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

https://arbital.com/p/566

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.