Nth Root of Integer is Integer or Irrational

Theorem
Let $n$ be a natural number.

Let $x$ be an integer.

If $x$ has a real $n$th root Which is not an integer, then that root must be irrational.

To put it another way, if $r$ is a rational number and $r^n$ is an integer, then $r$ is an integer.

Proof
Suppose that $x^{1/n}$ is not an integer.

Suppose for the sake of contradiction that the $n$th root of $x$ is rational.

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

Since $x$ is an integer, $a^n$ and $b^n$ must share a common factor if $b \ne 1$.

If $a^n$ and $b^n$ are coprime, and if $b \ne 1$, ${a^n}/{b^n}$ would not be an integer because ${a^n}/{b^n}$ in simplest terms would not have $1$ as a denominator.

However, since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime because no new prime factors are introduced.

Thus, $b$ must equal 1.

Thus, $x^{1/n}$ must be an integer, which is a contradiction.

Therefore the $n$th root of an integer must be irrational if it is not an integer.