Irrationality of roots

details show $\text{m,n>1}$
The minimize possible for that $\sqrt[n]{m}$ such that $\text{n}^{th}$ isn't perfect is $\sqrt[2]{2}$ $\text{=irrational}$ thus the mentioned statement is $\text{true}$ .

Assume that $\sqrt[n]{m} = \dfrac{p}{q}$ for positive integers $p,q$ . Then $m = \left( \dfrac{p}{q} \right) ^n$ .

Therefore, $mq^n = p^n$ . By the fundamental theorem of arithmetic, $p,q$ are multiples of a set of primes repeated $n$ times. But then the prime factors of $m$ do not repeat $n$ times (since $m$ is not a perfect $n^{\text{th}}$ power).

So $mq^n \ne p^n$ .