Solution manual for mathematical proofs a transition to advanced mathematics 3rd edition by chartrand

Exercises for Chapter 1Exercises for Section 1.1: Describing a Set 1.1 Only d and e are sets... In each case, not every two subsets are disjoint.. Furthermore, there is no set I such sub

Exercises for Chapter 1

Exercises for Section 1.1: Describing a Set

1.1 Only (d) and (e) are sets

− x = 0} = {0, 1}

(b) 5/2, 7/2, 4

Exercises for Section 1.2: Subsets

1.13 See Figure 1

6 2

7

U

B 8

5 4 3 1

A

r r

r r r r

r

r

.

.

Figure 1: Answer for Exercise 1.13

P(A) = {∅, {0}, {∅}, {{0}}, {0, ∅}, {0, {0}}, {∅, {0}}, A}

|A| = 2 and |C| = 5

(c) No For A =∅ and B = {1}, |P(A)| = 1 and |P(B)| = 2.

Exercises for Section 1.3: Set Operations

B 9

7 8 6

5 4 3 2

1 A

r r r

r r r r r r

.

.

Figure 2: Answer for Exercise 1.23

1.26 (a) and (b) are the same, as are (c) and (d)

1.28 See Figures 4(a) and 4(b)

7 8

6 3 4 1

U

5

r r

.

.

r r r Figure 3: Answer for Exercise 1.27

.

.

A C B A C B (C − B) ∪ A C ∩ (A − B) (a) (b)

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.

.

.

.

.

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Figure 4: Answers for Exercise 1.28

Exercises for Section 1.4: Indexed Collections of Sets

T

α∈ASα= S1∩ S3∩ S4={3}

U B

4 3 1 2

7

6 8 5

C

A

r r

r r

r

r

r r

.

.

.

.

.

Figure 5: Answer for Exercise 1.35 1.37 S X∈SX = A∪ B ∪ C = {0, 1, 2, , 5} andT X∈SX = A∩ B ∩ C = {2} 1.38 (a) S α∈SAα= A1∪ A2∪ A4={1} ∪ {4} ∪ {16} = {1, 4, 16} T α∈SAα= A1∩ A2∩ A4=∅ (b) S α∈SBα= B1∪ B2∪ B4= [0, 2]∪ [1, 3] ∪ [3, 5] = [0, 5] T α∈SBα= B1∩ B2∩ B4=∅ (c) S α∈SCα= C1∪ C2∪ C4= (1,∞) ∪ (2, ∞) ∪ (4, ∞) = (1, ∞) T α∈SCα= C1∩ C2∩ C4= (4,∞) 1.39 Since |A| = 26 and |Aα| = 3 for each α ∈ A, we need to have at least nine sets of cardinality 3 for their union to be A; that is, in order for S α∈SAα = A, we must have|S| ≥ 9 However, if we let S ={a, d, g, j, m, p, s, v, y}, then S α∈SAα= A Hence the smallest cardinality of a set S with S α∈SAα= A is 9 1.40 (a) S5 i=1A2 i= A2∪A4∪A6∪A8∪A10={1, 3}∪{3, 5}∪{5, 7}∪{7, 9}∪{9, 11} = {1, 3, 5, , 11}

i=1(Ai∩ Ai+1) =S5

i=1(A2 i−1∩ A2 i+1) =S5

r∈R +Ar=S

T

r∈R +Ar=T

case, we must have either two 5-element subsets of A or two 3-element subsets of A and a 4-element subset of A In each case, not every two subsets are disjoint Furthermore, there is no set I such

subset of A (which are not disjoint) or three 3-element subsets of A No 3-element subset of A

1.45 n∈NAn= n∈N(−n, 2−n) = (−1, 2);

T

n∈NAn=T

n∈N(−1

n, 2−1

Exercises for Section 1.5: Partitions of Sets

1.56 (a) Suppose that a collection S of subsets of A satisfies Definition 1 Then every subset is

to more than one subset, then the subsets in S would not be pairwise disjoint So the collec-tion satisfies Definicollec-tion 2

(b) Suppose that a collection S of subsets of A satisfies Definition 2 Then every subset is nonempty

Definition 2 So condition (2) in Definition 3 is satisfied Since every element of A belongs

to a (unique) subset in S, condition (3) in Definition 3 is satisfied Thus Definition 3 itself is satisfied

(c) Suppose that a collection S of subsets of A satisfies Definition 3 By condition (1) in Defi-nition 3, every subset is nonempty By condition (2), the subsets are pairwise disjoint By condition (3), every element of A belongs to a subset in S So Definition 1 is satisfied

Exercises for Section 1.6: Cartesian Products of Sets

P(A) × P(B) = {(∅, ∅), (∅, B), ({1}, ∅), ({1}, B), ({2}, ∅), ({2}, B), (A, ∅), (A, B)}

See Figure 6

P(A × B) = {∅, {(1, 1)}, {(2, 1)}, A × B}

x = 3, y = 2 and y = 6

(0, −3)

(0, 3)

(3, 0)

(2, −1)

(−1, 2) (1, 2)

(−3, 0)

(−2, −1)

(−1, −2) (1, −2)

.

r

r

r

r

r

r

r r

r

r

r

.

Figure 6: Answer for Exercise 1.63

Additional Exercises for Chapter 1

1.73 Let S ={{1}, {2}, {3, 4}, A} and let B = {3, 4}.

S

r∈IAr=∅

S

S

r∈ICr=∅

is impossible

1.83 (a) S ={(−3, 4), (0, 5), (3, 4), (4, 3)}.

2