Nth Root of Integer is Integer or Irrational

Theorem
Let $n$ be a natural number.

Let $x$ be an integer.

If the $n$th root of $x$ is not an integer, it must be irrational.

Proof
We prove the contrapositive: if the $n$th root of $x$ is rational, it must be an integer.

By Existence of Canonical Form of Rational Number, there exist an integer $a$ and a natural number $b$ which are coprime such that:

Since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime by Powers of Coprime Numbers are Coprime.

Hence $\dfrac {a^n} {b^n}$ is by definition in canonical form.

Suppose $b \ne 1$.

As the denominator of $\dfrac {a^n} {b^n}$ is not $1$, $x = \dfrac {a^n} {b^n}$ is not an integer.

This is a contradiction.

Thus $b = 1$, and thus:


 * $x^{1/n} = a$

which is an integer.