The definition of (real) exponent of b∈R, b>0 is give in Rudin’s Principles of Mathematical Analysis (3.ed), Chapter 1 Exercise 6 , which says that:
6. Fix b>1.
- If m,n,p,q are integers, n>0,q>0, and r=m/n=p/q, prove that
(bm)1/n=(bp)1/q.
Hence it makes sense to define br=(bm)1/n. - Prove that br+s=brbs if r and s are rational.
- If x is real, define B(x) to be the set of all numbers bt, where t is rational and t≤x. Prove that
br=supB(r)
when r is rational. Hence it makes sense to define
bx=supB(x)
for every real x. - Prove that bx+y=bxby for all real x and y.
It is a litter hard for beginners to solve this problem, so I will try to post my solution here.
For a. suppose α=(bm)1/n and β=(bp)1/q, then by the definition of nth roots, we know that
α=(bm)1/n⇔αn=bm,andβ=(bp)1/q⇔βq=bp.
Thus, by the meaning of notation bm (which says b multiples itself by m times), we know that
(bm)p=bmp=(bp)m,
thus
αnp=(αn)p=(bm)p=(bp)m=(βq)m=βqm,
since r=m/n=p/q, we have pn=mq, and
α=β,
by the uniqueness of nth roots of a real number. That is (bm)1/n=(bp)1/q.
For b. we can assume r=m/n and s=p/q for n,q>0 and m,n,p,q are integers. Then by a.,
br+s=bmn+pq=bmq+npnq=(bmq+np)1/nq=(bmqbnp)1/nq=(bmq)1/nq(bnp)1/nq=bm/nbp/q=brbs,
we employ the result (prove in the book) that:
(αβ)1/n=α1/nβ1/n,
for any α,β∈R and n be a positive integer.
Now, for c., by the definition, we can write B(x) as
B(x)={bt|t∈Q,t≤x},
and note that b>1 thus
bt≤br,
for all t,r∈Q with t≤r.
In fact, if bt>br. Set t=m/n, r=p/q, and t≤r then by the definition of ordered field,
mq≤np⇒bmq≤bnp,
thus by a.
(bt)nq=bmq≤bnp=(br)nq,
a contradiction to the assumption bt>br.
In conclusion, we know that
sup{bt|t∈Q,t≤r}≤br,
and apparently,
br≤supB(r)=sup{bt|t∈Q,t≤r},
thus br=supB(r).
For d., firstly, for any r,s∈Q, if r≤x, s≤y, then by the definition of bx,by, we have
br≤bx,bs≤by,
thus
br+s=brbs≤bxby,
since R is a ordered field.
Now, for any t∈Q, if t≤x+y, we can always find r,s∈Q such that r≤x,s≤y and r+s=t. In fact, by the definition of x+y, we know that x+y={r+s|r∈x,s∈y}, thus the assertion is certainly satisfied. Consequently,
bx+y=sup{bt|t∈Q,t≤x+y}≤bxby.
On the other hand, if bx+y<bxby, then set ε=bxby−bx+y>0. For any small enough δ>0, by the definition of bx,by, there exist r0,s0∈Q, with r0≤x, s0≤y, such that
br0>bx−δ>0,bs0>by−δ>0.
Since R is an ordered field, then
br0bs0>bxby−δ(bx+by−δ),
particularly, we can take δ be the positive roots of
δ(bx+by−δ)=ε/2,
then
2bx+y≥2br0bs0>2bxby−ε=bxby+bx+y,
that is
bx+y>bxby,
contradict to the assumption.
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