The n-th root of m is either an integer or irrational

by Joe Zeng Jul 6 2016

In other words, no power of a rational number that is not an integer is ever an integer.

There is an intuitive way to see that for any natural numbers $~$m$~$ and $~$n$~$, $~$\sqrt[n]m$~$ will always either be an integer or an irrational number.

Suppose that there was some $~$\sqrt[n]m$~$ that was a rational number $~$\frac{a}{b}$~$ that was not an integer. Suppose further that $~$\frac ab$~$ is written as a [-reduced] fraction, such that the Greatest common divisor of $~$a$~$ and $~$b$~$ is $~$1$~$. Then, since $~$\frac{a}{b}$~$ is not an integer, $~$b > 1$~$.

Since $~$\frac ab = \sqrt[n]m$~$, we have conversely that $~$(\frac ab)^n = m$~$, which is a natural number by our hypothesis. But let's take a closer look at $~$(\frac ab)^n$~$. It evaluates to $~$\frac{a^n}{b^n}$~$, which is still a reduced fraction. [todo: Proof of gcd(a^n, b^n) = 1.]

But since $~$b > 1$~$ before, we have that $~$b^n > 1$~$ as well, meaning that $~$(\frac ab)^n$~$ cannot be an integer, contradicting the fact that it equals $~$m$~$, a natural number.