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Discrete Mathematics Assignment # 5: Solutions Summer 03 Section 5.1 9. U = R A = {x R | -2 x 1} B = {x R | -1 < x < 3} a) b) c) d) e) f) g) h) AB = {x R | -2 x <3} AB = {x R | -1 < x 1} Ac = {x R | x<-2 or x > 1} Bc = {x R | x-1 or x 3} AcBc = {x R | x < -2 or x 3} AcBc = Ac = {x R | x-1 or x > 1} (AB)c = {x R | x-1 or x > 1} (AB)c = {x R | x < -2 or x 3} -2 -2 -1 -1 0 0 1 1 2 2 3 3 4 4 10. R Q Z Z a) Z+ Q b) R- Q c) Q Z d) Z-Z+ = Z e) QR = Q f) QZ = Q g) Z+R = Z+ h) ZQ = Z TRUE FALSE ( -2 R- but -2Q) FALSE (½ Q but ½ Z) FALSE (0Z but 0 Z- and 0 Z+) TRUE TRUE TRUE FALSE (½ Z, ZQ = Q) Discrete Mathematics Summer 03 15.a) AB d)A – ( BC) b)BC e)(AB)c c)Ac f)AcBc 17. A = {x, y, z, w}, B = {a, b} b) BxA = {(a,x), (a,y), (a,z), (a,w), (b,x), (b,y), (b,z), (b,w)} d) BxB = {(a,a), (a,b), (b,a), (b,b)} Discrete Mathematics Summer 03 Section 5.2 12. For all sets A, B, C; (A-B)(C-B) = A – (BUC) FALSE. Using Venn Diagrams: (A-B)(C-B) =? A – (BUC) Clearly, they are not equal. Using a concrete example: Let A = {a, e, f, g}, B ={b, d, e, g} and C={c, d, f, g} Then A-B = {a, f}, C-B = {f, c}, and therefore (A-B) (C-B) = {f}. On the other hand BUC = {b, e, g, f, d, c}, and A-BUC = {a} Clearly {a} {f} therefore (A-B) (C-B) A-BUC 19. Suppose AB. Claim: BcAc Proof: AB iff x, xA xB iff x, xB xA iff x, xBc xAc iff BcAc 33. For all sets A, B, C, show (A-B)U(B-A) = (AUB)-(AB) (A-B) (B-A) = (ABc) (BAc) = (AU(BAc)) (Bc (BAc)) by Thm. 5.2.2(10) by Thm 5.2.2(3) Discrete Mathematics Summer 03 = ((AB) (AAc)) ((Bc B) (Bc Ac)) by Thm. 5.2.2(3) c c = (AB) U U (B A ) by Thm 5.3.3(2b) c c = (AB) (B A ) by Thm. 5.2.2(4) = (AB) (BA)c by Thm. 5.2.2(7) = (AB) - (BA) by Thm. 5.2.2(10) = (AB) - (AB) by Thm. 5.2.2(1) 34. For all sets A,B,C, show (A-B)-C = A-(BUC) (A-B)-C = (A-B)Cc = (ABc) Cc = A (BcCc) = A (BUC)c = A – (BUC) by Thm. 5.2.2(10) by Thm. 5.2.2(10) by Thm. 5.2.2(2) by Thm. 5.2.2(7) by Thm 5.2.2(10) Section 5.3 14. For all sets A,B,C: if (BC)A then (A-B) (A-C) = . FALSE. Using Venn Diagrams: Consider the situation shown in the diagram. The shaded area shows (A-B)(A-C). Here we have that (BC)A, but the shaded area is not empty. Hence (BC)A does not imply that (A-B)(A-C). Using a concrete example: Let A = {d, b}, B = {c, d}, C={a,d}. (Note that BC = {d} A ) Then A-B = {b} and A-C = {b}.So (A-B)(A-C) = {b} . 19. For all sets A, Ax = Discrete Mathematics Summer 03 Ax = {(a,b) | a A and b } = ; since there’s no b such that b . 33. Defn. Symmetric difference: AB = (A-B) U (B-A) c) A = A A = (A-) (-A) = (Ac) (Ac) = (AU) (Ac) = (AU) = A = A d) A Ac = U AAc = (A-Ac) (Ac-A) = (AA) (AcAc) = A Ac = U e) A A = A A = (A-A) (A-A) = (AAc) (AAc) = = by Definition of by Thm. 5.2.2(10) by Thm. 5.3.3(2) by Thm. 5.3.3(3) by Definition of by Thm. 5.2.2(10) by Thm. 5.2.2(6) by Thm. 5.3.3(2) by Definition of by Thm. 5.2.2(10) by Thm. 5.3.3(4) by Thm. 5.3.3(3) by Thm. 5.2.2(4) by Thm. 5.3.3(1) f) If AC = BC then A=B. Suppose, AC = BC I.e., that (A-C) + (C-A) = (B-C) + (C-B) Claim that A=B. Proof. We show that AB. The proof that BA follows the same reasoning. Choose xA. We must show that xB. (1) xC . Then x(A-C). Hence either x (B-C) or x (C-B) But x(C-B) (since xC) Hence x (B-C) . Hence xB. Discrete Mathematics (2) xC. Then x AC. Hence x BC (since AC = BC ) Hence x C-B. Hence xB. Summer 03

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