$\bf Exercise\, 1.6$
Fix $b>1.$
Prove that if $m,n,p,q$ are integers, $n>0,q>0$ and $r=m/n=p/q$, then
$$
(b^m)^{1/n}=(b^p)^{1/q}.
$$
I'm not really sure what I can assume and what I can't assume, I think that all I need is $(x^y)^z=x^{yz}$ for integers $y,z$, but I'm not sure how to prove this (I don't even know what the expected definition of exponentiation is!).
Attempt
$$
\begin{align}
\left((b^m)^{1/n}\right)^n&=b^m\quad \text{By Theorem 1.21 (I think).}\\
\left((b^m)^{1/n}\right)^{nq}&=b^{mq}\quad \text{Here I use $(x^y)^z=x^{yz}$.}\\
\left((b^m)^{1/n}\right)^{nq}&=b^{np}\quad \text{As $mq=np$.}\\
\left((b^m)^{1/n}\right)^{qn}&=b^{pn}\quad
\end{align}
$$
Then, taking $n$-th and then $q$-th roots, we get our desired result (I think this is possible, again, by theorem $1.21$, but I'm not sure).
Could someone check my proof and tell me which facts about exponentiation we are allowed to assume and use for these kind of proofs?
Best Answer
In general, Rudin seems to assume we know about integers but not reals. Because of this, it looks like
(a) the definition of $b^i$, with $i$ a positive integer, is inferred from 1.13 ($i$ copies of $b$ multiplied together, just like for the case of $b$ an integer),
(b) $(b^i)^j=b^{ij}$ can be assumed known for positive integers $i, j$ ($j$ copies of $b^i$ multiplied together are the same as $ij$ copies of $b$ multiplied together, say "by associativity"),
(c) similar things can be done for $i$ and $j$ negative, with $b^i$ being $-i$ copies of $1/b$ multiplied together,
(d) from the Corollary to Theorem 1.21 we can also infer $(b^i)^{1/n}=(b^{1/n})^i$ for positive integers $i$ by induction on $i$: if we assume $(b^i)^{1/n}=(b^{1/n})^i$ already shown for some $i$, let $a=b^i$ in the Corollary; we then get $(b^{i+1})^{1/n}=(b^ib)^{1/n}=(b^i)^{1/n}b^{1/n} =(b^{1/n})^i b^{1/n} = (b^{1/n})^{i+1}$. From here we can also get the result for $i$ negative.
Assume wlog that $p$ and $q$ have no common divisor. From our background knowledge regarding integers, we know that $m=jp$, $n=jq$ for some integer $j$.
Our goal is $(b^m)^{1/n}=(b^p)^{1/q}$, that is $(b^{jp})^{1/jq}=(b^p)^{1/q}$. Because of the uniqueness of $n$th roots in Theorem 1.21, we just need to show that $((b^{jp})^{1/jq})^q=b^p$. Based on (d) above, $$((b^{jp})^{1/jq})^q=(b^{jpq})^{1/jq}=((b^p)^{jq})^{1/jq}=b^p$$ as desired.
It's so much fun to prove stuff with one hand tied behind your back!