Principles of Mathematics in Operations Research

266 Solutions

Problems of Chapter 9

9.1 Let a = inf A Then, Vx e A, a < x «=> -x < -a. Hence, (-A) in

bounded above. Also, —a is an upper bound of (—A). So,

sup(-.A) < -a <=> - sup(--A) > a = inf A.

Conversely, let /? = sup(—.4). Then, Vx G A, — x < (3 •£> x > —/?. Hence,

—/? is a lower bound of A. So,

inf A > -/J — -sup(-.A).

Thus, inf A = -sup(-A).

9.2

a) If m = 0, (bm)l'n = (6^0 )^1 /™ = l^1 /" = 1 (see (c)).

(b---b)1/n b1/n---b1/n

If m > 0, (ft"^1 )^1 /" = < ^-—' = ' » ' = (b^n)m.

m times m times

If m < 0, let m' = -m > 0. Then,

(vm\l/n _ /i.-m'\l/n _ / 1 \l/n _ 1 _ 1 _ 1 _ (ul/n\m

\U I — \U I — Vhm' ) — (hm'\l/n ~ /U/ti\m' — (hl/n)-m ~ (" ) •

b) If TO — 0, all terms are 1.
n
bm •••bm , ".
If m > 0, (bm)n = v ' = b...b...b...b =6mn.

m m
Similarly, (bm)n = bmn.
If TO < 0, let TO' = -TO > 0. Then,
(i.m\n (u—m'\n / 1 \n 1 1 1 umn
\U ) — \U ) — ~g^7) — (fcm')n — bm'n ~ (,-(mn) ~ "

c) Let l^1 /™ = x where x >- 0. Then, xn — 1. Also, l^1 /™ = 1. Since the positive
nth root of 1 is unique, we get x = 1.

d) Let 6^1 /™? = a, and (6^1 /")!/? = p where a,/3 X 0.
Then, b = anq and 6^1 /" = /?« => 6 = (/3^9 )n = /3"n = £"^9 = anq. Since the
positive nqth root of 6 is unique, we get a = /3, i.e. bl/nq = (6^1 /™)^1 /^9. Similarly,
frl/ng /bl/q\l/n

e) If p = 0, then W+q = b0+q = W = b°W = WW. Similarly, if q = 0,
6?+9 = bPW. So assume p ^ 0, g ^ 0.