 , a2 ∈ A such that f (a1 ) = b1 and f (a2 ) = b2 . If a1 = a2 then b1 = f (a1 ) = f (a2 ) = b2 , which is a contradiction, so a1 6= a2 . But (g ◦ f )(a1 ) = g(f (a1 )) = g(b1 ) = g(b2 ) = g(f (a2 )) = (g ◦ f )(a2 ). Therefore g ◦ f is not one-to-one. (b) Suppose f is not onto and g is one-to-one. Since f is not onto, we can choose some b ∈ B such that b ∈ / Ran(f ). Let c = g(b) ∈ C. Suppose g ◦ f is onto. Then we can choose some a ∈ A such that g(f (a)) = (g ◦ f )(a) = c = g(b). Since g is one-to-one, it follows that f (a) = b, which contradicts the fact that b ∈ / Ran(f ). Therefore g ◦ f is not onto.

12. Suppose f is onto. Suppose b1 , b2 ∈ B and g(b1 ) = g(b2 ). Since f is onto, we can choose some a ∈ A such that f (a) = b1 . Then a ∈ g(b1 ) = g(b2 ), so f (a) = b2 . Therefore b1 = b2 , so g is one-to-one.

If f is not onto, then g may fail to be one-to-one. Example: A = {1}, B = {2, 3, 4}, f = {(1, 2)}.

Then g(3) = g(4) = ∅.

13. (a) Suppose that f is one-to-one. Suppose that c1 ∈ C, c2 ∈ C, and (f C)(c1 ) = (f C)(c2 ). By exercise 7(a) of Section 5.1, it follows that f (c1 ) = f (c2 ), so since f is one-to-one, c1 = c2 . (b) Suppose that f C is onto. Suppose b ∈ B. Then since f C is onto, we can choose some c ∈ C such that (f C)(c) = b. But then c ∈ A, and by exercise 7(a) of Section 5.1, f (c) = b. (c) Let A = B = R and C = R+ . For (a), use f (x) = |x|, and for (b), use f (x) = x.

14. (a) Suppose A has more than one element. Let a1 and a2 be two distinct elements of A. Then f (a1 ) = b = f (a2 ), so f is not one-to-one.

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(b) Suppose B has more than one element. Then we can choose b0 ∈ B such that b0 6= b. Then for all a ∈ A, f (a) = b 6= b0 , so f is not onto.

15. (→) Suppose f ∪ g is one-to-one. Suppose Ran(f ) ∩ Ran(g) 6= ∅. Then we can choose some c ∈ Ran(f ) ∩ Ran(g). Since c ∈ Ran(f ), we can choose some a ∈ A such that f (a) = c. Similarly, since c ∈ Ran(g), we can choose some b ∈ B such that g(b) = c. Therefore (a, c) ∈ f ∪ g and (b, c) ∈ f ∪ g, so (f ∪ g)(a) = c = (f ∪ g)(b). But since A and B are disjoint, a 6= b. This contradicts the fact that f ∪ g is one-to-one. Therefore Ran(f ) and Ran(g) are disjoint. (←) Suppose Ran(f ) and Ran(g) are disjoint. Suppose x1 , x2 ∈ A∪B and (f ∪g)(x1 ) = (f ∪g)(x2 ).

Let c = (f ∪ g)(x1 ) = (f ∪ g)(x2 ) ∈ C. Case 1. x1 ∈ A. Then (x1 , f (x1 )) ∈ f ⊆ f ∪ g, so f (x1 ) = (f ∪ g)(x1 ) = c. Suppose x2 ∈ B. Then (x2 , g(x2 )) ∈ g ⊆ f ∪g, so g(x2 ) = (f ∪g)(x2 ) = c. But then c ∈ Ran(f )∩Ran(g), which contradicts the fact that Ran(f ) and Ran(g) are disjoint. Therefore x2 ∈ / B, so x2 ∈ A. Thus (x2 , f (x2 )) ∈ f ⊆ f ∪ g, so f (x2 ) = (f ∪ g)(x2 ) = c = f (x1 ). Since f is one-to-one, x1 = x2 .

Case 2. x1 ∈ B. Then a similar argument shows that x2 ∈ B and g(x1 ) = c = g(x2 ), so since g is one-to-one, x1 = x2 .

Thus x1 = x2 . Since x1 and x2 were arbitrary, f ∪ g is one-to-one.

16. Suppose S is one-to-one. Suppose a ∈ A. Let c = (S ◦ R)(a) ∈ C. Then (a, c) ∈ S ◦ R, so we can choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Now suppose b0 ∈ B and (a, b0 ) ∈ R. Since S : B → C, we can let c0 = S(b0 ). Then (a, b0 ) ∈ R and (b0 , c0 ) ∈ S, and therefore (a, c0 ) ∈ S ◦ R. Since (a, c) ∈ S ◦ R, (a, c0 ) ∈ S ◦ R, and S ◦ R is a function, c = c0 . Therefore S(b0 ) = c0 = c = S(b), so since S is one-to-one, b0 = b. Thus b is the only element of B such that (a, b) ∈ R. Since a was arbitrary, we conclude that R : A → B. 17. (a) Suppose R is reflexive and f is onto. Let x ∈ B be arbitrary. Since f is onto, we can choose some u ∈ A such that f (u) = x. Since R is reflexive, (u, u) ∈ R. Therefore (x, x) ∈ S. (b) Suppose R is transitive and f is one-to-one. Suppose that (x, y) ∈ S and (y, z) ∈ S. Since (x, y) ∈ S, we can choose some u and v in A such that f (u) = x, f (v) = y, and (u, v) ∈ R. Similarly, since (y, z) ∈ S we can choose p and q in A such that f (p) = y, f (q) = z, and (p, q) ∈ R. Since f (v) = y = f (p) and f is one-to-one, v = p. Therefore (v, q) = (p, q) ∈ R. Since we also have (u, v) ∈ R, by transitivity of R it follows that (u, q) ∈ R, so (x, z) ∈ S. 18. (a) Suppose X ∈ A/R. Then we can choose some x ∈ A such that X = [x]R . Therefore g(x) = [x]R = X, so g is onto. (b) (→) Suppose g is one-to-one. Since R is reflexive, by Theorem 4.3.4, iA ⊆ R. Now suppose (x, y) ∈ R. Then g(x) = [x]R = [y]R = g(y), so since g is one-to-one, x = y, and therefore (x, y) ∈ iA . Thus R ⊆ iA . Since iA ⊆ R and R ⊆ iA , R = iA . (←) Suppose R = iA . Suppose a1 , a2 ∈ A and g(a1 ) = g(a2 ). Then [a1 ]R = [a2 ]R . Since a1 ∈ [a1 ]R , it follows that a1 ∈ [a2 ]R , so (a1 , a2 ) ∈ R = iA , and therefore a1 = a2 . Since a1 and a2 were arbitrary, g is one-to-one.

19. (→) Suppose h is one-to-one. Suppose x, y ∈ A and f (x) = f (y). Then h([x]R ) = f (x) = f (y) = h([y]R ), so since h is one-to-one, [x]R = [y]R . Since x ∈ [x]R , x ∈ [y]R , so xRy. Since x and y were arbitrary, this shows that ∀x ∈ A∀y ∈ A(f (x) = f (y) → xRy). (←) Suppose ∀x ∈ A∀y ∈ A(f (x) = f (y) → xRy). Suppose X, Y ∈ A/R and h(X) = h(Y ). By definition of A/R, we can choose x, y ∈ A such that X = [x]R and Y = [y]R . Then f (x) = h([x]R ) = h(X) = h(Y ) = h([y]R ) = f (y), so by our assumption xRy. Therefore x ∈ [y]R , so by Lemma 4.5.5, X = [x]R = [y]R = Y . Since X and Y were arbitrary, this shows that h is one-to-one.

20. (a) Suppose that f is onto, g : B → C, h : B → C, and g ◦ f = h ◦ f . Let b ∈ B be arbitrary. Since f is onto, we can choose some a ∈ A such that f (a) = b. Therefore g(b) = g(f (a)) = (g ◦ f )(a) = (h ◦ f )(a) = h(f (a)) = h(b). Since b was arbitrary, this shows that ∀b ∈ B(g(b) = h(b)), so g = h.

66

SOLUTIONS (b) Suppose that C has at least two elements, and for all functions g and h from B to C, if g ◦f = h◦f then g = h. Let c1 and c2 be two distinct elements of C. Suppose b ∈ B. Let g and h be functions from B to C such that ∀x ∈ B(g(x) = c1 ), ∀x ∈ B \ {b}(h(x) = c1 ), and h(b) = c2 . (Formally, g = B × {c1 } and h = [(B \ {b}) × {c1 }] ∪ {(b, c2 )}.) Then g 6= h, so by assumption g ◦ f 6= h ◦ f , and therefore we can choose some a ∈ A such that g(f (a)) 6= h(f (a)). But by the way g and h were defined, the only x ∈ B for which g(x) 6= h(x) is x = b, so it follows that f (a) = b. Since b was arbitrary, this shows that f is onto.

21. (a) Suppose that f is one-to-one, g : A → B, h : A → B, and f ◦ g = f ◦ h. Let a ∈ A be arbitrary.

Then f (g(a)) = (f ◦ g)(a) = (f ◦ h)(a) = f (h(a)), and since f is one-to-one it follows that g(a) = h(a). Since a was arbitrary, we conclude that g = h. (b) Suppose that A 6= ∅, and for all functions g and h from A to B, if f ◦g = f ◦h then g = h. Suppose b1 , b2 ∈ B and f (b1 ) = f (b2 ). Let g and h be functions from A to B such that ∀a ∈ A(g(a) = b1 ) and ∀a ∈ A(h(a) = b2 ). (Formally, g = A × {b1 } and h = A × {b2 }.) Then for all a ∈ A, (f ◦ g)(a) = f (g(a)) = f (b1 ) = f (b2 ) = f (h(a)) = (f ◦ h)(a), so f ◦ g = f ◦ h, and therefore by assumption g = h. Since A 6= ∅, we can choose some a0 ∈ A. Then b1 = g(a0 ) = h(a0 ) = b2 .

Since b1 and b2 were arbitrary, f is one-to-one.

22. (a) Let j : R → R be defined by the formula j(x) = (x − 1)2 + 1. Then for all x ∈ R, (j ◦ f )(x) = j(f (x)) = j(x2 + 1) = (x2 + 1 − 1)2 + 1 = x4 + 1 = h(x). Therefore h = j ◦ f , so hRf .

Now suppose gRf . Then we can choose some function k : R → R such that g = k ◦ f . But then 2 = g(1) = (k ◦ f )(1) = k(f (1)) = k(2) = k(f (−1)) = (k ◦ f )(−1) = g(−1) = 0, which is a contradiction. Therefore it is not the case that gRf . (b) For every function f ∈ F, f = iR ◦ f , so f Rf . Therefore R is reflexive. Now suppose f, g, h ∈ F, f Rg, and gRh. Then we can choose functions j and k from R to R such that f = j ◦ g and g = k ◦ h. Therefore f = j ◦ (k ◦ h) = (j ◦ k) ◦ h, so f Rh. Thus f is transitive. (c) For every f ∈ F, f = f ◦ iR , so f RiR . (d) Let f ∈ F be arbitrary. (→) Suppose iR Rf . Then there is some function g : R → R such that g ◦ f = iR . See part 1 of Theorem 5.3.3 for a proof that f is one-to-one. (←) Suppose f is one-to-one. Let A = Ran(f ) and let h = f −1 ∪ ((R \ A) × {0}). To see that h : R → R, suppose x ∈ R. Case 1. x ∈ A. Then x ∈ Ran(f ), so we can choose some y ∈ R such that (y, x) ∈ f , and therefore (x, y) ∈ f −1 ⊆ h. Now suppose that for some y 0 ∈ R, (x, y 0 ) ∈ h. Clearly (x, y 0 ) ∈ / (R \ A) × {0}, since x ∈ A, so (x, y 0 ) ∈ f −1 . Therefore (y 0 , x) ∈ f , so f (y 0 ) = x = f (y).

Since f is one-to-one, y 0 = y.

Case 2. x ∈ / A. Then (x, 0) ∈ (R \ A) × {0} ⊆ h. Also, if (x, y) ∈ h then (x, y) ∈ / f −1 , since x∈ / A = Ran(f ), so (x, y) ∈ (R \ A) × {0}, and therefore y = 0.

Thus ∀x ∈ R∃!y ∈ R((x, y) ∈ h), so h : R → R. Finally, for any x ∈ R we have (f (x), x) ∈ f −1 ⊆ h, so (h ◦ f )(x) = h(f (x)) = x = iR (x). Therefore iR = h ◦ f , so iR Rf . (e) By part 1 of exercise 17 in Section 5.1, for all f ∈ R, g = g ◦ f , so gRf . (f) (→) Suppose f Rg. Then we can choose some function h : R → R such that f = h ◦ g. Since g is a constant function, we can choose some a ∈ R such that ∀x ∈ R(g(x) = a). Therefore for every x ∈ R, f (x) = (h ◦ g)(x) = h(g(x)) = h(a), so f is a constant function (with constant value h(a)). (←) Suppose f is a constant function. Then by part (e), f Rg. (g) We first show that [iR ]S = {f ∈ F | f is one-to-one}. Suppose f ∈ [iR ]S . Then (f, iR ) ∈ S = R ∩ R−1 , so iR Rf , and therefore, by part (d), f is one-to-one. Now suppose f ∈ F and f is one-to-one. Then by part (d), iR Rf , and by part (c), f RiR . Therefore (f, iR ) ∈ R ∩ R−1 = S, so f ∈ [iR ]S . This completes the proof that {f ∈ F | f is one-to-one} = [iR ]S ∈ F/S. To see that it is the largest element of F/S, let X ∈ F/S be arbitrary. Then we can choose some f ∈ F such

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that X = [f ]S . By part (c), f RiR , so [f ]S T [iR ]S , which means XR[iR ]S . Since X was arbitrary, this shows that [iR ]S is the largest element of F/S. The second half of the problem is very similar. Let g = R × {0}. Then g is the constant function whose constant value is 0. We first show that [g]S = {f ∈ F | f is a constant function}.

To prove this, suppose f ∈ [g]S . Then (f, g) ∈ S = R ∩ R−1 , so f Rg, and therefore, by part (f), f is a constant function. Now suppose f ∈ F and f is a constant function. Then by part (f), f Rg, and by part (e), gRf . Therefore (f, g) ∈ R ∩ R−1 = S, so f ∈ [g]S . This completes the proof that {f ∈ F | f is a constant function} = [g]S ∈ F/S. To see that it is the smallest element of F/S, let X ∈ F/S be arbitrary. Then we can choose some f ∈ F such that X = [f ]S . By part (e), gRf , so [g]S T [f ]S , which means [g]S T X. Since X was arbitrary, this shows that [g]S is the smallest element of F/S. 23. (a) Yes, because for every n ∈ N, f (n) = n, so n ∈ Ran(f ). (b) No, because, for example, −1 ∈ Z but −1 ∈ / Ran(f ).

Section 5.3 1. R−1 (p) = the person sitting immediately to the right of p.

2. F −1 (X) = A \ X. Thus, F −1 = F .

3. Let g(x) = (3x − 5)/2. Then for any x ∈ R, f (g(x)) =

3x − 5 + 5 3x 2(3x − 5)/2 + 5 = = =x 3 3 3

g(f (x)) =

3(2x + 5)/3 − 5 2x + 5 − 5 2x = = = x.

2 2 2

and

Therefore f ◦ g = iR and g ◦ f = iR , and by Theorems 5.3.4 and 5.3.5 it follows that f is one-to-one and onto and f −1 = g, so f −1 (x) = g(x) = (3x − 5)/2. p 4. Let g(x) = 3 (x + 3)/2. Then for any x ∈ R, r f (g(x)) = 2

3

x+3 2

!3 −3=2·

x+3 −3=x+3−3=x 2

and r g(f (x)) =

3

r 3 √ 2x3 − 3 + 3 3 3 2x = = x3 = x.

2 2

Therefore f ◦ g = iR and g ◦ f = iR , andpas in the previous exercise it follows that f is one-to-one and onto and f −1 = g, so f −1 (x) = g(x) = 3 (x + 3)/2.

5. Let g : R+ → R be defined by the formula g(x) = 2 − log x. Then for all x ∈ R+ , f (g(x)) = 102−(2−log x) = 10log x = x, and for all x ∈ R, g(f (x)) = 2 − log(102−x ) = 2 − (2 − x) = x.

Therefore f ◦ g = iR+ and g ◦ f = iR , and it follows that f is one-to-one and onto and f −1 = g, so f −1 (x) = g(x) = 2 − log x.

68

SOLUTIONS 6. (a) Let B = R \ {3}. Notice that for all x ∈ A, f (x) =

3x 3x − 6 6= = 3, x−2 x−2

so f (x) ∈ B. Therefore f : A → B. Define g : B → A by the formula g(x) = 2x/(x − 3). Then for all x ∈ B, f (g(x)) =

2x 3 · x−3 2x x−3 − 2

6x

= x−3 = 6 x−3

6x = x, 6

and for all x ∈ A, g(f (x)) =

3x 2 · x−2 3x x−2 − 3

6x

= x−2 = 6 x−2

6x = x.

6

Therefore f ◦ g = iB and g ◦ f = iA , so f is one-to-one and maps onto B. (b) It follows from part (a) that f −1 = g, so f −1 (x) = g(x) = 2x/(x − 3).

7. (a) For all x ∈ R, (f2 ◦ f1 )(x) = f2 (f1 (x)) = f2 (x + 7) = (x + 7)/5 = f (x). (b) It is easy to verify that f1−1 (x) = x − 7 and f2−1 (x) = 5x. Therefore (f1−1 ◦ f2−1 )(x) = f1−1 (f2−1 (x)) = f1−1 (5x) = 5x − 7 = f −1 (x).

8. (a) Let b be an arbitrary element of B. Let a = f −1 (b) ∈ A. Then (b, a) ∈ f −1 , so (a, b) ∈ f and therefore f (a) = b. Thus, (f ◦ f −1 )(b) = f (f −1 (b)) = f (a) = b = iB (b).

Since b was arbitrary, we have shown that ∀b ∈ B((f ◦ f −1 )(b) = iB (b)), so f ◦ f −1 = iB . (b) Since f −1 : B → A and (f −1 )−1 = f : A → B, by the first half of Theorem 5.3.2, iB = (f −1 )−1 ◦ f −1 = f ◦ f −1 .

9. Suppose that f : A → B, g : B → A, and f ◦ g = iB . Let b be an arbitrary element of B. Let a = g(b) ∈ A. Then f (a) = f (g(b)) = (f ◦ g)(b) = iB (b) = b. Since b was arbitrary, this shows that f is onto.

10. Let (b, a) be an arbitrary element of B × A. Suppose (b, a) ∈ g. Then g(b) = a, so f (a) = f (g(b)) = (f ◦ g)(b) = iB (b) = b. Therefore (a, b) ∈ f , so (b, a) ∈ f −1 . Now suppose (b, a) ∈ f −1 . Then (a, b) ∈ f , so f (a) = b. Therefore g(b) = g(f (a)) = (g ◦ f )(a) = iA (a) = a, so (b, a) ∈ g.

11. (a) Suppose that f is one-to-one and f ◦ g = iB . By part 2 of Theorem 5.3.3, f is also onto, so f −1 : B → A and f −1 ◦ f = iA . This gives us enough information to imitate the reasoning in the proof of Theorem 5.3.5: g = iA ◦ g = (f −1 ◦ f ) ◦ g = f −1 ◦ (f ◦ g) = f −1 ◦ iB = f −1 . (b) Suppose f is onto and g ◦f = iA . By part 1 of Theorem 5.3.3, f is also one-to-one, so f −1 : B → A and f ◦ f −1 = iB . Therefore g = g ◦ iB = g ◦ (f ◦ f −1 ) = (g ◦ f ) ◦ f −1 = iA ◦ f −1 = f −1 . (c) Suppose f ◦ g = iB and g ◦ f 6= iA . By part 2 of Theorem 5.3.3, f is onto. If f is one-to-one, then by part (a), g = f −1 , so f −1 : B → A. But then by Theorem 5.3.2, g ◦ f = f −1 ◦ f = iA , which is a contradiction. Therefore f is not one-to-one. Similarly, by part 1 of Theorem 5.3.3, g is one-to-one. If g is onto, then by part (b), f = g −1 , so g −1 : A → B. But then by Theorem 5.3.2, g ◦ f = g ◦ g −1 = iA , which is a contradiction. Therefore g is not onto.

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12. Let B 0 = Ran(f ) ⊆ B. Then f : A → B 0 , f is one-to-one, and f maps onto B 0 . Therefore by Theorem 5.3.1, f −1 : B 0 → A. 13. (a) It is clear from the definition of R that ∀x ∈ A∀y ∈ A(xRy → f (x) = f (y)), so in the terminology of exercise 21 in Section 5.1, f is compatible with R. Therefore, by part (a) of that exercise, there is a function h : A/R → B such that for all x ∈ A, h([x]R ) = f (x). (b) It is clear from the definition of R that ∀x ∈ A∀y ∈ A(f (x) = f (y) → xRy). Therefore, by exercise 19 in Section 5.2, h is one-to-one. To prove that h is onto, suppose b ∈ B. Since f is onto, there is some x ∈ A such that f (x) = b. Therefore h([x]R ) = f (x) = b. Since b was arbitrary, this proves that h is onto. (c) Suppose b ∈ B. Since f is onto, we can choose some y ∈ A such that f (y) = b. Therefore h([y]R ) = f (y) = b, so h−1 (b) = [y]R = {x ∈ A | xRy} = {x ∈ A | f (x) = f (y)} = {x ∈ A | f (x) = b}. (d) By part (c), f ◦ g = iB iff ∀b ∈ B(f (g(b)) = b) iff ∀b ∈ B(g(b) ∈ {x ∈ A | f (x) = b}) iff ∀b ∈ B(g(b) ∈ h−1 (b)).

14. (a) Suppose x ∈ A0 = Ran(g). Then we can choose some b ∈ B such that g(b) = x. Therefore (g ◦ f )(x) = g(f (g(b))) = g((f ◦ g)(b)) = g(iB (b)) = g(b) = x. (b) By the given information, (f A0 ) ◦ g = iB , and by part (a), g ◦ (f A0 ) = iA0 . Therefore

by Theorem 5.3.4, f A0 is a one-to-one, onto function from A0 to B, and by Theorem 5.3.5, g = (f A0 )−1 . √ 15. Note that Ran(g) = {x ∈ R | x ≥ 0} = B. Also, for all x ∈ B, (f ◦ g)(x) = f (g(x)) = ( x)2 = x, so f ◦ g = iB . Therefore by exercise 14, g = (f B)−1 .

16. (a) B = {x ∈ R | x ≤ 4}. To prove this, suppose first that y ∈ B = Ran(f ). Then we can choose some x ∈ R such that y = f (x) = 4x − x2 . Therefore 4 − y = x2 − 4x + 4 = (x − 2)2 ≥ 0, so y ≤ 4, and therefore√y ∈ {x ∈ R | x ≤ 4}. Now suppose y ∈ {x ∈ R | x ≤ 4}. Then y ∈ R and y ≤ 4.

Let x = 2 + 4 − y, which is defined since 4 − y ≥ 0. Then p p p p f (x) = 4(2 + 4 − y) − (2 + 4 − y)2 = 8 + 4 4 − y − (4 + 4 4 − y + 4 − y) = y, so y ∈ Ran(f ) = B. √ (b) Since B = Ran(f ), f : R → B. Define g : B → R by the formula g(x) = 2 + 4 − x. The calculation in part (a) shows that for all y ∈ B, f (g(y)) = y, which shows that f ◦ g = iB . By exercise 14, if we let A = √ Ran(g) then f A is a one-to-one, onto function from A to B, and (f A)−1 (x) = g(x) = 2 + 4 − x.

We now prove that A = {x ∈ R | x ≥ 2}. To prove this,√suppose first √ that y ∈ A = Ran(g).

Then we can choose some x ∈ R such that y = g(x) = 2 + 4 − x. Since 4 − x ≥ 0, y ≥ 2, so 2 y ∈ {x ∈ R | x ≥ 2}. Now suppose y ∈ {x ∈ R | x ≥ 2}. Then y ∈ R and y ≥ 2. √ Let x = 4y − y .

2 2 Then 4 − x√= y − 4y + 4 = (y − 2) , and since y ≥ 2, y − 2 ≥ 0. Therefore 4 − x = y − 2, so g(x) = 2 + 4 − x = 2 + y − 2 = y. This proves that y ∈ Ran(g) = A. 17. (a) Suppose f ∈ F. As we saw in Example 5.2.2, iA is one-to-one and onto, so iA ∈ P. Also, i−1 A ◦ f ◦ iA = iA ◦ f ◦ iA = f , so (f, f ) ∈ R. Therefore R is reflexive. Next, suppose (f, g) ∈ R. Then we can choose some h ∈ P such that f = h−1 ◦ g ◦ h. Since (h−1 )−1 = h : A → A, by Theorem 5.3.4, h−1 is one-to-one and onto, so h−1 ∈ P. Also, (h−1 )−1 ◦ f ◦ h−1 = h ◦ (h−1 ◦ g ◦ h) ◦ h−1 = (h ◦ h−1 ) ◦ g ◦ (h ◦ h−1 ) = iA ◦ g ◦ iA = g,

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SOLUTIONS so (g, f ) ∈ R. Therefore R is symmetric. Finally, suppose (f, g) ∈ R and (g, h) ∈ R. Then we can choose functions j, k ∈ P such that f = j −1 ◦ g ◦ j and g = k −1 ◦ h ◦ k. By Theorem 5.2.5, k ◦ j is one-to-one and onto, so k ◦ j ∈ P. Also, applying part 5 of Theorem 4.2.5, f = j −1 ◦ g ◦ j = j −1 ◦ (k −1 ◦ h ◦ k) ◦ j = (j −1 ◦ k −1 ) ◦ h ◦ (k ◦ j) = (k ◦ j)−1 ◦ h ◦ (k ◦ j), so (f, h) ∈ R. Therefore R is transitive. (b) Suppose f Rg. Then we can choose some h ∈ P such that f = h−1 ◦ g ◦ h. Therefore f ◦ f = (h−1 ◦ g ◦ h) ◦ (h−1 ◦ g ◦ h) = h−1 ◦ g ◦ (h ◦ h−1 ) ◦ g ◦ h = h−1 ◦ g ◦ iA ◦ g ◦ h = h−1 ◦ (g ◦ g) ◦ h, so (f ◦ f )R(g ◦ g). (c) Suppose f has a fixed point and f Rg. Then we can choose some a ∈ A such that f (a) = a, and we can also choose some h ∈ P such that f = h−1 ◦ g ◦ h. Therefore a = f (a) = h−1 (g(h(a))), so h(a) = g(h(a)). This means that h(a) is a fixed point of g.

18. Since g is one-to-one and onto, g −1 : C → B. Let h = g −1 ◦ f . Then h : A → B and g ◦ h = g ◦ (g −1 ◦ f ) = (g ◦ g −1 ) ◦ f = iC ◦ f = f .

Section 5.4 1. (a) No: f (0) = 1/2 ∈ / Z. (b) Yes. (c) Yes. (d) No: f (2) = 3/2 ∈ / {x ∈ R | 2 ≤ x < 4}.

2. (a) Yes. (b) Yes. (c) No: f ({100, 101, 102, . . . , 199}) = {17, 100, 101, 102, . . . , 199}, which has 101 elements. (d) Yes. 3. f (−1) = 2 and f (1) = 0, so we need to add those values to the set. It is easy to check that {−1, 0, 1, 2} is closed under f , so it is the closre.

4. By exercise 12 of Section 4.3, all three sets sets are closed under f .

5. Yes: the statement ∀x ∈ ∅(f (x) ∈ ∅) is vacuously true.

6. (a) Suppose Ran(f ) ⊆ C ⊆ A. Suppose x ∈ C. Then f (x) ∈ Ran(f ), so f (x) ∈ C. Therefore C is closed under f . (b) Clearly B ⊆ B ∪ Ran(f ) ⊆ A, and by part (a), B ∪ Ran(f ) is closed under f . Since the closure of B under f is the smallest subset of A that contains B and is closed under f , the closure is a subset of B ∪ Ran(f ).

7. Suppose C ⊆ A and C is closed under f . Suppose x ∈ A \ C, so x ∈ A and x ∈ / C. Then f −1 (x) ∈ A. −1 −1 Suppose f (x) ∈ C. Then since C is closed under f , x = f (f (x)) ∈ C, which is a contradiction.

Therefore f −1 (x) ∈ / C, so f −1 (x) ∈ A \ C. Since x was an arbitrary element of A \ C, this shows that A \ C is closed under f −1 .

8. Let D be the closure of C under f . (→) Suppose C is closed under f . Since D is the smallest subset of A that contains C and is closed under f , D ⊆ C. But also by definition of closure, C ⊆ D, so D = C. (←) Suppose D = C. By definition, D is closed under f , so C is closed under f .

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9. (a) Suppose x ∈ C1 ∪ C2 . Then either x ∈ C1 or x ∈ C2 .

Case 1. x ∈ C1 . Then since C1 is closed under f , f (x) ∈ C1 , so f (x) ∈ C1 ∪ C2 .

Case 2. x ∈ C2 . Then since C2 is closed under f , f (x) ∈ C2 , so f (x) ∈ C1 ∪ C2 .

Therefore f (x) ∈ C1 ∪ C2 . Since x was arbitrary, C1 ∪ C2 is closed under f . (b) Yes. Proof: Suppose x ∈ C1 ∩ C2 . Then x ∈ C1 and x ∈ C2 . Since x ∈ C1 and C1 is closed under f , f (x) ∈ C1 . Similarly, f (x) ∈ C2 . Therefore f (x) ∈ C1 ∩ C2 , so since x was arbitrary, C1 ∩ C2 is closed under f . (c) No. Here is a counterexample: A = {1, 2}, f = {(1, 2), (2, 2)}, C1 = {1, 2}, C2 = {2}.

10. (a) Suppose B1 ⊆ B2 . Then B1 ⊆ B2 ⊆ C2 ⊆ A and C2 is closed under f . Since C1 is the smallest subset of A that contains B1 and is closed under f , C1 ⊆ C2 . (b) Clearly B1 ∪ B2 ⊆ C1 ∪ C2 ⊆ A, and by part (a) of exercise 9, C1 ∪ C2 is closed under f . Now suppose B1 ∪ B2 ⊆ D ⊆ A and D is closed under f . Then since B1 ⊆ D and C1 is the smallest subset of A that contains B1 and is closed under f , C1 ⊆ D. Similarly, since B2 ⊆ D, C2 ⊆ D. Therefore C1 ∪ C2 ⊆ D. Since D was arbitrary, this shows that C1 ∪ C2 is the closure of B1 ∪ B2 under f . (c) No. Counterexample: A = {1, 2, 3}, f = {(1, 1), (2, 2), (3, 2)}, B1 = {1, 2}, B2 = {1, 3}. (d) No. The counterexample in part (c) is a counterexample for (d) as well.

11. Let F = {C ⊆ A | B ⊆ C and C is closed T under f }. You should be able to check that A ∈ F, and therefore F 6= ∅. Thus, we can let C = F, and by exercise 9 of Section 3.3, C ⊆ A. We will show that C is the closure of B under f by proving the three properties in Definition 5.4.8.

To prove the first property, suppose x ∈ B. Let D be an arbitrary element of F. Then by the definition of F, B ⊆ D, so x ∈ D. Since D was arbitrary, this shows that ∀D ∈ F(x ∈ D), so T x ∈ F = C. Thus, B ⊆ C. T Next, suppose x, y ∈ C and again let D be an arbitrary element of F. Then since x, y ∈ C = F, x ∈ D and y ∈ D. But since D ∈ F, D is closed under T f , so f (x, y) ∈ D. Since D was arbitrary, we can conclude that ∀D ∈ F(f (x, y) ∈ D), so f (x, y) ∈ F = C. Thus, we have shown that C is closed under f , which is the second property in Definition 5.4.8.

Finally, to prove the third property, suppose B ⊆ D ⊆ A and D is T closed under f . Then D ∈ F, and applying exercise 9 of Section 3.3 again we can conclude that C = F ⊆ D. 12. (a) Z. (b) {X ⊆ N | X is finite}.

13. (a) Let G = {C ⊆ A | B ⊆ C and C is closed under T F}. You should be able to check that A ∈ G, and therefore G = 6 ∅. Thus, we can let C = G, and by exercise 9 of Section 3.3, C ⊆ A. We will show that C is the closure of B under F. To prove that B ⊆ C, suppose x ∈ B. Let D be an arbitrary element of G. Then by the definition of G, B ⊆ D, so x ∈ D. Since D was arbitrary, this shows that ∀D ∈ G(x ∈ D), so T x ∈ G = C. Thus, B ⊆ C. Next,Tsuppose f ∈ F and x ∈ C and again let D be an arbitrary element of G. Then since x ∈ C = G, x ∈ D. But since D ∈ G, D is closed under T F, so f (x) ∈ D. Since D was arbitrary, we can conclude that ∀D ∈ G(f (x) ∈ D), so f (x) ∈ G = C. Thus, we have shown that C is closed under F. Finally, to prove that C is the smallest subset of A containing B that is closed under F, suppose B ⊆ D ⊆ A and D is closed T under F. Then D ∈ G, and applying exercise 9 of Section 3.3 again we can conclude that C = G ⊆ D. S (b) Suppose x ∈ f ∈F Cf . Then we can choose some f ∈ F such that x ∈ Cf . Since C is closed under F and f ∈ F, C is closed under f . Also, B ⊆ C ⊆ A. But Cf is the smallest subset of A that contains B and S is closed under f , so Cf ⊆ C. Since x ∈ Cf , x ∈ C. Since x was arbitrary, we conclude that f ∈F Cf ⊆ C.

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SOLUTIONS (c) No. Counterexample: A = {1, 2, 3}, f = {(1, 1), (2, 3), (3, 3)}, g = {(1, 2), (2, 2), (3, 3)}, F = {f, g}, B = {1}. (d) No. The counterexample in part (c) is a counterexample for (d) as well.

14. Z. 15. Q+ .

16. (a) Suppose X ⊆ N. Let E = {n ∈ N | n is even} and let O = {n ∈ N | n is odd}. Let Y = X ∪E and Z = X ∪ O. Clearly Y and Z are both infinite, so Y ∈ I and Z ∈ I. To prove that Y ∩ Z = X, suppose n ∈ Y ∩ Z. Then n ∈ Y and n ∈ Z. Case 1. n is even. Then n ∈ / O and n ∈ Z = X ∪ O, so n ∈ X. Case 2. n is odd. Then n ∈ / E and n ∈ Y = X ∪ E, so n ∈ X. Thus, n ∈ X. Next, suppose n ∈ X. Then n ∈ X ∪ E = Y and n ∈ X ∪ O = Z, so n ∈ Y ∩ Z. (b) P(N).

17. (a) Yes, by part 1 of Theorem 5.2.5. (b) Yes, by part 2 of Theorem 5.2.5. (c) Yes: If f and g are strictly increasing functions, then for all real numbers x and y, if x < y then g(x) < g(y), and therefore f (g(x)) < f (g(y)). This shows that f ◦ g is strictly increasing. (d) No. (The composition of two strictly decreasing functions is strictly increasing.)

18. (a) No. Let f (x) = x and g(x) = −x. Then f and g are one-to-one, but (f + g)(x) = 0, so f + g is not one-to-one. (b) No. Let f (x) = x and g(x) = −x. Then f and g are onto, but (f + g)(x) = 0, so f + g is not onto. (c) Yes: If f and g are strictly increasing functions, then for all real numbers x and y, if x < y then f (x) < f (y) and g(x) < g(y), so (f + g)(x) = f (x) + g(x) < f (y) + g(y) = (f + g)(y).. Therefore f + g is strictly increasing. (d) Yes. The proof is similar to the proof for part (c).

19. By exercise 16 in Section 4.3, the set of reflexive relations is closed under ◦. The set of symmetric relations need not be closed under ◦. For example, if A = {1, 2, 3}, R = {(1, 2), (2, 1)}, and S = {(2, 3), (3, 2)}, then R and S are symmetric, but R ◦ S = {(3, 1)}, which is not symmetric. The set of transitive relations also need not be closed under ◦. For example, if A = {1, 2, 3}, R = {(1, 2), (3, 3)}, and S = {(1, 1), (2, 3)}, then R and S are transitive, but R ◦ S = {(1, 2), (2, 3)}, which is not transitive.

20. (a) Not closed, because f (1, 0) = NaN ∈ / R. (b) Closed. (c) Not closed, because f (−1, −1) = 1 ∈ / R− . (d) Not closed, because f (1, 0) = NaN ∈ / Q. (e) Closed.

21. (a) Let G = {C ⊆ A | B ⊆ C and C is closed under T F}. You should be able to check that A ∈ G, and therefore G = 6 ∅. Thus, we can let C = G, and by exercise 9 of Section 3.3, C ⊆ A. We will show that C is the closure of B under F. To prove that B ⊆ C, suppose x ∈ B. Let D be an arbitrary element of G. Then by the definition of G, B ⊆ D, so x ∈ D. Since D was arbitrary, this shows that ∀D ∈ G(x ∈ D), so T x ∈ G = C. Thus, B ⊆ C. Next, suppose f ∈ F and x, y ∈ C and again let D be an arbitrary element of G. Then since T x, y ∈ C = G, x ∈ D and y ∈ D. But since D ∈ G, D is closed under F, soTf (x, y) ∈ D. Since D was arbitrary, we can conclude that ∀D ∈ G(f (x, y) ∈ D), so f (x, y) ∈ G = C. Thus, we have shown that C is closed under F.

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Finally, suppose B ⊆ D ⊆ A and D is closed T under F. Then D ∈ G, and applying exercise 9 of Section 3.3 again we can conclude that C = G ⊆ D. √ √ √ (b) We will use the notation Q( 2) = {a + b 2 | a, b ∈ Q}. Suppose x, y ∈ Q( 2).√ Then we can √ y = c + d 2. Therefore choose rational numbers √a, b, c, and√d such that x = a + b√ 2 and √ f (x, y) = x + y = (a + b 2) + (c + d 2) √ = (a + c)√ + (b + d) 2 ∈ Q( 2), since a√+ c ∈ Q√and 2) = (ac + 2bd) + (ad + bc) 2 ∈ Q( 2), b + d ∈ Q. Also, g(x, y) = xy = (a + b 2)(c + d √ since √ ac + 2bd ∈ Q and ad + bc ∈ Q. Therefore Q( 2) is closed √ under {f, g}. Now suppose Let x ∈ Q( 2)√be arbitrary. Then we can Q ∪ { 2} ⊆ D ⊆ R and D is closed under {f, g}. √ √ choose rational numbers a and b such that x = a + b 2 =√ f (a, g(b, 2)). Since a, b, 2 ∈ D and D is closed under {f, g}, x ∈ D. Since x was arbitrary, Q( 2) ⊆ D. √ √ (c) {a + b 3 2 + c( 3 2)2 | a, b, c ∈ Q}.

Section 5.5 1. (a) Yes. Proof: Suppose y ∈ f (W ∪ X). Then we can choose some x ∈ W ∪ X such that y = f (x).

Since x ∈ W ∪ X, either x ∈ W or x ∈ X. Case 1. x ∈ W . Then y = f (x) ∈ f (W ), so y ∈ f (W ) ∪ f (X).

Case 2. x ∈ X. Then y = f (x) ∈ f (X), so y ∈ f (W ) ∪ f (X).

Since y was arbitrary, f (W ∪ X) ⊆ f (W ) ∪ f (X). Now suppose y ∈ f (W ) ∪ f (X), so either y ∈ f (W ) or y ∈ f (X).

Case 1. y ∈ f (W ). Then we can choose some w ∈ W such that y = f (w). Since w ∈ W , w ∈ W ∪ X, so y = f (w) ∈ f (W ∪ X).

Case 2. y ∈ f (X). Then we can choose some x ∈ X such that y = f (x). Since x ∈ X, x ∈ W ∪ X, so y = f (x) ∈ f (W ∪ X).

Since y was arbitrary, f (W ) ∪ f (X) ⊆ f (W ∪ X). Therefore f (W ∪ X) = f (W ) ∪ f (X). (b) What is always true is that f (W ) \ f (X) ⊆ f (W \ X), and if f is one-to-one then f (W \ X) = f (W ) \ f (X). Proof: Suppose y ∈ f (W ) \ f (X). Then y ∈ f (W ) and y ∈ / f (X). Since y ∈ f (W ), we can choose some w ∈ W such that y = f (w). If w ∈ X then y = f (w) ∈ f (X), which is a contradiction, so w ∈ / X. Since w ∈ W and w ∈ / X, w ∈ W \ X, so y = f (w) ∈ f (W \ X). Since y was arbitrary, this proves that f (W ) \ f (X) ⊆ f (W \ X).

Now suppose f is one-to-one and y ∈ f (W \X). Then we can choose some w ∈ W \X such that y = f (w). Since w ∈ W \ X, w ∈ W and w ∈ / X. Therefore y = f (w) ∈ f (W ). Suppose y ∈ f (X).

Then we can choose some x ∈ X such that y = f (x). Therefore f (x) = y = f (w), so since f is one-to-one, w = x ∈ X, which is a contradiction. Therefore y ∈ / f (X), so y ∈ f (W ) \ f (X). Since y was arbitrary, f (W \ X) ⊆ f (W ) \ f (X), so f (W ) \ f (X) = f (W \ X).

Example in which f (W ) \ f (X) 6= f (W \ X): A = {1, 2, 3}, B = {4, 5, 6}, f = {(1, 4), (2, 5), (3, 5)}, W = {1, 2}, X = {1, 3}. (c) What is always true is that W ⊆ X → f (W ) ⊆ f (X), and if f is one-to-one then f (W ) ⊆ f (X) → W ⊆ X. Proof: Suppose W ⊆ X. Suppose y ∈ f (W ). Then we can choose some w ∈ W such that y = f (w). Since w ∈ W and W ⊆ X, w ∈ X, so y = f (w) ∈ f (X). Since y was arbitrary, this shows that f (W ) ⊆ f (X). Thus W ⊆ X → f (W ) ⊆ f (X).

Now suppose f is one-to-one and f (W ) ⊆ f (X). Suppose w ∈ W . Then f (w) ∈ f (W ), so since f (W ) ⊆ f (X), f (w) ∈ f (X). Therefore we can choose some x ∈ X such that f (w) = f (x).

But since f is one-to-one, this implies that w = x ∈ X. Since w was arbitrary, this proves that W ⊆ X. Thus, if f is one-to-one then f (W ) ⊆ f (X) → W ⊆ X. Example in which f (W ) ⊆ f (X) but W * X: Same as example in part (b).

2. (a) Yes. Proof: Let x ∈ A be arbitrary. Then x ∈ f −1 (Y ∩ Z) iff f (x) ∈ Y ∩ Z iff f (x) ∈ Y ∧ f (x) ∈ Z

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SOLUTIONS iff x ∈ f −1 (Y ) ∧ x ∈ f −1 (Z) iff x ∈ f −1 (Y ) ∩ f −1 (Z). (b) Yes. Proof: Let x ∈ A be arbitrary. Then x ∈ f −1 (Y ∪ Z) iff f (x) ∈ Y ∪ Z iff f (x) ∈ Y ∨ f (x) ∈ Z iff x ∈ f −1 (Y ) ∨ x ∈ f −1 (Z) iff x ∈ f −1 (Y ) ∪ f −1 (Z). (c) Yes. Proof: Let x ∈ A be arbitrary. Then x ∈ f −1 (Y \ Z) iff f (x) ∈ Y \ Z iff f (x) ∈ Y ∧ f (x) ∈ /Z iff x ∈ f −1 (Y ) ∧ x ∈ / f −1 (Z) iff x ∈ f −1 (Y ) \ f −1 (Z). (d) What is always true is that Y ⊆ Z → f −1 (Y ) ⊆ f −1 (Z), and if f is onto then f −1 (Y ) ⊆ f −1 (Z) → Y ⊆ Z. Proof: Suppose Y ⊆ Z. Suppose x ∈ f −1 (Y ). Then f (x) ∈ Y . Since f (x) ∈ Y and Y ⊆ Z, f (x) ∈ Z, so x ∈ f −1 (Z). Since x was arbitrary, this shows that f −1 (Y ) ⊆ f −1 (Z). Thus Y ⊆ Z → f −1 (Y ) ⊆ f −1 (Z).

Now suppose f is onto and f −1 (Y ) ⊆ f −1 (Z). Suppose y ∈ Y . Since f is onto, we can choose some x ∈ A such that f (x) = y. Since f (x) = y ∈ Y , x ∈ f −1 (Y ), and since f −1 (Y ) ⊆ f −1 (Z), it follows that x ∈ f −1 (Z). Therefore y = f (x) ∈ Z. Since y was arbitrary, this shows that Y ⊆ Z. Thus, if f is onto then f −1 (Y ) ⊆ f −1 (Z) → Y ⊆ Z. Example in which f −1 (Y ) ⊆ f −1 (Z) but Y * Z: A = {1, 2, 3}, B = {4, 5, 6}, f = {(1, 4), (2, 5), (3, 5)}, Y = {5, 6}, Z = {4, 5}.

3. What is always true is that X ⊆ f −1 (f (X)), and if f is one-to-one then f −1 (f (X)) = X. Proof: Suppose x ∈ X. Then f (x) ∈ f (X), so x ∈ f −1 (f (X)). Since x was arbitrary, this proves that X ⊆ f −1 (f (X)).

Now suppose f is one-to-one. Suppose x ∈ f −1 (f (X)). Then f (x) ∈ f (X), so we can choose some 0 x ∈ X such that f (x) = f (x0 ). Since f is one-to-one, x = x0 ∈ X. Since x was arbitrary, this shows that f −1 (f (X)) ⊆ X, so f −1 (f (X)) = X. An example in which f −1 (f (X)) 6= X was given in the text: f : R → R, f (x) = x2 , X = {x ∈ R | 0 ≤ x < 2}.

4. What is always true is that f (f −1 (Y )) ⊆ Y , and if f is onto then f (f −1 (Y )) = Y . Proof: Suppose y ∈ f (f −1 (Y )). Then we can choose some x ∈ f −1 (Y ) such that y = f (x). Since x ∈ f −1 (Y ), y = f (x) ∈ Y . Since y was arbitrary, this shows that f (f −1 (Y )) ⊆ Y .

Now suppose f is onto. Suppose y ∈ Y . Since f is onto, we can choose some x ∈ A such that f (x) = y. Since f (x) = y ∈ Y , x ∈ f −1 (Y ), so y = f (x) ∈ f (f −1 (Y )). Since y was arbitrary, this shows that Y ⊆ f (f −1 (Y )), so f (f −1 (Y )) = Y .

Example in which f (f −1 (Y )) 6= Y : f : R → R, f (x) = x2 , Y = {x ∈ R | −4 < x < 4}.

5. (a) → (b). Suppose C is closed under f . Suppose y ∈ f (C). Then we can choose some x ∈ C such that y = f (x). Since C is closed under f , y = f (x) ∈ C. Thus f (C) ⊆ C. (b) → (c). Suppose f (C) ⊆ C. Suppose x ∈ C. Then f (x) ∈ f (C), so since f (C) ⊆ C, f (x) ∈ C. Therefore x ∈ f −1 (C). Thus C ⊆ f −1 (C). (c) → (a). Suppose C ⊆ f −1 (C). Suppose x ∈ C. Then since C ⊆ f −1 (C), x ∈ f −1 (C), so f (x) ∈ C. Since x was arbitrary, this shows that C is closed under f .

6. Here are two interesting facts: (a) For all X ⊆ A, (g ◦ f )(X) = g(f (X)). (b) For all Z ⊆ C, (g ◦ f )−1 (Z) = f −1 (g −1 (Z)).

Proof of (a): Suppose X ⊆ A. Suppose z ∈ (g ◦ f )(X). Then we can choose some x ∈ X such that z = (g ◦ f )(x) = g(f (x)). Since x ∈ X, f (x) ∈ f (X), so z = g(f (x)) ∈ g(f (X)).

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75

Now suppose z ∈ g(f (X)). Then we can choose some y ∈ f (X) such that z = g(y). Since y ∈ f (X), we can choose some x ∈ X such that y = f (x). Therefore z = g(y) = g(f (x)) = (g ◦ f )(x) ∈ (g ◦ f )(X).

Thus (g ◦ f )(X) = g(f (X)).

Proof of (b): Suppose Z ⊆ C. Suppose x ∈ A. Then x ∈ (g ◦ f )−1 (Z) iff (g ◦ f )(x) ∈ Z iff g(f (x)) ∈ Z iff f (x) ∈ g −1 (Z) iff x ∈ f −1 (g −1 (Z)).

7. Let X1 be the inverse image of Y under f , and let X2 be the image of Y under f −1 . This means that X1 = {x ∈ A | f (x) ∈ Y }, X2 = {f −1 (y) | y ∈ Y } = {x ∈ A | ∃y ∈ Y (f −1 (y) = x)}.

We must prove X1 = X2 . Suppose x ∈ X1 . Then f (x) ∈ Y . Let y = f (x) ∈ Y . Then x = f −1 (y) ∈ X2 .

Now suppose x ∈ X2 . Then we can choose some y ∈ Y such that f −1 (y) = x. Therefore f (x) = y ∈ Y , so x ∈ X1 .

Chapter 6 Section 6.1 1. Base case: When n = 0, both sides of the equation are 0.

Induction step: Suppose that n ∈ N and 0 + 1 + 2 + · · · + n = n(n + 1)/2. Then 0 + 1 + 2 + · · · + (n + 1) = (0 + 1 + 2 + · · · + n) + (n + 1) n(n + 1) + (n + 1) 2 n (n + 1)(n + 2) = (n + 1) +1 = , 2 2

=

as required.

2. Base case: When n = 0, both sides of the equation are 0.

Induction step: Suppose n ∈ N and 02 + 12 + 22 + · · · + n2 = n(n + 1)(2n + 1)/6. Then 02 + 12 + 22 + · · · + (n + 1)2 = (02 + 12 + 22 + · · · + n2 ) + (n + 1)2 n(2n + 1) + 6(n + 1) n(n + 1)(2n + 1) + (n + 1)2 = (n + 1) · 6 6 (n + 1)(2n2 + 7n + 6) (n + 1)(n + 2)(2(n + 1) + 1) = = .

6 6 =

3. Base case: When n = 0, both sides of the equation are 0.

Induction step: Suppose n ∈ N and 03 + 13 + 23 + · · · + n3 = [n(n + 1)/2]2 . Then

03 + 13 + 23 + · · · + (n + 1)3 = 03 + 13 + 23 + · · · + n3 + (n + 1)3 2 2

n n(n + 1) + (n + 1)3 = (n + 1)2 +n+1 = 2 4

2 2 (n + 1)(n + 2) 2 n + 4n + 4 = (n + 1) · = .

4 2

76

SOLUTIONS 4. For all n ≥ 1, 1 + 3 + 5 + · · · + (2n − 1) = n2 . We prove this by mathematical induction.

Base case: When n = 1, both sides of the equation are 1.

Induction step: Suppose n ≥ 1 and 1 + 3 + 5 + · · · + (2n − 1) = n2 . Then 1 + 3 + 5 + · · · + (2(n + 1) − 1) = (1 + 3 + 5 + · · · + (2n − 1)) + (2n + 1) = n2 + 2n + 1 = (n + 1)2 .

5. Base case: When n = 0, both sides of the equation are 0.

Induction step: Suppose n ∈ N and 0 · 1 + 1 · 2 + 2 · 3 + · · · + n(n + 1) = n(n + 1)(n + 2)/3. Then 0 · 1 + 1 · 2 + 2 · 3 + · · · + (n + 1)(n + 2) = (0 · 1 + 1 · 2 + 2 · 3 + · · · n(n + 1)) + (n + 1)(n + 2) n(n + 1)(n + 2) + (n + 1)(n + 2) 3 n (n + 1)(n + 2)(n + 3) = (n + 1)(n + 2) +1 = .

3 3 =

6. For all n ∈ N, 0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + · · · + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4. We prove this by mathematical induction.

Base case: When n = 0, both sides of the equation are 0.

Induction step: Suppose n ∈ N and 0 · 1 · 2 + 1 · 2 · 3 + · · · + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4.

Then 0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + · · · + (n + 1)(n + 2)(n + 3) = (0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + · · · + n(n + 1)(n + 2)) + (n + 1)(n + 2)(n + 3) n(n + 1)(n + 2)(n + 3) + (n + 1)(n + 2)(n + 3) 4 (n + 1)(n + 2)(n + 3)(n + 4) n +1 = . = (n + 1)(n + 2)(n + 3) 4 4 =

7. For all n ∈ N, 30 + 31 + 32 + · · · + 3n = (3n+1 − 1)/2. We prove this by mathematical induction.

Base case: Setting n = 0, we get 30 = 1 = (31 − 1)/2.

Induction step: Suppose n ∈ N and 30 + 31 + 32 + · · · + 3n = (3n+1 − 1)/2. Then 30 + 31 + 32 + · · · + 3n+1 = (30 + 31 + 32 + · · · + 3n ) + 3n+1 =

3 · 3n+1 − 1 3n+2 − 1 3n+1 − 1 + 3n+1 = = .

2 2 2

8. Base case: When n = 1, the equation is 1 − 1/2 = 1/2, which is true.

Induction step: Suppose n ≥ 1 and 1−

1 1 1 1 1 1 1 1 + − + ··· + − = + + ··· + .

2 3 4 2n − 1 2n n+1 n+2 2n

Then 1−

1 1 1 1 1 + − + ··· + − 2 3 4 2(n + 1) − 1 2(n + 1)

1 1 1 1 1 1 1 = 1 − + − + ··· + − + − 2 3 4 2n − 1 2n 2n + 1 2n + 2

1 1 1 1 1 = + + ··· + + − n+1 n+2 2n 2n + 1 2n + 2

CHAPTER 6

77 1 1 1 = + + ··· + + n+2 n+3 2n + 1 1 1 1 = + + ··· + . n+2 n+3 2n + 2

1 1 − n + 1 2n + 2

9. (a) Base case: When n = 0, n2 + n = 0 = 0 · 2, so 2 | (n2 + n).

Induction step: Suppose n ∈ N and 2 | (n2 + n). Then there is some integer k such that 2 n +n = 2k. Therefore (n+1)2 +(n+1) = n2 +3n+2 = n2 +n+2n+2 = 2k+2n+2 = 2(k+n+1), so 2 | ((n + 1)2 + (n + 1)). (b) Base case: When n = 0, n3 − n = 0 = 0 · 6, so 6 | (n3 − n).

Induction step. Suppose n ∈ N and 6 | (n3 − n). Then there is some integer k such that 3 n − n = 6k. Also, by part (a), there is some integer j such that n2 + n = 2j. Therefore (n + 1)3 − (n + 1) = (n3 + 3n2 + 3n + 1) − (n + 1) = (n3 − n) + 3(n2 + n) = 6k + 3(2j) = 6(k + j), so 6 | ((n + 1)3 − (n + 1)).

10. Base case: When n = 0, 9n − 8n − 1 = 0 = 64 · 0, so 64 | (9n − 8n − 1).

Induction step: Suppose that n ∈ N and 64 | (9n − 8n − 1). Then there is some integer k such that n 9 − 8n − 1 = 64k. Therefore 9n+1 − 8(n + 1) − 1 = 9n+1 − 8n − 9 = 9n+1 − 72n − 9 + 64n = 9(9n − 8n − 1) + 64n = 9(64k) + 64n = 64(9k + n), so 64 | (9n+1 − 8(n + 1) − 1).

11. Base case: When n = 0, 4n + 6n − 1 = 0 = 0 · 9, so 9 | (4n + 6n − 1).

Induction step: Suppose n ∈ N and 9 | (4n + 6n − 1). Then there is some integer k such that n 4 + 6n − 1 = 9k. Therefore 4n+1 + 6(n + 1) − 1 = 4n+1 + 6n + 5 = 4n+1 + 24n − 4 + 9 − 18n = 4(4n + 6n − 1) + 9 − 18n = 4(9k) + 9(1 − 2n) = 9(4k + 1 − 2n), so 9 | (4n+1 + 6(n + 1) − 1).

12. (a) Base case: When n = 0, 7n − 5n = 0 = 2 · 0, so 7n − 5n is even.

Induction step: Suppose n ∈ N and 7n − 5n is even. Then there is some integer k such that n 7 − 5n = 2k. Therefore 7n+1 − 5n+1 = 7 · 7n − 5 · 5n = 2 · 7n + 5 · (7n − 5n ) = 2 · 7n + 5 · 2k = 2(7n + 5k), so 7n+1 − 5n+1 is even. (b) Base case: When n = 0, 2 · 7n − 3 · 5n + 1 = 0 = 0 · 24, so 24 | (2 · 7n − 3 · 5n + 1).

Induction step: Suppose n ∈ N and 24 | (2 · 7n − 3 · 5n + 1). Then there is some integer k such that 2 · 7n − 3 · 5n + 1 = 24k. Also, by part (a), there is some integer j such that 7n − 5n = 2j.

Therefore 2 · 7n+1 − 3 · 5n+1 + 1 = 14 · 7n − 15 · 5n + 1

78

SOLUTIONS = (2 · 7n − 3 · 5n + 1) + 12(7n − 5n ) = 24k + 12(2j) = 24(k + j), so 24 | (2 · 7n+1 − 3 · 5n+1 + 1).

13. Let a and b be arbitrary integers. We now use induction to prove that ∀n ∈ N[(a − b) | (an − bn )].

Base case: When n = 0, an − bn = 0 = 0 · (a − b), so (a − b) | (an − bn ).

Induction step: Suppose n ∈ N and (a − b) | (an − bn ). Then there is some integer k such that n a − bn = k(a − b). Therefore an+1 − bn+1 = an+1 − abn + abn − bn+1 = a(an − bn ) + bn (a − b) = ak(a − b) + bn (a − b) = (ak + bn )(a − b), so (a − b) | (an+1 − bn+1 ).

14. Let a and b be arbitrary integers. We now use induction to prove that ∀n ∈ N[(a + b) | (a2n+1 + b2n+1 )].

Base case: When n = 0, a2n+1 + b2n+1 = a + b, so (a + b) | (a2n+1 + b2n+1 ).

Induction step: Suppose n ∈ N and (a + b) | (a2n+1 + b2n+1 ). Then there is some integer k such that a2n+1 + b2n+1 = k(a + b). Therefore a2(n+1)+1 + b2(n+1)+1 = a2n+3 + a2 b2n+1 − a2 b2n+1 + b2n+3 = a2 (a2n+1 + b2n+1 ) − b2n+1 (a2 − b2 ) = a2 (k(a + b)) − b2n+1 (a − b)(a + b) = [a2 k − b2n+1 (a − b)](a + b), so (a + b) | (a2(n+1)+1 + b2(n+1)+1 ).

15. Base case: When n = 10, 2n = 1024 > 1000 = n3 .

Induction step: Suppose n ≥ 10 and 2n > n3 . Then 2n+1 = 2 · 2n > 2n3 3

=n +n

(inductive hypothesis) 3

≥ n3 + 10n2 3

2

= n + 3n + 7n

(since n ≥ 10) 2

≥ n3 + 3n2 + 70n 3

(since n ≥ 10)

2

= n + 3n + 3n + 67n > n3 + 3n2 + 3n + 1 = (n + 1)3 .

16. (a) Base case: 0 = 2 · 0, so 0 is even. If 0 is also odd, then there is some integer j such that 0 = 2j + 1.

But then j = −1/2, which is not an integer, so we have a contradiction. Therefore 0 is not odd.

Induction step: Suppose n ∈ N and n is either even or odd, but not both.

Case 1. n is even and not odd. Then there is some integer k such that n = 2k. Then n + 1 = 2k + 1, so n + 1 is odd. If n + 1 is also even, then there is some integer j such that n + 1 = 2j. But then n = 2j − 1 = 2(j − 1) + 1, so n is odd, which is a contradiction. Therefore n + 1 is not even.

Case 2. n is odd and not even. Then there is some integer k such that n = 2k + 1. Then n + 1 = 2k + 2 = 2(k + 1), so n + 1 is even. If n + 1 is also odd, then there is some integer j such that n + 1 = 2j + 1. But then n = 2j, so n is even, which is a contradiction. Therefore n + 1 is not odd.

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(b) Let n be an arbitrary integer. If n ≥ 0, then n is either even or odd, but not both, by part (a).

Now suppose n < 0. Then −n is a positive integer, so by part (a) −n is either even or odd, but not both.

Case 1. −n is even and not odd. Then for some integer k, −n = 2k. Then n = 2(−k), so n is even. If n is also odd, then there is some integer j such that n = 2j + 1. But then −n = −2j − 1 = 2(−j − 1) + 1, so −n is odd, which is a contradiction. Therefore n is not odd.

Case 2. −n is odd and not even. Then for some integer k, −n = 2k + 1. Then n = −2k − 1 = 2(−k − 1) + 1, so n is odd. If n is also even, then there is some integer j such that n = 2j. But then −n = 2(−j), so −n is even, which is a contradiction. Therefore n is not even.

17. Base case: When n = 1, both sides of the equation are equal to 4.

Inductions step: Suppose n ≥ 1 and 2 · 21 + 3 · 22 + · · · + (n + 1)2n = n2n+1 . Then 2 · 21 + 3 · 22 + · · · + (n + 2)2n+1 = (2 · 21 + 3 · 22 + · · · + (n + 1)2n ) + (n + 2)2n+1 = n2n+1 + (n + 2)2n+1 = (2n + 2)2n+1 = 2(n + 1)2n+1 = (n + 1)2n+2 .

18. (a) The proof doesn’t check the base case. In fact, when n = 0, the left side of the equation is 1 and the right side is 0, so the equation is false. (b) For all n ∈ N, 1 · 30 + 3 · 31 + 5 · 32 + · · · + (2n + 1)3n = n3n+1 + 1. The proof is by mathematical induction.

Base case: When n = 0, both sides of the equation are 1.

Induction step: Suppose n ∈ N and 1 · 30 + 3 · 31 + 5 · 32 + · · · + (2n + 1)3n = n3n+1 + 1. Then 1 · 30 + 3 · 31 + 5 · 32 + · · · + (2(n + 1) + 1)3n+1 = (1 · 30 + 3 · 31 + 5 · 32 + · · · + (2n + 1)3n ) + (2n + 3)3n+1 = n3n+1 + 1 + (2n + 3)3n+1 = (3n + 3)3n+1 + 1 = 3(n + 1)3n+1 + 1 = (n + 1)3n+2 + 1.

19. Base case: n = 0. Then n is even and an = a0 = 1 > 0.

Induction step: Suppose n ∈ N, if n is even then an > 0, and if n is odd then an < 0. We must prove two statements: if n + 1 is even then an+1 > 0, and if n + 1 is odd then an+1 < 0.

To prove the first statement, suppose that n + 1 is even. Then as in exercise 16, n is odd, so by inductive hypothesis, an < 0. Multiplying both sides of this inequality by a (and reversing the inequality, since a < 0), we get an+1 > 0.

The proof of the second statement is similar: Suppose n + 1 is odd. Then n is even, so by inductive hypothesis, an > 0. Multiplying by a, we get an+1 < 0.

20. (a) Base case: When n = 1, the statement to be proven is 0 < a < b, which was given.

Induction step: Suppose that n ≥ 1 and 0 < an < bn . Multiplying this inequality by the positive number a we get 0 < an+1 < abn , and multiplying the inequality a < b by the positive number bn gives us abn < bn+1 . Combining these inequalities, we can conclude that 0 < an+1 < bn+1 . √ √ √ √ (b) Let n ≥ 2 be arbitrary. Suppose it √ is not true that 0 < n a < n b. Since n a and n b are both √ positive, this√means√that 0 < n b ≤ b a.

Case 1. n b = n a. Raising both sides of this equation to the power n, we get b = a, which contradicts the fact √ that √ √ a < b. √ Case 2. 0 < n b < n a. Applying part (a), we get b = ( n b)n < ( n a)n = a, which contradicts the fact that a < b. √ √ Since we have reached a contradiction in both cases, we conclude that 0 < n a < n b.

80

SOLUTIONS (c) Suppose n ≥ 1. We are given that a < b, and by part (a), an < bn . Therefore b − a > 0 and bn − an > 0, so (b − a)(bn − an ) > 0. Multiplying this out gives us bn+1 − ban − abn + an+1 > 0.

Rearranging this, we get the required inequality abn + ban < an+1 + bn+1 . (d) Base case: n = 2. By part (c) with n = 1, 2ab < a2 + b2 . Adding a2 + b2 to both sides, we get a2 + 2ab + b2 < 2a2 + 2b2 , or equivalently (a + b)2 < 2(a2 + b2 ). Dividing by 4, we get (a + b)2 /4 < (a2 + b2 )/2, or equivalently ((a + b)/2)2 < (a2 + b2 )/2. This is the required inequality with n = 2.

Induction step: Suppose n ≥ 2 and ((a + b)/2)n < (an + bn )/2. Then

a+b 2

n+1

=

a+b 2

a+b 2

n

n

a+b a + bn 2 2 an+1 + abn + ban + bn+1 = 4 2an+1 + 2bn+1 < 4 an+1 + bn+1 . = 2

<

(inductive hypothesis)

(part (c))

Section 6.2 1. (a) We must prove that R0 is reflexive (on A0 ), transitive, and antisymmetric. For the first, suppose x ∈ A0 . Since R is reflexive (on A) and x ∈ A, (x, x) ∈ R, so (x, x) ∈ R ∩ (A0 × A0 ) = R0 . This shows that R0 is reflexive.

Next, suppose that (x, y) ∈ R0 and (y, z) ∈ R0 . Then (x, y) ∈ R, (y, z) ∈ R, and x, y, z ∈ A0 .

Since R is transitive, (x, z) ∈ R, so (x, z) ∈ R ∩ (A0 × A0 ) = R0 . Therefore R0 is transitive.

Finally, suppose that (x, y) ∈ R0 and (y, x) ∈ R0 . Then (x, y) ∈ R and (y, x) ∈ R, so since R is antisymmetric, x = y. Thus R0 is antisymmetric. (b) To see that T is reflexive, suppose x ∈ A. If x = a, then (x, x) = (a, a) ∈ {a} × A ⊆ T . If x 6= a, then x ∈ A0 , so since R0 is reflexive, (x, x) ∈ R0 ⊆ T 0 ⊆ T .

For transitivity, suppose that (x, y) ∈ T and (y, z) ∈ T . If x = a then (x, z) = (a, z) ∈ {a} × A ⊆ T . Now suppose x 6= a. Then (x, y) ∈ / {a} × A, so since (x, y) ∈ T = T 0 ∪ ({a} × A) we 0 0 0 0 must have (x, y) ∈ T . But T ⊆ A × A , so y ∈ A0 and therefore y 6= a. Similar reasoning now shows that (y, z) ∈ T 0 . Since T 0 is transitive, it follows that (x, z) ∈ T 0 ⊆ T .

To show that T is antisymmetric, suppose (x, y) ∈ T and (y, x) ∈ T . If x = a then (y, x) ∈ / T 0, so (y, x) ∈ {a} × A and therefore y = a = x. Similarly, if y = a then x = y. Now suppose x 6= a and y 6= a. Then as in the proof of transitivity it follows that (x, y) ∈ T 0 and (y, x) ∈ T 0 , so by antisymmetry of T 0 , x = y.

We now know that T is a partial order. To see that it is total, suppose x ∈ A and y ∈ A. If x = a then (x, y) ∈ {a} × A ⊆ T . Similarly, if y = a then (y, x) ∈ T . Now suppose x 6= a and y 6= a. Then x ∈ A0 and y ∈ A0 , so since T 0 is a total order, either (x, y) ∈ T 0 ⊆ T or (y, x) ∈ T 0 ⊆ T .

Finally, to see that R ⊆ T , suppose that (x, y) ∈ R. If x = a then (x, y) ∈ {a} × A ⊆ T . Now suppose x 6= a. If y = a then the fact that (x, y) ∈ R would contradict the R-minimality of a.

Therefore y 6= a. But then (x, y) ∈ R ∩ (A0 × A0 ) = R0 ⊆ T 0 ⊆ T .

2. Suppose R is a partial order on a set A. We now use induction to prove that ∀n ∈ N∀B ⊆ A[B has n elements → ∃T (T is a partial order on A ∧ R ⊆ T ∧ ∀x ∈ B∀y ∈ A(xT y ∨ yT x))].

Base case: n = 0. If B ⊆ A and B has 0 elements then B = ∅. We can let T = R, and the statement ∀x ∈ B∀y ∈ A(xT y ∨ yT x) will be true vacuously.

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81

Induction step: Suppose n ∈ N and ∀B ⊆ A[B has n elements → ∃T (T is a partial order on A ∧ R ⊆ T ∧ ∀x ∈ B∀y ∈ A(xT y ∨ yT x))]. Suppose B ⊆ A and B has n + 1 elements. Let b be any element of B and let B 0 = B \ {b}, a subset of A with n elements. By inductive hypothesis, we can let T 0 be a partial order on A such that R ⊆ T 0 and ∀x ∈ B 0 ∀y ∈ A(xT 0 y ∨ yT 0 x). Let A1 = {x ∈ A | (x, b) ∈ T 0 } and A2 = A \ A1 . Let T = T 0 ∪ (A1 × A2 ).

To see that T is reflexive, suppose x ∈ A. Since T 0 is a partial order on A, (x, x) ∈ T 0 , so (x, x) ∈ T . Thus T is reflexive. Next, to prove that T is transitive, suppose (x, y) ∈ T and (y, z) ∈ T .

Then (x, y) ∈ T 0 ∪ (A1 × A2 ), so either (x, y) ∈ T 0 or (x, y) ∈ A1 × A2 , and similarly either (y, z) ∈ T 0 or (y, z) ∈ A1 × A2 . We now consider three cases.

Case 1. (x, y) ∈ T 0 and (y, z) ∈ T 0 . Then since T 0 is transitive, (x, z) ∈ T 0 ⊆ T .

Case 2. (x, y) ∈ T 0 and (y, z) ∈ A1 × A2 . Then y ∈ A1 and z ∈ A2 . Since y ∈ A1 , by the definition of A1 , (y, b) ∈ T 0 . Since (x, y) ∈ T 0 , (y, b) ∈ T 0 , and T 0 is transitive, (x, b) ∈ T 0 , so x ∈ A1 . Since x ∈ A1 and z ∈ A2 , (x, z) ∈ A1 × A2 ⊆ T .

Case 3. (x, y) ∈ A1 × A2 . Then x ∈ A1 and y ∈ A2 . Since y ∈ A2 , y ∈ / A1 , so (y, z) ∈ / A1 × A2 , and therefore (y, z) ∈ T 0 . Suppose z ∈ A1 . Then by definition of A1 , (z, b) ∈ T 0 , so since (y, z) ∈ T 0 , by transitivity of T 0 , (y, b) ∈ T 0 . But then y ∈ A1 , which is a contradiction. Therefore z ∈ / A1 , so z ∈ A2 .

Since x ∈ A1 and z ∈ A2 , (x, z) ∈ A1 × A2 ⊆ T .

In call cases we have (x, z) ∈ T , so this proves that T is transitive.

To prove that T is antisymmetric, suppose (x, y) ∈ T and (y, x) ∈ T . As before, either (x, y) ∈ T 0 or (x, y) ∈ A1 × A2 , and either (y, x) ∈ T 0 or (y, x) ∈ A1 × A2 . Suppose (x, y) ∈ A1 × A2 . Then x ∈ A1 and y ∈ A2 . Thus (y, x) ∈ / A1 × A2 , so (y, x) ∈ T 0 . Since x ∈ A1 , (x, b) ∈ T 0 . Since (y, x) ∈ T 0 , 0 0 (x, b) ∈ T , and T is transitive, (y, b) ∈ T 0 . But this means y ∈ A1 , which is a contradiction. Therefore it cannot be the case that (x, y) ∈ A1 × A2 , so (x, y) ∈ T 0 . Similar reasoning shows that we can rule out (y, x) ∈ A1 × A2 , so (y, x) ∈ T 0 . Since (x, y) ∈ T 0 , (y, x) ∈ T 0 , and T 0 is antisymmetric, y = x.

Thus T is antisymmetric.

We have now shown that T is a partial order on A. Since R ⊆ T 0 and T 0 ⊆ T , R ⊆ T . Finally, let x ∈ B and y ∈ A be arbitrary. Once again we consider cases.

Case 1. x ∈ B 0 . Then by the choice of T 0 , either xT 0 y or yT 0 x. Since T 0 ⊆ T , it follows that either xT y or yT x.

Case 2. x ∈ / B 0 . Then x = b. We now consider two subcases.

Case 2a. y ∈ A1 . Then (y, x) = (y, b) ∈ T 0 ⊆ T , so yT x.

Case 2b. y ∈ / A1 , so y ∈ A2 . Notice that (b, b) ∈ T 0 , so b ∈ A1 . Therefore (x, y) = (b, y) ∈ A1 × A2 ⊆ T , so xT y.

Thus either xT y or yT x, as required.

3. We will prove by induction that ∀n ≥ 1∀B ⊆ A[B has n elements → B has an R-smallest element and an R-largest element].

Base case: n = 1. Suppose B ⊆ A and B has 1 element. Then B = {b}, for some b ∈ A. Since b is the only element of B and R is reflexive, ∀x ∈ B(bRx) and ∀x ∈ B(xRb), so b is both the R-smallest element of B and the R-largest element of B. Induction step. Suppose n ≥ 1 and ∀B ⊆ A[B has n elements → B has an R-smallest element and an R-largest element]. Now suppose B ⊆ A and B has n + 1 elements. Let b be any element of B, and let B 0 = B \ {b}. Then B 0 has n elements, so by inductive hypothesis we can let c be the R-smallest element of B 0 and d the R-largest element of B 0 .

Since R is a total order, either bRc or cRb. Case 1. bRc. Let x be an arbitrary element of B. If x = b, then since R is reflexive, bRx. If x 6= b, then x ∈ B 0 , so since c is the R-smallest element of B 0 , cRx. By transitivity of R, since bRc and cRx, bRx. Thus ∀x ∈ B(bRx), so b is the R-smallest element of B. Case 2. cRb. Let x be an arbitrary element of B. If x = b, then since cRb, cRx. If x 6= b, then x ∈ B 0 , so since c is the R-smallest element of B 0 , cRx. Thus ∀x ∈ B(cRx), so c is the R-smallest

82

SOLUTIONS element of B. Thus, B has an R-smallest element. A similar argument shows that either b or d is the R-largest element of B. 4. (a) We will prove the statement: ∀n ≥ 1∀B ⊆ A[B has n elements → ∃x ∈ B∀y ∈ B((x, y) ∈ R ◦ R)].

We proceed by induction on n.

Base case: Suppose n = 1. If B ⊆ A and B has one element, then for some x ∈ B, B = {x}.

Since R is reflexive, (x, x) ∈ R, and therefore (x, x) ∈ R ◦ R. But x is the only element in B, so ∀y ∈ B((x, y) ∈ R ◦ R), as required.

Induction step: Suppose that n ≥ 1 and for every B ⊆ A, if B has n elements then ∃x ∈ B∀y ∈ B((x, y) ∈ R ◦ R). Now suppose that B ⊆ A and B has n + 1 elements. Choose some b ∈ B, and let B 0 = B \ {b}. Then B 0 ⊆ A and B 0 has n elements, so by inductive hypothesis there is some x ∈ B 0 such that ∀y ∈ B 0 ((x, y) ∈ R ◦ R). We now consider two cases.

Case 1: (x, b) ∈ R ◦ R. Then ∀y ∈ B((x, y) ∈ R ◦ R), so we are done.

Case 2: (x, b) ∈ / R ◦ R. In this case, we will prove that ∀y ∈ B((b, y) ∈ R ◦ R). To do this, let y ∈ B be arbitrary. If y = b, then since R is reflexive, (b, b) ∈ R, and therefore (b, y) = (b, b) ∈ R ◦ R. Now suppose y 6= b. Then y ∈ B 0 , so by the choice of x we know that (x, y) ∈ R ◦ R. This means that for some z ∈ A, (x, z) ∈ R and (z, y) ∈ R. We have (x, z) ∈ R, so if (z, b) ∈ R then (x, b) ∈ R ◦ R, contrary to the assumption for this case. Therefore (z, b) ∈ / R, so by the hypothesis on R, (b, z) ∈ R. But then since (b, z) ∈ R and (z, y) ∈ R, we have (b, y) ∈ R◦R, as required. (b) Let A = the set of contestants and let R = {(x, y) ∈ A × A | x beats y} ∪ iA . By assumption, for all x and y in A, if x 6= y then either x beats y or y beats x, so xRy ∨ yRx. But also if x = y then (x, y) ∈ iA ⊆ R, so xRy. Thus R satisfies the hypothesis in part (a). We now apply part (a) with B = A to conclude that there is some x ∈ A such that for all y ∈ A, (x, y) ∈ R ◦ R. We claim that x is excellent. To prove this, suppose y ∈ A and y 6= x. Then (x, y) ∈ R ◦ R, so there is some z ∈ A such that (x, z) ∈ R and (z, y) ∈ R. If z = x then since (z, y) ∈ R, (x, y) ∈ R. But since x 6= y, (x, y) ∈ / iA , so by the definition of A, x beats y. Similarly, if z = y then since (x, z) ∈ R, (x, y) ∈ R, so x beats y. Finally, if z 6= x and z 6= y then x beats z and z beats y. This proves that x is excellent.

0

1

5. Base case: n = 1. F0 = 2(2 ) + 1 = 3 and F1 = 2(2 ) + 1 = 5, so F1 = F0 + 2.

Induction step: Suppose n ≥ 1 and Fn = (F0 · F1 · F2 · · · Fn−1 ) + 2. Then (F0 · F1 · F2 · · · Fn ) + 2 = ((F0 · F1 · F2 · · · Fn−1 ) · Fn ) + 2

n n = (Fn − 2) · Fn + 2 = 2(2 ) − 1 2(2 ) + 1 + 2 n 2 n = 2(2 ) − 1 + 2 = 2(2 ·2) + 1 = 2(2

n+1

)

+ 1 = Fn+1 .

6. Base case: If n = 1, then the statement to be proven is that for every real number a1 , |a1 | ≤ |a1 |, which is clearly true.

Induction step. Suppose n ≥ 1 and for all real numbers a1 , a2 , . . . , an , |a1 + a2 + · · · + an | ≤ |a1 | + |a2 | + · · · + |an |. Now suppose a1 , a2 , . . . , an+1 are real numbers. Then |a1 + a2 + · · · + an+1 | = |(a1 + a2 + · · · + an ) + an+1 | ≤ |a1 + a2 + · · · + an | + |an+1 |

(triangle inequality: ex. 13(c) in Sect. 3.5)

≤ |a1 | + |a2 | + · · · + |an | + |an+1 | (inductive hypothesis).

7. (a) Suppose a and b are positive real numbers. Then a2 − 2ab + b2 = (a − b)2 ≥ 0. Adding 2ab to both sides gives us a2 + b2 ≥ 2ab, and then dividing by ab, which is positive, we get a/b + b/a ≥ 2.

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83

(b) Since a ≤ b ≤ c, c − a ≥ 0 and c − b ≥ 0, so c2 − bc − ac + ab = (c − a)(c − b) ≥ 0. Adding ac to both sides gives us ab + c2 − bc ≥ ac, and dividing by ac, we get b/c + c/a − b/a ≥ 1. (c) We use induction.

Base case: If n = 2, the statement to be proven is that if 0 < a1 ≤ a2 then a1 /a2 + a2 /a1 ≥ 2.

This follows from part (a).

Induction step. Suppose n ≥ 2 and for all numbers a1 , a2 , . . . , an , if 0 < a1 ≤ a2 ≤ · · · ≤ an then a1 /a2 + a2 /a3 + · · · + an−1 /an + an /a1 ≥ n. Now suppose a1 , a2 , . . . , an+1 are real numbers and 0 < a1 ≤ a2 ≤ · · · ≤ an+1 . Then

a2 an an+1 a1 a2 an−1 an an an+1 an a1 + + ··· + + = + + ··· + + + + − a2 a3 an+1 a1 a2 a3 an a1 an+1 a1 a1

an an+1 an ≥n+ + − (inductive hypothesis) an+1 a1 a1 ≥n+1

(part (b)).

8. (a) Let m = (a + b)/2, the arithmetic mean of a and b, and let d = (a − b)/2. Then it is easy to check that m + d = a and m − d = b, so p √ p √ a+b . ab = (m + d)(m − d) = m2 − d2 ≤ m2 = m = 2 (b) We use induction on n.

Base case: n = 1. This case is taken care of by part (a).

Induction step: Suppose n ≥ 1, and the arithmetic mean–geometric mean inequality holds for lists of length 2n . Now let a1 , a2 , . . . , a2n+1 be a list of 2n+1 positive real numbers. Let a1 + a2 + · · · + a2n a2n +1 + a2n +2 + · · · + a2n+1 , m2 = . n 2 2n Notice that a1 + a2 + · · · + a2n = m1 2n , and similarly a2n +1 + a2n +2 + · · · + a2n+1 = m2 2n . Also, √ √ by inductive hypothesis, we know that m1 ≥ 2n a1 a2 · · · a2n and m2 ≥ 2n a2n +1 a2n +2 · · · a2n+1 .

Therefore √ a1 + a2 + · · · + a2n+1 m1 2n + m2 2n m1 + m2 = = ≥ m1 m2 n+1 n+1 2 2 2 q √ √ 2n a1 a2 · · · a2n 2n a2n +1 a2n +2 · · · a2n+1 ≥ √ n+1 = 2 a1 a2 · · · a2n+1 . m1 =

(c) We use induction on n.

Base case: If n = n0 , then by assumption the arithmetic mean–geometric mean inequality fails for some list of length n.

Induction step: Suppose n ≥ n0 , and there are positive real numbers a1 , a2 , . . . , an such that √ a1 + a2 + · · · + an < n a1 a2 · · · an . n

√ Let m = (a1 + a2 + · · · + an )/n, and let an+1 = m. Then we have m < n a1 a2 · · · an , so mn < a1 a2 · · · an . Multiplying both sides of this inequality by m gives us mn+1 < a1 a2 · · · an m = √ a1 a2 · · · an+1 , so m < n+1 a1 a2 · · · an+1 . But notice that we also have mn = a1 + a2 + · · · + an , so a1 + · · · + an+1 mn + m m(n + 1) √ = = = m < n+1 a1 a2 · · · an+1 . n+1 n+1 n+1 Thus, we have a list of length n + 1 for which the arithmetic mean–geometric mean inequality fails.

84

SOLUTIONS (d) Suppose that the arithmetic mean–geometric mean inequality fails for some list of positive real numbers. Let n0 be the length of this list, and choose an integer n ≥ 1 such that n0 ≤ 2n . (In fact, we could just let n = n0 , as you will show in exercise 12(a) in Section 6.3.) Then by part (b), the arithmetic mean–geometric mean inequality holds for all lists of length 2n , but by part (c), it must fail for some list of length 2n . This is a contradiction, so the inequality must always hold.

9. Suppose n ≥ 2 and a1 , a2 , . . . , an is a list of positive real numbers. By the arithmetic mean–geometric mean inequality (exercise 8) applied to the numbers 1/a1 , 1/a2 , . . . , 1/an , we have r 1 1 1 1 1 1 1 a1 + a2 + · · · + an ≥ n · ··· = √ > 0. n n a1 a2 an a1 a2 · · · an By exercise 6 in Section 3.1,

n 1 1 1 a1 + a2 + · · · + an

√ ≤ n a1 a2 · · · an .

10. (a) Suppose a1 , a2 , b1 , and b2 are real numbers, a1 ≤ a2 , and b1 ≤ b2 . Then (a1 b1 + a2 b2 ) − (a1 b2 + a2 b1 ) = (a2 − a1 )(b2 − b1 ) ≥ 0. Therefore a1 b2 + a2 b1 ≤ a1 b1 + a2 b2 . (b) We use induction on n.

Base case: n = 1. Suppose a1 and b1 are real numbers and f is a one-to-one, onto function from {1} to {1}. Then f (1) = 1, so a1 bf (1) = a1 b1 .

Induction step. Suppose n ≥ 1 and the result holds for sequences of real numbers of length n. Now suppose a1 , a2 , . . . , an+1 and b1 , b2 , . . . , bn+1 are real numbers, a1 ≤ a2 ≤ · · · ≤ an+1 , b1 ≤ b2 ≤ · · · ≤ bn+1 , and f is a one-to-one onto function from {1, 2, . . . , n+1} to {1, 2, . . . , n+1}.

We consider two cases.

Case 1. f (n + 1) = n + 1. Let f 0 = f \ {(n + 1, n + 1)}. Then f 0 is a one-to-one, onto function from {1, 2, . . . , n} to {1, 2, . . . , n}. (In the language of exercise 7 in Section 5.1, f 0 is the restriction of f to {1, 2, . . . , n}.) Therefore, by the inductive hypothesis, a1 bf (1) + a2 bf (2) + · · · + an+1 bf (n+1) = (a1 bf 0 (1) + a2 bf 0 (2) + · · · + an bf 0 (n) ) + an+1 bn+1 ≤ a1 b1 + a2 b2 + · · · + an bn + an+1 bn+1 .

Case 2. f (n + 1) 6= n + 1. Let k = f (n + 1) ≤ n. Since f is onto, there is some j ≤ n such that f (j) = n + 1. Now define g : {1, 2, . . . , n + 1} to {1, 2, . . . , n + 1} as follows:   if x = j, k, g(x) = n + 1, if x = n + 1,   f (x), otherwise.

In other words, g is the same as f , except that g(j) = k and g(n + 1) = n + 1. Notice that g is also a one-to-one, onto function from {1, 2, . . . , n + 1} to itself, and since g(n + 1) = n + 1, the reasoning in case 1 applies to g. Therefore, by case 1, a1 bg(1) + a2 bg(2) + · · · + an+1 bg(n+1) ≤ a1 b1 + a2 b2 + · · · an+1 bn+1 .

(∗)

Also, aj ≤ an+1 and bk ≤ bn+1 , so by part (a), aj bf (j) + an+1 bf (n+1) = aj bn+1 + an+1 bk ≤ aj bk + an+1 bn+1 = aj bg(j) + an+1 bg(n+1) , and therefore (aj bf (j) + an+1 bf (n+1) ) − (aj bg(j) + an+1 bg(n+1) ) ≤ 0.

Adding inequalities (∗) and (∗∗), we get the desired inequality, a1 bf (1) + a2 bf (2) + · · · + an+1 bf (n+1) ≤ a1 b1 + a2 b2 + · · · an+1 bn+1 .

(∗∗)

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85

11. We proceed by induction on n.

Base case: n = 0. If A has 0 elements, then A = ∅, so P(A) = {∅}, which has 1 = 20 elements.

Induction step: Suppose that for every set A with n elements, P(A) has 2n elements. Now suppose that A has n + 1 elements. Let a be any element of A, and let A0 = A \ {a}. Then A0 has n elements, so by inductive hypothesis P(A0 ) has 2n elements. There are two kinds of subsets of A: those that contain a as an element, and those that don’t. The subsets that don’t contain a are just the subsets of A0 , and there are 2n of these. Those that do contain a are the sets of the form X ∪ {a}, where X ∈ P(A0 ), and there are also 2n of these, since there are 2n possible choices for X. Thus the total number of elements of P(A) is 2n + 2n = 2n+1 .

12. We proceed by induction on n.

Base case: n = 0. If A has 0 elements, then A = ∅, so P2 (A) = ∅, which has 0 elements, and n(n − 1)/2 = 0.

Induction step: Suppose n ∈ N, and for every set A, if A has n elements then P2 (A) as n(n − 1)/2 elements. Now suppose A is a set with n+1 elements. Let a be some element of A, and let A0 = A\{a}.

Then A0 has n elements, so by inductive hypothesis P2 (A0 ) has n(n − 1)/2 elements. There are two kinds of two-element subsets of A: those that contain a as an element, and those that don’t. The subsets that don’t contain a are just the two-element subsets of A0 , and there are n(n − 1)/2 of these.

Those that do contain a are the sets of the form {a, x}, where x ∈ A0 , and there are n of these, since there are n possibilities for x. Thus the total number of elements of P2 (A) is n2 + n (n + 1)n n(n − 1) +n= = .

2 2 2 13. We use induction on n.

Base case: Suppose n = 1. Then the triangle is cut into 4 congruent equilateral triangles and one corner is removed. The remaining three triangles can be covered by one trapezoidal tile.

Induction step: Suppose n is a positive integer, and an equilateral triangle cut into a grid of 4n congruent equilateral triangles, with one corner removed, can be covered with trapezoidal tiles.

Now consider an equilateral triangle cut into a grid of 4n+1 congruent equilateral triangles, with one corner removed. The division lines include line segments connecting the centers of the sides of the triangle. These segments divide the triangle into four triangular subgrids, each of which is divided into 4n triangles. The corner that has been removed comes from one of these four subgrids. Place one trapezoidal tile at the point where the other three triangular subgrids meet, so that it covers one corner of each of these subgrids, as shown in the figure below. The area remaining to be covered now contains every triangle except for one corner in each of the triangular subgrids, so by applying the inductive hypothesis to each subgrid we see that this area can be covered with tiles.

14. Base case: n = 1. One chord cuts the circle into two regions, and (n2 + n + 2)/2 = 2.

Induction step: Suppose that when n chords are drawn, the circle is cut into (n2 + n + 2)/2 regions.

When another chord is drawn, it will intersect each of the first n chords exactly once. Therefore it will pass through n + 1 regions, cutting each of those regions in two. (Each time it crosses one of the first n chords, it passes from one region to another.) Therefore the number of regions after the next chord

86

SOLUTIONS is drawn is

n2 + n + 2 n2 + 3n + 4 (n + 1)2 + (n + 1) + 2 + (n + 1) = = , 2 2 2

as required.

15. Base case: When n = 1, there is one chord, cutting the circle into two regions. Color one region white and the other black.

Induction step: Suppose that the regions created by n chords can be colored as required, and consider a circle with n + 1 chords. Remove one chord and apply the inductive hypothesis to color the regions as required. Then add in the chord that was removed and reverse the colors of all regions on one side of that chord.

16. We use induction to prove that for every n ∈ N, if A is a set with n elements, f : A → A, and f is one-to-one, then f is onto.

Base case: n = 0. If A has 0 elements and f : A → A, then A = ∅, f = ∅, and f is both one-to-one and onto (vacuously).

Induction step: Suppose n ∈ N and for every set A with n elements and every function f : A → A, if f is one-to-one then f is onto. Now suppose A has n + 1 elements, f : A → A, and f is one-to-one but not onto. Since f is not onto, there is some a ∈ A such that a ∈ / Ran(f ). Let A0 = A \ {a} and 0 0 0 0 f = f ∩ (A × A ). Suppose x ∈ A . Then (x, f (x)) ∈ f and since a ∈ / Ran(f ), f (x) 6= a, so f (x) ∈ A0 .

0 0 0 Therefore (x, f (x)) ∈ A × A , so (x, f (x)) ∈ f . Now suppose (x, y1 ) ∈ f 0 and (x, y2 ) ∈ f 0 . Then (x, y1 ) ∈ f and (x, y2 ) ∈ f , so since f is a function, y1 = y2 . This proves that f 0 : A0 → A0 . Note that for every x ∈ A0 , (x, f 0 (x)) ∈ f 0 ⊆ f , so f (x) = f 0 (x). To see that f 0 is one-to-one, suppose x1 , x2 ∈ A0 and f 0 (x1 ) = f 0 (x2 ). Then f (x1 ) = f (x2 ), so since f is one-to-one, x1 = x2 . By inductive hypothesis, f 0 must map onto A0 . Since a ∈ / Ran(f ), f (a) 6= a, so f (a) ∈ A0 . Therefore, since f 0 is onto, there is 0 0 some x ∈ A such that f (x) = f (a). Since x ∈ A0 , x 6= a, but f (x) = f 0 (x) = f (a), which contradicts the fact that f is one-to-one.

17. To establish that every natural number is an element of A, you would have to let n be an arbitrary natural number and then prove that n ∈ A; the proof merely assumes that n ∈ A. This assumption applies only to the single (arbitrarily chosen) natural number n; it doesn’t apply to all natural numbers.

18. The mistake is the sentence “Now let a3 be an element of A that is different from both a1 and a2 .”

For such an element a3 to exist, A must have at least three elements. At this point in the proof, we know that A has n + 1 elements. If n ≥ 2, then A has at least three elements, and this step is correct.

But if n = 1 then A has only two elements. So the induction step doesn’t work when n = 1.

Section 6.3 1. We will prove by induction that for all n ≥ 1, n X

n 1 = . i(i + 1) n+1 i=1

Base case: When n = 1, both sides of Pnthe equation are equal to 1/2.

Induction step: Suppose n ≥ 1 and i=1 1/(i(i + 1)) = n/(n + 1). Then n+1 X

n X 1 1 1 = + i(i + 1) i(i + 1) (n + 1)(n + 2) i=1 i=1

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

(definition of summation) (inductive hypothesis)

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2. Base case: When n = 1, both sides of the equation are equal to 1/6.

Induction step: Suppose n ≥ 1 and n X

n2 + 3n 1 = . i(i + 1)(i + 2) 4(n + 1)(n + 2) i=1

Then n+1 X

n X 1 1 1 = + i(i + 1)(i + 2) i(i + 1)(i + 2) (n + 1)(n + 2)(n + 3) i=1 i=1

n2 + 3n 1 + 4(n + 1)(n + 2) (n + 1)(n + 2)(n + 3) n3 + 6n2 + 9n + 4 (n + 1)(n2 + 5n + 4) = = 4(n + 1)(n + 2)(n + 3) 4(n + 1)(n + 2)(n + 3) (n2 + 2n + 1) + (3n + 3) (n + 1)2 + 3(n + 1) = = .

4(n + 2)(n + 3) 4(n + 2)(n + 3) =

(inductive hypothesis)

3. Base case: When n = 2, both sides of the equation are equal to 1/3.

Induction step: Suppose n ≥ 2 and n X

1 3n2 − n − 2 = . (i − 1)(i + 1) 4n(n + 1) i=2

Then n+1 X

n X 1 1 1 = + (i − 1)(i + 1) (i − 1)(i + 1) n(n + 2) i=1 i=1

3n2 − n − 2 1 + 4n(n + 1) n(n + 2) 3n3 + 5n2 − 4n − 4 4n + 4 = + 4n(n + 1)(n + 2) 4n(n + 1)(n + 2) 3n2 + 5n 3(n + 1)2 − (n + 1) − 2 = = .

4(n + 1)(n + 2) 4(n + 1)(n + 2) =

(inductive hypothesis)

4. Base case: When n = 0, both sides of the equation are equal to 1.

Induction step: Suppose n ∈ N and n X

(2i + 1)2 =

i=0

(n + 1)(2n + 1)(2n + 3) .

3

Then n+1 X

n X (2i + 1)2 + (2n + 3)2

i=0

i=0

(2i + 1)2 =

(n + 1)(2n + 1)(2n + 3) + (2n + 3)2 3 (2n + 3)(2n2 + 9n + 10) (n + 2)(2n + 3)(2n + 5) = = .

3 3 =

(inductive hypothesis)

88

SOLUTIONS 5. Base case: When n = 0, both sides of the equation are equal to 1.

Induction step: Suppose n ∈ N and n X

ri =

i=0

rn+1 − 1 . r−1

Then n+1 X

ri =

n X

i=0

ri + rn+1

i=0 n+1

−1 + rn+1 r−1 rn+2 − 1 = . r−1

=

r

6. Base case: n = 1. Then

(inductive hypothesis)

n X 1

=1≤1=2−

i2 i=1

1 . n

Induction step: Suppose that n X 1 i=1

i2

≤2−

1 . n

Then n+1 X

n X 1 1 1 1 1 = + ≤2− + 2 2 2 i i (n + 1) n (n + 1)2 i=1 i=1

=2−

n2 + n 1 n2 + n + 1 < 2 − =2− . n(n + 1)2 n(n + 1)2 n+1

7. (a) We use induction on n.

Base case: When n = 0, the equation says a0 + b0 = a0 + b0 , which is true.

InductionPstep: Suppose nP∈ N, and P for any sequences of real numbers a0 , a1 , . . . , an and b0 , n n n b1 , . . . , bn , i=0 (ai + bi ) = i=0 ai + i=0 bi . Now suppose a0 , a1 , . . . , an+1 and b0 , b1 , . . . , bn+1 are sequences of real numbers. Then n+1 X

n X

i=0

i=0 n X

(ai + bi ) = =

=

(ai + bi ) + (an+1 + bn+1 )

i=0 n X

ai +

n X

! bi !

ai + an+1

i=0

=

n+1 X i=0

+ (an+1 + bn+1 )

(inductive hypothesis)

i=0

ai +

+

n X

! bi + bn+1

i=0 n+1 X

bi .

i=0

(b) We use induction on n.

Base case: When n = 0, both sides of the equation are equal to ca0 .

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89

P Induction step: Suppose n ∈ N, and for any real numbers c and a0 , a1 , . . . , an , c · i=0 (c · ai ). Then ! n+1 n X X c· ai = c · ai + an+1 i=0

Pn

i=0 ai =

i=0

=c·

n X

ai + c · an+1

i=0

=

n X (c · ai ) + c · an+1

(inductive hypothesis)

i=0

=

n+1 X

(c · ai ).

i=0

8. (a) We let m be arbitrary and then prove by induction that for all n ≥ m, Hn − Hm ≥ (n − m)/n.

Base case: n = m. Then Hn − Hm = 0 ≥ 0 = (n − m)/n.

Induction step: Suppose that n ≥ m and Hn − Hm ≥ (n − m)/n. Then n−m 1 1 − Hm ≥ + n+1 n n+1 n−m 1 n+1−m ≥ + = . n+1 n+1 n+1

Hn+1 − Hm = Hn +

(b) Base case: If n = 0 then H2n = H1 = 1 ≥ 1 = 1 + n/2.

Induction step: Suppose n ≥ 0 and H2n ≥ 1 + n/2. By part (a), H2n+1 − H2n ≥

2n+1 − 2n 1 = . n+1 2 2

Therefore

1 n 1 n+1 ≥1+ + =1+ .

2 2 2 2 (c) Since limn→∞ (1 + n/2) = ∞, by part (b) limn→∞ H2n = ∞. Clearly the Hn ’s form an increasing sequence, so limn→∞ Hn = ∞.

Pn−1 9. Base case: When n = 2, k=1 Hk = H1 = 1 = 2(3/2) − 2 = nHn − n.

Pn−1 Induction step: Suppose n ≥ 2 and k=1 Hk = nHn − n. Then H2n+1 ≥ H2n +

n X

Hk =

n−1 X

k=1

Hk + Hn

k=1

= nHn − n + Hn

(inductive hypothesis)

= (n + 1)Hn + 1 − (n + 1) = (n + 1) Hn +

1 n+1

− (n + 1)

= (n + 1)Hn+1 − (n + 1).

Pn 10. We will prove by induction that for all n ≥ 1, i=1 (i · (i!)) = (n + 1)! − 1.

Base case: When n = 1, both sides of Pnthe equation are equal to 1.

Induction step: Suppose n ≥ 1 and i=1 (i · (i!)) = (n + 1)! − 1. Then n+1 X

n X

i=1

i=1

(i · (i!)) =

(i · (i!)) + (n + 1) · (n + 1)!

90

SOLUTIONS = (n + 1)! − 1 + (n + 1) · (n + 1)!

(inductive hypothesis)

= (n + 2) · (n + 1)! − 1 = (n + 2)! − 1.

Pn 11. We will prove by induction that for all n ∈ N, i=0 (i/(i + 1)!) = 1 − 1/(n + 1)!.

Base case: When n = 0, both sides of equation are equal to 0.

Pthe n Induction step: Suppose n ∈ N and i=0 (i/(i + 1)!) = 1 − 1/(n + 1)!. Then n+1 X

n X i n+1 i = + (i + 1)! (i + 1)! (n + 2)! i=0 i=0

1 n+1 + (n + 1)! (n + 2)! n+2 n+1 1 =1− + =1− . (n + 2)! (n + 2)! (n + 2)!

=1−

(inductive hypothesis)

12. (a) We will prove by induction that for all n ∈ N, 2n ≥ n + 1, from which the desired conclusion follows.

Base case: When n = 0, 2n = 20 = 1 ≥ 0 + 1 = n + 1.

Induction step. Suppose n ∈ N and 2n ≥ n+1. Then 2n+1 = 2·2n ≥ 2·(n+1) = 2n+2 ≥ n+2. (b) Base case: n = 9. Then n! = 362880 ≥ 262144 = (2n )2 .

Induction step: Suppose that n ≥ 9 and n! ≥ (2n )2 . Then (n + 1)! = (n + 1) · n! ≥ (n + 1) · (2n )2 ≥ 10 · 22n ≥ 22 · 22n = 22n+2 = (2n+1 )2 .

2

(c) Base case: n = 0. Then n! = 1 ≤ 1 = 2(n ) .

2 Induction step: Suppose that n ∈ N and n! ≤ 2(n ) . Then 2

2

2

2

2((n+1) ) = 2n +2n+1 = 2(n ) · 22n+1 ≥ 2(n ) · 2n+1 > n! · (n + 1)

(by inductive hypothesis and part (a))

= (n + 1)!.

13. (a) Base case: When n = 0, (k 2 + n)! = (k 2 )! ≥ 1 = k 2n .

Induction step: Suppose n ∈ N and (k 2 +n)! ≥ k 2n . Then (k 2 +n+1)! = (k 2 +n+1)·(k 2 +n)! ≥ 2 k · k 2n = k 2n+2 = k 2(n+1) .

2 (b) Base case: When n = 2k 2 , by part (a), n! = (k 2 + k 2 )! ≥ k 2k = k n .

Induction step: Suppose n ≥ 2k 2 and n! ≥ k n . Then (n + 1)! = (n + 1) · n! ≥ (n + 1) · k n > k · k n = k n+1 .

14. Let a be an arbitrary real number and m an arbitrary natural number. We now prove by induction that for all n ∈ N, (am )n = amn .

Base case: When n = 0, (am )n = (am )0 = 1 = a0 = amn .

Induction step: Suppose n ∈ N and (am )n = amn . Then (am )n+1 = (am )n · am

(definition of exponentiation)

= amn · am

(inductive hypothesis)

= amn+m

(Example 6.3.2)

m(n+1)

=a

.

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91

15. Base case: n = 0. Then an = a0 = 0 = 20 − 0 − 1 = 2n − n − 1.

Induction step: Suppose that n ∈ N and an = 2n − n − 1. Then an+1 = 2an + n = 2(2n − n − 1) + n = 2n+1 − 2n − 2 + n = 2n+1 − n − 2 = 2n+1 − (n + 1) − 1. n

16. We will prove by induction that for all n ∈ N, an = 2(2 ) .

0 n Base case: n = 0. Then an = a0 = 2 = 2(2 ) = 2(2 ) . n+1 n n n Induction step: Suppose n ∈ N and an = 2(2 ) . Then an+1 = (an )2 = (2(2 ) )2 = 2(2 ·2) = 2(2 ) .

17. We will prove by induction that for all n ≥ 1, an = 1/n.

Base case: n = 1. Then an = a1 = 1 = 1/n.

Induction step: Suppose n ≥ 1 and an = 1/n. Then an+1 = 18. (a) (b)

n 0

an 1/n 1/n 1 = = = . an + 1 1/n + 1 (n + 1)/n n+1

= n!/(0!n!) = 1 and nn = n!/(n!0!) = 1.

n n! n n! + + = k−1 k!(n − k)! (k − 1)!(n − k + 1)! k n!(n − k + 1) n!k = + k!(n − k + 1)! k!(n − k + 1)!

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

(c) We follow the hint.

Base case: n = 0. Suppose A is a set with 0 elements. Then A = ∅, the only value of k we have to worry about is k = 0, P0 (A) = {∅}, which has 1 element, and 00 = 1.

Induction step: Suppose the desired conclusion holds for sets with n elements, and A is a set with n + 1 elements. Let a be an element of A, and let A0 = A \ {a}, which is a set with n elements. Now suppose 0 ≤ k ≤ n + 1. We consider three cases.

Case 1: k = 0. Then Pk (A) = {∅}, which has 1 element, and n+1 = 1. k

n+1 Case 2: k = n + 1. Then Pk (A) = {A}, which has 1 element, and k = 1.

Case 3. 0 < k ≤ n. There are two kinds of k-element subsets of A: those that contain a as an element, and those that don’t. The k-element subsets that don’t contain a are just the k-element

subsets of A0 , and by inductive hypothesis there are nk of these. Those that do contain a are the n sets of the form X ∪ {a}, where X ∈ Pk−1 (A0 ), and by inductive hypothesis there are k−1 of these, since this is the number of possibilities for X. Therefore by part (b), the total number of k-element subsets of A is

n n n+1 + = . k k−1 k (d) We let x and y be arbitrary and then prove the equation by induction on n.

Base case: n = 0. Then both sides of the equation are equal to 1.

Induction step: We will make use of parts (a) and (b). Suppose that n X n n−k k (x + y)n = x y . k k=0

Then (x + y)n+1 = (x + y)(x + y)n

92

SOLUTIONS n X n

xn−k y k (by inductive hypothesis) k k=0

n n n n−1 n n−2 2 x y+ x y = (x + y) x + 0 1 2

n n +··· + y n

n n+1 n n n n n n−1 2 = x + x y+ x y+ x y 0 0 1 1

n n n+1 + ··· + xy n + y n n

n n n n = xn+1 + + xn y + + xn−1 y 2 0 1 1 2

n n + ··· + + xy n + y n+1 n−1 n

n + 1 n+1 n+1 n n + 1 n−1 2 = x + x y+ x y 0 1 2

n+1 n + 1 n+1 + ··· + xy n + y n n+1 n+1 X n + 1 = xn+1−k y k . k

= (x + y)

k=0

19. (a) Proof 1: We apply part (d) of exercise 18, with x = y = 1: n

n

2 = (1 + 1) =

n X n k=0

k

n−k k

1

1 =

n X n k=0

k

.

Proof 2: Let A be a set with n elements. By part (c) of exercise 18, of elements in P(A). By exercise 11 of Section 6.2, this number is 2n . (b) We apply part (d) of exercise 18, with x = 1 and y = −1: 0 = (1 − 1)n =

n X n k=0

k

1n−k (−1)k =

n X

(−1)k

k=0

Pn

k=0

n k

is the number

n . k

20. We will prove by induction that for all n ≥ 1, 0 < an < 1/2, from which the desired conclusion follows.

Base case: n = 1. Then an = a1 = 1/4, so 0 < an < 1/2.

Induction step: Suppose n ≥ 1 and 0 < an < 1/2. Then by Example 3.1.2, a2n < (1/2)2 = 1/4, so an+1 = (an )2 + 1/4 < 1/4 + 1/4 = 1/2. It is also clear that an+1 = a2n + 1/4 ≥ 1/4 > 0, so 0 < an+1 < 1/2. x

(2x )

21. (a) f1 (x) = 2f0 (x) = 2x , f2 (x) = 2f1 (x) = 2(2 ) , f3 (x) = 2f2 (x) = 2(2 ) . (b) Let x and y be arbitrary positive integers with x < y. We now prove by induction that for all n ∈ N, fn (x) < fn (y). Base case: n = 0. Then fn (x) = f0 (x) = x < y = f0 (y) = fn (y).

Induction step: Suppose n ∈ N and fn (x) < fn (y). Then fn+1 (x) = 2fn (x) < 2fn (y) = fn+1 (y). (c) Let m be an arbitrary natural number and x an arbitrary positive integer. We now prove by induction that for all n ≥ m + 1, fm (x) < fn (x).

Base case: n = m + 1. Then by exercise 12(a), fm (x) < 2fm (x) = fm+1 (x) = fn (x).

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93

Induction step: Suppose n ≥ m + 1 and fm (x) < fn (x). As in the base case, by exercise 12(a), fn (x) < 2fn (x) = fn+1 (x). Therefore fm (x) < fn+1 (x). (d) Suppose n ∈ N. By part (c), for every x ∈ Z+ , fn (x) < fn+1 (x), so fn ∈ O(fn+1 ). Now suppose fn+1 ∈ O(fn ). Then we can choose a ∈ Z+ and c ∈ R+ such that for all x > a, fn+1 (x) ≤ cfn (x).

Let x be any positive integer such that x > a, x ≥ c, and x ≥ 5. Note that by part (c), fn (x) ≥ f0 (x) = x ≥ 5. Therefore fn+1 (x) = 2fn (x) > (fn (x))2

(Example 6.1.3, since fn (x) ≥ 5)

≥ cfn (x)

(since fn (x) ≥ f0 (x) = x ≥ c).

This contradicts the fact that for all x > a, fn+1 (x) ≤ cfn (x). Therefore fn+1 ∈ / O(fn ).

2

(23 )

(e) g(1) = f1 (1) = 21 = 2, g(2) = f2 (2) = 2(2 ) = 24 = 16, and g(3) = f3 (3) = 2(2 ) = 2256 = 115792089237316195423570985008687907853269984665640564039457584007913129639936. (f) Suppose n ∈ N. Then by part (c), for every x > n, fn (x) < fx (x) = g(x). Therefore for every n ∈ N, fn ∈ O(g). Now suppose n ∈ N and g ∈ O(fn ). Since fn+1 ∈ O(g), by exercise 19(b) in Section 5.1, fn+1 ∈ O(fn ), which contradicts part (d).

22. By changing the goal to a stronger statement, we made the inductive hypothesis stronger, which helped in the induction step.

Section 6.4 1. (a) (→) Suppose that ∀nQ(n). Let n be arbitrary. Then Q(n+1) is true, which means ∀k < n+1 P (k).

In particular, since n < n + 1, P (n) is true. Since n was arbitrary, this shows that ∀nP (n). (←) Suppose that ∀nP (n). Then for any n, it is clearly true that ∀k < n P (k), which means that Q(n) is true. (b) Base case: n = 0. Then Q(n) is the statement ∀k < 0 P (k), which is vacuously true.

Induction step: Suppose Q(n) is true. This means that ∀k < n P (k) is true, so by assumption, it follows that P (n) is true. Therefore ∀k < n + 1 P (k) is true, which means that Q(n + 1) is true. √ 2. Suppose q ∈ N, q > 0, and ∀k < q[0 < k → ¬∃j ∈ Z+ (j/k = 2)]. Now suppose there is some √ p ∈ Z+ such that p/q = 2. As in the proof of Theorem 6.4.5, we can show that p and q are both even, so there√are positive integers j and k such that p = 2j and q = 2k. Therefore √ 0 < k < q and j/k = p/q = 2, which contradicts the inductive hypothesis. Thus ¬∃p ∈ Z+ (p/q = 2). √ √ 3. Suppose 2 is rational, and let S = {q ∈ Z+ | ∃p ∈ Z+ (p/q = 2)} 6= ∅. By the well-ordering √ principle, let q be the smallest element of S, and let p be a positive integer such that p/q = 2. Then √ p2 /q 2 = 2, so p2 = 2q 2 . Since 1 < 2 = p/q < 2, q < p < 2q, and therefore p − q and 2q − p are positive integers. Since p2 = 2q 2 ,

2 2q − p 4q 2 − 4pq + p2 6q 2 − 4pq = 2 = = 2, p−q p − 2pq + q 2 3q 2 − 2pq √ so (2q − p)/(p − q) = 2. Therefore p − q ∈ S and p − q < 2q − q = q, which contradicts the fact that q is the smallest element of S. √ √ 4. (a) Suppose 6 is rational. Let S = {q ∈ Z+ | ∃p ∈ Z+ (p/q = 6)}. Then S 6= ∅, so we √ can let q be the smallest element of S, and we can choose a positive integer p such that p/q = 6. Therefore p2 = 6q 2 , so p2 is even, and hence p is even. This means that p = 2p, for some integer p. Thus 4p2 = 6q 2 , so 2p2 = 3q 2 and therefore 3q 2 is even. It is easy to check that if q is√odd then 3q 2 is odd, so q must be even, which means that q = 2q for some integer q. But then 6 = p/q and q < q, contradicting the fact that q is the smallest element of S.

94

SOLUTIONS √ √ √ √ (b) Suppose that 2 + 3 = p/q. Squaring both sides gives us 5 + 2 6 = p2 /q 2 , so 6 = (p2 − 5q 2 )/(2q 2 ), which contradicts part (a).

5. We use strong induction to prove that ∀n ∈ N(n ≥ 12 → some combination of red and blue beads is worth n Martian credits). Suppose n ∈ N, n ≥ 12, and ∀k < n(k ≥ 12 → some combination of red and blue beads is worth k Martian credits). We now consider several cases.

Case 1. n = 12. Then 4 blue beads is worth n credits.

Case 2. n = 13. Then 1 red bead and 2 blue beads is worth n credits.

Case 3. n = 14. Then 2 red beads is worth n credits.

Case 4. n ≥ 15. Then n − 3 ≥ 12, so by inductive hypothesis there are natural numbers b and r such that b blue beads and r red beads are worth n − 3 credits. Therefore b + 1 blue beads and r red beads are worth n − 3 + 3 = n credits.

6. We use strong induction. Suppose n ≥ 1 and for all k, if 1 ≤ k < n then xk + 1/xk is an integer.

Case 1. n = 1. Then xn + 1/xn = x + 1/x, which is an integer by assumption.

Case 2. n = 2. Then x2 + 1/x2 = (x + 1/x)2 − 2, which is an integer.

Case 3. n ≥ 3. Then by inductive hypothesis, xn−1 + 1/xn−1 and xn−2 + 1/xn−2 are integers, and therefore

1 1 1 1 x+ − xn−2 + n−2 xn + n = xn−1 + n−1 x x x x is also an integer.

7. (a) We use ordinary induction on n.

Base case: n = 0. Both sides of equation are equal to 0.

Pthe n Induction step: Suppose that i=0 Fi = Fn+2 − 1. Then n+1 X

Fi =

i=0

n X

Fi + Fn+1 = (Fn+2 − 1) + Fn+1 = Fn+3 − 1.

i=0

(b) We use ordinary induction on n.

Base case: n = 0. Both sides of equation are equal to 0.

Pthe n Induction step. Suppose that i=0 (Fi )2 = Fn Fn+1 . Then n+1 X

n X (Fi )2 + (Fn+1 )2 = Fn Fn+1 + (Fn+1 )2 = Fn+1 (Fn + Fn+1 ) = Fn+1 Fn+2 .

i=0

i=0

(Fi )2 =

(c) We use ordinary induction on n.

Base case: n = 0. Both sides of equation are equal to 1.

Pthe n Induction step: Suppose that i=0 F2i+1 = F2n+2 . Then n+1 X

F2i+1 =

i=0

n X

F2i+1 + F2n+3 = F2n+2 + F2n+3 = F2n+4 = F2(n+1)+2 .

i=0

Pn (d) We will use induction to prove that for all n, i=0 F2i = F2n+1 − 1.

Base case: n = 0. Both sides Pn of the equation are equal to 0.

Induction step: Suppose i=0 F2i = F2n+1 − 1. Then n+1 X i=0

F2i =

n X i=0

F2i + F2n+2 = F2n+1 − 1 + F2n+2 = F2n+3 − 1 = F2(n+1)+1 − 1.

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95

8. (a) Let m ≥ 1 be arbitrary. We now prove by strong induction that ∀n ∈ N(Fm+n = Fm−1 Fn + Fm Fn+1 ). Suppose that n ∈ N and ∀k < n(Fm+k = Fm−1 Fk + Fm Fk+1 ).

Case 1. n = 0. Then Fm−1 Fn + Fm Fn+1 = Fm−1 · 0 + Fm · 1 = Fm = Fm+n .

Case 2. n = 1. Then Fm−1 Fn + Fm Fn+1 = Fm−1 · 1 + Fm · 1 = Fm+1 = Fm+n .

Case 3. n ≥ 2. Then by inductive hypothesis, Fm+n−2 = Fm−1 Fn−2 + Fm Fn−1 , Fm+n−1 = Fm−1 Fn−1 + Fm Fn .

Adding these two equations, we get Fm+n = Fm+n−2 + Fm+n−1 = Fm−1 (Fn−2 + Fn−1 ) + Fm (Fn−1 + Fn ) = Fm−1 Fn + Fm Fn+1 . (b) Once again we let m ≥ 1 be arbitrary and then proceed by strong induction on n. Suppose that n ≥ 1 and ∀k(1 ≤ k < n → Fm+k = Fm+1 Fk+1 − Fm−1 Fk−1 ).

Case 1. n = 1. Then Fm+1 Fn+1 − Fm−1 Fn−1 = Fm+1 · 1 + Fm−1 · 0 = Fm+1 = Fm+n .

Case 2. n = 2. Then Fm+1 Fn+1 −Fm−1 Fn−1 = Fm+1 ·2−Fm−1 ·1 = Fm+1 +(Fm+1 −Fm−1 ) = Fm+1 + Fm = Fm+2 = Fm+n .

Case 3. n ≥ 3. Then by inductive hypothesis, Fm+n−2 = Fm+1 Fn−1 − Fm−1 Fn−3 , Fm+n−1 = Fm+1 Fn − Fm−1 Fn−2 .

Adding these two equations, we get Fm+n = Fm+n−2 + Fm+n−1 = Fm+1 (Fn−1 + Fn ) − Fm−1 (Fn−3 + Fn−2 ) = Fm+1 Fn+1 − Fm−1 Fn−1 . (c) We use ordinary induction.

Base case: n = 0. Then (Fn )2 + (Fn+1 )2 = 02 + 12 = 1 = F2n+1 and (Fn+2 )2 − (Fn )2 = 2 1 − 02 = 1 = F2n+2 .

Induction step: Suppose (Fn )2 + (Fn+1 )2 = F2n+1 , 2

2

(∗)

(Fn+2 ) − (Fn ) = F2n+2 .

(∗∗)

(Fn+1 )2 + (Fn+2 )2 = F2n+1 + F2n+2 = F2n+3 ,

(∗∗∗)

Adding (∗) and (∗∗) gives

which is the first equation we must prove. To prove the second, we compute (Fn+3 )2 − (Fn+1 )2 = (Fn+2 + Fn+1 )2 − (Fn+1 )2 = (Fn+2 )2 + 2Fn+2 Fn+1 + (Fn+1 )2 − (Fn+1 )2 = (Fn+2 )2 + 2(Fn+1 + Fn )Fn+1 = (Fn+2 )2 + 2(Fn+1 )2 + 2Fn Fn+1 = (Fn+1 )2 + (Fn+2 )2 + ((Fn+1 )2 + 2Fn Fn+1 + (Fn )2 ) − (Fn )2 = (Fn+1 )2 + (Fn+2 )2 + (Fn+1 + Fn )2 − (Fn )2 = (Fn+1 )2 + (Fn+2 )2 + (Fn+2 )2 − (Fn )2 = F2n+3 + F2n+2 = F2n+4 .

(by (∗∗) and (∗∗∗))

96

SOLUTIONS (d) Let m ∈ N be arbitrary. We first use induction to prove that ∀k ∈ N(Fm | Fkm ).

Base case: k = 0. Then Fkm = F0 = 0 = 0 · Fm , so Fm | Fkm .

Induction step: Suppose Fm | Fkm . Then there is some integer j such that Fkm = jFm .

Therefore by part (a), F(k+1)m = Fkm+m = Fkm−1 Fm + Fkm Fm+1 = Fkm−1 Fm + (jFm )Fm+1 = (Fkm−1 + jFm+1 )Fm , so Fm | F(k+1)m .

Finally, to complete the problem, suppose m | n. Then there is some natural number k such that n = km, so by the statement we have just proven, Fm | Fn . (e) We use ordinary induction.

Base case: If n = 1, then n−1 X

0 = 1 = F2n−1 , 0 i=0 n−1 X 2n − i − 1 1 = = 1 = F2n . i 0 i=0 2n − i − 2 i

=

Induction step: Suppose n ≥ 1 and F2n−1 =

n−1 X i=0

2n − i − 2 , i

F2n =

n−1 X i=0

2n − i − 1 . i

Then n−1 X 2n − i − 2 2n − i − 1 + i i i=0 i=0

2n − 1 2n − 2 n 2n − 2 2n − 3 n−1 = + + ··· + + + + ··· + 0 1 n−1 0 1 n−1

2n − 2 2n − 2 2n − 3 2n − 3 2n − 1 + + + + + ··· = 1 0 2 1 0

n n n−1 + + + n−1 n−2 n−1

2n − 1 2n − 2 n+1 =1+ + + ··· + +1 (by exercise 18 in Section 6.3) 1 2 n−1

X

n 2n 2n − 1 2n − 2 n 2n − i . = + + + ··· + = 0 1 2 n i i=0

F2n+1 = F2n + F2n−1 =

n−1 X

Similarly,

n X 2n − i

n−1 X

2n − i − 1 F2n+2 = F2n+1 + F2n = + i i i=0 i=0

2n 2n − 1 n 2n − 1 2n − 2 n = + + ··· + + + + ··· + 0 1 n 0 1 n−1

2n 2n − 1 2n − 1 2n − 2 2n − 2 = + + + + + ··· 0 1 0 2 1

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97

n n + + n n−1

2n 2n − 1 n+1 =1+ + + ··· + (by exercise 18 in Section 6.3) 1 2 n

X

n 2n + 1 2n 2n − 1 n+1 2n + 1 − i = + + + ··· + = .

0 1 2 n i i=0

9. (a) (→) Suppose a0 , a1 , a2 , . . . is a Gibonacci sequence. Then in particular a2 = a0 + a1 , which means c2 √ = 1+c. Solving this quadratic equation by the quadratic formula leads to the conclusion c = (1 ± 5)/2. √ √ (←) Suppose either c = (1 + 5)/2 or c = (1 − 5)/2. Then c2 = 1 + c, and therefore for every n ≥ 2, an = cn = cn−2 c2 = cn−2 (1 + c) = cn−2 + cn−1 = an−2 + an−1 . √ √ (b) It will be convenient to introduce the notation c1 = (1 + 5)/2 and c2 = (1 − 5)/2. Then for any n ≥ 2, an = scn1 + tcn2 = scn−2 c21 + tcn−2 c22 = scn−2 (1 + c1 ) + tc2n−2 (1 + c2 ) = (scn−2 + tcn−2 )+ 1 2 1 1 2 n−1 n−1 (sc1 + tc2 ) = an−2 + an−1 . √ √ (c) Let s = (5a0 + (2a1 − a0 ) 5)/10 and t = (5a0 − (2a1 − a0 ) 5)/10. For all n ∈ N let √ !n √ !n 1+ 5 1− 5 bn = s +t , 2 2 and notice that by part (b), b0 , b1 , . . . is a Gibonacci sequence.

We now prove by strong induction that for all n ∈ N, an = bn , which will solve the problem.

Suppose n ∈ N and for all k < n, ak = bk .

Case 1. n = 0. Then √ !0 √ !0 1+ 5 1− 5 b0 = s +t 2 2 √ √ 5a0 + (2a1 − a0 ) 5 5a0 − (2a1 − a0 ) 5 = + 10 10 10a0 = a0 . = 10 Case 2. n = 1. Then √ !1 1+ 5 +t 2

√ !1 1− 5 b1 = s 2 √ ! √ ! √ !

5a0 + (2a1 − a0 ) 5 1+ 5 5a0 − (2a1 − a0 ) 5 = + 10 2 10 √ √ 10a1 + (4a0 + 2a1 ) 5 10a1 − (4a0 + 2a1 ) 5 = + 20 20 20a1 = = a1 .

20

√ !

1− 5 2

Case 3. n ≥ 2. Then since both the a’s and the b’s are Gibonacci sequences, an = an−2 + an−1 = bn−2 + bn−1 = bn .

10. The Lucas numbers form a Gibonacci sequence, so we can apply our solution √ to exercise 9(c) √ to find a formula for them. According to that solution, for every n, Ln = s((1 + 5)/2)n + t((1 − 5)/2)n ,

98

SOLUTIONS where

√ √ √ √ 5L0 + (2L1 − L0 ) 5 10 + 0 · 5 5L0 − (2L1 − L0 ) 5 10 − 0 · 5 s= = = 1, t= = = 1.

10 10 10 10 Thus, √ !n √ !n 1− 5 1+ 5 + .

Ln = 2 2 11. Imitating exercise 9, we first consider sequence of the form an = cn , for some constant c. If such a sequence satisfies the recurrence an = 5an−1 − 6an−2 , then in particular a2 = 5a1 − 6a0 , so c2 = 5c − 6.

The solutions to this quadratic equation are c = 2, 3.

Now consider sequences of the form an = s2n + t3n , for any two numbers s and t. For any such sequence, if n ≥ 2 then 5an−1 − 6an−2 = 5(s2n−1 + t3n−1 ) − 6(s2n−2 + t3n−2 ) = 4s2n−2 + 9t3n−2 = s2n + t3n = an . In other words, any such sequence will satisfy the recurrence in the exercise. If we can choose s and t so that a0 = −1 and a1 = 0, then the sequence will be the sequence determined by the recursive definition in the exercise.

Since a0 = s20 + t30 = s + t and a1 = s21 + t31 = 2s + 3t, we want s and t satisfying the equations s + t = −1,

2s + 3t = 0.

The unique solution to this system of equations is s = −3, t = 2. So the sequence is given by the formula an = −3 · 2n + 2 · 3n .

12. We use strong induction. Suppose that n ∈ N, and for all k < n, ak = Fk .

Case 1. n = 0. Then an = a0 = 0 = F0 = Fn .

Case 2. n = 1. Then an = a1 = 1 = F1 = Fn .

Case 3. n = 2. Then an = a2 = 1 = F2 = Fn .

Case 4. n ≥ 3. Then 3 1 1 an = an−3 + an−2 + an−1 2 2 2 1 3 1 = Fn−3 + Fn−2 + Fn−1 (inductive hypothesis) 2 2 2 1 1 = (Fn−3 + Fn−2 ) + Fn−2 + Fn−1 2 2 1 1 = Fn−1 + Fn−2 + Fn−1 2 2 = Fn−2 + Fn−1 = Fn .

13. We use strong induction. Suppose n ∈ N and for all k < n, Pk has Fk+2 elements.

Case 1. n = 0. P0 = {∅}, which has 1 element, and Fn+2 = F2 = 1.

Case 2. n = 1. P1 = {∅, {1}}, which has 2 elements, and Fn+2 = F3 = 2.

Case 3. n ≥ 2. The elements of Pn that don’t contain n are the elements of Pn−1 , and by inductive hypothesis there are Fn+1 of them. If an element of Pn contains n, then since it cannot contain consecutive integers, it must not contain n − 1. Therefore it must have the form X ∪ {n}, where X ∈ Pn−2 , and every set of this form will be in Pn . The number of such sets is Fn , since by inductive hypothesis that is the number of possibilities for X. Therefore the number of elements of Pn is Fn+1 + Fn = Fn+2 .

14. (a) If n ≥ 0 then the result follows from the division algorithm, Theorem 6.4.1. Now suppose n < 0.

Then −n − 1 ≥ 0, so we can apply Theorem 6.4.1 to −n − 1 and m to conclude that there are natural numbers q 0 and r0 such that −n − 1 = q 0 m + r0 and 0 ≤ r0 < m, so 0 ≤ r0 ≤ m − 1.

Therefore n = −q 0 m − r0 − 1 = (−q 0 − 1)m + (m − 1 − r0 ). Let q = −q 0 − 1 and r = m − 1 − r0 .

Then q and r are integers, n = qm + r, and since 0 ≤ r0 ≤ m − 1, 0 ≤ r ≤ m − 1 < m.

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99

(b) Suppose n = qm + r = q 0 m + r0 , where q, r, q 0 , and r0 are integers, 0 ≤ r < m, and 0 ≤ r0 < m.

Then (q − q 0 )m = r0 − r and −m < r0 − r < m, so −m < (q − q 0 )m < m. Dividing by m, we get −1 < q − q 0 < 1. Since the only integer between −1 and 1 is 0, we must have q − q 0 = 0. Therefore q = q 0 , and r0 − r = (q − q 0 )m = 0, so r = r0 . (c) Suppose n is an integer. By part (a), there are integers q and r such that n = 3q + r ands 0 ≤ r < 3, so r ∈ {0, 1, 2}. Therefore n − r = 3q, so n ≡ r (mod 3). This proves that at least one of the three statements is true. To show that only one statement is true, suppose that n ≡ r1 (mod 3) and n ≡ r2 (mod 3), where r1 , r2 ∈ {0, 1, 2}. Then there are integers q1 and q2 such that n − r1 = 3q1 and n − r2 = 3q2 . Therefore n = 3q1 + r1 = 3q2 + r2 , and by part (b) it follows that r1 = r2 .

15. Let a be the larger of 5k and k(k + 1). Now suppose n > a, and by the division algorithm choose q and r such that n = qk + r and 0 ≤ r < k. Note that if q ≤ 4 then n = qk + r ≤ 4k + r < 5k ≤ a, which is a contradiction. Therefore q > 4, so q ≥ 5, and by Example 6.1.3 it follows that 2q ≥ q 2 .

Similar reasoning shows that q ≥ k + 1, so q 2 ≥ q(k + 1) = qk + q > qk + k > qk + r = n. Therefore 2n ≥ 2qk = (2q )k ≥ (q 2 )k ≥ nk .

16. (a) We use induction on k.

Base case: k = 1. Suppose f1 ∈ O(g) and a1 is a real number, and let f (n) = a1 f1 (n). We must prove that f ∈ O(g). Since f1 ∈ O(g), we can choose a ∈ Z+ and c ∈ R+ such that for all n > a, |f1 (n)| ≤ c|g(n)|. Let c0 = (|a1 | + 1)c ∈ R+ . Then for all n > a, |f (n)| = |a1 ||f1 (n)| ≤ (|a1 | + 1)c|g(n)| = c0 |g(n)|, so f ∈ O(g).

Induction step: Assume the statement holds for k, and suppose f1 , f2 , . . . , fk+1 are elements of O(g) and a1 , a2 , . . . , ak+1 are real numbers. Let h(n) = a1 f1 (n) + a2 f2 (n) + · · · + ak fk (n) and f (n) = a1 f1 (n) + a2 f2 (n) + · · · + ak+1 fk+1 (n) = h(n) + ak+1 fk+1 (n). Then by inductive hypothesis, h ∈ O(g), and therefore by exercise 19(c) of Section 5.1, f ∈ O(g).

17. We will prove by induction that for all n ∈ N, an = 2n .

Base case: When n = 0, an = a0 = 1 = 20 = 2n .

Induction step: Suppose n ∈ N and an = 2n . If n = 0 then an+1 = a1 = 1 + a0 = 1 + 1 = 2 = 21 = 2n+1 . Now suppose n ≥ 1. Then an+1 = 1 +

n X

ai = 1 +

n−1 X

i=0

ai + an = an + an = 2an = 2 · 2n = 2n+1 .

i=0

(Note: An alternative approach is to use strong induction and apply Example 6.1.1.)

18. We will prove by induction that for all n ∈ N, an = Fn+2 /Fn+1 .

Base case: When n = 0, an = a0 = 1 = 1/1 = F2 /F1 = Fn+2 /Fn+1 .

Induction step: Suppose n ∈ N and an = Fn+2 /Fn+1 . Then an+1 = 1 +

Fn+1 Fn+2 + Fn+1 Fn+3 1 =1+ = = . an Fn+2 Fn+2 Fn+2

19. (a) We first prove the following lemma.

Lemma. For every integer n, if n 6≡ 0 (mod 3) then n2 ≡ 1 (mod 3).

Proof. Suppose n is an integer and n 6≡ 0 (mod 3). Then by exercise 14(c), either n ≡ 1 (mod 3) or n ≡ 2 (mod 3).

Case 1. n ≡ 1 (mod 3). Then there is some integer k such that n − 1 = 3k, so n = 3k + 1.

Therefore n2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1, so n2 − 1 = 3(3k 2 + 2k) and n2 ≡ 1 (mod 3).

Case 2. n ≡ 2 (mod 3). Then there is some integer k such that n = 3k + 2, so n2 = 9k 2 + 12k + 4 = 3(3k 2 + 4k + 1) + 1 and therefore n2 ≡ 1 (mod 3).

100

SOLUTIONS Now, to solve the problem, suppose m and n are integers and 3 | (m2 + n2 ). Then m2 + n2 ≡ 0 (mod 3). Suppose m 6≡ 0 (mod 3). Then by the lemma, m2 ≡ 1 (mod 3), so there is some integer j such that m2 = 3j + 1.

Case 1. n ≡ 0 (mod 3). Then there is some integer k such that n = 3k. Therefore m2 + n2 = 3j + 1 + 9k 2 = 3(j + 3k 2 ) + 1, so m2 + n2 ≡ 1 (mod 3). By exercise 14(c), this contradicts the fact that m2 + n2 ≡ 0 (mod 3).

Case 2. n 6≡ 0 (mod 3). Then by the lemma n2 ≡ 1 (mod 3), so there is some integer k such that n2 = 3k + 1. Therefore m2 + n2 = 3j + 1 + 3k + 1 = 3(j + k) + 2, so m2 + n2 ≡ 2 (mod 3).

Once again, by exercise 14(c), this contradicts m2 + n2 ≡ 0 (mod 3).

Thus it is not possible that m 6≡ 0 (mod 3), so m ≡ 0 (mod 3) and therefore 3 | m. A similar argument shows that 3 | n. (b) Adding the two equations in (∗) we get c2 + d2 = 3a2 + 3b2 = 3(a2 + b2 ). Therefore 3 | (c2 + d2 ), so by part (a), 3 | c and 3 | d. (c) By part (b), there are positive integers j and k such that c = 3j and d = 3k. Therefore adding the equations in (∗) gives 3a2 + 3b2 = c2 + d2 = 9j 2 + 9k 2 . Dividing by 3, we get a2 + b2 = 3(j 2 + k 2 ), so 3 | (a2 + b2 ) and by part (a), 3 | a and 3 | b. (d) By part (c), there are positive integers s and t such that a = 3s and b = 3t. Substituting into (∗) we get 9s2 + 18t2 = 9j 2 , 18s2 + 9t2 = 9k 2 .

Dividing both equations by 9 gives us s2 + 2t2 = j 2 ,

2s2 + t2 = k 2 .

Therefore k ∈ S. But d = 3k > k, so this contradicts the choice of d as the smallest element of S. 20. (a) Let x = BC and y = BE. Then AE = EF = BC = x, so AB = AE + BE = x + y. The equality of ratios in the problem is then x+y y x = =1+ . y x x If we let r = x/y, then the equation is r = 1 + 1/r. Multiplying by r, we get r2 = r + 1, so √ 2 r − r − 1 = 0. The quadratic formula now gives us r = (1 ± 5)/2. Since r is clearly positive, √ we must have r = (1 + 5)/2 = ϕ. (b) We write all angles in degrees but drop the degree sign. Let x = cos(36). We use several trigonometric identities: cos(108) = cos(180 − 72) = − cos(72), cos(72) = cos(2 · 36) = 2 cos2 (36) − 1 = 2x2 − 1, cos(108) = cos(36 + 72) = cos(36) cos(72) − sin(36) sin(2 · 36) = x(2x2 − 1) − 2 sin2 (36) cos(36) = 2x3 − x − 2(1 − cos2 (36)) cos(36) = 2x3 − x − 2(1 − x2 )x = 4x3 − 3x.

The equation cos(108) = − cos(72) now gives us 4x3 − 3x = −2x2 + 1, so 4x3 + 2x2 − 3x − 1 = 0.

(∗)

Since 4x3 + 2x2 − 3x − 1 = (x + 1)(4x2 − 2x − 1), by the quadratic formula the solutions to (∗) are √ √ √ √ 1+ 5 2 − 20 1− 5 2 + 20 x = −1, x = = ≈ 0.809, x = = ≈ −0.309.

8 4 8 4 √ Since 0 < x < 1, the only possibility is x = (1 + 5)/4 = ϕ/2.

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101

(c) By Example 6.2.3, the angles in the pentagon ABCDE, measured in degrees, add up to 3 · 180 = 540, and therefore ∠ABC = 540/5 = 108. Since 4ABC is isosceles, ∠BAC = ∠BCA = 36.

Let M be the midpoint of AC. Then 4AM B is a right triangle, and AM = cos(36). Similarly, M C = cos(36), so by part (b), AC = 2 cos(36) = ϕ.

21. (a) For any numbers a, b, c, and d, (ab)(cd) = (cd)(ab)

(commutative law)

= c(d(ab))

(associative law)

= c((da)b)

(associative law)

= c((ad)b)

(commutative law).

(b) To simplify notation, we will assume that any product is the left-grouped product unless parentheses are used to indicate otherwise. We use strong induction on n. Assume the statement is true for products of fewer than n terms, and consider any product of a1 , a2 , . . . , an . If n = 1, then the only product is the left-grouped product, so there is nothing to prove. Now suppose n > 1.

Then our product has the form pq, where p is a product of a1 , . . . , ak−1 and q is a product of ak , . . . , an for some k with 2 ≤ k ≤ n. By inductive hypothesis, p = a1 · · · ak−1 and q = ak · · · an (where by our convention, these two products are left-grouped). Thus, it will suffice to prove (a1 · · · ak−1 )(ak · · · an ) = a1 · · · an . If k = n, then the left-hand side of this equation is already left-grouped, so there is nothing to prove. If k < n, then (a1 · · · ak−1 )(ak · · · an ) = (a1 · · · ak−1 )((ak · · · an−1 )an )

(definition of left-grouped)

= ((a1 · · · ak−1 )(ak · · · an−1 ))an

(associative law)

= (a1 · · · an−1 )an

(inductive hypothesis)

= a1 · · · an

(definition of left-grouped).

(c) By part (b), we may assume that the two products are left-grouped. Thus, we must prove that if b1 , b2 , . . . , bn is some reordering of a1 , a2 , . . . , an , then a1 · · · an = b1 · · · bn , where as in part (b) we assume products are left-grouped unless parentheses indicate otherwise. We use induction on n. If n = 1 then the products are clearly equal because b1 = a1 . Now suppose the statement is true for products of length n, and suppose that b1 , . . . , bn+1 is a reordering of a1 , . . . , an+1 .

Then bn+1 is one of a1 , . . . , an+1 . If bn+1 = an+1 then b1 · · · bn+1 = (b1 · · · bn )an+1

(definition of left-grouped)

= (a1 · · · an )an+1

(inductive hypothesis)

= a1 · · · an+1

(definition of left-grouped).

Now suppose bn+1 = ak for some k ≤ n. We will write a1 · · · a ck · · · an for the (left-grouped) product of the numbers a1 , . . . , an with the factor ak left out. Then b1 · · · bn+1 = (b1 · · · bn )ak

(definition of left-grouped)

= (a1 · · · a ck · · · an+1 )ak

(inductive hypothesis)

= ((a1 · · · a ck · · · an )an+1 )ak

(definition of left-grouped)

= (a1 · · · a ck · · · an )(an+1 ak )

(associative law)

= (a1 · · · a ck · · · an )(ak an+1 )

(commutative law)

= ((a1 · · · a ck · · · an )ak )an+1

(associative law)

= (a1 · · · an )an+1

(inductive hypothesis)

= a1 · · · an+1

(definition of left-grouped).

102

SOLUTIONS

Section 6.5 1. For all n, Bn = {n}. We prove this by induction.

Base case: If n = 0, then Bn = B0 = B = {0} = {n}.

Induction step: Suppose n ∈ N and Bn = {n}. Then Bn+1 = f (Bn ) = {f (n)} = {n + 1}.

2. For all n, Bn = {x ∈ Z | x ≥ −n}. We prove this by induction.

Base case: If n = 0, then Bn = B0 = B = N = {x ∈ Z | x ≥ 0}.

Induction step. Suppose n ∈ N and Bn = {x ∈ Z | x ≥ −n}. Suppose c ∈ Bn+1 = f (Bn ). Then there is some b ∈ Bn such that c = f (b) = b − 1. Since b ∈ Bn , b ∈ Z and b ≥ −n. Therefore b − 1 ∈ Z and b − 1 ≥ −n − 1 = −(n + 1), so c = b − 1 ∈ {x ∈ Z | x ≥ −(n + 1)}. Next, suppose c ∈ {x ∈ Z | x ≥ −(n + 1)}. Then c ∈ Z and c ≥ −(n + 1) = −n − 1. Let b = c + 1. Then b ∈ Z and b ≥ −n, so b ∈ {x ∈ Z | x ≥ −n} = Bn , and c = b − 1 = f (b) ∈ f (Bn ) = Bn+1 . This proves that Bn+1 = {x ∈ Z | x ≥ −(n + 1)}.

S 3. Let C = n∈N Bn . Since every element of F is a function from A to A, it is not hard to see that for every n, Bn ⊆ A (try proving it by induction), so C ⊆ A. It is also clear that BS= B0 ⊆ C. To see that C is closed under every function f ∈ F, suppose f ∈ F and x ∈ C. Since C = n∈N Bn , we can choose some n ∈ N such that x ∈ Bn . Therefore f (x) ∈ f (Bn ) ⊆ Bn+1 ⊆ C. Finally, suppose B ⊆ D ⊆ A and D is closed under every function f ∈ F. We claim that for every n ∈ N, Bn ⊆ D, from which it will follow that C ⊆ D. We prove this by induction on n. For the base case, B0 = B ⊆ D by assumption.

Now suppose Bn ⊆ D and x ∈ Bn+1 . Then by the definition of Bn+1 , there is some f ∈ F such that x ∈ f (Bn ), which means that x = f (b) for some b ∈ Bn . By the inductive hypothesis, Bn ⊆ D, so b ∈ D. Since D is closed under f , x = f (b) ∈ D. 4. B0 = {∅}, B1 = {X ∈ P(N) | X has exactly one element}, B2 = {X ∈ P(N) | X has either one or two elements}, . . . . In general, for every positive integer n, Bn = {X ∈ P(N) | X 6= ∅ and X has at most n elements}. We prove this last statement by induction on n.

Base case: When n = 1 we have Bn = B1 = {fk (∅) | k ∈ N} = {{k} | k ∈ N} = {X ∈ P(N) | X has exactly one element}.

Induction step: Suppose n ≥ 1 and Bn = {X ∈ P(N) | X 6= ∅ and X has at most n elements}.

Now suppose Y ∈ Bn+1 . Then for some k ∈ N and Z ∈ Bn , Y = fk (Z) = Z ∪ {k}. The number of elements of Y is either the same as the number of elements in Z (if k ∈ Z), or one larger (if k ∈ / Z).

Since Z ∈ Bn , by inductive hypothesis, Z 6= ∅ and Z has at most n elements. Therefore Y 6= ∅ and Y has at most n + 1 elements, so Y ∈ {X ∈ P(N) | X 6= ∅ and X has at most n + 1 elements}. Now suppose Y ∈ {X ∈ P(N) | X 6= ∅ and X has at most n + 1 elements}. Then Y 6= ∅, so we can choose some k ∈ Y . If Y has fewer than n + 1 elements, then Y ∈ Bn and Y = fk (Y ), so Y ∈ Bn+1 . If Y has n + 1 elements, then Y \ {k} ∈ Bn and Y = fk (Y \ {k}), and again we conclude that Y ∈ Bn+1 . Thus Bn+1 = {X ∈ P(N) | X 6= ∅ and X has at most n + 1 elements}.

5. The closure is the set C = {n ∈ Z | n ≥ 2}. It is clear that C is closed under f . By Theorem 6.4.2, every element of C is either prime or is a product of primes, so every element of C must be in any set that contains P and is closed under f .

6. The mistake is the sentence “Then by the definition of C, there is some m ∈ N such that x, y ∈ Bm .”

The correct conclusion is that there are m, n ∈ N such that x ∈ Bm and y ∈ Bn .

7. (a) B0 = {x ∈ R | −2 ≤ x ≤ 0}, B1 = {x ∈ R | 0 ≤ x ≤ 4}, B2 = {x ∈ R | 0 ≤ x ≤ 16}, . . . . In n general, for every positive integer n, Bn = {x ∈ R | 0 ≤ x ≤ 2(2 ) }.

S S S (b) Sn∈N Bn = {x ∈ R | x ≥ −2}. Therefore −1, 3 ∈ n∈N Bn but f (−1, 3) = −3 ∈ / n∈N Bn , so n∈N Bn is not closed under f . In other words, property 2 in Definition 5.4.8 does not hold. (c) The closure is R. To see why, suppose B ⊆ D ⊆ R and D is S closed under f . Then for all n ∈ N, Bn ⊆ D (that part of the proof in exercise 6 is correct), so n∈N Bn = {x ∈ R | x ≥ −2} ⊆ D.

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But now for any negative real number x, −x ∈ D and −1 ∈ D, so since D is closed under f , x = f (−x, −1) ∈ D. Thus R ⊆ D. 8. (a) Let m be an arbitrary natural number. We now prove by induction that for all n ≥ m, Bm ⊆ Bn .

Base case: n = m. Then Bm ⊆ Bm = Bn .

Induction step: Suppose n ≥ m and Bm ⊆ Bn . Clearly Bn ⊆ Bn ∪ f (Bn × Bn ) = Bn+1 , so Bm ⊆ Bn+1 .

S (b) Let C = n∈N Bn . It is not hard to see that each set Bn is a subset of A, so C ⊆ A, and B = B0 ⊆ C. To see that C is closed under f , suppose x, y ∈ C. Then by the definition of C, there are m, n ∈ N such that x ∈ Bm and y ∈ Bn . If m ≤ n then, by part (a), Bm ⊆ Bn , so x ∈ Bn .

Therefore f (x, y) ∈ f (Bn × Bn ) ⊆ Bn+1 , so f (x, y) ∈ C. A similar argument applies if m > n.

Finally, suppose B ⊆ D ⊆ A and D is closed under f . To prove that C ⊆ D, it will suffice to prove that ∀n ∈ N(Bn ⊆ D). We prove this by induction. The base case holds because B0 = B ⊆ D by assumption. For the induction step, suppose Bn ⊆ D and let x ∈ Bn+1 be arbitrary. By the definition of Bn+1 either x ∈ Bn or x = f (a, b) for some a, b ∈ Bn . In the first case, x ∈ D because by inductive hypothesis, Bn ⊆ D. In the second case, a, b ∈ D, so since D is closed under f , x = f (a, b) ∈ D. Therefore Bn+1 ⊆ D. 9. If B = ∅ then for all n ∈ N, Bn = ∅, so the closure is ∅. If B 6= ∅ then B0 = B and for all n ≥ 1, Bn = {c}; the closure is B ∪ {c}.

10. We use induction on n.

Base case: n = 1. Then x = 2! + 2 = 4. The only value of i we have to worry about is i = 0, and for this value of i we have i + 2 = 2 and x + i = 4. Since 2 | 4, we have (i + 2) | (x + i), as required.

Induction step: Suppose that n is a positive integer, and for every integer i, if 0 ≤ i ≤ n − 1 then (i + 2) | ((n + 1)! + 2 + i). Now let x = (n + 2)! + 2, and suppose that 0 ≤ i ≤ n. If i = n then we have x + i = (n + 2)! + 2 + i = (i + 2)! + (i + 2) = (i + 2)((i + 1)! + 1), so (i + 2) | (x + i). Now suppose 0 ≤ i ≤ n − 1. By inductive hypothesis, we know that (i + 2) | ((n + 1)! + 2 + i), so we can choose some integer k such that (n + 1)! + 2 + i = k(i + 2), and therefore (n + 1)! = (k − 1)(i + 2). Therefore x + i = (n + 2)! + 2 + i = (n + 2)(n + 1)! + (i + 2) = (n + 2)(k − 1)(i + 2) + (i + 2) = (i + 2)((n + 2)(k − 1) + 1), so (i + 2) | (x + i).

11. Let m be an arbitrary positive integer. We now prove by induction that for all n ∈ Z+ , Rm+n = Rm ◦ Rn .

Base case: n = 1. By definition, Rm+n = Rm+1 = Rm ◦ R = Rm ◦ R1 = Rm ◦ Rn .

Induction step: Suppose n ∈ Z+ and Rm+n = Rm ◦ Rn . Then Rm+n+1 = Rm+n ◦ R

(definition of Rm+n+1 )

= (Rm ◦ Rn ) ◦ R

(inductive hypothesis)

= Rm ◦ (Rn ◦ R)

(associativity of ◦: Theorem 4.2.5, part 4)

m

=R ◦R

n+1

(definition of Rn+1 ).

12. (a) We use induction.

Base case: If n = 1, then f n = f , which is a function from A to A by assumption.

Induction step: Suppose n ∈ Z+ and f n : A → A. Then by Theorem 5.1.5, since f n+1 = f n ◦f , f n+1 : A → A.

104

SOLUTIONS (b) We use induction.

Base case: If n = 1, then f n (B) = f (B0 ) = B1 = Bn .

Induction step: Suppose n ∈ Z+ and f n (B) = Bn . Then by exercise 11, f n+1 = f 1+n = f 1 ◦ f n = f ◦ f n . Therefore, by our solution to exercise 6 in Section 5.5, f n+1 (B) = (f ◦ f n )(B) = f (f n (B)) = f (Bn ) = Bn+1 .

13. (a) Suppose a is a periodic point for f . Then for some positive integer n, f n (a) = a. Let D = {f (a), f 2 (a), . . . , f n (a)}, which is a finite set. Notice that a = f n (a) ∈ D, so {a} ⊆ D. Suppose x ∈ D. Then for some k ∈ {1, 2, . . . , n}, x = f k (a). If k < n then f (x) = f (f k (a)) = (f 1 ◦f k )(a) = f k+1 (a) ∈ D. If k = n then x = f n (a) = a, so f (x) = f (a) ∈ D. Thus, D is closed under f . By definition, the closure of {a} under f must be a subset of D, so it is finite. (b) No. Counterexample: f : R → R, f (x) = |x|, a = −1.

14. Clearly T is a relation on A and R = R1 ⊆ T . To see that T is transitive, suppose (x, y) ∈ T and n (y, z) ∈ T . Then by the definition of T , we can choose positive integers n and S m suchnthat (x, y) ∈ R m m n m+n and (y, z) ∈ R . Thus by exercise 11, (x, z) ∈ R ◦ R = R , so (x, z) ∈ n∈Z+ R = T . Therefore T is transitive.

Finally, suppose R ⊆ S ⊆ A × A and S is transitive. We must show that T ⊆ S, and clearly by the definition of T it suffices to show that ∀n ∈ Z+ (Rn ⊆ S). We prove this by induction on n. We have assumed R ⊆ S, so when n = 1 we have Rn = R1 = R ⊆ S. For the induction step, suppose n is a positive integer and Rn ⊆ S. Now suppose (x, y) ∈ Rn+1 . Then by definition of Rn+1 we can choose some z ∈ A such that (x, z) ∈ R and (z, y) ∈ Rn . By assumption R ⊆ S, and by inductive hypothesis Rn ⊆ S. Therefore (x, z) ∈ S and (z, y) ∈ S, so since S is transitive, (x, y) ∈ S. Since (x, y) was an arbitrary element of Rn+1 , this shows that Rn+1 ⊆ S. 15. We use induction.

Base case: n = 1. Then Rn = R ⊆ S = S n .

Induction step: Suppose n ∈ Z+ and Rn ⊆ S n . Suppose (x, y) ∈ Rn+1 = Rn ◦ R. Then we can choose some z ∈ A such that (x, z) ∈ R and (z, y) ∈ Rn . Since R ⊆ S and Rn ⊆ S n , (x, z) ∈ S and (z, y) ∈ S n . Therefore (x, y) ∈ S n ◦ S = S n+1 . Since (x, y) was arbitrary, Rn+1 ⊆ S n+1 .

16. (a) R ∩ S ⊆ R and R ∩ S ⊆ S. Therefore by exercise 15, for every positive integer n, (R ∩ S)n ⊆ Rn and (R ∩ S)n ⊆ S n , so (R ∩ S)n ⊆ Rn ∩ S n . However, the two need not be equal. For example, if A = {1, 2, 3, 4}, R = {(1, 2), (2, 4)}, and S = {(1, 3), (3, 4)}, then (R ∩ S)2 = ∅ but R2 ∩ S 2 = {(1, 4)}. (b) R ⊆ R ∪ S and S ⊆ R ∪ S. Therefore by exercise 15, for every positive integer n, Rn ⊆ (R ∪ S)n and S n ⊆ (R ∪S)n , so Rn ∪S n ⊆ (R ∪S)n . However, the two need not be equal. Counterexample: A = {1, 2, 3}, R = {(1, 2)}, S = {(2, 3)}.

17. (a) Let m = d(a, b) and n = d(b, c). Then (a, b) ∈ Rm and (b, c) ∈ Rn , so by exercise 11, (a, c) ∈ Rn ◦ Rm = Rm+n . Therefore d(a, c) ≤ m + n = d(a, b) + d(b, c). (b) Let n = d(a, c). Then (a, c) ∈ Rn = R(n−m)+m = Rn−m ◦ Rm , so there is some b ∈ A such that (a, b) ∈ Rm and (b, c) ∈ Rn−m . Therefore d(a, b) ≤ m and d(b, c) ≤ n − m. If either d(a, b) < m or d(b, c) < n − m then d(a, b) + d(b, c) < m + n − m = n = d(a, c), which would contradict part (a). Therefore d(a, b) = m and d(b, c) = n − m = d(a, c) − m.

18. (a) We use induction.

Base case: n = 1. Suppose (a, b) ∈ R1 = R. Let f = {(0, a), (1, b)}. Then f is an R-path from a to b of length 1. For the other direction, suppose f is an R-path from a to b of length 1.

By the definition of R-path, this means that f (0) = a, f (1) = b, and (f (0), f (1)) ∈ R. Therefore (a, b) ∈ R = R1 .

Induction step: Suppose n is a positive integer and Rn = {(a, b) ∈ A × A | there is an R-path from a to b of length n}. Now suppose (a, b) ∈ Rn+1 = R1 ◦ Rn by exercise 11. Then there

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is some c such that (a, c) ∈ Rn and (c, b) ∈ R. By inductive hypothesis, there is an R-path f from a to c of length n. Then f ∪ {(n + 1, b)} is an R-path from a to b of length n + 1. For the other direction, suppose f is an R-path from a to b of length n + 1. Let c = f (n). Then f \ {(n + 1, b)} is an R-path from a to c of length n, so by inductive hypothesis (a, c) ∈ Rn . But also (c, b) = (f (n), f (n + 1)) ∈ R, so (a, b) ∈ R1 ◦ Rn = Rn+1 . (b) This follows from part (a) and exercise 14.

19. (a) Suppose f is an R-path from a to b of length n, but f is not one-to-one. Then we can choose i and j such that 0 ≤ i < j ≤ n and f (i) = f (j). Let g = {(0, f (0)), (1, f (1)), . . . , (i, f (i)) = (i, f (j)), (i + 1, f (j + 1)), . . . , (i + n − j, f (n))}.

Notice that since a 6= b, it cannot be the case that i = 0 and j = n, so either i ≥ 1 or j ≤ n − 1, and therefore i + n − j ≥ 1. It is not hard to check that g is an R-path from a to b of length i+n−j < n, so by exercise 18(a), (a, b) ∈ Ri+n−j . But this contradicts the definition of n = d(a, b) as the smallest positive integer m such that (a, b) ∈ Rm . (b) Suppose f is an R-path from a to a of length n and it is not the case that ∀i < n∀j < n(f (i) = f (j) → i = j). Then we can choose i and j so that 0 ≤ i < j < n and f (i) = f (j). Now the same method as in part (a) can be used to construct an R-path from a to a of length i + n − j, contradicting the definition of d(a, a).

S S S 20. By exercise 14, {Rn | 1 ≤ n ≤ m} ⊆ n∈Z+ Rn = T , so we just need to prove T ⊆ {Rn | 1 ≤ n ≤ m}.

Suppose (a, b) ∈ T . Let n = d(a, b). Then (a, b) ∈ Rn , so by exercise 18(a), we can let f be an R-path from a to b of length n. By exercise 19, f (0), f (1),S. . . , f (n − 1) are n distinct elements of A. But A has only m elements, so n ≤ m. Therefore (a, b) ∈ {Rn | 1 ≤ n ≤ m}.

Chapter 7 Section 7.1 1. (a) D(a) = D(57) = {1, 3, 19, 57}, D(b) = D(36) = {1, 2, 3, 4, 6, 9, 12, 18, 36}, D(a) ∩ D(b) = {1, 3}. (b) The Euclidean algorithm shows that gcd(57, 36) = 3. Here are the steps: n qn rn Division 0 57 1 36 57 = 1 · 36 + 21 2 1 21 36 = 1 · 21 + 15 3 1 15 21 = 1 · 15 + 6 4 1 6 15 = 2 · 6 + 3 5 2 3 6=2·3+0 6 2 0 2. (a) The extended Euclidean algorithm shows that gcd(775, 682) = 31 = −7 · 775 + 8 · 682: n qn rn sn tn Division 0 775 1 0 1 682 0 1 775 = 1 · 682 + 93 2 1 93 1 − 1 · 0 = 1 0 − 1 · 1 = −1 682 = 7 · 93 + 31 3 7 31 0 − 7 · 1 = −7 1 − 7 · (−1) = 8 93 = 3 · 31 + 0 4 3 0 (b) The extended Euclidean algorithm shows that gcd(562, 243) = 1 = 16 · 562 − 37 · 243:

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SOLUTIONS n 0 1 2 3 4 5

qn

2 3 5 15

rn 562 243 76 15 1 0

sn 1 0 1−2·0= 1 0 − 3 · 1 = −3 1 − 5 · (−3) = 16

tn 0 1 0 − 2 · 1 = −2 1 − 3 · (−2) = 7 −2 − 5 · 7 = −37

Division 562 = 2 · 243 + 76 243 = 3 · 76 + 15 76 = 5 · 15 + 1 15 = 15 · 1 + 0

3. (a) The extended Euclidean algorithm shows that gcd(2790, 1206) = 18 = 16 · 2790 − 37 · 1206: n qn rn sn tn Division 0 2790 1 0 1 1206 0 1 2790 = 2 · 1206 + 378 2 2 378 1−2·0= 1 0 − 2 · 1 = −2 1206 = 3 · 378 + 72 3 3 72 0 − 3 · 1 = −3 1 − 3 · (−2) = 7 378 = 5 · 72 + 18 4 5 18 1 − 5 · (−3) = 16 −2 − 5 · 7 = −37 72 = 4 · 18 + 0 5 4 0 (b) The extended Euclidean algorithm shows that gcd(191, 156) = 1 = −49 · 191 + 60 · 156: n qn rn sn tn Division 0 191 1 0 1 156 0 1 191 = 1 · 156 + 35 2 1 35 1−1·0= 1 0 − 1 · 1 = −1 156 = 4 · 35 + 16 3 4 16 0 − 4 · 1 = −4 1 − 4 · (−1) = 5 35 = 2 · 16 + 3 4 2 3 1 − 2 · (−4) = 9 −1 − 2 · 5 = −11 16 = 5 · 3 + 1 5 5 1 −4 − 5 · 9 = −49 5 − 5 · (−11) = 60 3=3·1+0 6 3 0 4. Note that a = 1 · a + 0 · b ∈ L, so L is not the empty set, and L ⊆ Z+ ⊆ N. Therefore, by the well-ordering principle, L has a smallest element. Let d be the smallest element of L. Since d ∈ L, there are integers s and t such that d = sa + tb. Let q and r be the quotient and remainder when a is divided by d. Thus, a = qd + r and 0 ≤ r < d. Suppose r 6= 0. Then r ∈ Z+ and r = a − qd = a − q(sa + tb) = (1 − qs)a − (qt)b, so r ∈ L. But r < d, so this contradicts the fact that d is the smallest element of L. Therefore r = 0, so a = qd, which implies that d | a. A similar argument shows that d | b, so d ∈ D(a) ∩ D(b). To show that it is the largest element of D(a) ∩ D(b), suppose c ∈ D(a) ∩ D(b). Then there are integers j and k such that a = jc and b = kc, so d = sa + tb = sjc + tkc = (sj + tk)c. Therefore c | d, which implies that c ≤ d.

5. Let n be an arbitrary integer. (→) Suppose n is a linear combination of a and b. Then there are integers s and t such that n = sa + tb. Since d = gcd(a, b), d | a and d | b, so there are integers j and k such that a = jd and b = kd. Therefore n = sa + tb = sjd + tkd = (sj + tk)d, so d | n. (←) Suppose d | n. Then there is some integer k such that n = kd. By Theorem 7.1.4, there are integers s and t such that d = sa + tb. Therefore n = kd = k(sa + tb) = (ks)a + (kt)b, so n is a linear combination of a and b.

6. We will prove that D(a) ∩ D(b) = D(a + bc) ∩ D(b), from which the desired conclusion follows.

Suppose d ∈ D(a) ∩ D(b). Then there are integers j and k such that a = jd and b = kd. Therefore a + bc = jd + kdc = (j + kc)d, so d ∈ D(a + bc). Since we also have d ∈ D(b), we conclude that d ∈ D(a + bc) ∩ D(b).

Now suppose that d ∈ D(a + bc) ∩ D(b). Then there are integers j and k such that a + bc = jd and b = kd. Therefore a = jd − bc = jd − kdc = (j − kc)d, so d ∈ D(a). Since we also have d ∈ D(b), we

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conclude that d ∈ D(a) ∩ D(b).

7. (a) No. Counterexample: a = b = 2, a0 = 3, b0 = 4. (b) Yes. Suppose a | a0 and b | b0 . Let d = gcd(a, b). Then d | a and d | b. Since d | a and a | a0 , by Theorem 3.3.7, d | a0 . Similarly, d | b0 . Therefore, by Theorem 7.1.6, d | gcd(a0 , b0 ).

8. Let d = gcd(5a + 2, 13a + 5). Note that 13(5a + 2) − 5(13a + 5) = (65a + 26) − (65a + 25) = 1. Thus, 1 is a linear combination of 5a + 2 and 13a + 5, and by exercise 5 it follows that d | 1, so d = 1.

9. We use strong induction on the maximum of a and b. In other words, we prove the following statement by strong induction: ∀k ∈ Z+ [∀a ∈ Z+ ∀b ∈ Z+ (max(a, b) = k → gcd(2a − 1, 2b − 1) = 2gcd(a,b) − 1)], where max(a, b) denotes the maximum of a and b.

Let k ∈ Z+ be arbitrary and assume that for every positive integer k 0 < k, ∀a ∈ Z+ ∀b ∈ Z+ (max(a, b) = k 0 → gcd(2a − 1, 2b − 1) = 2gcd(a,b) − 1).

Now let a and b be arbitrary positive integers and assume that max(a, b) = k. We may assume that a ≥ b, since otherwise we can swap the values of a and b. We consider two cases.

Case 1. a = b. Then gcd(2a − 1, 2b − 1) = gcd(2a − 1, 2a − 1) = 2a − 1 = 2gcd(a,a) − 1 = 2gcd(a,b) − 1.

Case 2. a > b. Let c = a − b > 0, so that a = c + b. Let k 0 = max(c, b). Since b < a and c < a, k < a = max(a, b) = k. Therefore 0

gcd(2a − 1, 2b − 1) = gcd(2c − 1 + 2a − 2c , 2b − 1) = gcd(2c − 1 + 2c (2b − 1), 2b − 1) = gcd(2c − 1, 2b − 1) gcd(c,b)

=2

−1

gcd(c+b,b)

=2

gcd(a,b)

=2

−1

(exercise 6) (inductive hypothesis) (exercise 6)

− 1.

10. Let d = gcd(a, b) and e = gcd(na, nb). Then d | a and d | b, so there are integers j and k such that a = jd and b = kd. Therefore na = j(nd) and nb = k(nd), so nd | na and nd | nb. Since e is the greatest common divisor of na and nb, nd ≤ e.

Since n | na and n | nb, by Theorem 7.1.6, n | e, so we can choose an integer r such that e = nr.

Since e | na and e | nb, we can choose integers s and t such that na = se = snr and nb = te = tnr.

Dividing these equations by n, we conclude that a = sr and b = tr. Therefore r | a and r | b. Since d is the greatest common divisor of a and b, r ≤ d, so e = nr ≤ nd. Combining this with our earlier conclusion that nd ≤ e, we have e = nd; in other words, gcd(na, nb) = n gcd(a, b).

11. (a) Let d = gcd(a, b). Suppose n ∈ D(d). Then n | d. Since d | a and d | b, it follows by Theorem 3.3.7 that n | a and n | b, so n ∈ D(a) ∩ D(b).

Now suppose n ∈ D(a) ∩ D(b). Then n | a and n | b, so by Theorem 7.1.6, n | d, and therefore n ∈ D(d). Thus D(gcd(a, b)) = D(d) = D(a) ∩ D(b). (b) By definition, gcd(gcd(a, b), c) is the largest element of D(gcd(a, b)) ∩ D(c). But by part (a), D(gcd(a, b)) ∩ D(c) = D(a) ∩ D(b) ∩ D(c). Thus, gcd(gcd(a, b), c) is the largest element of D(a) ∩ D(b) ∩ D(c).

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12. (a) The Euclidean algorithm shows that gcd(55, 34) = 1: n qn rn Division 0 55 1 34 55 = 1 · 34 + 21 2 1 21 34 = 1 · 21 + 13 3 1 13 21 = 1 · 13 + 8 4 1 8 13 = 1 · 8 + 5 5 1 5 8=1·5+3 6 1 3 5=1·3+2 7 1 2 3=1·2+1 8 1 1 2=2·1+0 9 2 0 The numbers ri are the Fibonacci numbers. There are 8 divisions. (b) There are n − 1 division steps, and gcd(Fn+1 , Fn ) = 1. To prove this, let r0 , r1 , . . . be the sequence of numbers produced by the Euclidean algorithm. We claim first that for every k ≤ n−1, rk = Fn+1−k . We prove this by strong induction. Suppose that k ≤ n − 1 and for all k 0 < k, rk0 = Fn+1−k0 . We consider three cases.

Case 1. k = 0. Then rk = r0 = Fn+1 = Fn+1−0 .

Case 2. k = 1. Then rk = r1 = Fn = Fn+1−1 .

Case 3. k ≥ 2. Then rk is the remainder when dividing rk−2 by rk−1 . By inductive hypothesis, rk−2 = Fn+3−k and rk−1 = Fn+2−k . By the definition of the Fibonacci numbers, Fn+3−k = 1 · Fn+2−k + Fn+1−k , and since k ≤ n − 1, n + 1 − k ≥ 2, so Fn+1−k < Fn+2−k . Therefore the quotient when dividing rk−2 by rk−1 is 1, and the remainder is Fn+1−k , so rk = Fn+1−k .

This completes the proof by induction that for all k ≤ n − 1, rk = Fn+1−k . In particular, we have rn−2 = F3 = 2 and rn−1 = F2 = 1. To compute rn , we divide by rn−2 by rn−1 , which gives a quotient of 2 and remainder of 0. Thus rn = 0, so there are n − 1 division steps, and gcd(Fn+1 , Fn ) = rn−1 = 1.

13. (a) We use strong induction. Suppose k < m, and for all k 0 < k, rm−k0 ≥ Fk0 +2 . We consider three cases.

Case 1. k = 0. Then rm−k = rm 6= 0, so rm−k ≥ 1 = F2 = Fk+2 .

Case 2. k = 1. Then m ≥ 2, so rm−1 > rm ≥ 1. Therefore rm−k = rm−1 ≥ 2 = F3 = Fk+2 .

Case 3. k ≥ 2. Then rm−k+2 is the remainder when dividing rm−k by rm−k+1 . Let qm−k+2 be the quotient. Since rm−k > rm−k+1 , the quotient is at least 1, so rm−k = qm−k+2 · rm−k+1 + rm−k+2 ≥ 1 · rm−(k−1) + rm−(k−2) ≥ Fk+1 + Fk

(inductive hypothesis)

= Fk+2 . √ (b) Let α = (1 − 5)/2 ≈ −0.618. Then −1 < α < 0, and you can prove by induction that for every positive integer k, −1 < αk < 1.

Let k be an arbitrary positive integer. Then by Theorem 6.4.3, √ n √ n 1+ 5 − 1−2 5 2 ϕk − αk ϕk − 1 ϕk 1 ϕk √ Fk = = √ > √ = √ − √ > √ − 1.

5 5 5 5 5 5 (c) Applying first part (a) with k = m − 1 and then part (b), we get ϕm+1 b = r1 = rm−(m−1) ≥ Fm+1 > √ − 1.

5

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Rearranging this inequality, we get

√ ϕm+1 < (b + 1) 5.

Taking the logarithm of both sides and applying properties of logarithms, we get √ log 5 (m + 1) log ϕ = log(ϕm+1 ) < log((b + 1) 5) = log(b + 1) + , 2 and therefore log(b + 1) log 5 m< + − 1. log ϕ 2 log ϕ (d) We apply part (c), using base-10 logarithms. Suppose b has at most 100 digits. Then b+1 ≤ 10100 , so log(b + 1) ≤ 100, and the number of divisions when using the Euclidean algorithm is m<

log(b + 1) log 5 100 log 5 + −1≤ + − 1 ≈ 479.169. log ϕ 2 log ϕ log ϕ 2 log ϕ

Since m is an integer, m ≤ 479.

14. (a) Suppose a and b are positive integers. By the division algorithm (Theorem 6.4.1), there are natural numbers q0 and r0 such that a = q0 b + r0 and r0 < b. We now consider two cases.

Case 1. r0 ≤ b/2. Let q = q0 and r = r0 . Then a = qb + r and r ≤ b/2.

Case 2. r0 > b/2. Let q = q0 + 1 and r = b − r0 > 0. Then a = q0 b + r0 = (q0 + 1)b − (b − r0 ) = qb − r and r = b − r0 < b − b/2 = b/2.

Thus, there are natural numbers q and r such that r ≤ b/2 and either a = qb + r or a = qb − r. (b) We will prove that D(a) ∩ D(b) = D(b) ∩ D(r), from which the desired conclusion follows. If a = qb + r then the proof is exactly the same as in the text. Now consider the case a = qb − r.

Suppose d ∈ D(a) ∩ D(b). Then there are integers j and k such that a = jd and b = kd. Therefore r = qb − a = qkd − jd = (qk − j)d, so d ∈ D(r). Since we also have d ∈ D(b), d ∈ D(b) ∩ D(r).

Now suppose d ∈ D(b) ∩ D(r). Then there are integers j and k such that b = jd and r = kd.

Therefore a = qb − r = qjd − kd = (qj − k)d, so d ∈ D(a). Since we also have d ∈ D(b), d ∈ D(a) ∩ D(b). (c) The proof is almost exactly the same as in the text. If there is no n such that rn = 0, then the sequence r0 , r1 , . . . is an infinite sequence, and for every n ≥ 1, rn+1 ≤ rn /2 < rn . Therefore {r0 , r1 , . . .} is a nonempty set of natural numbers with no smallest element, which contradicts the well-ordering principle. Thus there must be some positive integer m such that rm 6= 0 and rm+1 = 0. Applying part (b) repeatedly, we have gcd(a, b) = gcd(r0 , r1 ) = gcd(r1 , r2 ) = · · · = gcd(rm−1 , rm ) = rm . (d) The least absolute remainder Euclidean algorithm takes fewer steps, as these calculations show: Euclidean Algorithm Least Absolute Remainder Euclidean Algorithm n qn rn Division n qn rn Division 0 1515 0 1515 1 555 1515 = 2 · 555 + 405 1 555 1515 = 3 · 555 − 150 2 2 405 555 = 1 · 405 + 150 2 3 150 555 = 4 · 150 − 45 3 1 150 405 = 2 · 150 + 105 3 4 45 150 = 3 · 45 + 15 4 2 105 150 = 1 · 105 + 45 4 3 15 45 = 3 · 15 + 0 5 1 45 105 = 2 · 45 + 15 5 3 0 6 2 15 45 = 3 · 15 + 0 7 3 0

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SOLUTIONS

Section 7.2 1. 650 = 2 · 52 · 13, 756 = 22 · 33 · 7, 1067 = 11 · 97.

2. lcm(1495, 650) = 1495 · 650/ gcd(1495, 650) = 971750/65 = 14950.

3. lcm(1953, 868) = 1953 · 868/ gcd(1953, 868) = 1695204/217 = 7812.

4. Let e and f be arbitrary real numbers. We consider two cases.

Case 1. e ≤ f . Then min(e, f ) = e and max(e, f ) = f , so min(e, f ) + max(e, f ) = e + f .

Case 2. e > f . Then min(e, f ) = f and max(e, f ) = e, so min(e, f ) + max(e, f ) = f + e = e + f .

5. We prove the contrapositives of both directions. (→) Suppose some prime number p appears in the prime factorizations of both a and b. Then p | a and p | b, so gcd(a, b) ≥ p > 1, and therefore a and b are not relatively prime. (←) Suppose a and b are not relatively prime. Let d = gcd(a, b) > 1. Let p be any prime number in the prime factorization of d. Then since d | a and d | b, p must occur in the prime factorizations of both a and b.

6. (→) Suppose a and b are relatively prime. Then by Theorem 7.1.4, there are integers s and t such that sa + tb = gcd(a, b) = 1. (←) Suppose there are integers s and t such that sa + tb = 1. By exercise 5 in Section 7.1, gcd(a, b) | 1, so gcd(a, b) = 1.

7. Suppose d is a positive integer such that d | a0 and d | b0 . Since a0 | a and b0 | b, by the transitivity of the divisibility relation, d | a and d | b. Since a and b are relatively prime, d = 1. Thus, the only common divisor of a0 and b0 is 1, so a0 and b0 are relatively prime.

8. Let d = gcd(a, b) and x = ab/gcd(a, b) = ab/d. (a) Since d = gcd(a, b), d | b, so there is some integer k such that b = kd. Therefore x = akd/d = ak, so x is an integer and a | x. A similar argument shows that b | x, so x is a common multiple of a and b. Since m is the least common multiple, m ≤ x. (b) Suppose r > 0. Since a | m, there is some integer t such that m = ta. Therefore r = ab − qm = ab − qta = (b − qt)a, so a | r. Similarly, b | r. But r < m, so this contradicts the definition of m as the least positive integer that is divisible by both a and b. Therefore r = 0. (c) With t defined as in part (b), ab = qm = qta. Dividing both sides by a, we get b = qt, so q | b.

The proof that q | a is similar. (d) Since q | a and q | b, q ≤ gcd(a, b). Therefore ab = qm ≤ gcd(a, b)m, so m ≥ ab/gcd(a, b).

9. By Theorem 7.1.4, there are integers s and t such that d = sa + tb = sjd + tkd. Dividing this equation by d, we get 1 = sj + tk, so by exercise 6, j and k are relatively prime.

10. Suppose a, b, and d are positive integers and d | ab. Let d2 = gcd(d, b). Then there are positive integers d1 and k such that d = d1 d2 and b = kd2 , and by exercise 9, d1 and k are relatively prime. Since d | ab, we can choose an integer j such that ab = jd. Therefore akd2 = jd1 d2 , and dividing by d2 we get ak = jd1 . Thus d1 | ak. Since gcd(d1 , k) = 1, by Theorem 7.2.2, d1 | a. Thus we have d = d1 d2 , d1 | a, and d2 | b.

11. Suppose a, b, and m are positive integers, a | m, and b | m. Then we can choose integers j and k such that m = ja = kb. Also, a | lcm(a, b) and b | lcm(a, b), so we can choose integers s and t such that lcm(a, b) = sa = tb. Let q and r be the quotient and remainder when m is divided by lcm(a, b). Then m = q lcm(a, b) + r and 0 ≤ r < lcm(a, b). Suppose r > 0. Then r = m − q lcm(a, b) = ja − qsa = (j − qs)a, so a | r, and also r = m − q lcm(a, b) = kb − qtb = (k − qt)b, so b | r. This contradicts the fact that lcm(a, b) is the smallest positive integer that is divisible by both a and b. Thus r = 0, so m = q lcm(a, b) and lcm(a, b) | m.

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12. Let n = lcm(lcm(a, b), c). Since a | m and b | m, by exercise 11, lcm(a, b) | m. Since n is the smallest positive integer that is divisible by both lcm(a, b) and c, n ≤ m.

Since n = lcm(lcm(a, b), c), lcm(a, b) | n and c | n. But also a | lcm(a, b) and b | lcm(a, b), so by Theorem 3.3.7, a | n and b | n. Since m is the smallest positive integer that is divisible by a, b, and c, m ≤ n. Combining this with our earlier conclusion that n ≤ m, we have m = n = lcm(lcm(a, b), c).

13. Suppose a and b are positive integers and a2 | b2 . Let the prime factorization of b be b = pe11 pe22 · · · pekk .

2ek 2 2 1 2e2 Then the prime factorization of b2 is b2 = p2e 1 p2 · · · pk . Since a | b , every prime factor of a must f1 f2 fk be one of p1 , p2 , . . . , pk , so a = p1 p2 · · · pk for some natural numbers f1 , f2 , . . . , fk . Therefore 2fk 2 2 2 a2 = p12f1 p2f 2 · · · pk . Since a | b , for every i we must have 2fi ≤ 2ei , and therefore fi ≤ ei . Thus a | b.

14. (a) Suppose p is prime and 5p + 9 ∈ {n2 | n ∈ N}. Then there is some natural number n such that 5p + 9 = n2 . Note that n2 = 5p + 9 > 9, so n ≥ 4. Rearranging the equation 5p + 9 = n2 , we get 5p = n2 − 9 = (n − 3)(n + 3), so p | ((n − 3)(n + 3)). Since p is prime, by Theorem 7.2.3 it follows that either p | (n − 3) or p | (n + 3).

If p | (n−3), then there is some integer k such that n−3 = kp, so 5p = (n−3)(n+3) = kp(n+3).

Dividing by p, we conclude that 5 = k(n + 3), so (n + 3) | 5. But since n ≥ 4, n + 3 ≥ 7, so this is impossible. Thus p - (n − 3), so we must have p | (n + 3). Therefore there is an integer k such that n + 3 = kp, so 5p = (n + 3)(n − 3) = kp(n − 3). Dividing by p, we get 5 = k(n − 3), so (n − 3) | 5. But the only divisors of 5 are 1 and 5, so either n − 3 = 1 or n − 3 = 5. In other words, either n = 4 or n = 8.

We now check each of these values of n to see if it leads to a solution for p. If n = 4, then 5p + 9 = n2 = 16, so p = 7/5, which is not an integer. If n = 8, then 5p + 9 = n2 = 64, so p = 55/5 = 11, which is a prime number. Thus the only solution is p = 11. (b) Suppose p is a prime and 15p + 4 ∈ {n2 | n ∈ N}. Then there is some natural number n such that 15p + 4 = n2 . Since n2 > 4, n ≥ 3. Rearranging the equation 15p + 4 = n2 , we get 15p = n2 − 4 = (n − 2)(n + 2). Therefore p | ((n − 2)(n + 2)), so since p is prime, either p | (n − 2) or p | (n + 2).

If p | (n−2) then there is some positive integer k such that n−2 = kp, so 15p = (n−2)(n+2) = kp(n+2), and dividing by p we get 15 = k(n+2). Therefore (n+2) | 15. Since D(15) = {1, 3, 5, 15}, this implies that n + 2 must be 1, 3, 5, or 15. But we also have n ≥ 3, so n + 2 ≥ 5. Therefore either n + 2 = 5 or n + 2 = 15, which means that either n = 3 or n = 13.

If p | (n + 2) then similar reasoning leads to the conclusion that (n − 2) | 15, so n − 2 must be 1, 3, 5, or 15, which means n is one of 3, 5, 7, or 17.

We now have n narrowed down to five possible values: 3, 5, 7, 13, or 17. We can now check each to see whether it leads to a solution for p: n = 3: p = 1/3, which is not an integer; n = 5: p = 7/5, which is not an integer; n = 7: p = 3, which is prime; n = 13: p = 11, which is prime; n = 17: p = 19, which is prime.

Thus, there are three solutions for p: 3, 11, and 19. (c) Suppose p is a prime and 5p + 8 ∈ {n3 | n ∈ N}. Then there is some natural number n such that 5p + 8 = n3 . Since n3 > 8, n ≥ 3. From 5p + 8 = n3 we get 5p = n3 − 8 = (n − 2)(n2 + 2n + 4), and imitating the reasoning used in parts (a) and (b) we can conclude that either (n − 2) | 5 or (n2 + 2n + 4) | 5. But since n ≥ 3, n2 + 2n + 4 ≥ 19, so (n2 + 2n + 4) - 5. Therefore (n − 2) | 5, so

112

SOLUTIONS either n − 2 = 1 or n − 2 = 5, which means that either n = 3 or n = 7. If n = 3 then p = 19/5, which is not an integer. If n = 7, then p = 67, which is prime. So p = 67 is the only solution.

15. (a) Suppose x, y ∈ H. Then we can choose natural numbers m and n such that x = 4m + 1 and y = 4n + 1. Therefore xy = (4m + 1)(4n + 1) = 16mn + 4m + 4n + 1 = 4(4mn + m + n) + 1 ∈ H. (b) We prove by strong induction that for every natural number m, if m > 1 and m ∈ H then m is either a Hilbert prime or a product of two or more Hilbert primes. So suppose that m ∈ H and m > 1, and for every k < m, if k > 1 and k ∈ H then k is either a Hilbert prime or a product of two or more Hilbert primes. Of course, if m is a Hilbert prime then there is nothing to prove, so suppose that m is not a Hilbert prime. Then we can choose Hilbert numbers a and b such that m = ab, a < m, and b < m. Note that since a < m = ab, it follows that b > 1, and similarly we must have a > 1. Thus, by inductive hypothesis, each of a and b is either a Hilbert prime or a product of Hilbert primes. But then since m = ab, m is a product of Hilbert primes. (c) First note that 441 = 4 · 110 + 1, so 441 ∈ H. We have 441 = 9 · 49 = 21 · 21. Each of the numbers 9, 49, and 21 can be written as a product of smaller natural numbers in only one way: 9 = 3 · 3, 49 = 7 · 7, and 21 = 3 · 7. However, 3 and 7 are not Hilbert numbers, so 9, 49, and 21 are Hilbert primes.

16. Let p1 , p2 , . . . , pk be a list of all primes that occur in the prime factorization of either a or b, so that a = pe11 pe22 · · · pekk ,

b = pf11 pf22 · · · pfkk

for some natural numbers e1 , e2 , . . . , ek and f1 , f2 , . . . , fk . For i = 1, 2, . . . , k, let ( ( 0, if ei ≥ fi , ei , if ei ≥ fi , hi = gi = fi , if ei < fi .

0, if ei < fi , Let c = p1g1 pg22 · · · pgkk ,

d = ph1 1 ph2 2 · · · phk k .

Then for all i, gi ≤ ei and hi ≤ fi , and therefore c | a and d | b. Also, c and d have no prime factors in common, so by exercise 5, c and d are relatively prime. Finally, max(e1 ,f1 )

cd = p1g1 +h1 · · · pgkk +hk = p1

max(ek ,fk )

· · · pk

= lcm(a, b).

17. (a) Let s, t, u, and v be integers such that gcd(a, b) = sa + tb,

gcd(a, c) = ua + vc.

Then gcd(a, b) · gcd(a, c) = (sa + tb)(ua + vc) = (sua + svc + tbu)a + (tv)bc.

Thus, gcd(a, b) · gcd(a, c) is a linear combination of a and bc, and by exercise 5 in Section 7.1 it follows that gcd(a, bc) | (gcd(a, b) · gcd(a, c)). (b) Let d = gcd(a, b). Then d | a and d | b. Since d | b and b | bc, d | bc. Since d | a and d | bc,

by Theorem 7.1.6, d | gcd(a, bc); in other words, gcd(a, b) | gcd(a, bc). A similar argument shows that gcd(a, c) | gcd(a, bc). Therefore, by exercise 11, lcm(gcd(a, b), gcd(a, c)) | gcd(a, bc). (c) Since b and c are relatively prime, gcd(a, b) | b, and gcd(a, c) | c, by exercise 7, gcd(a, b) and gcd(a, c) are relatively prime. Therefore lcm(gcd(a, b), gcd(a, c)) =

gcd(a, b) · gcd(a, c) gcd(a, b) · gcd(a, c) = = gcd(a, b) · gcd(a, c). gcd(gcd(a, b), gcd(a, c)) 1

Substituting into part (b), we get (gcd(a, b) · gcd(a, c)) | gcd(a, bc), and thus gcd(a, b) · gcd(a, c) ≤ gcd(a, bc). And by part (a) we have gcd(a, bc) ≤ gcd(a, b) · gcd(a, c). Combining these facts, we conclude that gcd(a, bc) = gcd(a, b) · gcd(a, c).

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18. Suppose m is a positive integer and 2m + 1 is prime. Suppose m is not a power of two. Then the prime factorization of m must include some odd prime number p, so there is some positive integer r such that m = pr. Since p is odd, there is some positive integer n such that p = 2n + 1. Therefore 2m + 1 = 2(2n+1)r + 1 = (2r )2n+1 + 12n+1 . By exercise 14 in Section 6.1, (2r + 1) | ((2r )2n+1 + 12n+1 ), so (2r + 1) | (2m + 1). Since r < m, 1 < 2r + 1 < 2m + 1, so this contradicts the fact that 2m + 1 is prime. Therefore m is a power of two.

19. (a) Since x is a positive rational number, there are positive integers m and n such that x = m/n. Let d = gcd(m, n). By exercise 9, we can let a and b be positive integers such that m = da, n = db, and gcd(a, b) = 1. Then da a m x= = = . n db b (b) Since a/b = c/d, ad = bc. Therefore a | bc. Since gcd(a, b) = 1, by Theorem 7.2.2, a | c. A similar argument shows c | a, so a = c. Therefore ad = bc = ba, and dividing both sides by a we conclude that b = d. (c) By part (a), we have x = a/b, where a and b are relatively prime positive integers. Let the prime factorizations of a and b be g

a = r1g1 r2g2 · · · rj j ,

b = sh1 1 sh2 2 · · · shl l .

Note that by exercise 5, these factorizations have no primes in common. Then g

x=

r1g1 r2g2 · · · rj j sh1 1 sh2 2 · · · shl l

g

= r1g1 r2g2 · · · rj j s1−h1 s2−h2 · · · sl−hl .

Rearranging the primes r1 , . . . , rj , s1 , . . . , sl into increasing order gives the required product pe11 · · · pekk . (d) We begin by reversing the steps of part (c). Let r1 , r2 , . . . , rj be those primes in the product pe11 pe22 · · · pekk whose exponents are positive, listed in increasing order, and s1 , s2 , . . . , sl those whose exponents are negative. Rewriting each prime raised to a negative power as the prime to a positive power in the denominator, we get g

x = pe11 pe22 · · · pekk =

r1g1 r2g2 · · · rj j sh1 1 sh2 2 · · · shl l

,

where all the exponents gi and hi are positive integers. The numerator and denominator have no fm can prime factors in common, so they are relatively prime. Similarly, the product q1f1 q2f2 · · · qm be rewritten as a fraction with all exponents positive: fm x = q1f1 q2f2 · · · qm =

v1y1 v2y2 · · · vtyt . w1z1 w2z2 · · · wuzu

g

By part (b), r1g1 · · · rj j = v1y1 · · · vtyt and sh1 1 · · · shl l = w1z1 · · · wuzu . By the uniqueness of prime factorizations, j = t and for all i ∈ {1, . . . , j}, ri = vi and gi = yi , and also l = u and for all i ∈ {1, . . . , l}, si = wi and hi = zi . Rewriting the primes in the denominator as primes raised to fm negative powers, we find that the original two products pe11 · · · pekk and q1f1 · · · qm are the same.

20. Suppose the prime factorizations of a and b are a = 2c pe11 pe22 · · · pekk ,

fm b = 2d q1f1 q2f2 · · · qm ,

where p1 , p2 , . . . , pk and q1 , q2 , . . . , qm are odd primes. Substituting into the equation a2 = 2b2 , we get 2ek 2fm 1 2e2 22c p2e = 22d+1 q12f1 q22f2 · · · qm .

1 p2 · · · pk By uniqueness of prime factorizations, the exponents of 2 in these two products must be the same, so 2c = 2d + 1. But 2c is even and 2d + 1 is odd, so this is a contradiction.

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Section 7.3 1.

+ [0]6 [1]6 [2]6 [3]6 [4]6 [5]6

[0]6 [0]6 [1]6 [2]6 [3]6 [4]6 [5]6

[1]6 [1]6 [2]6 [3]6 [4]6 [5]6 [0]6

[2]6 [2]6 [3]6 [4]6 [5]6 [0]6 [1]6

[3]6 [3]6 [4]6 [5]6 [0]6 [1]6 [2]6

[4]6 [4]6 [5]6 [0]6 [1]6 [2]6 [3]6

[5]6 [5]6 [0]6 [1]6 [2]6 [3]6 [4]6

· [0]6 [1]6 [2]6 [3]6 [4]6 [5]6

[0]6 [0]6 [0]6 [0]6 [0]6 [0]6 [0]6

[1]6 [0]6 [1]6 [2]6 [3]6 [4]6 [5]6

[2]6 [0]6 [2]6 [4]6 [0]6 [2]6 [4]6

[3]6 [0]6 [3]6 [0]6 [3]6 [0]6 [3]6

[4]6 [0]6 [4]6 [2]6 [0]6 [4]6 [2]6

[5]6 [0]6 [5]6 [4]6 [3]6 [2]6 [1]6

2. Let P = [ab]m . Now let x ∈ X and y ∈ Y be arbitrary. Then x ∈ [a]m and y ∈ [b]m , so x ≡ a (mod m) and y ≡ b (mod m). By Lemma 7.3.4 it follows that xy ≡ ab (mod m), so xy ∈ [ab]m = P . Since x and y were arbitrary, we conclude that ∀x ∈ X∀y ∈ Y (xy ∈ P ).

To prove that P is unique, suppose P 0 is another equivalence class such that ∀x ∈ X∀y ∈ Y (xy ∈ 0 P ). Since a ∈ X and b ∈ Y , ab ∈ P and ab ∈ P 0 . Therefore P and P 0 are not disjoint, and since Z/≡m is pairwise disjoint, this implies that P = P 0 .

3. Parts 5–9 of Theorem 7.3.6 follow from these calculations: 5.

X · Y = [a]m · [b]m = [ab]m = [ba]m = [b]m · [a]m = Y · X,

6. (X · Y ) · Z = ([a]m · [b]m ) · [c]m = [ab]m · [c]m = [(ab)c]m = [a(bc)]m = [a]m · [bc]m = [a]m · ([b]m · [c]m ) = X · (Y · Z), 7.

X · [1]m = [a]m · [1]m = [a · 1]m = [a]m = X,

8.

X · [0]m = [a]m · [0]m = [a · 0]m = [0]m ,

9.

X · (Y + Z) = [a]m · ([b]m + [c]m ) = [a]m · [b + c]m = [a(b + c)]m = [ab + ac]m = [ab]m + [ac]m = ([a]m · [b]m ) + ([a]m · [c]m ) = (X · Y ) + (X · Z).

4. (a) Since Z1 is an additive identity element, Z1 + Z2 = Z2 . And since Z2 is an additive identity element, Z1 + Z2 = Z1 . Therefore Z1 = Z1 + Z2 = Z2 . (b) Since X10 is an additive inverse for X, X10 + X + X20 = [0]m + X20 = X20 . Similarly, since X20 is an additive inverse for X, X10 + X + X20 = X10 + [0]m = X10 . Therefore X10 = X20 . (c) Suppose O1 and O2 are multiplicative identity elements. Then O1 = O1 · O2 = O2 . (d) Suppose X10 and X20 are multiplicative inverses of X. Then X10 = X10 · [1]m = X10 · X · X20 = [1]m · X20 = X20 .

5. Suppose X ∈ Z/≡p and X 6= [0]p . Then there is some integer r such that 1 ≤ r ≤ p − 1 and X = [r]p .

Since p - r and p is prime, gcd(r, p) = 1, so by Theorem 7.3.7, X = [r]p has a multiplicative inverse.

6. No. Counterexample: m = 6, a = 2, b = 3.

7. Since d = gcd(m, a), we can choose integers j and k such that m = jd and a = kd. Suppose b is an integer and d - b. Suppose there is some integer x such that ax ≡ b (mod m). Then we can choose an integer t such that b − ax = tm, and therefore b = ax + tm = kdx + tjd = (kx + tj)d.

Thus d | b, which is a contradiction. Thus, there is no integer x such that ax ≡ b (mod m).

8. Let a and b be arbitrary integers. Then na ≡ nb (mod nm)

iff

∃k ∈ Z(nb − na = knm)

iff

∃k ∈ Z(b − a = km)

iff

a ≡ b (mod m).

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9. Let x be the number of packages of file cards the teacher bought. Then he handed out 20x + 2 cards, giving the same number to each of 26 students, so 26 | (20x + 2), and therefore 20x ≡ −2 (mod 26).

By Theorem 7.3.11, this is equivalent to 10x ≡ −1 (mod 13). To solve this congruence, we first note that [10]−1 13 = [4]13 , since [10]13 · [4]13 = [40]13 = [1]13 , and then we compute: 10x ≡ −1 (mod 13)

iff

[10]13 · [x]13 = [−1]13 = [12]13

iff

[x]13 = [10]−1 13 · [12]13 = [4]13 · [12]13 = [48]13 = [9]13

iff x ∈ [9]13 .

Thus, the possible values of x are 9, 22, 35, 48, . . . . But we have one more piece of information: each student got between 10 and 20 file cards. The number of file cards received by each student is (20x + 2)/26, so 10 ≤ (20x + 2)/26 ≤ 20, and therefore 12.9 ≤ x ≤ 25.9. Thus x = 22.

10. (a) The extended Euclidean algorithm shows that [40]−1 237 = [160]237 , so 40x ≡ 8 (mod 237) iff

[40]237 · [x]237 = [8]237

iff

[x]237 = [40]−1 237 · [8]237 = [160]237 · [8]237 = [1280]237 = [95]237

iff

x ∈ [95]237 .

(b) We first rewrite the congruence as 4 · 10x ≡ 4 · 2 (mod 4 · 59), which is equivalent to 10x ≡ 2 (mod 59). We then compute [10]−1 59 = [6]59 , so 10x ≡ 2 (mod 59)

iff

[10]59 · [x]59 = [2]59

iff

[x]59 = [10]−1 59 · [2]59 = [6]59 · [2]59 = [12]59

iff

x ∈ [12]59 .

11. (a) The extended Euclidean algorithm shows that [31]−1 384 = [223]384 , so 31x ≡ 24 (mod 384)

iff

[31]384 · [x]384 = [24]384

iff

[x]384 = [31]−1 384 · [24]384 = [223]384 · [24]384 = [5352]384 = [360]384

iff x ∈ [360]384 . (b) The Euclidean algorithm shows that gcd(384, 32) = 32 and 32 - 24, so by Theorem 7.3.10, the congruence 32x ≡ 24 (mod 384) has no solutions.

12. (a) Suppose Alice bought x chairs without arms and y chairs with arms. Then since she spent $720, 35x + 50y = 720. Therefore 20 − 35x = 50y − 700 = 50(y − 14), so 50 | (20 − 35x), which means that 35x ≡ 20 (mod 50). (b) The congruence in part (a) can be written as 5 · 7x ≡ 5 · 4 (mod 5 · 10), which is equivalent to 7x ≡ 4 (mod 10). Since [7]−1 10 = [3]10 , we can solve the congruence as follows: 7x ≡ 4 (mod 10)

iff

[7]10 · [x]10 = [4]10

iff

[x]10 = [7]−1 10 · [4]10 = [3]10 · [4]10 = [12]10 = [2]10

iff

x ∈ [2]10 .

(c) The solutions to the congruence in part (b) are the elements of the set [2]10 = {2, 12, 22, 32, . . . , −8, −18, −28, . . . }. However, not all of these values make sense in the problem. We must have x ≥ 0 and 35x ≤ 720, so x ≤ 720/35 ≈ 20.57. Therefore the only possible solutions are x = 2, 12.

If x = 2 then from 35x + 50y = 720 we get y = 13, and if x = 12 then y = 6. So she bought either 2 chairs without arms and 13 chairs with arms, or 12 chairs without arms and 6 chairs with arms.

13. Let a and b be arbitrary integers. Suppose first that a ≡ b (mod m). Then [a]m = [b]m , so [na]m = [n]m · [a]m = [n]m · [b]m = [nb]m , and therefore na ≡ nb (mod m).

Now suppose that na ≡ nb (mod m), so [n]m · [a]m = [na]m = [nb]m = [n]m · [b]m . Since m and n are relatively prime, [n]m has a multiplicative inverse. Multiplying both sides of the equation [n]m · [a]m = [n]m · [b]m by [n]−1 m , we get [a]m = [b]m , so a ≡ b (mod m).

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SOLUTIONS

14. Suppose a and b are integers, a ≡ b (mod m1 ), and a ≡ b (mod m2 ). Then m1 | (b−a) and m2 | (b−a).

Therefore, by exercise 11 in Section 7.2, lcm(m1 , m2 ) | (b − a), so a ≡ b (mod lcm(m1 , m2 )).

15. Suppose a ≡ b (mod m). Then m | (b − a), so we can choose an integer n such that b − a = nm, so b = a + nm. Therefore by exercise 6 in Section 7.1, gcd(m, b) = gcd(m, a + nm) = gcd(m, a).

16. We use mathematical induction.

Base case: If n = 0 then an = a0 = 1 = b0 = bn , so an ≡ bn (mod m).

Induction step: Suppose n is a natural number and an ≡ bn (mod m). We also have a ≡ b (mod m), so by Lemma 7.3.4, an · a ≡ bn · b (mod m), or in other words, an+1 ≡ bn+1 (mod m).

17. (a) First note that 10 ≡ 1 (mod 3). Therefore, by exercise 16, for every i ∈ N, 10i ≡ 1 (mod 3), so [10i ]3 = [1]3 . Thus [n]3 = [d0 + 10d1 + · · · + 10k dk ]3 = [d0 ]3 + [10]3 · [d1 ]3 + · · · + [10k ]3 · [dk ]3 = [d0 ]3 + [1]3 · [d1 ]3 + · · · + [1]3 · [dk ]3 = [d0 + d1 + · · · + dk ]3 .

In other words, n ≡ (d0 + d1 + · · · + dk ) (mod 3). (b) 3 | n iff [n]3 = [0]3 iff [d0 + · · · + dk ]3 = [0]3 iff 3 | (d0 + · · · + dk ).

18. (a) First note that 10 ≡ −1 (mod 11). Therefore, by exercise 16, for every i ∈ N, 10i ≡ (−1)i (mod 11), so [10i ]11 = [(−1)i ]11 . Thus [n]11 = [d0 + 10d1 + 102 d2 + · · · + 10k dk ]11 = [d0 ]11 + [10]11 · [d1 ]11 + [102 ]11 · [d2 ]11 + · · · + [10k ]11 · [dk ]11 = [d0 ]11 + [−1]11 · [d1 ]11 + [1]11 · [d2 ]11 + · · · + [(−1)k ]11 · [dk ]11 = [d0 − d1 + d2 − d3 + · · · + (−1)k dk ]11 .

In other words, n ≡ (d0 − d1 + d2 − d3 + · · · + (−1)k dk ) (mod 11). (b) 11 | n iff [n]11 = [0]11 iff [d0 −d1 +d2 −d3 +· · ·+(−1)k dk ]11 = [0]11 iff 11 | (d0 −d1 +· · ·+(−1)k dk ). (c) 2 − 7 + 1 − 5 + 3 − 5 = −11, which is divisible by 11. Therefore 535172 is divisible by 11.

19. (a) Suppose n ≥ 10. First note that 10f (n) = (dk · · · d1 0)10 + 50d0 = (dk · · · d1 d0 )10 + 49d0 = n + 49d0 .

Therefore 3f (n) − n = 49d0 − 7f (n) = 7(7d0 − f (n)), so n ≡ 3f (n) (mod 7), or equivalently [n]7 = [3]7 · [f (n)]7 . Since [3]−1 7 = [5]7 , it follows that [f (n)]7 = [5]7 · [n]7 , so f (n) ≡ 5n (mod 7). (b) Suppose n ≥ 10. If 7 | n then [n]7 = [0]7 , so [f (n)]7 = [5n]7 = [5]7 · [0]7 = [0]7 , and therefore 7 | f (n). Similarly, if 7 | f (n) then [f (n)]7 = [0]7 , so [n]7 = [3f (n)]7 = [3]7 · [0]7 = [0]7 and 7 | n. (c) f (627334) = 62733 + 5 · 4 = 62753; f (62753) = 6275 + 5 · 3 = 6290; f (6290) = 629 + 5 · 0 = 629; f (629) = 62 + 5 · 9 = 107; f (107) = 10 + 5 · 7 = 45; f (45) = 4 + 5 · 5 = 29. Since 7 - 29, 7 - 627334.

20. (a) One example is m = 5, a = a0 = 2, b = 1, b0 = 6. (b) Suppose we could define exponentiation on equivalence classes in such a way that for all positive integers m, a, and b, ([a]m )[b]m = [ab ]m . Then [2]5 = [21 ]5 = ([2]5 )[1]5 = ([2]5 )[6]5 = [26 ]5 = [64]5 .

But 2 6≡ 64 (mod 5), so this is a contradiction.

21. (a) Suppose x1 , x2 , y1 , and y2 are integers, x1 ≡m y1 , and x2 ≡m y2 . Then f (x1 , x2 ) = [x1 + x2 ]m = [x1 ]m + [x2 ]m = [y1 ]m + [y2 ]m = [y1 + y2 ]m = f (y1 , y2 ). (b) Suppose x1 and x2 are integers. Then h([x1 ]m , [x2 ]m ) = [x1 ]m + [x2 ]m = [x1 + x2 ]m = f (x1 , x2 ).

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117

Section 7.4 1. (Z/≡20 )∗ = {[1]20 , [3]20 , [7]20 , [9]20 , [11]20 , [13]20 , [17]20 , [19]20 }.

2. (a) ϕ(539) = ϕ(72 · 11) = ϕ(72 ) · ϕ(11) = 71 (7 − 1) · (11 − 1) = 420. (b) ϕ(540) = ϕ(22 · 33 · 5) = ϕ(22 ) · ϕ(33 ) · ϕ(5) = 21 (2 − 1) · 32 (3 − 1) · (5 − 1) = 144. (c) 541 is prime, so ϕ(541) = 541 − 1 = 540.

3. (a) ϕ(18) = ϕ(2 · 32 ) = (2 − 1) · 31 (3 − 1) = 6; 56 − 1 = 15624 = 868 · 18, so 56 ≡ 1 (mod 18). (b) ϕ(19) = 19 − 1 = 18; 218 − 1 = 262143 = 13797 · 19, so 218 ≡ 1 (mod 19). (c) ϕ(20) = ϕ(22 · 5) = 21 (2 − 1) · (5 − 1) = 8; 38 − 1 = 6560 = 328 · 20, so 38 ≡ 1 (mod 20).

4. (a) We follow the calculations in the proof of Lemma 7.4.7. First we use the extended Euclidean algorithm to show that gcd(8, 5) = 1 = 2 · 8 + (−3) · 5. In other words, sm + tn = 1, where s = −3 and t = 2. Next, we compute x = tna + smb = 2 · 8 · 4 + (−3) · 5 · 1 = 49. Finally, we compute mn = 40, and we find r such that 1 ≤ r ≤ 40 and r ≡ 49 (mod 40). The answer is r = 9; it is easy to verify that 9 ≡ 4 (mod 5) and 9 ≡ 1 (mod 8). (b) We do the same calculations as in part (a): 3 · 7 + (−2) · 10 = 1, so s = 3 and t = −2; x = tna + smb = (−2) · 10 · 6 + 3 · 7 · 4 = −36; mn = 70, −36 ≡ 34 (mod 70), so r = 34.

5. Base case: If n = 1, then [a]nm = [a]1m = [a]m = [a1 ]m = [an ]m . = [a]nm · [a]m = Induction step: Suppose n is a positive integer and [a]nm = [an ]m . Then [a]n+1 m n n n+1 [a ]m · [a]m = [a · a]m = [a ]m .

6. Suppose a ≡ b (mod mn). Then mn | (b − a), so for some integer k, b − a = kmn. Therefore m | (b − a) and n | (b − a), so a ≡ b (mod m) and a ≡ b (mod n).

Now suppose a ≡ b (mod m) and a ≡ b (mod n). Since a ≡ b (mod n), n | (b − a), so there is some integer j such that b − a = jn. Since a ≡ b (mod m), m | (b − a), so m | jn. But gcd(m, n) = 1, so by Theorem 7.2.2 it follows that m | j. Let k be an integer such that j = km. Then b − a = jn = kmn.

Therefore mn | (b − a), so a ≡ b (mod mn). (An alternative approach is to use exercise 14 in Section 7.3.)

7. Let a, b, and c be arbitrary positive integers. (→) Suppose gcd(ab, c) = 1. Then there are integers s and t such that sab + tc = 1. We can write this equation as either (sb)a + tc = 1 or (sa)b + tc = 1, so by exercise 6 in Section 7.2, gcd(a, c) = gcd(b, c) = 1. (←) Suppose gcd(a, c) = gcd(b, c) = 1. Then there are integers s, t, u, and v such that sa + tc = 1 and ub + vc = 1. Therefore 1 = (sa + tc)(ub + vc) = (su)ab + (sav + tub + tcv)c, so be exercise 6 in Section 7.2, gcd(ab, c) = 1.

8. The first half of the solution to exercise 6 does not use the hypothesis that m and n are relatively prime, so the left-to-right direction of the “iff” statement is correct even if this hypothesis is dropped.

Here is a counterexample for the other direction: a = 0, b = 12, m = 4, n = 6.

9. No, the lemma would be incorrect. Counterexample: m = 4, n = 6, a = 1, b = 2. Suppose there is an integer r such that r ≡ 1 (mod 4) and r ≡ 2 (mod 6). Then 4 | (r − 1) and 6 | (r − 2), so we can choose integers j and k such that r − 1 = 4j and r − 2 = 6k. Therefore r = 4j + 1 = 2(2j) + 1, so r is odd, and also r = 6k + 2 = 2(3k + 1), so r is even. This is a contradiction.

10. Suppose p is prime and a is a positive integer. We consider two cases.

Case 1. p - a. Then p and a are relatively prime, so by Theorem 7.4.2, [a]pp−1 = [1]p . Therefore [ap ]p = [a]p−1 · [a]p = [1]p · [a]p = [a]p , so ap ≡ a (mod p). p Case 2. p | a. Then [a]p = [0]p , so [ap ]p = [0]pp = [0]p = [a]p and therefore ap ≡ a (mod p).

118

SOLUTIONS

11. Suppose m and a are relatively prime positive integers. Then by Euler’s theorem, [a]m · [aϕ(m)−1 ]m = ϕ(m)−1 [aϕ(m) ]m = [1]m , so [a]−1 ]m . m = [a 12. Suppose m, a, p, and q are positive integers, m and a are relatively prime, and p ≡ q (mod ϕ(m)). By Euler’s theorem, aϕ(m) ≡ 1 (mod m). It follows, by exercise 16 in Section 7.3, that for every natural number n, anϕ(m) = (aϕ(m) )n ≡ 1 (mod m), so [anϕ(m) ]m = [1]m . We now consider two cases: Case 1. p ≥ q. Then since p ≡ q (mod ϕ(m)), there is some natural number n such that p − q = nϕ(m), so p = q + nϕ(m). Therefore [ap ]m = [aq+nϕ(m) ]m = [aq · anϕ(m) ]m = [aq ]m · [anϕ(m) ]m = [aq ]m · [1]m = [aq ]m , so ap ≡ aq (mod m).

Case 2. p < q. Similar to case 1, but with the roles of p and q reversed.

13. Let a be an arbitrary positive integer. We now use mathematical induction to prove the following statement: for every k ≥ 1, if b1 , b2 , . . . , bk is any list of k positive integers and gcd(a, b1 ) = gcd(a, b2 ) = · · · = gcd(a, bk ) = 1, then gcd(a, b1 b2 · · · bk ) = 1.

Base case: If k = 1, then the statement to be proven is that if b1 is any positive integer and gcd(a, b1 ) = 1, then gcd(a, b1 ) = 1. This is clearly true.

Induction step: Suppose the statement holds for lists of k positive integers, and suppose b1 , b2 , . . . , bk+1 is a list of k + 1 positive integers such that gcd(a, b1 ) = gcd(a, b2 ) = · · · = gcd(a, bk+1 ) = 1.

By the inductive hypothesis, gcd(a, b1 b2 · · · bk ) = 1. But now since gcd(a, bk+1 ) = 1, by Lemma 7.4.6, gcd(a, b1 b2 · · · bk+1 ) = gcd(a, (b1 b2 · · · bk )bk+1 ) = 1.

14. We use induction on k.

Base case: If k = 1, then the statement to be proven is that for any positive integer m1 and any integers a and b, a ≡ b (mod m1 ) iff a ≡ b (mod m1 ). This is clearly true.

Induction step: Suppose the statement holds for lists of k pairwise relatively prime positive integers, and suppose that m1 , m2 , . . . , mk+1 is a list of k + 1 positive integers that are pairwise relatively prime. Let M 0 = m1 m2 · · · mk and M = m1 m2 · · · mk+1 = M 0 mk+1 . Note that by exercise 13, gcd(M 0 , mk+1 ) = 1. Let a and b be arbitrary integers. Then a ≡ b (mod M )

iff

a ≡ b (mod M 0 mk+1 )

iff

a ≡ b (mod M 0 ) and a ≡ b (mod mk+1 )

iff

for every i ∈ {1, 2, . . . , k}, a ≡ b (mod mi ), and a ≡ b (mod mk+1 )

(Lemma 7.4.5) (inductive hypothesis)

iff

for every i ∈ {1, 2, . . . , k + 1}, a ≡ b (mod mi ).

15. (a) We proceed by induction on k.

Base case: When k = 1, the statement to be proven is that for every positive integer m1 and every integer a1 , there is an integer r such that 1 ≤ r ≤ m1 and r ≡ a1 (mod m1 ). This is true because {1, 2, . . . , m1 } is a complete residue system modulo m1 .

Induction step: Suppose that the statement holds for lists of k pairwise relatively prime positive integers, and let m1 , m2 , . . . , mk+1 be a list of k + 1 pairwise relatively prime positive integers. Let M 0 = m1 m2 · · · mk and M = m1 m2 · · · mk+1 = M 0 mk+1 . Let a1 , a2 , . . . , ak+1 be arbitrary integers. By inductive hypothesis, there is an integer r0 such that for all i ∈ {1, 2, . . . , k}, r0 ≡ ai (mod mi ). By exercise 13, gcd(M 0 , mk+1 ) = 1, so by Lemma 7.4.7 there is some integer r such that 1 ≤ r ≤ M , r ≡ r0 (mod M 0 ), and r ≡ ak+1 (mod mk+1 ). By exercise 14, for every i ∈ {1, 2, . . . , k}, r ≡ r0 (mod mi ), and therefore r ≡ ai (mod mi ). (b) Suppose that 1 ≤ r1 , r2 ≤ M and for all i ∈ {1, 2, . . . , k}, r1 ≡ ai (mod mi ) and r2 ≡ ai (mod mi ). Then for all i ∈ {1, 2, . . . , k}, r1 ≡ r2 (mod mi ), so by exercise 14, r1 ≡ r2 (mod M ).

Therefore r1 = r2 .

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119

16. (a) Suppose a ∈ D(m) and b ∈ D(n). Then we can choose integers j and k such that m = ja and n = kb. Therefore mn = jakb = (jk)ab, so ab ∈ D(mn). (b) Suppose a1 , a2 ∈ D(m), b1 , b2 ∈ D(n), and f (a1 , b1 ) = f (a2 , b2 ). Then a1 b1 = a2 b2 , so a1 | a2 b2 .

Since m and n are relatively prime, a1 | m, and b2 | n, by exercise 7 in Section 7.2, a1 and b2 are relatively prime. Since a1 | a2 b2 and gcd(a1 , b2 ) = 1, by Theorem 7.2.2, a1 | a2 . A similar argument shows that a2 | a1 , so a1 = a2 . Therefore a1 b1 = a2 b2 = a1 b2 , and dividing by a1 we get b1 = b2 . This proves that f is one-to-one.

To prove that f is onto, suppose d ∈ D(mn). Then by exercise 10 in Section 7.2, there are positive integers a and b such that d = ab, a | m, and b | n. Therefore (a, b) ∈ D(m) × D(n) and f (a, b) = ab = d. This prove that f is onto. (c) Since f is one-to-one and onto, D(m) × D(n) and D(mn) have the same number of elements. But the number of elements in D(m) × D(n) is τ (m) · τ (n) and the number of elements in D(mn) is τ (mn), so τ (mn) = τ (m) · τ (n).

17. Suppose m and n are relatively prime. Let the elements of D(m) be a1 , a2 , . . . , as , and let the elements of D(n) be b1 , b2 , . . . , bt . Then σ(m) = a1 + a2 + · · · + as and σ(n) = b1 + b2 + · · · + bt . Using the function f from part (b) of exercise 16, we see that the elements of D(mn) are all products of the form ai bj , where 1 ≤ i ≤ s and 1 ≤ j ≤ t. Thus we can arrange the elements of D(mn) in a table with s rows and t columns, where the entry in row i, column j of the table is ai bj ; every element of D(mn) appears exactly once in this table. To compute σ(mn), we must add up all entries in this table. We will do this by first adding up each row of the table, and then adding these row sums.

For 1 ≤ i ≤ s, let ri be the sum of row i of the table. Then ri = ai b1 + ai b2 + · · · + ai bt = ai (b1 + b2 + · · · + bt ) = ai σ(n).

Therefore σ(mn) = r1 + r2 + · · · + rs = a1 σ(n) + a2 σ(n) + · · · + as σ(n) = (a1 + a2 + · · · + as )σ(n) = σ(m)σ(n).

18. Suppose p is a positive integer and 2p − 1 is prime. Since 2p − 1 is prime, D(2p − 1) = {1, 2p − 1}, and D(2p−1 ) = {1, 2, 22 , . . . , 2p−1 }. Therefore D(2p−1 ) ∩ D(2p − 1) = {1}, so 2p−1 and 2p − 1 are relatively prime. Applying exercise 17 and Example 6.1.1, we get σ(2p−1 (2p − 1)) = σ(2p−1 ) · σ(2p − 1) = (1 + 2 + 22 + · · · + 2p−1 ) · (1 + (2p − 1)) = (2p − 1) · 2p = 2 · (2p−1 (2p − 1)).

Therefore 2p−1 (2p − 1) is perfect.

19. (a) Since n is even, one of the primes in the prime factorization of n is 2. Suppose the prime e factorization of n is n = 2k pe11 pe22 · · · pj j , where k and e1 , e2 , . . . , ej are positive integers, p1 , p2 , e . . . , pj are prime numbers, and 2 < p1 < p2 < · · · < pj . Let m = pe11 pe22 · · · pj j , so n = 2k m. Since 2 is the only even prime number, m is a product of odd numbers, so it is odd (see exercise 9 in Section 3.4). (b) The set of divisors of 2k is D(2k ) = {1, 2, 22 , . . . , 2k }, and since m is odd, the only element of this set that divides m is 1. Therefore 2k and m are relatively prime, so by exercise 17, σ(2k m) = σ(2k ) · σ(m). By Example 6.1.1, σ(2k ) = 1 + 2 + 22 + · · · + 2k = 2k+1 − 1, so σ(2k m) = (2k+1 − 1)σ(m). But since n is perfect, we also have σ(2k m) = σ(n) = 2n = 2k+1 m.

Therefore 2k+1 m = (2k+1 − 1)σ(m). (c) By part (b), 2k+1 | ((2k+1 − 1)σ(m)). But 1 · 2k+1 + (−1) · (2k+1 − 1) = 1, so by exercise 6 in Section 7.2, 2k+1 and 2k+1 − 1 are relatively prime. Therefore, by Theorem 7.2.2, 2k+1 | σ(m).

120

SOLUTIONS (d) Combining parts (b) and (c), we have 2k+1 m = (2k+1 − 1)σ(m) = (2k+1 − 1)2k+1 d. Dividing both sides by 2k+1 , we get m = (2k+1 − 1)d. (e) By part (d), d | m. Suppose d > 1. Since m = (2k+1 − 1)d > d, 1, d, and m are distinct divisors of m. Therefore σ(m) ≥ 1 + d + m = 1 + d + (2k+1 − 1)d = 1 + 2k+1 d > 2k+1 d = σ(m), which is a contradiction. Therefore d = 1. (f) Combining parts (d) and (e), we have 2p − 1 = 2k+1 − 1 = m and σ(2p − 1) = σ(m) = 2k+1 = 2p .

If 2p − 1 is not prime, then there is some integer c such that 1 < c < 2p − 1 and c | (2p − 1).

Therefore σ(2p − 1) ≥ 1 + c + (2p − 1) = 2p + c > 2p , which is a contradiction. Therefore 2p − 1 is prime.

Section 7.5 1. (a) n = pq = 5 · 11 = 55, ϕ(n) = (p − 1)(q − 1) = 4 · 10 = 40. To find d, we use the extended Euclidean algorithm to find [7]−1 40 : n qn rn sn tn Division 0 40 1 0 1 7 0 1 40 = 5 · 7 + 5 2 5 5 1−5·0= 1 0 − 5 · 1 = −5 7=1·5+2 3 1 2 0 − 1 · 1 = −1 1 − 1 · (−5) = 6 5=2·2+1 4 2 1 1 − 2 · (−1) = 3 −5 − 2 · 6 = −17 2=2·1+0 5 2 0 Therefore [7]−1 40 = [−17]40 = [23]40 , so d = 23. (b) To encrypt the message, Alice computes [9]755 = [4782969]55 = [4]55 , so c = 4. To decrypt the message, Bob computes [4]23 55 = [70368744177664]55 = [9]55 , thus recovering the original message m = 9.

2. (a) n = 71·83 = 5893, ϕ(n) = 70·82 = 5740. To compute d, we use the extended Euclidean algorithm −1 to find [e]−1 ϕ(n) = [1369]5740 . Here are the steps: n qn rn sn tn 0 5740 1 0 1 1369 0 1 2 4 264 1 −4 3 5 49 −5 21 4 5 19 26 −109 5 2 11 −57 239 6 1 8 83 −348 7 1 3 −140 587 8 2 2 363 −1522 9 1 1 −503 2109 10 2 0 Therefore d = 2109. (b) To encrypt the message, Alice computes [m]en = [1001]1369 5893 = [3421]5893 . Thus, c = 3421. To decrypt the message, Bob computes [c]dn = [3421]2109 5893 = [1001]5893 , recovering the original message 1001. Here are the steps for both calculations:

CHAPTER 7

121 Computing [m]en = [c]n : k 2 5 10 21 42

[m]kn [191]n [4453]n [5157]n [5087]n [1406]n

k 85 171 342 684 1369

[m]kn [2366]n [437]n [2393]n [4346]n [3421]n

Computing [c]dn = [m]n : k 2 4 8 16 32 65

[c]kn [5636]n [1226]n [361]n [675]n [1864]n [4993]n

k 131 263 527 1054 2109

[c]kn [3540]n [945]n [1251]n [3356]n [1001]n

3. If p = 71 and q = 83 then n = 71 · 83 = 5893 and ϕ(n) = 70 · 82 = 5740. If e = 1368 then gcd(ϕ(n), e) = gcd(5740, 1368) = 4, so e and ϕ(n) are not relatively prime, and therefore [e]−1 ϕ(n) does not exist.

4. (a) n = 17389· 14947 = 259913383, ϕ(n) = 17388· 14946 = 259881048. To find d, we use the extended −1 Euclidean algorithm to find [e]−1 ϕ(n) = [35824631]259881048 . Here are the steps: n qn rn sn tn 0 259881048 1 0 1 35824631 0 1 2 7 9108631 1 −7 3 3 8498738 −3 22 4 1 609893 4 −29 5 13 570129 −55 399 6 1 39764 59 −428 7 14 13433 −881 6391 8 2 12898 1821 −13210 9 1 535 −2702 19601 10 24 58 66669 −483634 11 9 13 −602723 4372307 12 4 6 2477561 −17972862 13 2 1 −5557845 40318031 14 6 0 Thus, d = 40318031. (b) To encrypt the message, Alice computes [m]en = [123456789]35824631 259913383 . Here are the steps: k [m]kn k [m]kn 2 [136994585]n 17492 [257205644]n 4 [56583995]n 34984 [213784457]n 8 [23493482]n 69969 [84216864]n 139939 [60180746]n 17 [227859358]n 34 [43323921]n 279879 [63604562]n 68 [234613571]n 559759 [206839246]n 136 [116149500]n 1119519 [170936373]n 273 [86505049]n 2239039 [226170104]n 4478078 [82842549]n 546 [178247596]n 1093 [160153466]n 8956157 [242033287]n 2186 [230571701]n 17912315 [4821172]n 4373 [239533299]n 35824631 [61227739]n 8746 [259805247]n Thus, c = 61227739. To decrypt the message, Bob computes [c]dn = [61227739]40318031 259913383 . Here are the steps:

122

SOLUTIONS k [c]kn k [c]kn 2 [35123006]n 19686 [22383387]n 39373 [138315399]n 4 [220223434]n 9 [126967424]n 78746 [219737906]n 19 [212058828]n 157492 [85999571]n 314984 [189148374]n 38 [83815709]n 76 [44143053]n 629969 [82986766]n 1259938 [61251000]n 153 [191858627]n 307 [7156952]n 2519876 [102479822]n 5039753 [159955387]n 615 [32302294]n 1230 [66892573]n 10079507 [231960641]n 20159015 [66795572]n 2460 [23335695]n 4921 [35600742]n 40318031 [123456789]n 9843 [155385804]n Thus, Bob recovers the original message m = 123456789.

5. (a) n = 493 = 17 · 29. (b) ϕ(n) = 16 · 28 = 448, [129]−1 448 = [257]448 , so d = 257.

257 (c) [149]493 = [183]493 , so m = 183.

6. (a) The fact that [s]en = [m]n follows from Theorem 7.5.1, with the roles of e and d reversed. Now suppose 0 ≤ s0 < n and [s0 ]en = [m]n . Then by Theorem 7.5.1, [s0 ]n = [m]dn = [s]n , so s0 = s. (b) Because the impostor doesn’t know the exponent d.

7. (a) [123]95 315 = [72]315 , so c = 72. (b) [95]−1 272 = [63]272 , so Bob computes d = 63. (c) [72]63 315 = [288]315 , so Bob gets 288. (d) ϕ(315) = ϕ(32 · 5 · 7) = 31 (3 − 1) · (5 − 1) · (7 − 1) = 144; [95]−1 144 = [47]144 , so Bob would have gotten d = 47. [72]47 315 = [18]315 , so Bob would have gotten 18.

8. (a) X 1 = X, and for every positive integer n, X n+1 = X n · X. (b) We let a be an arbitrary positive integer and then proceed by induction on b.

Base case: b = 1. Then X a · X b = X a · X 1 = X a · X = X a+1 = X a+b .

Induction step: Suppose b is a positive integer and X a · X b = X a+b . Then X a · X b+1 = X a · (X b · X) = (X a · X b ) · X =X

a+b

=X

a+b+1

·X

(definition of X b+1 ) (associativity of ·) (inductive hypothesis) (definition of X a+b+1 ).

(c) We let a be an arbitrary positive integer and then proceed by induction on b.

Base case: b = 1. Then (X a )b = (X a )1 = X a = X a·1 .

Induction step: Suppose b is a positive integer and (X a )b = X ab . Then (X a )b+1 = (X a )b · X a = X ab · X a =X

ab+a

(definition of (X a )b+1 ) (inductive hypothesis) (part (b))

= X a(b+1) .

9. We use strong induction. Suppose that a is a positive integer, and for every positive integer k < a, the computation of X k uses at most 2 log2 k multiplications.

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123

Case 1. a = 1. Then X a = X 1 = X, so no multiplications are needed, and 2 log2 a = 2 log2 1 = 0.

Case 2. a is even. Then a = 2k for some positive integer k < a, and to compute X a we use the formula X a = X k · X k . Let m be the number of multiplications used to compute X k . By inductive hypothesis, m ≤ 2 log2 k. To compute X a we use one additional multiplication (to multiply X k by itself), so the number of multiplications is m + 1 ≤ 2 log2 k + 1 < 2(log2 k + 1) = 2 log2 (2k) = 2 log2 a.

Case 3. a > 1 and a is odd. Then a = 2k + 1 for some positive integer k < a, and to compute X a we use the formula X a = X k · X k · X. As in case 2, if we let m be the number of multiplications used to compute X k then we have m ≤ 2 log2 k. To compute X a we use two additional multiplications, so the number of multiplications is m + 2 ≤ 2 log2 k + 2 = 2(log2 k + 1) = 2 log2 (2k) < 2 log2 (2k + 1) = 2 log2 a.

10. (a) 414 − 1 = 268435455 = 17895697 · 15, so 414 ≡ 1 (mod 15). But 314 = 4782969 ≡ 9 (mod 15). (b) Suppose that n is a Fermat pseudoprime to the base a. Then an−1 ≡ 1 (mod n). Therefore [a]n · [an−2 ]n = [an−1 ]n = [1]n , so [an−2 ]n = [a]−1 n . By Theorem 7.3.7, it follows that n and a are relatively prime. n

n

n

n

11. (a) 2(2 ) − (−1) = 2(2 ) + 1 = Fn , so Fn | (2(2 ) − (−1)). Therefore 2(2 ) ≡ −1 (mod Fn ). n n n n n n (b) By part (a) and Theorem 7.3.4, 2(2 ) · 2(2 ) ≡ (−1) · (−1) (mod Fn ). But 2(2 ) · 2(2 ) = 2(2 +2 ) = n+1 n+1 2(2 ) and (−1) · (−1) = 1. Therefore 2(2 ) ≡ 1 (mod Fn ). (c) By exercise 12(a) in Section 6.3, 2n ≥ n + 1. Let k = 2n − (n + 1) ∈ N, so 2n = k + n + 1. Then n Fn − 1 = 2(2 ) = 2k+n+1 = 2k · 2n+1 . Therefore 2n+1 | (Fn − 1). (d) By part (c), we can choose some positive integer j such that Fn − 1 = j · 2n+1 . Therefore n+1 n+1 j n+1 2Fn −1 = 2(j·2 ) = 2(2 ) . By part (b), 2(2 ) ≡ 1 (mod Fn ), so by exercise 16 in Section 7.3, n+1 j 2(2 ) ≡ 1j (mod Fn ). Thus 2Fn −1 ≡ 1 (mod Fn ).

6= [1]n . And since gcd(n, a) = 1, [a]n has a multiplicative inverse.

12. Since a ∈ R2 , [a]n−1 n = [1]n . Since {0, 1, . . . , n − 1} is a complete (a) Suppose x ∈ R1 . Then 2 ≤ x ≤ n − 1 and [x]n−1 n residue system modulo n, there is a unique y such that 0 ≤ y ≤ n − 1 and ax ≡ y (mod n), so −1 [a]n · [x]n = [y]n . We must prove that y ∈ R2 . If y = 0 then [x]n = [a]−1 n · [y]n = [a]n · [0]n = [0]n , = which contradicts the fact that 2 ≤ x ≤ n − 1. Therefore 1 ≤ y ≤ n − 1. And [y]n−1 n 6= [1]n . Therefore y n−1 6≡ 1 (mod n). It follows that y 6= 1, · [1]n = [a]n−1 = [a]n−1 · [x]n−1 [a]n−1 n n n n so 2 ≤ y ≤ n − 1. (b) Suppose f (x1 ) = f (x2 ) = y. Then [a]n · [x1 ]n = [y]n = [a]n · [x2 ]n , so [x1 ]n = [a]−1 n · [y]n = [x2 ]n , and therefore x1 = x2 . (c) By part (b), R1 has the same number of elements as Ran(f ). Since Ran(f ) ⊆ R2 , R2 has at least as many elements as R1 . So at least half the elements of R are in R2 .

13. (a) Since gcd(561, a) = 1 and 3 | 561, by exercise 7 in Section 7.2, gcd(3, a) = 1. Since 3 is prime, ϕ(3) = 3 − 1 = 2. Thus, by Euler’s theorem, a2 ≡ 1 (mod 3). It follows, by exercise 16 in Section 7.3, that a560 = (a2 )280 ≡ 1 (mod 3). (b) The steps are similar to the steps in part (a): Since gcd(561, a) = 1 and 11 | 561, gcd(11, a) = 1.

Since 11 is prime, ϕ(11) = 10. Thus, by Euler’s theorem, a10 ≡ 1 (mod 11). Therefore a560 = (a10 )56 ≡ 1 (mod 11). (c) As in parts (a) and (b), we have gcd(561, a) = 1 and 17 | 561, so gcd(17, a) = 1. By Euler’s theorem, a16 ≡ 1 (mod 17). Therefore a560 = (a16 )35 ≡ 1 (mod 17). (d) Since 3, 11, and 17 are pairwise relatively prime and 3 · 11 · 17 = 561, it follows from parts (a)–(c), by exercise 14 in Section 7.4, that a560 ≡ 1 (mod 561).

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14. (a) This is essentially the same as part (a) of exercise 19 in Section 7.4, but this time we’ll give a different proof. By exercise 12(a) in Section 6.3, 2n−1 > n − 1, so 2n−1 - (n − 1). Therefore {k ∈ N | 2k - (n − 1)} 6= ∅, so by the well-ordering principle, we can let k be the least natural number such that 2k - (n − 1). Since n is odd, n − 1 is even, so k ≥ 2. Let s = k − 1, which is a positive integer. By the definition of k, 2s | (n − 1), so we can choose a positive integer d such that n − 1 = 2s d. If d is even then there is some integer j such that d = 2j, so n − 1 = 2s · 2j = 2s+1 j = 2k j which contradicts the fact that 2k - (n − 1). Therefore d is odd. (b) Suppose n is prime, b is a positive integer, and b2 ≡ 1 (mod n). Then n | (b2 − 1), which means n | (b − 1)(b + 1). Since n is prime, by Theorem 7.2.3, either n | (b − 1) or n | (b + 1). In other words, either b ≡ 1 (mod n) or b ≡ −1 (mod n). s (c) Suppose a is a Miller-Rabin witness for n and n is prime. Then by Euler’s theorem, a2 d = an−1 ≡ 1 (mod n). Therefore, by the well-ordering principle, we can let k be the smallest natural number k such that a2 d ≡ 1 (mod n). If k = 0 then ad ≡ 1 (mod n), which contradicts the definition of i i+1 k Miller-Rabin witness. Therefore k > 0. Let i = k − 1 ∈ N. Then (a2 d )2 = a2 d = a2 d ≡ 1 i i (mod n), so by part (b), either a2 d ≡ 1 (mod n) or a2 d ≡ −1 (mod n). But both of these i i possibilities lead to contradictions: a2 d ≡ 1 (mod n) contradicts the definition of k, and a2 d ≡ −1 (mod n) contradicts the definition of Miller-Rabin witness. Thus, if there is a Miller-Rabin witness for n then n is not prime. (d) In the notation of earlier parts of this exercise, n = 85 and n − 1 = 84 = 22 · 21, so s = 2 and 1 d = 21. We have 132 ·21 = 1342 ≡ −1 (mod 85), so 13 is not a Miller-Rabin witness for 85. But 1 1421 ≡ 29 (mod 85) and 142 ·21 = 1442 ≡ 76 (mod 85), so 14 is a Miller-Rabin witness.

Chapter 8 Section 8.1 1. (a) Define f : Z+ → N by the formula f (n) = n − 1. It is easy to check that f is one-to-one and onto. (b) Let E = {n ∈ Z | n is even}, and define f : Z → E by the formula f (n) = 2n. It is easy to check that f is one-to-one and onto, so Z ∼ E. But we already know that Z+ ∼ Z, so by Theorem 8.1.3, Z+ ∼ E, and therefore E is denumerable.

2. (a) By Theorem 8.1.6, Z+ ∼ Q. Therefore, by Theorem 8.1.2, Z+ ×Z+ ∼ Q×Q. Since Z+ ∼ Z+ ×Z+ , it follows that Z+ ∼ Q × Q. In other words, Q × Q is denumerable. √ √ √ (b) Define f : Q×Q → Q( 2) by the formula f (a, b) = a+b 2. By the definition of Q( 2), f is√onto.

By part (a), there is some function g : Z+ → Q × Q that is onto. Therefore f ◦ g : Z+ → Q( 2) is √ onto, so by Theorem 8.1.5, Q( 2) is countable. It is clearly not finite, so it must be denumerable.

3. (a) Define f : [0, 1] → [0, 2] by the formula f (x) = 2x. Then f is one-to-one and onto. (b) Define f : (−π/2, π/2) → R by the formula f (x) = tan x. Then f is one-to-one and onto. (c) Define f : (0, 1) → (−π/2, π/2) by the formula f (x) = πx − π/2. Then f is one-to-one and onto, so (0, 1) ∼ (−π/2, π/2). But then part (b) implies that (0, 1) ∼ R. (d) Let A = {1, 1/2, 1/4, 1/8, . . .} = {1/2n | n ∈ N}, B = (0, 1] \ A, and A0 = A \ {1}. Then B is disjoint from both A and A0 , A ∪ B = (0, 1], and A0 ∪ B = (0, 1). Define f : A → A0 by the formula f (x) = x/2. Then f is one-to-one and onto, so A ∼ A0 . By part 1 of Theorem 8.1.3, B ∼ B, so by part 2 of Theorem 8.1.2, (0, 1] = A ∪ B ∼ A0 ∪ B = (0, 1).

4. (a) No. Counterexample: Let A = B = C = Z+ and D = {1}. (b) No. Counterexample: Let A = B = N, C = Z− , and D = ∅.

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5. Suppose A ∼ B. Let f : A → B be one-to-one and onto. For every set X ⊆ A, the image of X under f is the set f (X) = {f (x) | x ∈ X} ⊆ B. Define a function g : P(A) → P(B) by the formula g(X) = f (X). We claim that g is one-to-one and onto. To see that g is one-to-one, suppose g(X) = g(X 0 ). Then f (X) = f (X 0 ). Suppose x ∈ X. Then f (x) ∈ f (X) = f (X 0 ), so there is some x0 ∈ X 0 such that f (x) = f (x0 ). But f is one-to-one, so it follows that x = x0 ∈ X 0 . Since x was arbitrary, this shows that X ⊆ X 0 . A similar argument shows that X 0 ⊆ X, so X = X 0 . To see that g is onto, suppose Y ⊆ B. Let X = f −1 (Y ) ⊆ A. By exercise 4 in Section 5.5, g(X) = f (f −1 (Y )) = Y .

Thus g is one-to-one and onto, so P(A) ∼ P(B).

6. (a) We prove that ∀n ∈ N∀m ∈ N(In ∼ Im → n = m) by induction on n.

Base case: n = 0. Suppose that m ∈ N and there is a one-to-one, onto function f : In → Im .

Since n = 0, In = ∅. But then since f is onto, we must also have Im = ∅, so m = 0 = n.

Induction step: Suppose that n ∈ N, and for all m ∈ N, if In ∼ Im then n = m. Now suppose that m ∈ N and In+1 ∼ Im . Let f : In+1 → Im be a one-to-one, onto function. Let k = f (n + 1), and notice that 1 ≤ k ≤ m, so m is positive. Using the fact that f is onto, choose some j ≤ n + 1 such that f (j) = m.

We now define a function g with domain In as follows: ( f (i), if i 6= j, g(i) = k, if i = j.

We first verify that for all i ∈ In , g(i) ∈ Im−1 . Suppose i ∈ In . If i 6= j then g(i) = f (i) ∈ Im .

But also since f is one-to-one, f (i) 6= f (j) = m, so g(i) = f (i) ∈ Im \ {m} = Im−1 . If i = j then g(i) = k = f (n + 1) ∈ Im . But also since f is one-to-one and j = i ≤ n, k = f (n + 1) 6= f (j) = m, so g(i) = k ∈ Im \ {m} = Im−1 . Thus g : In → Im−1 .

Next we verify that g is one-to-one and onto. To see that g is one-to-one, suppose i1 , i2 ∈ In and g(i1 ) = g(i2 ). If i1 = j and i2 6= j then f (n + 1) = k = g(i1 ) = g(i2 ) = f (i2 ), which contradicts the fact that f is one-to-one. Thus, this possibility can be ruled out. Similarly, we can rule out i1 6= j and i2 = j. If i1 = j and i2 = j then of course i1 = i2 . Finally, suppose that i1 6= j and i2 6= j. Then f (i1 ) = g(i1 ) = g(i2 ) = f (i2 ), so since f is one-to-one, i1 = i2 .

This proves that g is one-to-one. To see that g is onto, suppose that t ∈ Im−1 . Then t ∈ Im and t 6= m. If t = k then f (n + 1) = k = t 6= m = f (j), so j 6= n + 1, and therefore j ∈ In and g(j) = k = t. Now suppose t 6= k. Since f is onto, there is some i ∈ In+1 such that f (i) = t.

Then f (i) = t 6= m = f (j), so i 6= j, and also f (i) = t 6= k = f (n + 1), so i 6= n + 1. Therefore i ∈ In and g(i) = f (i) = t. This proves that g is onto.

Thus In ∼ Im−1 . By inductive hypothesis, it follows that n = m − 1, so n + 1 = m. (b) Suppose A is finite. Then by definition of “finite,” we know that there is at least one n ∈ N such that In ∼ A. To see that it is unique, suppose that n and m are natural numbers, In ∼ A, and Im ∼ A. Then by Theorem 8.1.3, In ∼ Im , so by part (a), n = m.

7. Since A is finite, we can let n = |A|, and In ∼ A. (→) Suppose A ∼ B. Then since In ∼ A, In ∼ B. Therefore B is finite and |B| = n = |A|. (←) Suppose B is finite and |B| = |A| = n. Then In ∼ B, so since In ∼ A, A ∼ B. 8. (a) We use induction on n.

Base case: n = 0. Suppose A ⊆ In = ∅. Then A = ∅, so |A| = 0.

Induction step: Suppose that n ∈ N, and for all A ⊆ In , A is finite, |A| ≤ n, and if A 6= In then |A| < n. Now suppose that A ⊆ In+1 . If A = In+1 then clearly A ∼ In+1 , so A is finite and |A| = n + 1. Now suppose that A 6= In+1 . If n + 1 ∈ / A, then A ⊆ In , so by inductive hypothesis, A is finite and |A| ≤ n. If n + 1 ∈ A, then there must be some k ∈ In such that k ∈ / A. Let

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SOLUTIONS A0 = (A ∪ {k}) \ {n + 1}, and define f : A0 → A as follows: ( x, if x 6= k, f (x) = n + 1, if x = k.

Then f is one-to-one and onto, so A0 ∼ A. Also, A0 ⊆ In , so by inductive hypothesis, A0 is finite and |A0 | ≤ n. Therefore by exercise 7, A is finite and |A| ≤ n. (b) Suppose A is finite and B ⊆ A. Let n = |A|, and let f : A → In be one-to-one and onto. Then f (B) ⊆ In , so by part (a), f (B) is finite, |f (B)| ≤ n, and if B 6= A then f (B) 6= In , so |f (B)| < n.

Also, f B is a one-to-one, onto function from B to f (B) (see exercise 7 in Section 5.1 for the meaning of the notation used here). Therefore B ∼ f (B), so by exercise 7, |B| = |f (B)| and the desired conclusion follows.

9. If A is finite then by exercise 8, |B| < |A|, and by exercise 7 this contradicts B ∼ A. Therefore A is infinite.

10. Define g : B → In by the formula g(x) = the smallest i ∈ In such that f (i) = x.

Suppose x, y ∈ B and g(x) = g(y). Let i = g(x) = g(y) ∈ In . Then x = f (i) = y. This shows that g is one-to-one. Therefore B ∼ Ran(g) ⊆ In , so by exercises 7 and 8, B is finite and |B| = | Ran(g)| ≤ n.

11. (a) We will prove the contrapositive. Suppose f is onto. Let n = |A|, and let g : In → A be one-to-one and onto. Then f ◦ g : In → B is onto, so by exercise 10, |B| ≤ n = |A|. (b) We will prove the contrapositive. Suppose f is one-to-one. Then A ∼ Ran(f ) ⊆ B, so by exercises 7 and 8, |A| = | Ran(f )| ≤ |B|. (c) Suppose |A| = |B|. (→) Suppose f is one-to-one but not onto. Then A ∼ Ran(f ) ⊆ B and Ran(f ) 6= B, so by exercises 7 and 8, |A| = | Ran(f )| < |B|, which is a contradiction. (←) Suppose f is onto but not one-to-one. Then there are x, y ∈ A such that x 6= y but f (x) = f (y). Let A0 = A \ {y} and f 0 = f \ {(y, f (y))}. Then f 0 : A0 → B and f 0 is onto. But by exercise 8, |A0 | < |A| = |B|, so this contradicts part (a).

12. Notice first that either i + j − 2 or i + j − 1 is even, so f (i, j) is a positive integer, and therefore f is a function from Z+ × Z+ to Z+ , as claimed. It will be helpful to verify two facts about the function f .

Both of the facts below can be checked by straightforward algebra: (a) For all j ∈ Z+ , f (1, j + 1) − f (1, j) = j. (b) For all i ∈ Z+ and j ∈ Z+ , f (1, i + j − 1) ≤ f (i, j) < f (1, i + j). It follows that i + j is the smallest k ∈ Z+ such that f (i, j) < f (1, k).

To see that f is one-to-one, suppose that f (i1 , j1 ) = f (i2 , j2 ). Then by fact (b) above, i1 + j1 = the smallest k ∈ Z+ such that f (i1 , j1 ) < f (1, k) = the smallest k ∈ Z+ such that f (i2 , j2 ) < f (1, k) = i2 + j2 .

Using the definition of f , it follows that (i1 + j1 − 2)(i1 + j1 − 1) 2 (i2 + j2 − 2)(i2 + j2 − 1) = f (i2 , j2 ) − = i2 .

2

i1 = f (i1 , j1 ) −

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But then since i1 = i2 and i1 + j1 = i2 + j2 , we must also have j1 = j2 , so (i1 , j1 ) = (i2 , j2 ). This shows that f is one-to-one.

To see that f is onto, suppose n ∈ Z+ . It is easy to verify that f (1, n + 1) > n, so we can let k be the smallest positive integer such that f (1, k) > n. Notice that f (1, 1) = 1 ≤ n, so k ≥ 2. Since k is smallest, f (1, k − 1) ≤ n, and therefore by fact (a), 0 ≤ n − f (1, k − 1) < f (1, k) − f (1, k − 1) = k − 1.

Adding 1 to all terms, we get 1 ≤ n − f (1, k − 1) + 1 < k.

Thus, if we let i = n − f (1, k − 1) + 1 then 1 ≤ i < k. Let j = k − i, and notice that i ∈ Z+ and j ∈ Z+ .

With this choice for i and j we have (i + j − 2)(i + j − 1) +i 2 (k − 2)(k − 1) + n − f (1, k − 1) + 1 = 2

(k − 2)(k − 1) (k − 2)(k − 1) = +n− + 1 + 1 = n.

2 2

f (i, j) =

13. Suppose f (m1 , n1 ) = f (m2 , n2 ). Then 2m1 −1 (2n1 − 1) = 2m2 −1 (2n2 − 1).

(∗)

Suppose m1 < m2 . Then we can divide both sides of (∗) by 2m1 −1 to get 2n1 − 1 = 2m2 −m1 (2n2 − 1).

But the left side of this equation is odd and the right side is even, so we have a contradiction. Thus it cannot be the case that m1 < m2 . A similar argument rules out m2 < m1 , so m1 = m2 . But then from (∗) we get 2n1 − 1 = 2n2 − 1, so n1 = n2 . This proves that f is one-to-one.

To see that f is onto, suppose k ∈ Z+ . By exercise 12(a) in Section 6.3, 2k > k, so 2k - k.

Thus, by the well-ordering principle, we can let m be the least natural number such that 2m - k.

Since 20 = 1 | k, m > 0, so m ∈ Z+ and 2m−1 | k. Thus there is some positive integer j such that k = 2m−1 j. If j is even then there is some positive integer s such that j = 2s, so k = 2m s, which contradicts the fact that 2m - k. Therefore j is odd, so we can let n = (j + 1)/2 ∈ Z+ . Then f (m, n) = 2m−1 (2n − 1) = 2m−1 j = k.

14. Suppose f : A → B, g : C → D, f and g are both one-to-one and onto, A and C are disjoint, and B and D are disjoint. Clearly f ∪ g ⊆ (A ∪ C) × (B ∪ D). Suppose x ∈ A ∪ C. Then either x ∈ A or x ∈ C. Case 1. x ∈ A. Then since f : A → B, there is some y ∈ B such that (x, y) ∈ f ⊆ f ∪ g. Now suppose that for some z, (x, z) ∈ f ∪ g. Since x ∈ A and A and C are disjoint, x ∈ / C, so (x, z) ∈ / C ×D and therefore (x, z) ∈ / g. So (x, z) ∈ f , and since (x, y) ∈ f and f is a function, z = y.

Case 2. x ∈ C. Then a similar argument shows that there is a unique y ∈ B ∪ D such that (x, y) ∈ f ∪ g.

Thus (f ∪ g) : A ∪ C → B ∪ D. To see that f ∪ g is one-to-one, suppose (f ∪ g)(x1 ) = (f ∪ g)(x2 ).

Let y = (f ∪ g)(x1 ) = (f ∪ g)(x2 ) ∈ B ∪ D. Then (x1 , y) ∈ f ∪ g and (x2 , y) ∈ f ∪ g.

Case 1. y ∈ B. Then since B and D are disjoint, y ∈ / D. Therefore (x1 , y) ∈ / g and (x2 , y) ∈ / g, so (x1 , y) ∈ f and (x2 , y) ∈ f . This means that f (x1 ) = y = f (x2 ), so since f is one-to-one, x1 = x2 .

Case 2. y ∈ D. Then a similar argument shows that g(x1 ) = y = g(x2 ), so since g is one-to-one, x1 = x2 .

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SOLUTIONS Finally, to see that f ∪ g is onto, suppose y ∈ B ∪ D. Case 1. y ∈ B. Then since f is onto, there is some x ∈ A such that (x, y) ∈ f ⊆ f ∪ g.

Case 2. y ∈ D. Then since g is onto, there is some x ∈ B such that (x, y) ∈ g ⊆ f ∪ g.

15. (a) If B \ {f (m) | m ∈ Z+ , m < n} = ∅ then B ⊆ {f (m) | m ∈ Z+ , m < n}, so by exercises 8 and 10, B is finite. But we assumed that B was infinite, so this is impossible. (b) We use strong induction. Suppose that ∀m < n, f (m) ≥ m. Now suppose that f (n) < n.

Let m = f (n). Then by inductive hypothesis, f (m) ≥ m. Also, by the definition of f (n), m = f (n) ∈ B \ {f (k) | k ∈ Z+ , k < n} ⊆ B \ {f (k) | k ∈ Z+ , k < m}. But since f (m) is the smallest element of this last set, it follows that f (m) ≤ m. Since we have f (m) ≥ m and f (m) ≤ m, we can conclude that f (m) = m. But then m ∈ / B \ {f (k) | k ∈ Z+ , k < n}, so we have a contradiction. (c) Suppose that i ∈ Z+ , j ∈ Z+ , and i 6= j. Then either i < j or j < i. Suppose first that i < j.

Then according to the definition of f (j), f (j) ∈ B \ {f (m) | m ∈ Z+ , m < j}, and clearly f (i) ∈ {f (m) | m ∈ Z+ , m < j}. It follows that f (i) 6= f (j). A similar argument shows that if j < i then f (i) 6= f (j). This shows that f is one-to-one.

To see that f is onto, suppose that n ∈ B. By part (b), f (n + 1) ≥ n + 1 > n. But according to the definition of f , f (n + 1) is the smallest element of B \ {f (m) | m ∈ Z+ , m < n + 1}. It follows that n ∈ / B \ {f (m) | m ∈ Z+ , m < n + 1}. But n ∈ B, so it must be the case that also n ∈ {f (m) | m ∈ Z+ , m < n + 1}. In other words, for some positive integer m < n + 1, f (m) = n.

16. (a) One possibility is to define f : Z+ → Z \ {0} by the formula ( (n + 1)/2, if n is odd, f (n) = −n/2, if n is even. (b) Suppose g(n1 ) = g(n2 ). Let the prime factorizations of n1 and n2 be n1 = pe11 pe22 · · · pekk and fm . Then since g(n1 ) = g(n2 ), n2 = q1f1 q2f2 · · · qm f (e1 ) f (e2 ) f (e ) f (f ) f (f ) f (fm ) p2 · · · pk k = q1 1 q2 2 · · · qm .

p1

By exercise 19(d) in Section 7.2, k = m and for all i ∈ {1, 2, . . . , k}, pi = qi and f (ei ) = f (fi ).

Since f is one-to-one, it follows that ei = fi , so fm n1 = pe11 pe22 · · · pekk = q1f1 q2f2 · · · qm = n2 .

This proves that g is one-to-one. To prove that g is onto, suppose x ∈ Q+ . By exercise 19(c) in Section 7.2, there are prime numbers p1 , p2 , . . . , pk and nonzero integers f1 , f2 , . . . , fk such that p1 < p2 < · · · < pk and x = pf11 pf22 · · · pfkk .

Since f is onto, there are positive integers e1 , e2 , . . . , ek such that for all i ∈ {1, 2, . . . , k}, f (ei ) = fi . Therefore f (e1 ) f (e2 ) f (e ) · · · pk k = pf11 pf22 · · · pfkk = x. p2

g(pe11 pe22 · · · pkek ) = p1 (c) Define h : Z → Q by the formula

  if n > 0, g(n), h(n) = 0, if n = 0,   −g(−n), if n < 0.

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17. Suppose B ⊆ A and A is countable. Then by Theorem 8.1.5, there is a one-to-one function f : A → Z+ .

By exercise 13 of Section 5.2, f B is a one-to-one function from B to Z+ , so B is countable. (See exercise 7 of Section 5.1 for the definition of the notation used here.)

18. Suppose B ⊆ A, A is infinite, and B is finite. Since B and A \ B are disjoint, if A \ B is finite then by Theorem 8.1.7, A = B ∪ (A \ B) is finite, which is a contradiction. Therefore A \ B is infinite.

19. Following the hint, we recursively define partial orders Rn , for n ∈ N, so that R = R0 ⊆ R1 ⊆ R2 ⊆ · · · and ∀i ∈ In ∀j ∈ Z+ ((ai , aj ) ∈ Rn ∨ (aj , ai ) ∈ Rn ). (∗) Let R0 = R. Given Rn , to define Rn+1 we apply exercise 2 of Section 6.2, with B = {ai | i ∈ In+1 }.

S Finally, let T = n∈N Rn . Clearly T is reflexive, because every Rn is. To see that T is transitive, suppose that (a, b) ∈ T and (b, c) ∈ T . Then for some natural numbers m and n, (a, b) ∈ Rm and (b, c) ∈ Rn . If m ≤ n then Rm ⊆ Rn , and therefore (a, b) ∈ Rn and (b, c) ∈ Rn . Since Rn is transitive, it follows that (a, c) ∈ Rn ⊆ T . A similar argument shows that if n < m then (a, c) ∈ T , so T is transitive. The proof that T is antisymmetric is similar. Finally, to see that T is a total order, suppose x ∈ A and y ∈ A. Since we have numbered the elements of A, we know that for some positive integers m and n, x = am and y = an . But then by (∗) we know that either (am , an ) or (an , am ) is an element of Rn , and therefore also an element of T .

20. Since B and A \ B are disjoint and A = B ∪ (A \ B), by Theorem 8.1.7, |A| = |B| + |A \ B|, so |A \ B| = |A| − |B|.

21. We proceed by induction.

Base case: n = 1. Then the statement to be proven is that A1 is finite and |A1 | = |A1 |, which is clearly true.

Induction and for any pairwise disjoint finite sets A1 , A2 , S step: Suppose n isSa positive integer, Pn . . . , An , i∈In S Ai is finite and | i∈In Ai | = i=1 |Ai |. Now S suppose A1 , AS 2 , . . . , An+1 are pairwise disjoint. Then i∈In Ai and An+1 are disjoint finite sets, so i∈In+1 Ai = ( i∈In Ai ) ∪ An+1 is finite and [ [

Ai = Ai ∪ An+1 i∈In+1

i∈In

=

=

[

Ai + |An+1 |

i∈In n X

|Ai | + |An+1 |

(Theorem 8.1.7)

(inductive hypothesis)

i=1

=

n+1 X

|Ai |.

i=1

22. (a) We follow the hint.

Base case: n = 0. Suppose A and B are finite sets and |B| = 0. Then B = ∅, so A × B = ∅, which is finite, and |A × B| = 0 = |A| · 0.

Induction step: Let n be an arbitrary natural number, and suppose that for all finite sets A and B, if |B| = n then A × B is finite and |A × B| = |A| · n. Now suppose A and B are finite sets and |B| = n + 1. Choose an element b ∈ B, and let B 0 = B \ {b}, a set with n / B0, A × B0 elements. Then A × B = A × (B 0 ∪ {b}) = (A × B 0 ) ∪ (A × {b}), and since b ∈ 0 0 and A × {b} are disjoint. By inductive hypothesis, A × B is finite and |A × B | = |A| · n. Also, it is not hard to see that A ∼ A × {b} – just match up each x ∈ A with (x, b) ∈ A × {b} – so A × {b} is finite and |A × {b}| = |A|. By Theorem 8.1.7, it follows that A × B is finite and |A × B| = |A × B 0 | + |A × {b}| = |A| · n + |A| = |A| · (n + 1).

130

SOLUTIONS (b) To order a meal, you name an element of A × B, where A = {steak, chicken, pork chops, shrimp, spaghetti} and B = {ice cream, cake, pie}. So the number of meals is |A×B| = |A|·|B| = 5·3 = 15.

23. (a) Suppose A ∼ B and C ∼ D. Let f : A → B and g : C → D be one-to-one and onto. By Theorem 5.3.1, f −1 : B → A and g −1 : D → C. For h ∈ A C and j ∈ B D we now define F (h) = g ◦ h ◦ f −1 ∈ B D,

G(j) = g −1 ◦ j ◦ f ∈ A C.

Then F : A C → B D and G : B D → A C. Also, for all h ∈ A C and j ∈ B D, G(F (h)) = g −1 ◦ g ◦ h ◦ f −1 ◦ f = iC ◦ h ◦ iA = h and F (G(j)) = g ◦ g −1 ◦ j ◦ f ◦ f −1 = iD ◦ j ◦ iB = j.

Therefore, by Theorem 5.3.3, F is one-to-one and onto, so A C ∼ B D. (b) Suppose A, B, and C are sets and A ∩ B = ∅. For f ∈ A∪B C, g ∈ A C, and h ∈ B C, define F (f ) = (f ∩ (A × C), f ∩ (B × C)),

G(g, h) = g ∪ h.

Then by exercise 7 in Section 5.1, F (f ) = (f A, f B) ∈ A C × B C, and by exercise 12 in Section 5.1, G(g, h) ∈ A∪B C. Thus, F : A∪B C → A C × B C and G : A C × B C → A∪B C. If f ∈ A∪B C then f ⊆ (A ∪ B) × C, so G(F (f )) = (f ∩ (A × C)) ∪ (f ∩ (B × C)) = f ∩ ((A × C) ∪ (B × C)) = f ∩ ((A ∪ B) × C) = f.

Also, if (g, h) ∈ A C × B C then g ⊆ A × C, h ⊆ B × C, and (A × C) ∩ (B × C) = ∅, so F (G(g, h)) = ((g ∪ h) ∩ (A × C), (g ∪ h) ∩ (B × C)) = (g, h).

Therefore, by Theorem 5.3.3, F is one-to-one and onto, so A∪B C ∼ A C × B C. (c) We use induction to prove that ∀n ∈ N∀A∀B(A and B are finite sets and |A| = n → |A B| = |B||A| ).

Base case: n = 0. Suppose A and B are finite sets and |A| = 0. Then A = ∅, so A B = {∅} and |A B| = 1 = |B|0 = |B||A| .

Induction step: Suppose n ∈ N and for all finite sets A and B with |A| = n, |A B| = |B||A| = n |B| . Now suppose A and B are finite sets and |A| = n + 1. Let a be any element of A and 0 let A0 = A \ {a}, a set with n elements. Then A = A0 ∪ {a}, so by part (b), A B = A ∪{a} B ∼ 0 0 A B × {a} B. By inductive hypothesis, |A B| = |B|n . We claim now that |{a} B| = |B|. To see why, define F : B → {a} B by the formula F (b) = {(a, b)}; it is not hard to check that F is one-to-one and onto, so B ∼ {a} B. Therefore by exercise 22(a), 0

0

|A B| = |A B × {a} B| = |A B| · |{a} B| = |B|n · |B| = |B|n+1 = |B||A| . (d) Let S be the set of students in the class, and let G = {A, B, C, D, F}. Then a grade assignment is a function from S to G, so the set of all possible grade assignments is S G. Since |S| = 20 and |G| = 5, by part (c), the number of possible grade assignments is |S G| = |G||S| = 520 = 95367431640625.

24. (a) Base case: n = 0. If |A| = 0 then A = ∅, so F = {∅}, and |F | = 1 = 0!.

Induction step: Suppose n is a natural number, and the desired conclusion holds for n. Now let A be a set with n + 1 elements, and let F = {f | f is a one-to-one, onto function from In+1 to A}. Let g : In+1 → A be a one-to-one, onto function. For each i ∈ In+1 , let Ai = A \ {g(i)}, a set with n elements, and let Fi = {f | f is a one-to-one, onto function from In to Ai }. By inductive hypothesis, Fi is finite and |Fi | = n!. Now let Fi0 = {f ∈ F | f (n + 1) = g(i)}. Define

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a function h : Fi → Fi0 by the formula h(f ) = f ∪ {(n + 1, g(i))}. It is not hard to check S that h is one-to-one and onto, so Fi0 is finite and |Fi0 | = |Fi | = n!. Finally, notice that F = i∈In+1 Fi0 and ∀i ∈ In+1 ∀j ∈ In+1 (i 6= j → Fi0 ∩ Fj0 = ∅). It follows, by exercise 21, that F is finite and Pn+1 |F | = i=1 |Fi0 | = (n + 1) · n! = (n + 1)!. (b) Notice that if f ∈ F then f −1 : A → In . Let h(f ) = {(a, b) ∈ A × A | f −1 (a) ≤ f −1 (b)}. We first check that h(f ) is a total order on A. For any a ∈ A, f −1 (a) ≤ f −1 (a), so (a, a) ∈ h(f ). Therefore h(f ) is reflexive on A. Suppose (a, b) ∈ h(f ) and (b, c) ∈ h(f ). Then f −1 (a) ≤ f −1 (b) and f −1 (b) ≤ f −1 (c). Therefore f −1 (a) ≤ f −1 (c), so (a, c) ∈ h(f ). This proves that h(f ) is transitive.

Next, suppose (a, b) ∈ h(f ) and (b, a) ∈ h(f ). Then f −1 (a) ≤ f −1 (b) and f −1 (b) ≤ f −1 (a), so f −1 (a) = f −1 (b). Therefore a = f (f −1 (a)) = f (f −1 (b)) = b, so h(f ) is antisymmetric. Finally, if a, b ∈ A then either f −1 (a) ≤ f −1 (b) or f −1 (b) ≤ f −1 (a), so either (a, b) ∈ h(f ) or (b, a) ∈ h(f ).

Thus, h(f ) is a total order on A. In other words, h : F → L. Next, we show that h is one-to-one. Suppose f, g ∈ F and f 6= g. By the well-ordering principle, let i be the smallest element of In such that f (i) 6= g(i). Let j = f −1 (g(i)), so f (j) = g(i). Suppose j < i. Then by the definition of i, f (j) = g(j) 6= g(i) = f (j), which is a contradiction. Therefore j ≮ i. Also, f (j) = g(i) 6= f (i), so j 6= i. Thus i < j. But i = f −1 (f (i)) and j = f −1 (g(i)), so this shows that f −1 (f (i)) < f −1 (g(i)), which implies that (f (i), g(i)) ∈ h(f ). A similar argument shows that g −1 (g(i)) < g −1 (f (i)), so (f (i), g(i)) ∈ / h(g).

Thus h(f ) 6= h(g), which proves that h is one-to-one.

Finally, to show that h is onto, suppose R ∈ L, which means that R is a total order on A. Notice that for all a ∈ A, a ∈ {x ∈ A | xRa} ⊆ A, so 1 ≤ |{x ∈ A | xRa}| ≤ |A| = n.

Therefore we can define g : A → In by the formula g(a) = |{x ∈ A | xRa}|. We claim now that ∀a ∈ A∀b ∈ A((a, b) ∈ R ↔ g(a) ≤ g(b)). To prove this, let a, b ∈ A be arbitrary. (→) Suppose (a, b) ∈ R. Then by transitivity of R, for every x ∈ A, if (x, a) ∈ R then (x, b) ∈ R. In other words, {x ∈ A | xRa} ⊆ {x ∈ A | xRb}, and therefore g(a) = |{x ∈ A | xRa}| ≤ |{x ∈ A | xRb}| = g(b). (←) We will prove the contrapositive. Suppose (a, b) ∈ / R. Then since R is a total order, (b, a) ∈ R, and as in the (→) proof this implies that {x ∈ A | xRb} ⊆ {x ∈ A | xRa}. But also since (a, b) ∈ / R, a ∈ / {x ∈ A | xRb}, whereas since R is reflexive, a ∈ {x ∈ A | xRa}. Therefore g(b) = |{x ∈ A | xRb}| < |{x ∈ A | xRa}| = g(a), so g(a) g(b).

This completes the proof that ∀a ∈ A∀b ∈ A((a, b) ∈ R ↔ g(a) ≤ g(b)). Now we claim that g is one-to-one. To prove this, suppose a, b ∈ A and g(a) = g(b). Then g(a) ≤ g(b) and g(b) ≤ g(a), so (a, b) ∈ R and (b, a) ∈ R. By antisymmetry of R, a = b. Thus, g is one-to-one.

By exercise 11(c), it follows that g is also onto. Therefore g −1 : In → A and g −1 is one-to-one and onto; in other words, g −1 ∈ F . Finally, we claim that h(g −1 ) = R. To see this, note that for all a, b ∈ A, (a, b) ∈ h(g −1 )

iff

(g −1 )−1 (a) ≤ (g −1 )−1 (b)

iff g(a) ≤ g(b)

iff (a, b) ∈ R.

−1

Since h(g ) = R and R was an arbitrary element of L, we conclude that h is onto. Thus h : F → L and h is one-to-one and onto, so F ∼ L and |L| = |F | = n!. (c) Each way of seating the people corresponds to a total order on the set of five people. Therefore, by part (b), the number of ways to seat them is 5! = 120.

25. Let m = |A|, and let f : Im → A be one-to-one and onto. Define g : A → A/R by the formula g(x) = [x]R . Then g is onto, so g ◦ f : Im → A/R is onto. By exercise 10, A/R is finite.

S Let k = |A/R| and let h : Ik → A/R be one-to-one and onto. Because A/R is a partition of A, i∈Ik h(i) = A and ∀i ∈ Ik ∀j ∈ Ik (i 6= j → h(i) ∩ h(j) = ∅). Therefore, by exercise 21, |A| =

[ i∈Ik

h(i) =

k X i=1

|h(i)|.

132

SOLUTIONS But by assumption, for every i ∈ Ik , |h(i)| = n. So m = |A| =

k X i=1

n = n + n + · · · + n = nk, {z } | k terms

so |A/R| = k = m/n = |A|/n.

26. (a) The sets A and B \ (A ∩ B) are disjoint, A ∪ B = A ∪ (B \ (A ∩ B)), and A ∩ B ⊆ B. Therefore,

by Theorem 8.1.7 and exercise 20, |A ∪ B| = |A ∪ (B \ (A ∩ B))| = |A| + |B \ (A ∩ B)| = |A| + |B| − |A ∩ B|. (b) We use part (a) repeatedly: |A ∪ B ∪ C| = |A ∪ (B ∪ C)| = |A| + |B ∪ C| − |A ∩ (B ∪ C)| = |A| + |B| + |C| − |B ∩ C| − |(A ∩ B) ∪ (A ∩ C)| = |A| + |B| + |C| − |B ∩ C| − (|A ∩ B| + |A ∩ C| − |(A ∩ B) ∩ (A ∩ C)|) = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.

S 27. Base case: n = 1. Then In = {1}, P = {{1}}, and A{1} = A1 . Therefore | i∈ln Ai | = |A1 | and P |S|+1 |AS | = (−1)2 |A{1} | = |A1 |.

S∈P (−1) Induction step: Suppose the inclusion-exclusion principle holds for n sets, and suppose A1 , A2 , . . . , An+1 are finite sets. Let Pn = P(In ) \ {∅} and Pn+1 = P(In+1 ) \ {∅}. By exercise 26(a), exercise 23(a) of Section 3.4, and the inductive hypothesis, [

[ Ai = Ai ∪ An+1 i∈In+1

i∈In

=

[

=

X

Ai + |An+1 | −

[

Ai ∩ An+1

i∈In

i∈In

[

(−1)|S|+1 |AS | + |An+1 | −

(Ai ∩ An+1 ) .

i∈In

S∈Pn

Now notice that for every S ∈ Pn , \ \ (Ai ∩ An+1 ) = Ai ∩ An+1 = AS∪{n+1} . i∈S

i∈S

Therefore, by another application of the inductive hypothesis, [ X (Ai ∩ An+1 ) = (−1)|S|+1 |AS∪{n+1} |. i∈In

S∈Pn

Thus [

Ai =

i∈In+1

X

(−1)|S|+1 |AS | + |An+1 | −

S∈Pn

=

X

X

(−1)|S|+1 |AS∪{n+1} |

S∈Pn |S|+1

(−1)

2

|AS | + (−1) |A{n+1} |

S∈Pn

+

X

(−1)|S∪{n+1}|+1 |AS∪{n+1} |.

S∈Pn

Finally, notice that there are three kinds of elements of Pn+1 : those that are elements of Pn , the set {n of the form S ∪ {n + 1}, where S ∈ Pn . It follows that the last formula above is just P + 1}, and sets |S|+1 |AS |, as required.

S∈Pn+1 (−1)

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133

28. First note that A \ B = A \ (A ∩ B) and B \ A = B \ (A ∩ B), and these sets are disjoint. Therefore

by Theorem 8.1.7 and exercise 20, |A 4 B| = |(A \ B) ∪ (B \ A)| = |(A \ (A ∩ B)) ∪ (B \ (A ∩ B)))| = |A \ (A ∩ B)| + |B \ (A ∩ B)| = (|A| − |A ∩ B|) + (|B| − |A ∩ B|) = 2(|A| − |A ∩ B|).

Therefore |A 4 B| is even.

29. Let D = {0, 1, 2, . . . , 9}. We can think of a PIN as a function f : I4 → D. (For each i ∈ I4 , f (i) is the ith digit of the PIN.) So the set of all possible PIN numbers is I4 D, and by exercise 23, |I4 D| = 104 = 10,000. Now let C be the set of all bank customers, and define the function F : C → I4 D by the formula f (c) = c’s PIN number. If the bank has more than 10,000 customers, then |C| > |I4 D|, and therefore by the pigeonhole principle (exercise 11(b)), F is not one-to-one. Therefore there are c1 , c2 ∈ C such that c1 6= c2 and F (c1 ) = F (c2 ). In other words, there are two different customers who have the same PIN number.

30. Let C be the set of all dates on which the class met, A the set of dates on which Alice attended class, and B the set of dates on which Bob attended class. Then A and B are disjoint (since Bob never saw Alice in class) and A ∪ B ⊆ C. Therefore by Theorem 8.1.7 and exercise 8(b), |A| + |B| = |A ∪ B| ≤ |C|.

If neither Alice nor Bob missed at least half of the classes, then |A| > |C|/2 and |B| > |C|/2, so |A| + |B| > |C|/2 + |C|/2 = |C|, which is a contradiction. Therefore either Alice or Bob missed at least half of the classes.

Section 8.2 1. (a) By Theorem 8.1.6, Q is countable. If R\Q were countable then, by Theorem 8.2.1, Q∪(R\Q) = R would be countable, contradicting Theorem 8.2.6. Thus, R \ Q must be uncountable. √ √ (b) Let A = { 2 + n | n ∈ Z+ }, and define f : Z+ → A by the formula f (n) = 2 + n. It is easy to check that f is one-to-one and onto, so A is denumerable. √ If A and Q are not disjoint, then there is some √ positive integer n such √ that 2 + n ∈ Q, so we can choose positive integers p and √ q such that 2 + n = p/q. But then 2 = p/q − n = (p − nq)/q, which contradicts the fact that 2 is irrational. Therefore A and Q are disjoint.

By Theorems 8.1.6 and 8.2.1, A ∪ Q is denumerable, and therefore A ∪ Q ∼ A. Finally, observe that R = (R \ (A ∪ Q)) ∪ (A ∪ Q) and R \ Q = (R \ (A ∪ Q)) ∪ A. By part 2 of Theorem 8.1.2, it follows that R ∼ R \ Q. 2. Suppose f, g ∈ Sn , a, b ∈ A, and F (f, a) = F (g, b) = h ∈ Sn+1 . Then for all i ∈ In , f (i) = h(i) = g(i), and therefore f = g. Also, a = h(n + 1) = b. Thus (f, a) = (g, b), so F is one-to-one.

To see that F is onto, suppose g ∈ Sn+1 . Let a = g(n + 1) ∈ A, and let f = g \ {(n + 1, a)}. Then f ∈ Sn , so (f, a) ∈ Sn × A and F (f, a) = (g \ {(n + 1, a)}) ∪ {(n + 1, a)} = g. Thus F is onto.

3. Suppose f1 , f2 ∈ S and F (f1 ) = F (f2 ). Let n1 and n2 be the lengths of f1 and f2 , respectively. Then g(f (1)) g(f (2)) g(f1 (1)) g(f1 (2)) 1 (n1 )) 2 (n2 )) · · · png(f . p2 · · · pg(f = F (f1 ) = F (f2 ) = p1 2 p2 2 n1 2

p1

By the uniqueness of prime factorizations it follows that n1 = n2 and for all i ∈ In1 , g(f1 (i)) = g(f2 (i)).

Since g is one-to-one, it follows that for all i ∈ In1 , f1 (i) = f2 (i), so f1 = f2 . Therefore F is one-to-one.

4. Let S be the set of all finite sequences of positive integers. By Theorem 8.2.4, S is countable. Let g : Z+ → S be a function that is onto. By exercise 10 in Section 8.1, for each f ∈ S, Ran(f ) is a finite subset of Z+ . Therefore we can define a function F : S → P by the formula F (f ) = Ran(f ).

We claim that F is onto. To see why, suppose X ∈ P . Then X ⊆ Z+ and X is finite. Let n = |X|.

Then there is some function f : In → X such that f is one-to-one and onto. But then f ∈ S and

134

SOLUTIONS F (f ) = Ran(f ) = X. Since X was an arbitrary element of P , this shows that F is onto. Since g and F are both onto, F ◦ g : Z+ → P is onto. Thus, by Theorem 8.1.5, P is countable. It is clearly infinite. so it is denumerable.

5. Suppose that A ∼ P(A). Then there is a function f : A → P(A) that is one-to-one and onto. Let X = {a ∈ A | a ∈ / f (a)} ∈ P(A). Since f is onto, there must be some a ∈ A such that f (a) = X. But then according to the definition of X, a ∈ X iff a ∈ / f (a), so X 6= f (a), which is a contradiction.

6. (a) Let A, B, and C be arbitrary sets. Define p : B × C → B and q : B × C → C by the formulas p(b, c) = b and q(b, c) = c. For any function f : A → B × C, p ◦ f : A → B and q ◦ f : A → C, so we can define F : A (B × C) → A B × A C by the formula F (f ) = (p ◦ f, q ◦ f ). Note that if a ∈ A and f (a) = (b, c) ∈ B × C, then (p ◦ f )(a) = b and (q ◦ f )(a) = c, so f (a) = ((p ◦ f )(a), (q ◦ f )(a)).

To see that F is one-to-one, suppose F (f ) = F (g). Then p ◦ f = p ◦ g and q ◦ f = q ◦ g, so for all a ∈ A, f (a) = ((p ◦ f )(a), (q ◦ f )(a)) = ((p ◦ g)(a), (q ◦ g)(a)) = g(a).

Therefore f = g, so F is one-to-one. To see that F is onto, suppose (g, h) ∈ A B × A C, so g : A → B and h : A → C. Define f : A → (B × C) by the formula f (a) = (g(a), h(a)). Then for all a ∈ A, (p ◦ f )(a) = p(g(a), h(a)) = g(a) and (q ◦ f )(a) = q(g(a), h(a)) = h(a), so p ◦ f = g and q ◦ f = h. Therefore F (f ) = (p ◦ f, q ◦ f ) = (g, h). This proves that F is onto. (b) Let A, B, and C be arbitrary sets. If f ∈ A (B C) then f : A → B C, so for any a ∈ A, f (a) ∈ B C. But that means that f (a) : B → C, so for any b ∈ B, f (a)(b) ∈ C. That means we can define a function F (f ) : A × B → C by the formula F (f )(a, b) = f (a)(b). Then F (f ) ∈ (A×B) C, so F : A (B C) → (A×B) C. To see that F is one-to-one, suppose F (f ) = F (g). Let a ∈ A be arbitrary. Then for every b ∈ B, f (a)(b) = F (f )(a, b) = F (g)(a, b) = g(a)(b), so by Theorem 5.1.4, f (a) = g(a). But since a was arbitrary, it follows by another application of Theorem 5.1.4 that f = g. To see that F is onto, suppose g ∈ (A×B) C, so g : A × B → C. Define a function f with domain A by the formula f (a) = {(b, c) ∈ B × C | g(a, b) = c}.

It is not hard to see that f (a) is a function from B to C, and for all b ∈ B, f (a)(b) = g(a, b).

Therefore f (a) ∈ B C, so f : A → B C, which means that f ∈ A (B C). For all (a, b) ∈ A × B, F (f )(a, b) = f (a)(b) = g(a, b), so F (f ) = g. (c) Let A be an arbitrary set. Define F : P(A) → A {yes, no} by the formula F (X) = (X × {yes}) ∪ ((A \ X) × {no}). It is not hard to see that F (X) : A → {yes, no}, so F (X) ∈ A {yes, no} as required, and for all a ∈ A, ( F (X)(a) =

yes, if a ∈ X, no, if a ∈ / X.

Define G : A {yes, no} → P(A) by the formula G(f ) = {a ∈ A | f (a) = yes}.

Suppose X ∈ P(A). Then G(F (X)) = {a ∈ A | F (X)(a) = yes} = {a ∈ A | a ∈ X} = X. This proves that G ◦ F = iP(A) . Now suppose f ∈ A {yes, no}. Then for every a ∈ A, F (G(f ))(a) = yes

iff a ∈ G(f )

iff

f (a) = yes.

Thus for all a ∈ A, F (G(f ))(a) = f (a), so F (G(f )) = f . This proves that F ◦ G = iA {yes,no} .

Thus, by Theorem 5.3.4, F is one-to-one and onto.

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(d) We apply the transitivity of ∼ (part 3 of Theorem 8.1.3): Z+

+

+

P(Z+ ) ∼ Z (Z {yes, no}) +

(part (c) and exercise 23(a) in Section 8.1)

+

∼ (Z ×Z ) {yes, no} ∼

(part (b))

Z+

(Z+ × Z+ ∼ Z+ , exercise 23(a) in Section 8.1)

{yes, no}

∼ P(Z+ )

(part (c)). +

Therefore, by the transitivity of ∼, Z P(Z+ ) ∼ P(Z+ ).

7. Since A is denumerable and Z+ × Z+ is denumerable, there is a function f : A → Z+ × Z+ that is one-to-one and onto. For each n ∈ Z+ , let Zn = Z+ × {n}, and let P = {f −1 (Zn ) | n ∈ Z+ }. It is easy to verify that P is a partition of A, P is denumerable, and every element of P is denumerable.

8. Define f : P(A) × P(B) → P(A ∪ B) by the formula f (X, Y ) = X ∪ Y . Since A and B are disjoint, if X ⊆ A and Y ⊆ B then f (X, Y ) ∩ A = (X ∪ Y ) ∩ A = X and f (X, Y ) ∩ B = (X ∪ Y ) ∩ B = Y . To see that f is one-to-one, suppose f (X1 , Y1 ) = f (X2 , Y2 ). Then X1 = f (X1 , Y1 ) ∩ A = f (X2 , Y2 ) ∩ A = X2 and Y1 = f (X1 , Y1 ) ∩ B = f (X2 , Y2 ) ∩ B = Y2 . To see that f is onto, suppose Z ⊆ A ∪ B. Let X = Z ∩ A ⊆ A and Y = Z ∩ B ⊆ B. Then f (X, Y ) = X ∪ Y = (Z ∩ A) ∪ (Z ∩ B) = Z ∩ (A ∪ B) = Z. 9. (a) Suppose B ⊆ S R and B is uncountable.

By the proof in exercise 23(a) in Section 3.4, B = S B ∩ R = B ∩ ( n∈Z+ An ) = n∈Z+ (B ∩ An ). If B ∩ An is countable for every n ∈ Z+ then by Theorem 8.2.2, B is countable, which is a contradiction. Therefore there is some positive integer n such that B ∩ An is uncountable.

S (b) For every positive integer n, An ⊆ n∈Z+ An = Z+ . Suppose that for every positive integer n, An 6= Z+ . Then we can let an be the least element of Z+ \An . Let B = {an | n ∈ Z+ }. Notice that for all positive integers m and n, if n ≤ m then, since An ⊆ Am and am ∈ / Am , am ∈ / An . Thus, if am ∈ An then m < n. It follows that for every positive integer n, B ∩ An ⊆ {a1 , a2 , . . . , an−1 }, so B ∩ An is finite. Therefore, by the assumption in the problem, B is finite, so for some positive integer N , B = {a1 , a2 , . . . , aN }. In other words, for every n > N , an is equal to one of a1 , a2 , . . . , aN .

S Since a1 ∈ Z+ = n∈Z+ An , we can choose some positive integer k1 such that a1 ∈ Ak1 .

Similarly, we can choose positive integers k2 , k3 , . . . , kN such that a2 ∈ Ak2 , a3 ∈ Ak3 , . . . , aN ∈ AkN . Let K be the largest of k1 , k2 , . . . , kN . Then Ak1 ⊆ AK , Ak2 ⊆ AK , . . . , AkN ⊆ AK , and therefore B = {a1 , a2 , . . . , aN } ⊆ AK . But aK ∈ / AK and aK ∈ B, so this is a contradiction.

Therefore, for some positive integer n, An = Z+ .

S 10. For each positive integer n, let An = {x ∈ A | x ≥ 1/n}. Clearly n∈Z+ An ⊆ A. Now suppose x ∈ A. Then x ∈ R+ , so x > 0. Let nSbe a positive integer large S enough that n ≥ 1/x. Then x ≥ 1/n, so x ∈ An . We conclude that A ⊆ n∈Z+ An , and therefore n∈Z+ An = A. Suppose a1 , a2 , . . . , ak are distinct elements of An . Then b ≥ a1 + a2 + · · · + ak ≥

1 1 k 1 + + ··· + = , n n n n

so k ≤ bn. Therefore An is finite, and in fact |An | ≤ bn. By Theorem 8.2.2, it follows that A = S n∈Z+ An is countable.

11. Suppose R/E is countable and there is some x ∈ R such that [x]E is countable. Since for all real numbers x andSy, [x]E ∼ [y]E , it follows that for every x ∈ R, [x]E is countable. Since R/E is a partition of R, (R/E). But then by Theorem 8.2.2, R is countable, which contradicts Theorem 8.2.6.

12. (a) Suppose x ∈ Q. Then there are integers p and q, with q 6= 0, such that x = p/q. Therefore qx − p = 0, so x ∈ A.

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SOLUTIONS √ (b) Let x = 2. Then x2 − 2 = 0, so x ∈ A. (c) Let C = {f | for some integer n > 0, f : In → Z and f (n) 6= 0}. Then C is a subset of the set of all finite sequences of elements of Z, so by Theorem 8.2.4 and exercise 17 in Section 8.1, C is countable. If f ∈ C and f : In → Z, let Sf = {x ∈ R | f (0) +Sf (1)x + f (2)x2 + · · · + f (n)xn = 0}; as stated in the problem, Sf is finite. By definition, A = f ∈C Sf , so by Theorem 8.2.2, A is countable.

13. First note that if F = ∅ then g can be any function. If F 6= ∅, then since F is countable, we can write its elements in a list: F = {f1 , f2 , . . .}. Now define g : Z+ → R by the formula g(n) = max{|f1 (n)|, |f2 (n)|,. . . , |fn (n)|}. Consider any function fk ∈ F. By the definition of g, ∀n ≥ k(|fk (n)| ≤ |g(n)|), so fk ∈ O(g). Therefore F ⊆ O(g).

14. First suppose ∅ ∈ / F. Then by the well-ordering principle, every element of F has a smallest element.

Define f : F → Z+ by the formula f (X) = the smallest element of X. Then for every X ∈ F, f (X) ∈ X. If X, Y ∈ F and f (X) = f (Y ) then f (X) = f (Y ) ∈ X ∩ Y . Therefore X ∩ Y 6= ∅, so since F is pairwise disjoint, X = Y . This shows that f is one-to-one, so by Theorem 8.1.5, F is countable.

Finally, if ∅ ∈ F then the argument of the previous paragraph proves that F \ {∅} is countable.

Since F = (F \ {∅}) ∪ {∅}, it follows by Theorem 8.2.1 that F is countable.

15. (a) If Q is countable, then by part 2 of Theorem 8.2.1, P ∪ Q is countable. But P ∪ Q = P(Z+ ), which is uncountable by Cantor’s theorem. Therefore Q is uncountable. (b) Suppose A ∈ Q. For every n ∈ Z+ , A ∩ In ⊆ In , so by exercise 8(a) in Section 8.1, A ∩ In is finite.

Therefore SA ⊆ P . Now suppose SA is finite. Then there is some positive integer n such that SA = {A∩I1 , A∩I2 , . . . , A∩In }. We claim now that A ⊆ In ; this will complete the proof, because it implies that A is finite, contradicting our assumption that A ∈ Q. To prove this claim, suppose that m ∈ A. Then A ∩ Im ∈ SA , so there is some k ≤ n such that A ∩ Im = A ∩ Ik ⊆ Ik ⊆ In .

But m ∈ A ∩ Im , so we conclude that m ∈ In , as required. (c) Suppose A ∈ Q, B ∈ Q, and A 6= B. Then there is some positive integer n such that either n ∈ A and n ∈ / B or n ∈ B and n ∈ / A. We will assume n ∈ A and n ∈ / B; the proof for the other case is similar. We claim now that SA ∩ SB ⊆ {A ∩ I1 , A ∩ I2 , . . . , A ∩ In−1 }; this will complete the proof, because it implies that SA ∩ SB is finite. To prove the claim, suppose that X ∈ SA ∩ SB .

Then there are positive integers nA and nB such that X = A ∩ InA and X = B ∩ InB . If nA ≥ n then n ∈ A ∩ InA = X = B ∩ InB ⊆ B, which is a contradiction. Therefore nA < n, so X = A ∩ InA ∈ {A ∩ I1 , . . . , A ∩ In−1 }, as required. (d) If A ∈ Q then SA ⊆ P , so since g : P → Z+ , g(SA ) ⊆ Z+ . Also, since SA is infinite and g is one-to-one, g(SA ) is also infinite. This proves that F ⊆ P(Z+ ) and every element of F is infinite.

To see that F is pairwise almost disjoint, suppose X, Y ∈ F and X 6= Y . Then there are sets A, B ∈ Q such that X = g(SA ) and Y = g(SB ). Since X 6= Y , A 6= B, so by part (c), SA ∩ SB is finite, and therefore g(SA ∩ SB ) is finite. By Theorem 5.5.2, g(SA ∩ SB ) = g(SA ) ∩ g(SB ) = X ∩ Y , so X and Y are almost disjoint. Finally, define h : Q → F by the formula h(A) = g(SA ). It is easy to check that h is one-to-one and onto, so F ∼ Q and therefore, by part (a), F is uncountable.

16. By two applications of Theorem 8.2.1, (Z+ × Z+ ) × Z+ is countable, so we can write its elements in a list: (Z+ × Z+ ) × Z+ = {((a1 , b1 ), c1 ), ((a2 , b2 ), c2 ), . . .}. For each i ∈ Z+ , let Xi = {ai n + bi | n ∈ Z+ }; then Xi is an infinite subset of Z+ . Define g : Z+ → Z+ recursively by the formula g(i) = the smallest element of Xi \ {g(j) | j < i}; note that since Xi is infinite and {g(j) | j < i} is finite, Xi \ {g(j) | j < i} is a nonempty subset of Z+ , and therefore by the well-ordering principle it has a smallest element. Since ∀j < i(g(i) 6= g(j)),

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g is one-to-one. Let R = Ran(g) ⊆ Z+ . Then g is a one-to-one, onto function from Z+ to R, so g −1 : R → Z+ . Finally, we define f : Z+ → Z+ as follows: ( cg−1 (x) , if x ∈ R, f (x) = 0, if x ∈ / R. To see that f has the required property, suppose a, b, and c are positive integers. Then ((a, b), c) ∈ (Z+ × Z+ ) × Z+ , so there is some positive integer i such that ((a, b), c) = ((ai , bi ), ci ); in other words, ai = a, bi = b, and ci = c. Let x = g(i) ∈ R ⊆ Z+ . Then by the definition of g, x ∈ Xi , so there is some positive integer n such that x = ai n + bi = an + b. Also, g −1 (x) = i, so f (an + b) = f (x) = cg−1 (x) = ci = c, as required.

17. Let W be the set of all words of English. This is a finite set (a complete list of its elements can be found in a dictionary). Therefore by Theorem 8.2.4 the set S of all finite sequences of elements of W is countable. The set of English sentences is a subset of S, so it is also countable. To see that it is infinite, consider the following list of infinitely many sentences: My mother had red hair; My mother’s mother had red hair; My mother’s mother’s mother had red hair; . . . . Thus, the set of English sentences is denumerable.

18. (a) As in exercise 17, let S be the set of all finite sequences of words of English, which is a countable set. The set P of phrases that define numbers is a subset of S, so it is also countable. Let f : Z+ → P be onto. Let D be the set of all numbers that can be defined by an English phrase, and let g : P → D be defined by the formula g(p) = the number defined by the phrase p. Then g ◦ f maps Z+ onto D, so D is countable. To see that it is infinite, consider the following list of infinitely many phrases defining numbers: the smallest positive integer; the smallest positive integer that is larger than some positive integer; the smallest positive integer that is larger than some positive integer that is larger than some positive integer; . . . . This shows that Z+ ⊆ D, so D is infinite. (b) Since D is denumerable and R is uncountable, there must be elements of R that are not in D; that is, there are real numbers that cannot be defined by English phrases.

Section 8.3 1. (a) The function iA : A → A is one-to-one. (b) Suppose A - B and B - C. Then there are one-to-one functions f : A → B and g : B → C. By part 1 of Theorem 5.2.5, g ◦ f : A → C is one-to-one, so A - C. 2. (a) Let A be an arbitrary set. Then A ∼ A, so A 6≺ A. (b) Suppose A ≺ B and B ≺ C. Then A - B and B - C, so by exercise 1(b), A - C. Now suppose A ∼ C. Then C - A, so since A - B, by exercise 1(b), C - B. Since B - C and C - B, by the Cantor-Schröder-Bernstein theorem, B ∼ C. But this contradicts the fact that B ≺ C. Therefore A C. Since A - C and A C, A ≺ C. 3. Since A ⊆ B ⊆ C, A - B and B - C. Since A ∼ C, C - A. Since C - A and A - B, by exercise 1(b), C - B. Since B - C and C - B, by the Cantor-Schröder-Bernstein theorem, B ∼ C. 4. Since A - B and C - D, there are one-to-one functions f : A → B and g : C → D. (a) Define h : A × C → B × D by the formula h(a, c) = (f (a), g(c)). To see that h is one-toone, suppose h(a1 , c1 ) = h(a2 , c2 ). Then (f (a1 ), g(c1 )) = (f (a2 ), g(c2 )), so f (a1 ) = f (a2 ) and g(c1 ) = g(c2 ). Since f and g are one-to-one, it follows that a1 = a2 and c1 = c2 , and therefore (a1 , c1 ) = (a2 , c2 ). Thus h is one-to-one, so A × C - B × D.

138

SOLUTIONS (b) By exercise 12(a) in Section 5.1, f ∪ g : A ∪ C → B ∪ D. To see that f ∪ g is one-to-one, suppose (f ∪ g)(x) = (f ∪ g)(y).

Case 1. x ∈ A. Then (f ∪ g)(x) = f (x) ∈ B. If y ∈ C then (f ∪ g)(y) = g(y) ∈ D. But since B and D are disjoint, this would contradict the assumption that (f ∪ g)(x) = (f ∪ g)(y). Therefore y ∈ / C, so y ∈ A. Therefore f (x) = (f ∪ g)(x) = (f ∪ g)(y) = f (y). But f is one-to-one, so it follows that x = y.

Case 2. x ∈ C. A similar argument shows that x = y.

Therefore f ∪ g is one-to-one, so A ∪ C - B ∪ D. (c) Define h : P(A) → P(B) by the formula h(X) = f (X). (Note: This equation should be read as saying that the result of applying h to X is equal to the image of X under f .) To see that h is one-to-one, suppose h(X) = h(Y ). Suppose x ∈ X. Then f (x) ∈ f (X) = h(X) = h(Y ) = f (Y ), so there is some y ∈ Y such that f (x) = f (y). Since f is one-to-one, it follows that x = y ∈ Y .

Since x was arbitrary, we conclude that X ⊆ Y . A similar argument shows Y ⊆ X, so X = Y .

Therefore h is one-to-one, so P(A) - P(B).

5. Let g : A → B and h : C → D be one-to-one functions. (a) Since A 6= ∅, we can choose some a0 ∈ A. Notice that g −1 : Ran(g) → A. Define j : B → A as follows: ( g −1 (b), if b ∈ Ran(g), j(b) = a0 , otherwise.

We claim that j is onto. To see why, suppose a ∈ A. Then g(a) ∈ Ran(g), so j(g(a)) = g −1 (g(a)) = a.

Now define F : A C → BD by the formula F (f ) = h ◦ f ◦ j. To see that F is one-to-one, suppose that f1 ∈ A C, f2 ∈ A C, and F (f1 ) = F (f2 ), which means h ◦ f1 ◦ j = h ◦ f2 ◦ j.

Let a ∈ A be arbitrary. Since j is onto, there is some b ∈ B such that j(b) = a. Therefore h(f1 (a)) = (h ◦ f1 ◦ j)(b) = (h ◦ f2 ◦ j)(b) = h(f2 (a)), and since h is one-to-one, it follows that f1 (a) = f2 (a). Since a was arbitrary, this shows that f1 = f2 . Thus F is one-to-one, so A C - B D. (b) Yes. Suppose A = C = D = ∅ but B 6= ∅. Then A - B, C - D, A C = {∅}, and B D = ∅, so A C 6- B D. 6. (a) Suppose A - B and B is finite. Let f : A → B be one-to-one, and let R = Ran(f ) ⊆ B. Then by exercise 8 in Section 8.1, R is finite and |R| ≤ |B|. Also, f is a one-to-one, onto function from A to R, so A ∼ R. Therefore A is finite and |A| = |R| ≤ |B|. (b) Suppose A ≺ B and B is finite. Then A - B, so by part (a), A is finite and |A| ≤ |B|. Suppose |A| = |B|, and let n = |A| = |B|. Then A ∼ In and B ∼ In , so A ∼ B, which contradicts A ≺ B. Therefore |A| = 6 |B|, so |A| < |B|.

7. Let A be an arbitrary set. Define f : A → P(A) by the formula f (a) = {a}. Then f is one-to-one, so A - P(A). But by exercise 5 of Section 8.2, A P(A). Therefore A ≺ P(A).

8. (a) We let n be arbitrary and then proceed by induction on m.

Base case: If m = n + 1, then by exercise 7, An ≺ P(An ) = An+1 = Am .

Induction step: Suppose m > n and An ≺ Am . By exercise 7, Am ≺ P(Am ) = Am+1 , so by exercise 2(b), An ≺ Am+1 .

S S S (b) For everySpositive integer n, An ⊆ m∈Z+ Am , so An - m∈Z+ Am . If An ∼ m∈Z+ Am then since An+1 - m∈Z+ Am , An+1 - An . But also An - An+1 , so by theSCantor-Schröder-Bernstein theS orem, A ∼ A , which contradicts part (a).

Therefore A

+ Am , so An ≺ + Am . n n+1 n m∈Z n∈Z S S So m∈Z+ Am is a larger size of infinity than all of the sets An . Of course, P( m∈Z+ Am ) is an even larger size of infinity, and applying the power set operation repeatedly we can continue to generate larger sizes of infinity.

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9. Using the notation of the proof of the Cantor-Schröder-Bernstein theorem, we have A1 = {1}, A2 = S {1/2}, A3 = {1/4}, and so on, so X = n∈Z+ An = {1, 1/2, 1/4, . . .} = {1/2n | n ∈ N}. The function h : (0, 1] → (0, 1) is defined by the formula ( x/2, if x ∈ X, h(x) = x, if x ∈ / X. 10. (a) Note that E ⊆ P(Z+ × Z+ ). It follows, using exercise 5 of Section 8.1, that E - P(Z+ × Z+ ) ∼ P(Z+ ), and therefore E - P(Z+ ). (b) Suppose f (X) = f (Y ). Then X ∪ {1} ∈ f (X) = f (Y ) = {Y ∪ {1}, (A \ Y ) ∪ {2}}, so either X ∪ {1} = Y ∪ {1} or X ∪ {1} = (A \ Y ) ∪ {2}. But clearly 2 ∈ / X ∪ {1}, so the second possibility can be ruled out. Therefore X ∪ {1} = Y ∪ {1}. Since neither X nor Y contains 1, it follows that X =Y. (c) Clearly A is denumerable (define h : Z+ → A by the formula f (n) = n + 2), and we showed at the end of Section 5.3 that P ∼ E. It follows that P(Z+ ) ∼ P(A) - P ∼ E, so P(Z+ ) - E. Combining this with part (a) and applying the Cantor-Schröder-Bernstein theorem gives the desired conclusion.

11. For every R ∈ T , R ⊆ Z+ × Z+ . Therefore T ⊆ P(Z+ × Z+ ), so by exercise 5 of Section 8.1, T - P(Z+ × Z+ ) ∼ P(Z+ ), and therefore T - P(Z+ ).

Now let A = Z+ \ {1}. For every X ⊆ A, let f (X) = {(a, b) ∈ X × X | a ≤ b} ∪ {(a, b) ∈ (Z+ \ X) × (Z+ \ X) | a ≤ b} ∪ (X × (Z+ \ X)).

It is tedious but not hard to verify that f (X) is a total order on Z+ , so f : P(A) → T . To see that f is one-to-one, suppose that X, Y ∈ P(A) and f (X) = f (Y ). Suppose x ∈ X. Since X ⊆ A = Z+ \ {1}, 1∈ / X, so x 6= 1 and therefore x > 1. Since 1 ∈ Z+ \ X, (x, 1) ∈ (X × (Z+ \ X)) ⊆ f (X) = f (Y ).

But since x > 1, the only way this can happen is if (x, 1) ∈ Y × (Z+ \ Y ), so x ∈ Y . Since x was arbitrary, this shows that X ⊆ Y , and a similar argument shows that Y ⊆ X. Therefore X = Y .

This proves that f is one-to-one, so P(A) - T . Clearly Z+ ∼ A (define h : Z+ → A by the formula h(n) = n + 1), so P(Z+ ) ∼ P(A), and therefore P(Z+ ) - T . Since T - P(Z+ ) and P(Z+ ) - T , by the Cantor-Schröder-Bernstein theorem, T ∼ P(Z+ ).

12. (a) Suppose A has at least two elements and A × A ∼ A. Define f : P(A) → P(A) × P(A) by the formula f (X) = (X, ∅). Clearly f is one-to-one, so P(A) - P(A) × P(A).

Since A × A ∼ A, P(A × A) ∼ P(A). Let a and b be distinct element of A, and define g : P(A) × P(A) → P(A × A) by the formula g(X, Y ) = (X × {a}) ∪ (Y × {b}). Suppose g(X1 , Y1 ) = g(X2 , Y2 ). Then X1 = {z ∈ A | (z, a) ∈ g(X1 , Y1 )} = {z ∈ A | (z, a) ∈ g(X2 , Y2 )} = X2 and Y1 = {z ∈ A | (z, b) ∈ g(X1 , Y1 )} = {z ∈ A | (z, b) ∈ g(X2 , Y2 )} = Y2 .

Therefore (X1 , Y1 ) = (X2 , Y2 ), so g is one-to-one. Thus P(A) × P(A) - P(A × A) ∼ P(A), so P(A) × P(A) - P(A). Since P(A) - P(A) × P(A) and P(A) × P(A) - P(A), by the Cantor-Schröder-Bernstein theorem, P(A) × P(A) ∼ P(A). (b) We apply part (a) with A = Z+ and also Theorems 8.1.2 and 8.3.3: R × R ∼ P(Z+ ) × P(Z+ ) ∼ P(Z+ ) ∼ R. 13. Since Q is denumerable, we can let f : Z+ → Q be one-to-one and onto. If I ∈ F then since I is nondegenerate, there are numbers x, y ∈ I such that x < y. By Lemma 8.3.4, there is some rational

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SOLUTIONS number q such that x < q < y, and since I is an interval, q ∈ I. Since f is onto, there is some n ∈ Z+ such that f (n) = q ∈ I. Thus, we have proven that ∀I ∈ F∃n ∈ Z+ (f (n) ∈ I). Define g : F → Z+ as follows: g(I) = the smallest n ∈ Z+ such that f (n) ∈ I. Suppose g(I1 ) = g(I2 ), and let n = g(I1 ) = g(I2 ). Then f (n) ∈ I1 and f (n) ∈ I2 . Since F is pairwise disjoint, this implies that I1 = I2 . Thus g is one-to-one. By Theorem 8.1.5, F is countable.

14. (a) According to the definition of function, R R ⊆ P(R × R), and therefore by exercise 12(b) and exercise 5 of Section 8.1, R R - P(R × R) ∼ P(R).

Clearly {yes, no} - R, so by exercise 6(c) of Section 8.2 and exercise 5, P(R) ∼ R {yes, no} R R. Since we have both R R - P(R) and P(R) - R R, by the Cantor-Schröder-Bernstein theorem, R R ∼ P(R). (b) By Theorems 8.1.6 and 8.3.3, exercise 23(a) of Section 8.1, and exercise 6(d) of Section 8.2, + Q R ∼ Z P(Z+ ) ∼ P(Z+ ) ∼ R. (c) Define F : C → Q R by the formula F (f ) = f Q. (See exercise 7 of Section 5.1 for the meaning of the notation used here.) Suppose f ∈ C, g ∈ C, and F (f ) = F (g). Then f Q = g Q, which means that for all x ∈ Q, f (x) = g(x). Now let x be an arbitrary real number. Use Lemma 8.3.4 to construct a sequence x1 , x2 , . . . of rational numbers such that limn→∞ xn = x. Then since f and g are continuous, f (x) = limn→∞ f (xn ) = limn→∞ g(xn ) = g(x). Since x was arbitrary, this shows that f = g. Therefore F is one-to-one, so C - Q R. Combining this with part (b), we can conclude that C - R. Now define G : R → C by the formula G(x) = R × {x}. In other words, G(x) is the constant function whose value at every real number is x. Clearly G is one-to-one, so R - C. By the Cantor-Schröder-Bernstein theorem, it follows that C ∼ R.