proof that $b^{\frac{m}{n}}=b^{\frac{p}{q}}$ if $\frac{m}{n}=\frac{p}{q}$ So I just solved this problem, however I'm not sure if my proof works, can someone tell me if it's correct or not, also I suspect there is a much simpler way to prove it so don't hesitate to post your own proof. 

The problem is: if $m,n,p,q$ are integers, $n\gt 0,q\gt 0,$ and $r=\frac{m}{n}=\frac{p}{q},$ prove that $$(b^m)^\frac{1}{n}=(b^p)^\frac{1}{q}.$$ 

First, I had to show that $b^\frac{1}{an}=(b^\frac{1}{n})^\frac{1}{a}$, which can be done by simply noting that $((b^\frac{1}{n})^\frac{1}{a})^{an}=(((b^\frac{1}{n})^\frac{1}{a})^a)^n=(b^\frac{1}{n})^n)=b,$ and since there exists a unique positive nth root for every integer we must have $b^\frac{1}{an}=(b^\frac{1}{n})^\frac{1}{a}$. 

Now since $r=\frac{m}{n}=\frac{p}{q}$, without loss of generality we can write $m=ap, $ and $n=aq$ for some integer $a$ thus $(b^m)^\frac{1}{n}=(b^{ap})^\frac{1}{aq}=(((b^p)^a)^\frac{1}{a})^\frac{1}{q}=(b^p)^\frac{1}{q}.$

My main problem with this proof is that I feel like I have cheated a bit with adding the factor $a$ since it's a bit more number theoretic, and I just started doing analysis. Also I think there should be a much easier way to do it.



From the comments For a positive real $b$ and a positive integer $n$, $b^{1/n}$ is defined to be the unique positive real such that $(b^{1/n})^n = b$.
Try $\dfrac{m}{n}=\dfrac{p}{q}=r$.

Then $b^{\frac{m}{n}}=b^r=b^{\frac{p}{q}}$.
As I said in the comments, the factor $a$ doesn't necessarily exist: consider the example of $3/3 = 2/2.$

However, if you let $d = \gcd(m,n)$ such that $m = dm'$ and $n = dn'$, then $m/n = m'/n'$ and you can apply your argument to show that $(b^m)^{1/n} = (b^{m'})^{1/n'}$. Similarly, if you let $e = \gcd(p,q)$, $p = ep'$, $q = eq'$, then $(b^p)^{1/q} = (b^{p'})^{1/q'}$.

But of course $m/n = p/q \implies m'/n' = p'/q'$, and besides $\gcd(p',q') = \gcd(m',n') = 1$. But it's a standard fact every (positive) fraction has a unique representation with coprime (positive) numerator and denominator (use Euclid's lemma four times and the fact $x \mid y \text{ and } y \mid x \implies x = y$ for positive $x$, $y$), so $m' = p'$ and $n' = q'$ and the claim is proven.