Proof Verification: Chapter 1 Exercise 6(a) of Baby Rudin I have an idea for a proof, but I'm not sure it's valid, so I would like to hear your opinion. 
The Question
Fix $b>1$.
(a) If $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = \frac{m}{n} = \frac{p}{q}$, prove that 
$$(b^m)^\frac{1}{n} = (b^p)^\frac{1}{q}$$
Hence it makes sense to define $b^r = (b^m)^\frac{1}{n}$.
My Proof 
From Theorem 1.21 we know that there exists $y_1,y_2\in\mathbb{R}$, such that $y_1^n = b^m$ and $y_2^q = b^p$. We show $y_1=y_2$. $y_1^n = b^{rn}$ and $y_2^q = b^{rq}$. Hence $b^{rn} = (b\cdot b \cdot b \cdot ...)\cdot (b\cdot b \cdot b \cdot ...) \cdot (b\cdot b \cdot b \cdot ...) ...$ where the number of $b$'s in each $(b\cdot b \cdot b \cdot ...)$ is $r$ and the number of $(b\cdot b \cdot b \cdot ...)$ is $n$. Thus $b^{rn} = (b^r)^n$ and in the same way, we get $b^{rq} = (b^r)^q$. From the uniqueness of $y_1$ and $y_2$ it follows that $y_1 = b^r = y_2$.   
 A: from "I'm not sure if my argument holds when m or p are negative."
It doesn't entirely.  Also you claim that it makes sense to define $b^r = (b*...*b)$ a number $r$ times.  That does not make any sense if $r$ is not an integer. 
One subtle thing is Rudin introduces $b^n$ as shorthand for $b*.... *b$.  He doesn't bother to mention that $b^0$ and $b^{-n}; n\in \mathbb N$ are extensions of that shorthand.  He also doesn't state or think it's worth proving that $b^{n+m} = b^nb^m$.  However as, at this stage, as $b^n$ is just shorthand for $b*.... *b$ and writing numbers on paper is basic addition and addition is associative this doesn't actual require proof.  It's immediate.
I suppose if I were teaching this and I had all the time in the world to bore my students out of their skulls, and make comments as we introduce the field axioms I'd apply them to integers and exponents:
$b$ is in some field $F$ but the exponents $b^n$ or $b^m$ are so that $n,m$ are integers which are a subset of the field of rationals.
Multiplication in $F$ is a associative and addition in $\mathbb Z\subset Q$ is associative so: Therefore $b^nb^m = \underbrace{\underbrace{(b*b*...*b)}_{n times}\underbrace{(b*b*...*b)}_{m times}=(\underbrace{b*b*...*b}_{n times}\underbrace{b*b*...*b}_{m times})}_{\text{Associativity of Multiplication in F}}=$
$\underbrace{(\underbrace{b*b*...*b}_{n times}\underbrace{b*b*...*b}_{m times})=(\underbrace{b*b*....*b}_{n+mtimes})}_{\text{Associativity of addition in }\mathbb Z} = b^{n+m}$.
There exists a multiplicative identity $1$ in $F$ and additive indentity $0$ in $\mathbb Z\subset Q$.  Therefore $b^0*b^m = b^m$ and $b^0 = 1$.
There exist multiplicative inverses in $F$ and additive inverses in $\mathbb Z\subset Q$: Therefore $b^{-n}b^{n} = b^0 =1$ so $b^{-n}=\frac 1{b^n};$ the multiplicative inverse.
Distributive axiom in $\mathbb Z\subset \mathbb Q$; $m(n+p) = mn + mp$: Therefore $(b^m)^n=b^{mn}$ (because if $m = 1+1+1...+1$ then $m*n = n+n+....+n$).
According to Rudin all this can be taken as a given and is still just "shorthand".  But this changes with the idea of $b^r; r \in \mathbb Q$ which is not our shorthand.
Now as $(b^{\frac 1n})^n =  b^{\frac 1n*n} = b^1 = b$ it might make sense to claim $b^{\frac 1n} = \sqrt[n]{b}$ as $\mathbb Q$ is a field.  But in the field $F$ we have absolutely no reason to believe that such a number $y; y^n = b$ exists. (They certainly do not exist in the field $\mathbb Q$ as $\sqrt 2 \not \in \mathbb Q$ attests).  Nor if we thought such numbers existed we could we assume that if $\frac mn = \frac pq$ then the numbers $w,v$ so that $w^n = b^m$ and $v^q = b^p$ would in anyway imply that $w = v$.
Th. 1.21 proves that such positive numbers exist for positive members of the field $\mathbb R$ and that they are unique.
But Ex.  6a) is what we need to do show that such values would be consistent.
Now when we write $r = \frac mn = \frac pq$ we assume $m,n,p,q$ are integers but also $n>0;q>0$ (so if $r< 0$ then it is the integers $m$ and $p$ that are negative).
And $\frac mn =\frac pq \iff mq = np$.  Let $k = mq = np$
Let $y_1$ and $y_2$ be the unique numbers so that $y_1^n = b^m$ and $y_2^q = b^p$.  Then $y_1^k = y_1^{np}  =(b^m)^p= b^{mp}$ and $y_2^{k} = y_2^{mq} = (b^p)^m = b^{mp}$.  So $y_1^k = y_2^k = b^{mp}$ but by theorem $1.21$ such $y_1, y_2$ are unique so $y_1=y_2$.

Note:  the introduction and explanation as to why we are doing the exercise and what the exercise signifies was over 7 times as long as doing the exercise itself.  
To mmy mind that was a singular oversight on Rudin's part.
