How can you prove that a value raised to $\frac{1}{n}$ is the n'th root of $x$?

Question

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.

Answer 1

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.

Answer 2

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?

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After the discussion below this comes to mind:

$$\displaystyle \log(2) = \lim_{k\to \infty } \, \sum\limits_{n=k}^{2 k} \frac{1}{n}$$

$$\displaystyle \log(2) = \lim_{k\to \infty } \, \int\limits_k^{2 k} \frac{1}{n} \, dn$$