Does $\left\{\frac{p^a}{q^b}:p,q\in\mathbb{Z}\right\}=\left\{\frac{j^{\gcd(a,b)}}{k^{\gcd(a,b)}}:j,k\in\mathbb{Z}\right\}$ for $a,b\in\mathbb{N}$? How to prove $$\left\{\frac{p^a}{q^b}:p,q\in\mathbb{Z}\right\}=\left\{\frac{j^{\gcd(a,b)}}{k^{\gcd(a,b)}}:j,k\in\mathbb{Z}\right\}$$ for $a,b\in\mathbb{N}$?
First, is this identity correct?
For example if $a=1$ 
$$\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}=\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}$$ 
If we set $j=p$ and $k=q$
$$\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}=\left\{\frac{j(k^{b-1})}{k(k^{b-1})}\right\}=\left\{\frac{j}{k^b}:j,k\in\mathbb{Z}\right\}=\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}$$
However, this is not a formal way of proving this case for $a=1$. I have to show 
$$\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}\subseteq\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\} \land \left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}\subseteq\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}$$ 
How do I do this for $a=1$? How do I do this for $a\in\mathbb{N}$?

Yes, it is true.
The containment $\subseteq$ is easy to show.
For $\supseteq$, take any $a,b,i,j$. Let $x_1,x_2,y_1,y_2$ be such that $ax_1+by_1=ax_2+by_2=d=\operatorname{gcd}(a,b)$ and $x_1,y_2\geq 0\geq y_1,x_2$. Take $p=i^{x_1}j^{-x_2}$, $q=i^{-y_1}j^{y_2}$. Then $p^aq^{-b}=i^{ax_1+by_1}j^{-ax_2-by_2}=i^dj^{-d}$ and you are done.
