User:Byte Me/Sandbox

Theorem
The nth root of any integer is irrational if the nth root is not an integer.

Proof
Let $p$ be an integer.

Suppose that the nth root of $p$ is rational.

Then there exist $a$ and $b$ which are natural numbers and coprime such that:

Since $p$ is an integer, $a$ and $b$ must share a common factor. However, since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime because no new factors are introduced.

Thus, $b$ must equal 1.

Since $n-1$ and $a$ are both integers, $a^{n-1}$ must be an integer.

Thus, $\frac p a$ must be an integer, which is a contradiction because this implies that $a$ divides a prime factor of $p$.

So, the Nth root of an integer must be irrational.