TEST BANK FOR A Discrete Transition to Advanced Mathematics By Bettina Richmond and Thomas Richmond

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}}
5. (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 2n subsets.
1
2 CHAPTER 1. SETS AND LOGIC
8. (a) 3, 4, 5, and 7: |S3| = |{t, h, r, e}| = 4 = |S4| = |{f, o, u, r}| = |S5| = |{f, i, v, e}| =
|S7| = |{s, e, v, n}|.
(b) S21 = S22 or S2002 = S2000, for example.
(c) a ∈ S1000 and a 6∈ Sk for k = 1, 2, . . . , 999.
(d) (i) True (ii) True (iii) True (iv) False (v) False (vi) True (vii) True
(viii) False (ix) True (x) True (xi) True: {n, i, e} = S9 ∈ S. (xii) True
(xiii) False (xiv) True (xv) True (xvi) False
9. (a) D1 = ∅,D2 = {2},D10 = {2, 5},D20 = {2, 5}
(b) (i) True (ii) False (iii) False (iv) True (v) True (vi) False (vii) True
(viii) False (ix) True (x) True (xi) False (xii) True
(c) |D10| = |{2, 5}| = 2; |D19| = |{19}| = 1.
(d) Observe that D2 = D4 = D8 = D16, D6 = D12 = D18, D3 = D9, D10 = D20.
Thus |D| = |{D1,D2, . . . ,D20}| = |{D1,D2,D3,D5, D6,D7,D10,D11,D13,D14,
D15,D17,D19}| = 13.
10. For example, let S1 = S2 = S3 = {1, 2, 3}, S4 = {4}, and S5 = {5}. Now S =
{Sk}5
k=1 = {{1, 2, 3}, {4}, {5}}, so |S| = 3.
1.2 Set Operations
1. (a) S ∩ T = {1, 3, 5}
(b) S ∪ T = {1, 2, 3, 4, 5, 7, 9}
(c) S ∩ V = {3, 9}
(d) S ∪ V = {1, 3, 5, 6, 7, 9}
(e) (T ∩ V ) ∪ S = {3} ∪ S = S = {1, 3, 5, 7, 9}
(f) T ∩ (V ∪ S) = T ∩ {1, 3, 5, 6, 7, 9} = {1, 3, 5}.
(g) V × T = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (9, 1),
(9, 2), (9, 3), (9, 4), (9, 5)}
(h) U × (T ∩ S) = {(3, 1), (3, 3), (3, 5), (6, 1), (6, 3), (6, 5), (9, 1), (9, 3), (9, 5)}.
2. (a) A ∩ D = {A♦}; cardinality 1
(c) A ∩ (S ∪ D) = {A♠,A♦}; cardinality 2
(e) (A ∩ S) ∪ (K ∩ D) = {A♠,K♦}; cardinality 2
(g) K ∩ Sc = {K♣,K♦,K♥}; cardinality 3
(i) (A ∪ K)c ∩ S = {2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠,Q♠}; cardinality 11
(n) K \ S = {K♥,K♣,K♦}; cardinality 3