Show that $f(x) = x^{m}$ is continuous for $m$ rational. Consider the function $f(x) = x^{m}$ with $m$ rational. 
Show that $f$ is continuous on all nonnegative numbers when $m$ is positive and that $f$ is continuous on all positive numbers when $m$ is negative.
Could we use the fact that $m = p(1/q)$ in order to show that $f$ is the inverse of a function that is continuous on the desired interval? 
Overall, I'm not sure.
 A: Note that
$$|x^p - c^p| = |x - c||x^{p-1} + x^{p-2}c + \ldots + xc^{p-2} + c^{p-1}|, \\ |x - c| = |x^{1/q} - c^{1/q}||x^{(q-1)/q} + x^{(q-2)/q}c^{1/q} + \ldots + x^{1/q}c^{(q-2)/q} +  c^{(q-1)/q}|. $$
To prove $f:x \mapsto x^p$ (p > 0) is continuous, we want to find $\delta$ such that $|x-c|< \delta \implies |x^p - c^p| < \epsilon$. 
We have $|x^{p-1} + x^{p-2}c + \ldots + xc^{p-2} + c^{p-1}| \leqslant |x|^{p-1} + |x|^{p-2}|c| + \ldots + |x||c|^{p-2} + |c|^{p-1} $ by the triangle inequality.
If $|x - c| < 1$, then $|x| < 1 + |c|$ and $|x|^{p-1-j}|c|^j < (1 + |c|)^{p-1}.$
Hence,
$$|x^p - c^p| = |x - c||x^{p-1} + x^{p-2}c + \ldots + xc^{p-2} + c^{p-1}| < |x-c|p(1 +|c|)^{p-1},$$
and if $|x-c| < \min(1, \epsilon/[p(1 +|c|)^{p-1}])$, then $|x^p - c^p| < \epsilon.$ 
Similarly, we can prove $g:x \mapsto x^{1/q}$  using the second identity with a lower bound on the second factor on the RHS. (See If $(x_n )\rightarrow a $ show that $ \sqrt[3]{x_n} \rightarrow \sqrt[3]{a}$.)
Finally, $h: x \mapsto f(g(x)) = x^{p/q}$ is the composition of continuous functions and, therefore, is continuous.
A: Here’s a general fact that proves that all sorts of reasonable functions are continuous:
Lemma.Let $I$ and $J$ be open intervals in $\Bbb R$, and let $f:I\to J$ be surjective and strictly monotone. Then $f$ is continuous.
Proof: Without loss of generality, we can suppose that $f$ is increasing. Let $x\in I$, $y=f(x)$ its image in $J$. Start with a positive $\varepsilon$, I’ll show you how to get your $\delta$. Since $J$ is open, we can assume without loss of generality that $y-\varepsilon$ and $y+\varepsilon$ both are in $J$, and therefore there are $x_-\in I$ and $x_+\in I$ with $f(x_-)=y-\varepsilon$, $f(x_+)=y+\varepsilon$ (because $f$ is surjective) and satisfying $x_-<x<x_+$ (because $f$ is increasing). Now let $\delta=\min(x-x_-,x_+-x)$. That’s it, ’cause if $|x-\xi|<\delta$, then $\xi$ is between $x_-$ and $x_+$, and $f(\xi)$ is between $f(x_-)=y-\varepsilon$ and $f(x_+)=y+\varepsilon$, in other words, $|f(\xi)-y|<\varepsilon$. q.e.d.
Your function $x\mapsto x^{p/q}$ is strictly monotone on $\langle0,\infty\rangle$ (unless $p/q=0$) and surjective, so continuous by the Lemma.
