Principles of Mathematics in Operations Research

268 Solutions

Then, r + s G Q, r + s < x + y =$• br+s = bsbr G B(x + y) =^> brbs < bx+y.

Keep s fixed. 6r < ^-, V6r € -B(x). Thus, ^j- is an upper bound for B(x).

Hence, 6* = sup(£(x)) < ^ <£> bs < ^. Similarly, 6r < ^.

Now vary s. W € -B(y), &s < ^js~- Thus, ^j- is an upper bound for

B(y).

by = sup{B(y)) < h^- => bxby < bx+y.

bx

Claim: bxby > bx+y.

Proof: Suppose not. bxby < bx+y for some x, y 6 K. 3a G Q C K 3 6x 6 y <

a < 6a!+!/, by Archimedean property, b*,^ > 0 =>• a > 0. Since a < 6x+!/,

a is NOT an upper bound of B(x + y). So, 3br G B(x + y) 3 a > br. Let

t = £ > 1. If n > ffi (see problem 9.4-c)) 61/n < i = £ =>• a < fc^-^1 /" =

^r-i/n ^rue for rationals). Also r — 1/n < r < x + y. So, r — A — x < y.

3v€<Q>3r-^-x<v<y. Then, v < y and r - ^ -1> < x. Thus, 6W € -B(y)

and 6 r-i-" 6 B(x). That is, bv < by and

br-*-v < bx «• 6r-" = br~"-vbv < bxby <a< br~".

We have a contradiction from the first and the last terms of the above relation.
9.4
a) bn-l = (b-l)(bn-^1 +bn-^2 + --- + b+l) > (6- 1)(1H hi) > (b-l)n.

b) Let t = b^1 /". Apply part a) for t: tn - 1 > n(t - 1) => b - 1 > n{bxln - 1).

c)£fA<n=>^<t-l=>^± + l<i. We have ^=i > blln - 1. Thus,

bl'n < t.

d) Let t = $• > 1. Use part c), 6^1 /" < t = $• =4> fc^+V" = fc^^1 /" < y if

e) y > 0 => t = ^ > 1. If n > bjE$, use c), 6^1 /" < t = % => y < •& =
lAV — l/n

f)
Claim: A is bounded above.
Proof: If not, V/3 > 0, 3w G A 3 w > j3. In particular, Vn G N, 3w G
A 9 w > n. Hence, Vn G N, 3w G A 3 6" < 6W < y, i.e. Vn G N, 6" < y.
If 0 < y < 1, we have a Contradiction since bn > 1. Assume y > 1, use
(c) \fn 3 n > fEj, V1/n < b => y < bn. Hence, Vn 3 n > |5j we have
bn <y <bn, Contradiction.
Let x = sup(yl) = sup {w G R : bw -< y).