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

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. 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.