Discrete Mathematics with Graph Theory (3rd Edition) 52

# Discrete Mathematics with Graph Theory (3rd Edition) 52 - A...

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50 Solutions to Review Exercises Chapter 2 Review 1. Since A = {I, 2, 3, 4,5, 6} and B = {3, 4, 5, 6, 7}, we have A EB B = {I, 2, 7} and (A EB B) , C = {1,7}. 2. (a) A = {-I, 0,1, 2}, B ={ -5, -3, -1, I}, C = {-1, -~, 0, ±1, ±2, ±i, ±!}; (b) An B = {±1}, so (A n B) x B = {(I, -5), (1, -3), (1, -1), (1, 1), (-1, -5), (-1, -3), (-1, -1), (-1, I)}; (c) B, C = {-5, -3}; (d) A EB C = {-1, -~, -2, ±i, ±i}· 3. (a) True. (----+) Suppose A n B = A and let a E A. Then a E An B, so a E B. Thus A ~ B. (f-- ) Suppose A ~ B. To prove A n B = A, we prove each of the two sets, A n B and A, is a subset of the other. Let x E An B. By definition of n, x is in both A and B, in particular, x E A. Thus A n B ~ A. Conversely, let x E A. Since A ~ B, x E B. Since x is in both A and B, x E An B. Thus A ~ An B. (b) This is false. If A = 0, (A n B) U C = C while A n (B U C) = 0, so any C =F 0, any B, and A = 0 provides a counterexample. (c) False. Take A = B = 0. 4. Let bE B and let a be any element
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Unformatted text preview: A. Then (a, b) E A x B, so (a, b) E A x C. Thus b E C. This shows tht B ~ C and a similar argument shows C ~ B, so B = C. 5. (a) Region 2: (A n C) , IJ Region 3: An B n C Region 4: (A n B) , C Region 5: (B n C) , A Region 6: B , (A U C) Region 7: C'- (A U B) (b) Region 2,3,4,5,7 is (A n B) U C; region 2,3,4 is A n (B U C) (c) B, (C, A) consists of regions 3, 4, and 6. (B' C) ,A consists of region 6. C 6. (a) A B (b) i. (A U B) n C = {2, 3, 8, 9} ii. A, (B '-C) = {2, 3, 7, 9} iii. AEB B = {2,6, 7,8,9} iv. (A, B) x (B n C) = {(2, 3), (2,8), (7,3), (7,8), (9,3), (9, 8)} 7. P(A) = {0, A}, so P(P(A)) = {0, {0}, {A}, {0, A}}. 8. (a) Take A = B = C = {3}. Then B ,C = 0 soAEB (B ,C) = A.On the other hand, AEBB = 0 so (A EB B) , C = 0 =F A. (b) Let (a, b) E A x B. Then a E Aandb E B. Since a E AandA ~ C, a E C. Sinceb E Band B ~ D, bED. So (a, b) E C x D. Hence A x B ~ C x D. 9. Take B = C = 0, A = {I} = D. Then A x B = 0 = C x D, so A x B ~ C x D. On the other hand,A g C....
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