Discrete Mathematics
Summer 03
Assignment # 5: Solutions
Section 5.1
9. U = R
A = {x  R | -2  x  1}
-2
-1
0
1
2
3
4
-2
-1
0
1
2
3
4
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}
10.
R
Q
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
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)
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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
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)
d) A Ac = U
AAc = (A-Ac)  (Ac-A)
= (AA)  (AcAc)
=
A  Ac
=
U
by Definition of 
by Thm. 5.2.2(10)
by Thm. 5.2.2(6)
by Thm. 5.3.3(2)
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)
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.
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(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