My purpose is to prove that the function $f:Q\to R$ given by $x\mapsto x^a$ ($a$ is real) can be continuously extended to $R$.
For that I want to prove that this function is Cauchy-continuous. I'm able to prove by induction that $x\mapsto x^n$ is Cauchy-continuous when $n$ is integer.
My question is about the next step: when $n$ is a positive integer, how to prove that $x\mapsto x^{1/n}$ is Cauchy continuous ?
EDIT1: $x^n$ is defined as $x\cdot x^{n-1}$ and $x^1=x$.
EDIT2: If $q>0$ is rational and if $n\geq 1$ is integer, the number $q^{1/n}$ is defined as the unique real $y>0$ such that $y^n=q$. (I assume the existence and unicity to be already done)
You want to show $f$ is Cauchy continuous, and $f$ is a function defined on $\mathbb{Q}$ and $x$ is a fixed real number; this will prove the extensibility.
Why then are you considering functions that vary in $x$, in the next paragraph. $x$ is fixed from the beginning (and must be $>0$ IIRC).
So what you have to show (not by induction) is that when $q_n$ is a rational Cauchy sequence then $x^{q_n}$ is Cauchy (in $\Bbb R$) too.
If $q_n = \frac{a_n}{b_n}$ in lowest terms and $b_n >0$, $x^{q_n}$ is defined as $(\sqrt[b_n]{x})^{a_n}$, where negative integral powers are fractions ($x^{-k}=\frac{1}{x^k}$ etc.) Now start estimating with epsilons, distinguishing cases etc. It's a messy thing, I'm afraid.