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....
View
Full Document
 Summer '10
 any
 Graph Theory, Region, DVD region code, i. ii. iii.

Click to edit the document details