How can you prove that a value raised to $\frac{1}{n}$ is the n'th root of $x$? I know that if you raise a value to $\frac{1}{2}$ for example, you take the square root, but that is not what I am asking, what I am asking is; what are you actually doing when raising a value to $\frac{1}{2}$ apart from figuring out which number multiplied by it self gives you that first value?
For example, if you raise $4$ to the power of $2$ you multiply $4$ by itself, obviously, but what do you do when you have $\frac{1}{2}$ as an exponent for example?
This is something that all through out my school years(this far, I am in High School) has not been explained.
 A: The best way to see it, before calculus, is to just realize that if we want to keep the "nice" property of integer exponentiation:
$$\left(x^n\right)^m = x^{nm}$$ for rational exponentiation, we want to define $x^{p/q}$ so that $$\left(x^{p/q}\right)^q=x^p$$
That this works is due to something deeper going on that requires some knowledge of calculus, and, in fact, even more deeply, complex numbers. And even then, it is mysterious - it is the kind of thing a mathematician gets deeper and deeper understanding of as one gets older.
Numerically, we can define a function:
$$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$
Sorry for the notation - I realize that it is a bit confusing if you haven't covered calculus, and perhaps even if you have. $n!$ is the product of the first $n$ positive integers $n! = n\cdot(n-1)\cdots2\cdot 1$ with $0!=1$ by definition, and this notation means that we add up all possible values of $\frac{x^n}{n!}$. So this formula looks like:
$$\exp(x) = 1 + \frac{x^1}{1} + \frac{x^2}{2\cdot 1} + \frac{x^3}{3\cdot2\cdot 1} + \frac{x^4}{4\cdot 3\cdot 2\cdot 1}\cdots$$
Then we prove that this is defined for all real (or even complex) $x$, and then show that $\exp(x+y)=\exp(x)\exp(y)$, that $\exp(0)=1$, and define $e=\exp(1)$ and show that $e\neq 1$.
This means that $\exp(n)=e^n$ for any integer $n$, so we often write $\exp(x)=e^x$.
We can also define, in several different ways, the natural logarithm of positive $x$, $\ln x$ so that $\ln(\exp(x))= x$.
This in turn lets us define $x^y$ for any $y$ and $x>0$ - namely, by defining $$x^y = \exp(y\ln x)$$
This gets more fun when you get to complex numbers. In general, $x^y$ has one possible complex value if $y$ is an integer, it as $q$ possible values if $y=p/q$ is a reduced rational number with $q\geq 1$, and $x^y$ has infinitely many possible values when $y$ is irrational.
A: I find this question interesting.
Take the cuberoot of 2 as an example which can be written as:
$$\sqrt[3]{2} = 1.25992...$$
$$2{}^{1/3} = 1.25992...$$
The conventional way to define this operation is as pointed out in Thomas Andrews answer:
$$\exp \left(\frac{\log (2)}{3}\right) = 1.25992...$$
That is the exponential of the logarithm of 2 divided by 3. Division by three as in cube root. Logarithm of 2 as in the number to take the cube root of. But what then is a logarithm?
The exponential function is easy to understand, but the logarithm? One way to find a logarithm is to take a limit:
$$1 \left(1-\frac{1}{2^{1/1}}\right) = 0.5$$
$$10 \left(1-\frac{1}{2^{1/10}}\right) = 0.66967...$$
$$100 \left(1-\frac{1}{2^{1/100}}\right) = 0.69075...$$
$$1000 \left(1-\frac{1}{2^{1/1000}}\right) = 0.692907...$$
$$10000 \left(1-\frac{1}{2^{1/10000}}\right) = 0.693123...$$
$$100000 \left(1-\frac{1}{2^{1/100000}}\right) = 0.693145...$$
$$1000000 \left(1-\frac{1}{2^{1/1000000}}\right) = 0.693147...$$
which converges to logarithm of 2:
$$\text{log}(2) = 0.693147...$$
But the problem persists, what do you do when you raise a number to a fraction?

After the discussion below this comes to mind:
$$\displaystyle \log(2) = \lim_{k\to \infty } \, \sum\limits_{n=k}^{2 k} \frac{1}{n}$$
$$\displaystyle \log(2) = \lim_{k\to \infty } \, \int\limits_k^{2 k} \frac{1}{n} \, dn$$
