If $m$ and $n$ are positive integers, prove that $\sqrt[m]{n}$ is either a positive integer or irrational.
I just need someone to verify my proof, which is as follows.
Clearly, if $n = k^m$ for some positive integer $k$, then $\sqrt[m]{n}$ is a positive integer, and hence the first part of the proof. Now suppose that $n$ $\neq$ $k^m$ for all positive integer $k$. We can prove by contradiction.
Assume $\sqrt[m]{n}$ = $\frac{a}{b}$, where $\frac{a}{b}$ is not an integer and $a$ is coprime relative to $b$.
$n = \frac{a^m}{b^m}$, $nb^m = a^m$
$b^m$ and $a^m$ are in the form $(p_1p_2...p_i)^m$, where p is a prime factor. $n$ can also be expressed as a product of prime factors. However, in order for $a$ to be an integer, n would have to have the form $(p_1p_2...p_i)^m$. However, this forms a contradiction as the condition above states that $\sqrt[m]{n}$ cannot be an integer. Hence, since all deductions were correct, then what is wrong must have been the argument itself, which is that $\sqrt[m]{n}$ is a rational non-integer. Therefore, $\sqrt[m]{n}$ is either a positive integer or is irrational.
Any comments or changes would be appreciated
