Does nth-roots with non-natural indexes exists? My high school math teacher stated that roots with non-natural indexes are meaningless, just like $\frac{\infty}{\infty}$ or $0^0$, because "mathematicians decided so and so it is unless we change axioms".
It doesn't make sense to me.
Isn't $\sqrt[\frac 1a]{x}$ just $x^{a}$ for every $a\in\mathbb{R}$ (or even $\mathbb{C}$) or am I missing something?
 A: You are correct.  Natural powers are easy to define (repeated multiplication).  If negative powers are to work nicely with positive integer powers, they must represent reciprocals.  If rational powers are to work nicely with integer powers, they must represent natural roots.  The way to define irrational powers is trickier, but perfectly well-defined with a bit of calculus (continuous extension of a function defined on a dense subset).
As an easy counter-example to "no non-natural roots", $\sqrt[\frac{1}{2}]{2} = 2^\frac{1}{\frac{1}{2}} = 2^2 = 4$.  However, even irrational roots make sense.  If you have a sequence of rational numbers ($(q_n)\in\mathbb{Q}^\mathbb{N}$) that converges to $\pi$, then $\sqrt[\pi]{a} = \lim_{n\rightarrow\infty} \sqrt[q_n]{a} = \lim_{n\rightarrow\infty} a^{\frac{1}{q_n}}$.  Admittedly, proving that statement makes sense is complicated, but I think it makes intuitive sense.
From Wolfram|Alpha: http://www.wolframalpha.com/input/?i=2%5E%5B1%2FPi%5D .
A: We can define exponentiation to every real number as follows:
Let $\alpha\in\mathbb{R}$ and $x\in \mathbb{R}_+$ be a non negative real number. 
We define $x^{\alpha}$ as $$x^{\alpha}:=\text{exp}(\alpha\ln(x))$$
with this definition one can prove all the 'exponent laws', for example $$x^{\alpha}x^{\beta}=x^{\alpha+\beta}$$

Another approach is to start with exponentiation of natural numbers, defined recursively as follows:
For $n\in\mathbb{N}$, and $x\in\mathbb{R}_+$; $$x^n=x^{n-1}\cdot x$$ $$x^1=x$$ 
Then we can extend this definition, allowing the exponent to be an integer number putting $$x^{-1}=\frac{1}{x}$$
To allow rational exponents, we define the $n$-th root as the (unique positive) solution to $y^n=x$ and we denote it as $$x^{\frac{1}{n}}=y$$
The final step needs the supreme axiom;
Let $\alpha\in\mathbb{R}$ and $x\in\mathbb{R}_+$, $$x^{\alpha}:=\text{sup}\lbrace x^r \mid r\in\mathbb{Q} \,\,\text{   and   }\,\, r\leq\alpha\rbrace$$
One can prove that the definition given at the beginning  of this post and this last one are equivalent.
