Let $f(x)=x^\alpha$ where $\alpha\in(0,1)$
I want to show to that $f(x)$ is continuous on the interval $I=[0,+\infty)$
Let $x_{0}\in[0,+\infty)$
$|f(x)-f(x_{0}|=|x^\alpha-{x_{0}}^\alpha|$
For $\alpha\in\mathbb{Q}\cap(0,1)$, we can write $\alpha=\frac{a}{b}$ where $a,b\in\mathbb{N}$ and $a<b$
I know how to prove that $x^{\frac{1}{b}}$ is continuous on $I$ and that $x^a$ is continous on $I$. So by composition, so is $x^{\frac{a}{b}}$.
However, what do I do in the case where $\alpha\in\mathbb{Q^\complement}\cap(0,1)$?