Assignment # 3 Solutions

							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) AB = {x  R | -2  x <3} AB = {x  R | -1 < x  1} Ac = {x  R | x<-2 or x > 1} Bc = {x  R | x-1 or x  3} AcBc = {x  R | x < -2 or x 3} AcBc = Ac = {x  R | x-1 or x > 1} (AB)c = {x  R | x-1 or x > 1} (AB)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) QR = Q f) QZ = Q g) Z+R = Z+ h) ZQ = Z

TRUE FALSE ( -2  R- but -2Q) FALSE (½  Q but ½  Z) FALSE (0Z but 0 Z- and 0 Z+) TRUE TRUE TRUE FALSE (½  Z, ZQ = Q)

Discrete Mathematics

Summer 03

15.a) AB

d)A – ( BC)

b)BC

e)(AB)c

c)Ac

f)AcBc

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 AB. Claim: BcAc Proof: AB iff x, xA  xB iff x, xB  xA iff x, xBc  xAc iff BcAc

33. For all sets A, B, C, show (A-B)U(B-A) = (AUB)-(AB) (A-B)  (B-A) = (ABc)  (BAc) = (AU(BAc)) (Bc (BAc)) by Thm. 5.2.2(10) by Thm 5.2.2(3)

Discrete Mathematics

Summer 03 = ((AB)  (AAc))  ((Bc B)  (Bc Ac)) by Thm. 5.2.2(3) c c = (AB)  U  U (B  A ) by Thm 5.3.3(2b) c c = (AB)  (B  A ) by Thm. 5.2.2(4) = (AB)  (BA)c by Thm. 5.2.2(7) = (AB) - (BA) by Thm. 5.2.2(10) = (AB) - (AB) 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 = (ABc) Cc = A  (BcCc) = 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 (BC)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 (BC)A, but the shaded area is not empty. Hence (BC)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 BC = {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: AB = (A-B) U (B-A) c) A = A A = (A-)  (-A) = (Ac)  (Ac) = (AU)  (Ac) = (AU)   = A = A d) A Ac = U AAc = (A-Ac)  (Ac-A) = (AA)  (AcAc) = A  Ac = U e) A A =  A A = (A-A)  (A-A) = (AAc)  (AAc) =  =  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 AC = BC then A=B. Suppose, AC = BC I.e., that (A-C) + (C-A) = (B-C) + (C-B) Claim that A=B. Proof. We show that AB. The proof that BA follows the same reasoning. Choose xA. We must show that xB. (1) xC . Then x(A-C). Hence either x  (B-C) or x  (C-B) But x(C-B) (since xC) Hence x (B-C) . Hence xB.

Discrete Mathematics (2) xC. Then x  AC. Hence x  BC (since AC = BC ) Hence x  C-B. Hence xB.

Summer 03