prove that $p^{m/n}$, with $p$ prime and $\gcd(m, n)=1$, is irrational I need to prove that $p^{m/n}$, $m$ and $n$ naturals, $n &gt; 1$, with $p$ prime and $\gcd(m, n)=1$, is irrational. It's suggested that this proof should be by contradiction or contraposition.
Suppose $\,p^{m/n}$ is rational. It's a  root of $\,x^n\! = p^m$ so by the Rational Root Test $\,x\,$ is an integer $\,a.$ Therefore, $ $ comparing powers of $\,p\,$ in $\,a^n = p^m$ $\Rightarrow$ $\,n\alpha = m\,$ so $\,n\mid m\,$ contra $\,n,m\,$ coprime.

Remark $ $ Without RRT: $\ $ if $\ p^{m/n}\! = a/b\ $ is rational then $\, p^m b^n = a^n\,$ hence comparing powers of $\,p\,$ on both sides of the prior equation yields  $\, m+n\beta = n\alpha\, $ so $\, n\mid m\, $ contra $\,n,m\,$ coprime
Assume the negation of the statement and set out for a contradiction. That is $$\frac{k}{l}^{\frac{n}{m}}=p$$

But the greatest common denominator is one so

$$\frac{k^{\frac{n}{m}}}{l^{\frac{n}{m}}}$$

Which must be a multiple of $l^{\frac{n}{m}}$. So $p=j^{\frac{n}{m}}$.

$p=({g^{\frac{1}{m}}})^n$ which contradicts the fact that $p$ was prime.