368 ANSWERS AND SOLUTIONS TO SELECTED EXERCISES
- (d) Proof Let A be an arbitrary subset of U. Then, A A U = (A - U) u
(U - A) = (25 u A' = A'. - (a) (ii) Proof Let A, B, and C be arbitrary sets. Then (A u B) - C =
(A u B) n C' = (A n C') u (B n C') = (A - C) u (B - C). - (a) Proof Assume n, k E N with k I n. Then (,lf ,) + C;) =
[n!/((k - l)!(n - k - I)!)] + [n!/k!(n - k)!] =
[kn! + (n - k + l)n!]/[k!(n - k + I)!] = n!(k + n - k + l)/k!(n - k + I)! =
(n + l)n!/k!(n + 1 - k)! = (n + l)!/k!(n + 1 - k)! = (n + l)!/k!(n + 1 - k)! =
("tl). - (b) Proof Assume f and g are odd functions. To show f + g is odd, we
must show that (f + g)(-x) = -( f + g)(x) for all x E R. So let x E R be
arbitrarily chosen. Then (f + g)( - x) = f ( - x) + g( - x) = - f (x) +
( - (g(x))) = -( f (x) + g(x)), as desired. Similarly, (f - g)( - x) = f ( - x) -
g(- x) = - f (x) - ( - g(x)) = - (f(x) - g(x)) = - (( f - g)(x)), as desired. Also,
(f O g)( - 4 = f (d - x)) = f ( - dx)) = -f (g(x)) = - (f O g)(x). Finally,
(fg)( - x) = f ( - x)g( - 4 = ( -f (XI)( - dx)) = f (x)g(x) = (fg)(x), so that fg is
even, as claimed. - (b) Proof Rewrite 60 + 14x - 2x2 in the form -2(x2 - 7x + y) +
(60 + 9) = -2(x - $I2 + (y). Clearly this expression assumes its maximum
value of at x = 3. 0 - (d) We show the statement is false by displaying a counterexample. Let
U={1,2,3,4,5),A={1,2,5),B={3,4),C={1,2,4). ThenAn
(B u C) = {1,2), whereas (A n B) u C = (1,2,4). - (a) Proof Given A = (aij), ,, and B = (bij),. ., to prove A = B, we need
only show that aij = bij for all i, j satisfying 1 I i I m, 1 I j 5 n. So
choose arbitrary integers h and k satisfying 1 5 h I m, 1 I k I n. Now
A' = (cji), xm and B' = (dji),,. ,, where cji = aij and dji = bij for all i, j with
1 I i 5 m, 1 I j I n. Since A' = B', we know that ckh = dkh. Hence
ahk = ckh = dkh = bhk, and we have a, = b,, as desired.
(e) Proof Let A and B be arbitrary symmetric square matrices, so that
A = A' and B = B'. To prove A + B is symmetric, we need only show
(A + B)'= A + B. But (A + B)'= At+ Bt= A + B.
Article 5.2
- (a) Proof Let A and B be sets such that A c B. To prove @(A) G @(B),
let X be an arbitrary element of B(A). To prove X E S(B), we must show
X c B. Now since X E @(A), we know X c A. Since X E A and A E B, then
X c B (by Example 7, Article 4.1), our desired conclusion. - (b) Apply the criterion in (a) with h = k = 1, f(x) = 1 + (l/(x - 1)). Assume
x # 1, x # 0. Then f(-x + 1) = 1 + (l/((-x + 1) - 1)) = 1 - (llx) =
2 - [l + ((l/(x + 1) - I))] = 2 - f(x + 1). - (a) Proof To prove [O,l] is convex, let x, y E [0, 11 and let t be a real
number satisfying 0 I; t 5 1. Then tx 2 0, 1 - t 2 0, and y 2 0, so that
tx + (1 - t)y 2 0. Also, since 0 I x I. 1, then tx I t and since 0 I y < 1, then
(1 - t)y I 1 - t. Hence tx + (1 - t)y I t + (1 - t) = 1. Thus
0 5 tx + (1 - t)y I; 1, as desired.