Definition of $x^{-a}$ and $x^{1/a}$ So, the other night I was wondering why $x^{-a}=\dfrac{1}{x^a}$ and $x^{1/a}= \sqrt[a]{x}$, because in school the teacher (at least for me), gave this formulas without any explanation whatsoever.
This is how I tried to justify them:

$x^{-a}=\dfrac{1}{x^a}$:
Let $x \in \Bbb R$. As $\Bbb R$ is a field there exists an multiplicative inverse for $x$: $x^{-1}$ such that $xx^{-1}=1$, $x\neq 0$. So we get $x^{-1}=1/x$. Then, for any $n \in \Bbb N$:
$$x^{-n}=\left(x^{-1}\right)^n = \prod_1^n \frac{1}{x}= \frac{1}{x^n}$$
And then for any $a \in \Bbb R$ it's just a generalization of this.

$x^{1/a}= \sqrt[a]{x}$:
For any $n \in \Bbb N$, then
$$\prod_1^n x^{1/n} = x$$
So if we solve the equation we'd have: $x^{1/n} = \sqrt[n] x$.
But the thing is that, this is not quite true. For example: if $n = 2$ we'd have:
$$x^{1/2}x^{1/2}=x$$
and if we'd solve this we would get $x^{1/2} = \pm \sqrt{x}$. In fact, for each $n$ there are $n$ values that $x^{1/n}$ can take.
So why do we define this to be just the positive nth root? Or is this just not the way that lead to the definition of $x^{1/n}$? And, also, is my deduction for $x^{-a}$ also right?
 A: For any $a>0$ and reals $x$ and $y$ we have:
$$\frac{a^x}{a^y}=a^{x-y},$$ which for $x=y$ gives:
$$1=a^0.$$
Now, let $x=0$.
Thus,we obtain:
$$\frac{a^0}{a^y}=a^{-y}$$ or
$$a^{-y}=\frac{1}{a^y}$$ and we got the first property:
For $x>0$ we have $x^{-a}=\frac{1}{x^a}.$
I think, the second property it's just a definition for any natural $n\geq2$:
$$x^{\frac{1}{n}}=\sqrt[n]{x}.$$
For $n=2$ we don't write the degree of the radical.
We have here the following definition.
For any $x>0$, natural $n\geq2$ and integer $m$ we define:
$$x^{\frac{m}{n}}=\sqrt[n]{x^m}.$$
A: My explanation:
From the definition of the powers as repeated products, it is elementary to establish
$$x^{m+n}=x^mx^n$$ and $$(x^m)^n=x^{mn}.$$
It seems reasonable to extend the definition of the powers to signed, rational and even real exponents in a way that preserves these properties.
A reasoning can be
$$1=x^0=x^{m+(-m)}=x^mx^{-m}$$ so that
$$x^{-m}:=\dfrac1{x^m}.$$
And
$$x=x^1=x^{n\cdot 1/n}=(x^{1/n})^n$$
so that $x^{1/n}$ is the solution of the equation $$(x^{1/n})^n=x,$$ which we denote by $$x^{1/n}:=\sqrt[n]x.$$

The generalization to rationals is immediate,
$$x^{m/n}:=\sqrt[n]{x^m}=\left(\sqrt[n]x\right)^m.$$

These are mere definitions, and it remains to check that they form a coherent set. We know that this is the case, though powers of negative numbers require special care.
