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Assignment # 3 Solutions

By Melissa Reed,2014-05-07 10:38
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Assignment # 3 Solutions

Discrete Mathematics Summer 03

    Assignment # 5: Solutions

Section 5.1

9. U = R -2 -1 0 1 2 3 4

     A = {x ? R | -2 ? x ? 1}

    -2 -1 0 1 2 3 4

     B = {x ? R | -1 < x < 3}

a) A?B = {x ? R | -2 ? x <3}

    b) A?B = {x ? R | -1 < x ? 1} cc) A = {x ? R | x<-2 or x > 1} cd) B = {x ? R | x?-1 or x ? 3} cce) A?B = {x ? R | x < -2 or x ?3} cccf) A?B = A = {x ? R | x?-1 or x > 1} cg) (A?B) = {x ? R | x?-1 or x > 1} ch) (A?B) = {x ? R | x < -2 or x ?3}

10. R

     Q

     Z Z

     + a) Z ? Q TRUE -- b) R ? Q FALSE ( -?2 ? R but -?2?Q)

     c) Q ? Z FALSE (? ? Q but ? ? Z) -+-+ d) Z?Z = Z FALSE (0?Z but 0? Z and 0? Z)

     e) Q?R = Q TRUE

     f) Q?Z = Q TRUE ++ g) Z?R = Z TRUE

     h) Z?Q = Z FALSE (? ? Z, Z?Q = Q)

Discrete Mathematics Summer 03

    15.a) A?B d)A ( B?C)

     c b)B?C e)(A?B)

     c ccc)A f)A?B

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. cc Claim: B?A

     Proof:

     A?B iff

     ?x, x?A ? x?B iff

     ?x, x?B ? x?A iff cc?x, x?B ? x?A iff cc B?A

    33. For all sets A, B, C, show (A-B)U(B-A) = (AUB)-(A?B)

     cc (A-B) ? (B-A) = (A?B) ? (B?A) by Thm. 5.2.2(10) ccc = (AU(B?A)) ?(B? (B?A)) by Thm 5.2.2(3)

Discrete Mathematics Summer 03

    cccc)) ? ((B? B) ? (B? A)) by Thm. 5.2.2(3) = ((A?B) ? (A?Acc = (A?B) ? U ? U? (B? A) by Thm 5.3.3(2b) cc = (A?B) ? (B? A) by Thm. 5.2.2(4) c = (A?B) ? (B?A) 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)

     c (A-B)-C = (A-B)?C by Thm. 5.2.2(10) cc = (A?B) ?C by Thm. 5.2.2(10) cc = A ? (B?C) by Thm. 5.2.2(2) c = A ? (BUC) by Thm. 5.2.2(7)

     = A (BUC) 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) by Definition of ? cc) ? (??A) by Thm. 5.2.2(10) = (A??c = (A?U) ? (??A) by Thm. 5.3.3(4)

     = (A?U) ? ? by Thm. 5.3.3(3)

     = A ? ? by Thm. 5.2.2(4)

     = A by Thm. 5.3.3(1)

     cd) A? A = U

     ccc A?A = (A-A) ? (A-A) by Definition of ? cc = (A?A) ? (A?A) by Thm. 5.2.2(10) c = A ? A by Thm. 5.2.2(6)

     = U by Thm. 5.3.3(2)

     e) A? A= ?

     A? A= (A-A) ? (A-A) by Definition of ? cc = (A?A) ? (A?A) 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.

Discrete Mathematics Summer 03

    (2) x?C.

    Then x ? A?C. Hence x ? B?C (since A?C = B?C )

    Hence x ? C-B. Hence x?B.

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