Exam (elaborations) TEST BANK FOR A Discrete Transition to Advanced Mathematics By Bettina Richmond and Thomas Richmond (Student's Solution Manual)

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