Solutions 267
b---b b---bb---b
Case 1 : p > 0, q > 0, W+q = ^~" = ^-v-'^-' = ^69;
P + <? p q
Case 2 : p < 0, q > 0, Let p' = -p > 0. So, lf+q = b-P'+q;
' Case 2a: p' = q ^ &-"'+« = 6° = l = j£ = ^ = fcPfc*.,, 6 • • • b b- • -b W ,„
Case 26 : p' < q =• b~" +q = s~^-' = ^^ -r^-r = &= WW.
q-p q-p &"3
P'
_ Case 2c : p' > a =» b-r>'+q = 6"CP'-«) = ^ = ^ = £ = *W.
Case 3 : p > 0, q < 0, similar to Case 2;
I Case 4 : p < 0, a < 0, then, p + a < 0 => &»+« = p^, = ^^ = WW.9.3
a) Let a = (bm)lln, (3 = (&P)^1 /^9 where a,/3 > 0.' Case 1 : m — 0, => p = 0. So, a = /3 = 1.
Case 2: m > 0, =» p > 0. a = ft™)^1 /" =>. a» = ft™ =-, ft = (an)£.
Similarly, 6 = (/3«)* =>• 6 mP = a"? = /?«m.
Thus, np = qm=> anp = /3np.
Since the positive (np)th root is unique, a = 0.
Case 3 : m < 0, =>• p < 0. Let m' = —m, p' = —p => m',p' > O.Case 2!
(um\l/n _ /L-m'\l/n _ / 1 U/n _ 1 _
= 1 = (bp)^1 /g
(b-p'y/i ^ >So, 6 r, r £ Q are well defined.
b) Let r = ^, s = | where n,g > 0.
br+s _ (pmq+np^i-q _ ^mg^p^ _ ^mq^ ^np^ _= ((bm)")^ ((W)n)^ = (((bm)q)^1 ^)^n(((bp)n)^n)^1 ^q =
{bm)^1 ln{bp)l/q = brbs.c) Let 6 e B{r). Then, <eQ,t<r^r-t>0,r-eQ. Since 6 > 1 and
r — t is a nonnegative rational number, we get br~l > 1.
Claim: Let & > 1, s e Q+. 6s > 1.
Proof: If s = 0 =>• 6 s = b° = 1. Assume s > 0. Then, s = E where p, 0 > 0.
ftS = (&P)l/«. 6>1=i.a = 6P>1=>. 6 »= al/« > L
Hence, 1 > 6r-' = Wb'^1 = ^ =• 6 < br. That is V6' £ B(r), b < br; i.e.
br is an upper bound for B(r). Then, sup(£(r)) < br. If r € Q, br £ B(r). So,
br < sup(5(r)). Thus, br = sup(j5(r)).
Now, we can safely define bx = sup(B(x)), Vx £ R.
d) Fix 6r arbitrary in -B(a;) and fix 6s arbitrary in B(y): r, s £ Q, r < x, s < y.