Bridge to Abstract Mathematics: Mathematical Proof and Structures

(Dana P.) #1
ANSWERS AND SOLUTIONS TO SELECTED EXERCISES 365

Article 3.2



  1. (d) All easy problems are solvable. (i) Some unsolvable problems are not
    mathematics problems.
    2- (4 (3x)(p(x) A -W) (0 WMx) -, (2-44 A -4w.
    3. (c) (i) F (ii) T (iii) T (iu) F (v) T (vi) F (d) (i) T
    (ii) T (iii) T (iv) F (v) T (vi) T.


9. (a) (Vx)(p(x) - q(x)) (b) P = Q (c) P = Q if and only if


(P' u Q) n (P u Q') = U, for any sets P and Q.



  1. (a) (b) (d) (e) p implies q; s and t are equivalent.


Article 3.3



  1. (c) All women are either not young or not athletes. (g) Some athletes are
    either not young or are men and no men are athletes.

  2. (4 (34 -2-44 A -- q(4) (4 (34 (PW A - dx)) v t -- 2-4~) A dx) 1.

  3. (4 (3mo) A - q(x) A - 44).

  4. (d) Prove that there exists a function f such that f has a relative maximum or
    minimum at x = 0, but f (0) does not equal zero.

  5. (b) O= x2 + 8x + 16 = (x + 4)2. Hence ifx2 + 8x + 16 = 0, then x = -4, so
    that there is at most one solution. Substituting -4 for x in x2 + 8x + 16 yields
    (-4)2 + 8(-4) + 16 = 0, so that, in fact, there is exactly one solution, x = -4,
    as desired.


Article 3.4



  1. (b) proposition (f) propositional function of two variables.

  2. (e) There exists x such that, for every y and z, fix, y, z).

  3. (c) Everyone has someone to whom they are a friend, who is not a friend in
    return. (g) Everyone is his own friend.

  4. (a) (v) - 8 divides 0 (true) (xi) For any integers m and n, if m divides n,
    then n divides m (false).

  5. (b) There exists a real number x such that xy = x for every real number y
    (true-let x = 0). (g) To every positive real number p, there corresponds
    at least one positive integer n such that lln < p.
    lo. (a) (3x)(Vy)(x I y) (false). (d) (Vx E Q')(Vy E Q')(3z E Q')(x < z < y) (true).


Article 3.5



  1. dx): x is a good citizen, q(x): x registers to vote, r(x): x does community service,
    $x); x is lazy; the argument has the form: P E Q, Q n R # 0, R n S = 0,
    therefore P n S' # @. This is invalid, as demonstrated by the substitutions
    U=(1,2,3,4,5),P=(l,2),Q=(l,2,3),R=(3,4),S={l,2,5).

Free download pdf