Student Manual for A Discrete Transition to Advanced Mathematics Richmond, Exercises of Discrete Mathematics

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A Discrete

Transition

to Advanced

Mathematics

Bettina Richmond

Thomas Richmond

Students' Solutions Manual for

Contents

This solution manual accompanies A Discrete Transition to Advanced Mathematics by Bettina Richmond and Tom Richmond. The text contains over 650 exercises. This manual includes solutions to parts of 210 of them. These solutions are presented as an aid to learning the material, and not as a substitute for learning the material. You should attempt to solve each problem on your own and consult the solutions manual only as a last resort. It is important to note that there are many different ways to solve most of the exercises. Looking up a solution before following through with your own approach to a problem may stifle your creativity. Consulting the solution manual after finding your own solution might reveal a different approach. There is no claim that the solutions presented here are the “best” solutions. These solutions use only techniques which should be familiar to you.

Chapter 1

Sets and Logic

1.1 Sets

1. (a) True (b) The elements of a set are not ordered, so there is no “first” element of a set.
2. |{M, I, S, S, I, S, S, I, P, P, I}| = |{M, I, S, P }| = 4 < 7 = |{F, L, O, R, I, D, A}|.
3. (a) { 1 , 2 , 3 } ⊆ { 1 , 2 , 3 , 4 } (b) 3 ∈ { 1 , 2 , 3 , 4 } (c) { 3 } ⊆ { 1 , 2 , 3 , 4 } (d) {a} ∈ {{a}, {b}, {a, b}} (e) ∅ ⊆ {{a}, {b}, {a, b}} (f) {{a}, {b}} ⊆ {{a}, {b}, {a, b}}
4. (a) A 0-element set ∅ has 20 = 1 subset, namely ∅.

(b) A 1-element set { 1 } has 21 = 2 subsets, namely ∅ and { 1 }.

(c) A 2-element set has 22 = 4 subsets. A 3-element set has 23 = 8 subsets. A 4-element set { 1 , 2 , 3 , 4 } should have 24 = 16 subsets

(d) The 16 subsets of { 1 , 2 , 3 , 4 } are: ∅, { 1 }, { 2 }, { 3 }, { 4 }, { 1 , 2 }, { 1 , 3 }, { 1 , 4 }, { 2 , 3 }, { 2 , 4 }, { 3 , 4 },

{ 1 , 2 , 3 }, { 1 , 2 , 4 }, { 1 , 3 , 4 }, { 2 , 3 , 4 }, { 1 , 2 , 3 , 4 } (e) A 5-element set has 25 = 32 subsets. A 6-element set has 26 = 64 subsets. An n-element set has 2 n^ subsets.

1

1.3. PARTITIONS 3

7. (^) (x, y) ∈ A × (B ∩ C) ⇐⇒ x ∈ A, y ∈ B ∩ C ⇐⇒ x ∈ A, y ∈ B and y ∈ C ⇐⇒ x ∈ A and y ∈ B and x ∈ A and y ∈ C ⇐⇒ (x, y) ∈ (A × B) ∩ (A × C).

This shows that the elements of A × (B ∩ C) are precisely those of (A × B) ∩ (A × C), and thus the two sets are equal.

9. The conditions are not equivalent. For example, the collection {S 1 , S 2 } where S 1 = S 2 6 = ∅ satisfies (Si ∩ Sj 6 = ∅ ⇒ Si = Sj ), but not (Si ∩ Sj 6 = ∅ ⇒ i = j). However, if the sets of the collection {Si|i ∈ I} are distinct, the statements will be equivalent.
10. Let A be the set of students taking Algebra and let S be the set of students taking Spanish. Now |A ∪ S| = |A| + |S| − |A ∩ S| = 43 + 32 − 7 = 68. Thus, there are 68 students taking Algebra or Spanish.
11. A tree diagram for the outcomes will have 2 branches for the choice of meat, each stem of which has 7 branches for the possible choices for vegetables, and each of these stems has 5 branches for the choice of dessert. Thus, 2 choices for meat, 7 choices for vegetable, and 5 choices for dessert give 2 · 7 · 5 = 70 choices for the special.
12. Observe that there are not 4 · 3 options, for Luis can not take both physics and chemistry at 2:00. There are only 11 scheduling options, as shown in the tree diagram below.

PP

PP

PP

PP

PP

PP

PP

PP

1.3 Partitions

3. (a) Not necessarily. Some Bi may be empty. (b) Yes (S 6 = ∅ and L 6 = ∅), S ∪ L = B, and S ∩ L = ∅. (c) No. S and P partition A, but D has nonempty intersection with S or P yet D 6 = S and D 6 = P. (d) No. X = ∅. (e) No. R ∩ S = S 6 = ∅, but R 6 = S.

4 CHAPTER 1. SETS AND LOGIC

5. (a) Yes. (b) No. L 3 6 = L 4 even though L 3 ∩ L 4 = {(0, 0)} 6 = ∅. Also, (0, 1) 6 ∈

S

D.

(c) Yes. (d) Yes. (e) No. (0, 1) 6 ∈

S

G. Also, P 3 6 = P 4 yet P 3 ∩ P 4 = {(0, 0)} 6 = ∅. (f) No. (π, π) 6 ∈

S

H.

8. Each Ci is nonempty: Given i ∈ I, Bi 6 = ∅, so ∃b ∈ Bi, and

b ∈ Ci. C is a mutually disjoint collection: If Ci ∩ Cj 6 = ∅, then ∃z ∈ Ci ∩ Cj , and from the definition of Ci and Cj , we have z^2 ∈ Bi ∩ Bj. Since Bi ∩ Bj 6 = ∅ and {Bi|i ∈ I} is a partition, it follows that Bi = Bj , so {x ∈ R|x^2 ∈ Bi} = {x ∈ R|x^2 ∈ Bj }, that is, Ci = Cj. S C = R: Clearly

S

C ⊆ R, so it suffices to show R ⊆

S

S C. Given^ x^ ∈^ R,^ x^2 ∈^ [0,^1 ) = B, so x^2 ∈ Bi for some i ∈ I, which shows x ∈ Ci. Thus, x ∈ R ⇒ x ∈

S

C, as needed.

11. (a) Given any partition P of S, each block of P may be partitioned into singleton sets (that is, into blocks of D), so D is finer than any partition P of S. (b) The coarsest partition of a set S is the one-block partition I = {S}. Given any partition P of S, the block S of I is further partitioned by the blocks of P, so ever partition P of S is finer than I. (c) Since each block of the coarser partition Q is the union of one or more blocks of the finer partition P, we have |P| ≥ |Q|. (d) No. P = {(−1, 0], (0, 1 )} and Q = {(−1, 5], (5, 6), [6, 1 )} are partitions of R with |P| ≤ |Q|, but neither partition is a refinement of the other.

1.4 Logic and Truth Tables

1. (a) S∧ ∼ G (b) H∨ ∼ S (c) ∼ (S ∧ G) (d) (S ∧ G) ∨ (∼ H) (e) (S∨ ∼ S) ∧ G (f) S ∧ H ∧ G (g) (S ∧ H) ∨ (∼ G)
2. (a) P Q P ∧ Q ∼ (P ∧ Q) ∼ Q ∼ (P ∧ Q)∧ ∼ Q T T T F F F T F F T T T F T F T F F F F F T T T ∼ (P ∧ Q)∧ ∼ Q = ∼ Q since the columns for these two statements are identical. (b) Note that if Q fails, then (P ∧ Q) fails, so that Q fails and (P ∧ Q) fails. On the other hand, if Q fails and some other conditions occur (namely, (P ∧ Q) fails), then Q fails.

6 CHAPTER 1. SETS AND LOGIC

(b) False. For a = 0, b a is not even defined. Negation: ∃a ∈ Z such that ∀b ∈ Z, (^) ab 6 ∈ Z. (c) True. ∀u ∈ N, take v = 2u. Negation: ∃u ∈ N such that ∀v ∈ N \ {u}, v u 6 ∈ N. (d) False. For u = 1, (^1) v 6 ∈ N ∀v ∈ N. Negation: ∃u ∈ N such that ∀v ∈ N \ {u}, u v 6 ∈ N. (e) True. ∀a ∈ N, take b = a^2 and c = a. Negation: ∃a ∈ N such that ∀b, c ∈ N, ab 6 = c^3.

6. (a) ∃a, b ∈ S such that ∀n ∈ N, na ≤ b. (b) (i) No. (ii) Yes. (iii) No. (iv) Yes.

1.6 Implications

4. (a) S ⇒ U is false if and only if the stock market goes up but unemployment does not go up. (b) The converse of S ⇒ U is false if and only if unemployment goes up but the stock market does not go up. (c) The contrapositive of ∼ I ⇒ U is false if and only if unemployment does not go up and interest rates do not go down.
5. (a) x^2 = 4 only if x = 2. False. Converse: x^2 = 4 if x = 2. True. (b) If 2x ≤ x, then x^2 > 0. False (consider x = 0). Converse: If x^2 > 0, then 2x ≤ x. False. (c) If 2 is a prime number, then 2^2 is a prime number. False. Converse: If 2^2 is a prime number, then 2 is a prime number. True. (d) If x is an integer then

x is an integer. False. Converse: If

x is an integer, then x is an integer. True. (e) If every line has a y-intercept, then every line contains infinitely many points. True. Converse: If every line contains infinitely many points then every line has a y-intercept. False. (f) A line has undefined slope only if it is vertical. True. Converse: A line has undefined slope if it is vertical. True. (g) x = −5 only if x^2 − 25 = 0. True. Converse: x = −5 if x^2 − 25 = 0. False. (h) x^2 is positive only if x is positive. (Assume x ∈ R.) False. Converse: x^2 is positive if x is positive. True.

7. (a) “m is a multiple of 8” is sufficient but not necessary for m 2 ∈ Z. (b) “m ∈ Z” is necessary but not sufficient for m 2 ∈ Z.

1.6. IMPLICATIONS 7

(c) “m is a multiple of 2” is a necessary and sufficient condition on m for m 2 ∈ Z.

10. (a) P Q P ⇒ Q ∼ P ⇒ Q ∼ P ∨ Q P ∨ Q ∼ (P ⇒∼ Q) P ∧ Q T T T T T T T T T F F T F T F F F T T T T T F F F F T F T F F F (b) (iii) and (v): (P ⇒ Q) = (∼ P ∨ Q); (iv) and (vi): (∼ P ⇒ Q) = (P ∨ Q); (vii) and (viii): ∼ (P ⇒∼ Q) = (P ∧ Q).
11. (c) P Q S P ⇒ S Q ⇒ S (P ⇒ S)∨(Q ⇒ S) P ∨ Q (P ∨ Q) ⇒ S T T T T T T T T T T F F F F T F T F T T T T T T T F F F T T T F F T T T T T T T F T F T F T T F F F T T T T F T F F F T T T F T Because the columns corresponding to (P ⇒ S)∨(Q ⇒ S) and (P ∨ Q) ⇒ S are not identical, the statements are not equivalent.

Chapter 2

Proofs

2.1 Proof Techniques

2. Partition the set L of lattice points inside a given circle C into blocks B(x, y) where, for (x, y) ∈ L, B(x, y) contains (x, y) and (x, y) rotated around the origin by 90 ◦, 180 ◦, and 270 ◦. Now for each (x, y) ∈ L{(0, 0)}, the block B(x, y) contains 4 elements, and B(0, 0) contains one element. Since |L| is the sum of |B(x, y)| taken over all distinct blocks, we have |L| = 4 k + 1 where k + 1 is the number of blocks in this partition.
3. (b) Suppose x, y ≥ 0 are given. Then

0 ≤ bxc ≤ x and 0 ≤ byc ≤ y. Multiplying these equations gives bxcbyc ≤ xy,

so bxcbyc is an integer which is ≤ xy. By definition, bxyc is the largest integer which is ≤ xy, so bxcbyc ≤ bxyc. This argument holds for any x, y ∈ [0, 1 ).

6. Suppose p(x) = ax^2 + bx + c and p(1) = p(−1). The equation p(1) = p(−1) becomes a + b + c = a − b + c, and subtracting (a + c) from both sides gives b = −b, so b = 0. Thus, p(x) = ax^2 + c, so p(2) = 22 a + c = (−2)^2 a + c = p(−2). Conversely, Suppose p(x) = ax^2 +bx+c and p(2) = p(−2). The equation p(2) = p(−2) becomes 4 a + 2 b + c = 4 a − 2 b + c, and again we find that b = 0. Thus, p(x) = ax^2 + c, so p(1) = 12 a + c = (−1)^2 a + c = p(−1).
7. Note that n^3 + n = n(n^2 + 1). Since n and n^2 have the same parity, n and n^2 + 1 have opposite parities (that is, one is even and the other is odd). Since any multiple of an even number is even, it follows that n(n^2 + 1) = n^3 + n is even.
8. (a) Suppose a is a multiple of 3, say a = 3 n where n ∈ Z. Then a = (n−1)+n+(n+1), the sum of three consecutive integers. Conversely, suppose a = k+(k+1)+(k+2)

9

10 CHAPTER 2. PROOFS

is the sum of three consecutive integers. Then a = 3(k + 1), so a is a multiple of

(b) No. The sum 1 + 2 + 3 + 4 = 10 of four consecutive integers is not a multiple of 4, and 8, a multiple of 4, cannot be written as a sum of four consecutive integers: 0 + 1 + 2 + 3 = 6 < 8 < 10 = 1 + 2 + 3 + 4. (c) The sum of k consecutive integers has form (n + 1) + (n + 2) + · · · + (n + k) = kn + (1 + 2 + · · · + k). Since kn is a multiple of k, the sum will be a multiple of k if and only if 1 + 2 + · · · + k is a multiple of k. Thus, a is a multiple of k if and only if a may be written as a sum of k consecutive integers is true if and only if 1 + 2 + · · · + k is a multiple of k. We will see later that 1 + 2 + · · · + k is the kth^ triangular number and is given by the formula k(k 2 +1). Thus, 1 + 2 + · · · + k is a multiple of k if and only if k+1 2 ∈ Z, that is, if and only if k is odd.

12. Direct proof: For any k ∈ Z, xk = π 2 + 2πk is a solution to sin x = 1, so sin x = 1 has infinitely many solutions. Indirect proof: Suppose to the contrary that sin x = 1 has only finitely many solutions. The solution set is nonempty since sin( π 2 ) = 1. Let xm be the largest member of the solution set. Now sin(xm + 2π) = sin xm = 1, so xm + 2π is an element of the solution set which is larger than xm, contrary to the choice of xm as the largest solution. Assuming that there were only finitely many solutions gave a contradiction, so there must be infinitely many solutions.
13. Suppose k and l are distinct lines that intersect. Suppose A and B are points of intersection of k and l. If A 6 = B, then the two distinct points A and B determine a unique line, contrary to k and l being distinct lines through A and B. Thus, A = B. That is, k and l intersect in a unique point.
14. Moving a knight out and back to his original position on the first move effectively gave the other player the first move in the double move chess game. In initial double move chess, moving a knight out and back does not exchange the roles of first player and second player, since the first player was playing initial double move chess and the second player is left with a different game—one in which only one player gets an initial double move.

2.2 Mathematical Induction

2. (b) For n = 1, the statement is 1^3 = 1

(^2) (1+1) 2 4 , which is true. Suppose the statement holds for n = k, that is, suppose 1^3 + 2^3 + · · · + k^3 = k

(^2) (k+1) 2 4.^ We wish to show that the statement holds for n = k + 1, that is, we wish to show that 13 + 2^3 + · · · + k^3 + (k + 1)^3 = (k+1)

(^2) (k+2) 2

4. Adding (k^ + 1)

(^3) to both sides of the

12 CHAPTER 2. PROOFS

9. Suppose α > − 1 , α 6 = 0. We wish to show (1 + α)n^ > 1 + nα for n ≥ 2. Observe that α > −1 guarantees that the powers (1 + α)n^ are all positive. For n = 2, the statement is (1 + α)^2 > 1 + 2α, which is true since (1 + α)^2 = a + 2α + α^2 , and α^2 > 0 for α 6 = 0. Now suppose (1 + α)k^ > 1 + kα for n = k ≥ 2. We wish to show (1 + α)k+1^ > 1 + (k + 1)α. But

(1 + α)k+1^ = (1 + α)k(1 + α)

(1 + kα)(1 + α) (Induction hypothesis) = 1 + kα + α + kα^2 = 1 + (k + 1)α + kα^2 1 + (k + 1)α since kα^2 > 0 for α 6 = 0.

Now the statement holds for n = 2 and for n = k + 1 whenever it holds for n = k, so by mathematical induction, (1 + α)n^ > 1 + nα for every natural number n ≥ 2.

14. (b) Any combination of m 4-cent stamps and n 10-cent stamps gives (4m+10n)-cents postage. Since 4m + 10n is always even, 4-cent and 10-cent stamps can never be combined to give any odd amount.
15. Assuming that all horses of any n-element set have the same color, the induction step argues that all horses of an n + 1-element set H = {h 1 ,... , hn+1} have the same color since all horses of the n-element set H \ {h 1 } have the came color C, all horses of the n-element set H \ {hn+1} have the came color D, and C = D since H \ {h 1 } ∩ H \ {hn+1} 6 = ∅. However, H \ {h 1 } ∩ H \ {hn+1} = ∅ if n = 1. Thus, the first induction step (if true for n = 1, then true for n = 2) fails.

2.3 The Pigeonhole Principle

3. (a) 24. Worst case: First 8 nickels, 10 dimes, 3 quarters, then 3 pennies. (b) 9. The pigeonhole principle applies. Worst case: 2 of each of the 4 types, then one more. (c) All 33. Worst case: the last coin drawn is a quarter. (d) 25. Worst case: First 12 pennies, 8 nickels, 3 quarters, then 2 dimes. (e) 16. Worst case: First 12 pennies, then 4 more coins to get a second pair.
4. Given 5 lattice points (ai, bi) i = 1, 2 , 3 , 4 , 5, the pigeonhole principle implies that at least three of the integers a 1 ,... , a 5 have the same parity. Without loss of generality, assume a 1 , a 2 , and a 3 have the same parity. Now by the pigeonhole principle, at least two of the points b 1 , b 2 , b 3 must have the same parity. Without loss of generality, assume b 1 and b 2 have the same parity. Now the midpoint of the segment from (a 1 , b 1 ) to (a 2 , b 2 ) is ( a^1 + 2 a^2 , b^1 + 2 b^2 ), and this is a lattice point since a 1 and a 2 have the same parity and b 1 and b 2 have the same parity.
5. In the worst case, each of the six cameras would receive 23 exposures before the next exposure would give one camera 24 exposures. Thus, 23 × 6 + 1 = 139 exposures are needed to guarantee that one camera has 24 exposures.

2.3. THE PIGEONHOLE PRINCIPLE 13

9. (a) Partition the balls into “50-sum” sets { 1 , 49 }, { 2 , 48 },... , { 24 , 26 } and two un- paired singleton { 25 } and { 50 }. This gives 26 sets. If balls are drawn and assigned to the appropriate set, to insure that one set receives two balls, we must draw 27 balls.

Chapter 3

Number Theory

3.1 Divisibility

4. If d | n^2 , then it need not be true that d | n. For example, 4 | 62 but 4 6 | 6.
5. (a) Suppose a|b. Then b = na for some n ∈ Z. If c ∈ Da, then c|a, so a = mc for some m ∈ Z, so b = na = n(mc) = (nm)c where nm ∈ Z, so c|b, that is, c ∈ Db. Thus, Da ⊆ Db. Conversely, suppose Da ⊆ Db. Now a ∈ Da so a ∈ Db, and thus a|b. (b) Suppose a|b. Then b = na for some n ∈ Z. If c ∈ Mb, then c = mb for some m ∈ Z, so c = mb = m(na) = (mn)a where mn ∈ Z, so a|c. Thus, c ∈ Ma. This shows that Mb ⊆ Ma. Conversely, suppose Mb ⊆ Ma. Since b ∈ Mb, we have b ∈ Ma, so a|b. (c) (^) Da = Db ⇐⇒ Da ⊆ Db and Db ⊆ Da ⇐⇒ a|b and b|a by part (a) ⇐⇒ a = ±b (Theorem 3.1.7) ⇐⇒ |a| = |b|

(d) (^) Ma = Mb ⇐⇒ Ma ⊆ Mb and Mb ⊆ Ma ⇐⇒ b|a and a|b by part (b) ⇐⇒ a = ±b (Theorem 3.1.7) ⇐⇒ |a| = |b|

9. (a) a = 73, b = 25: q = 0 , r = 25. (b) a = 25, b = 73: q = 2 , r = 23. (c) a = −73, b = −25: q = 1 , r = 48. (d) a = −25, b = −73: q = 3 , r = 2. (e) a = 79, b = −17: q = − 1 , r = 62.

15

16 CHAPTER 3. NUMBER THEORY

(f) a = −17, b = 79: q = − 4 , r = 11. (g) a = −37, b = 13: q = 0, r = 13. (h) a = 13, b = −37: q = − 3 , r = 2.

17. (a) If a and b leave a remainder of 2 when divided by 7, then a = 7q +2 and b = 7s+ for some integers q and s, and thus a − b = 7q + 2 − (7s + 2) = 7(q − s) where q − s ∈ Z, so 7|(a − b). (b) If a = 7q + 2, then 10a = 70q + 20 = 70q + 14 + 6 = 7(10q + 2) + 6. Thus, by uniqueness of the quotient and remainder when 10a is divided by 7, we have a quotient of 10q + 2 and a remainder of 6.
18. We will show 4|(13n^ −1) ∀n ∈ N by mathematical induction. If n = 1, then 4|(13^1 −1) since 4|12. Suppose 4|(13k^ − 1). We wish to show that 4|(13k+1^ − 1). Now

13 k+1^ − 1 = 13(13k^ − 1 + 1) − 1 = 13(13k^ − 1) + 13 − 1 = 13(13k^ − 1) + 12

Now 4|(13k^ − 1) by the induction hypothesis and 4|12, so 4|(13k+1^ − 1). Now by mathematical induction, 4|(13n^ − 1) ∀n ∈ N. Alternatively, the result of Exercise 23 shows that 12|(13n^ − 1) ∀n ∈ N, and since 4 |12, we have 4|(13n^ − 1) ∀n ∈ N.

3.2 The Euclidean Algorithm

5. If a, b ∈ Z and z and w are linear combinations of a and b using integer coefficients, say z = ja + kb and w = la + mb (j, k, l, m ∈ Z) then a linear combination of z and w with integer coefficients has form sz + tw (s, t ∈ Z). Now

sz + tw = s(ja + kb) + t(la + mb) = (sj + tl)a + (ks + tm)b

is a linear combination of a and b with integer coefficients sj + tl and ks + tm.

7. gcd(15, 39) = 3, so 3 should divide 15s + 39t for any s, t ∈ Z. The bill should be of form 15s + 39t (s, t ∈ N ∪ { 0 }), so the bill should be a multiple of 3 cents. It is not.
8. Suppose a, b, q, r ∈ Z \ { 0 } and a = bq + r.

(a) gcd(a, b) = gcd(b, r) is true. Proof: If d is any common divisor of a and b, then d is a divisor of a − bq = r. Thus, any common divisor of a and b is a common divisor of b and r. Conversely, any common divisor of b and r must also divide b and bq + r = a, and therefore must be a common divisor of a and b. This shows that the common divisors of a and b are precisely the common divisors of b and r, so gcd(a, b) = gcd(b, r). (c) In general, gcd(q, r) does not divide b. For example, with a = 45, b = 7, q = 6, and r = 3, we have gcd(q, r) = 3 but 3 does not divide 7.