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, $n > 1$, with $p$ prime and $\gcd(m, n)=1$, is irrational. It's suggested that this proof should be by contradiction or contraposition. 
 A: 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
A: Since you want to prove that
$p^{m/n}$
is irrational,
you have to assume that
it is rational
and derive a contradiction.
So,
assume that
$p^{m/n}
= \frac{a}{b}
$
where
$(a, b) = 1$.
Then
$p^{m}
= \frac{a^n}{b^n}
$
or
$b^np^{m}
= a^n
$.
$p$ must divide $a$,
so let
$a = p^kq$
where
$p \not\mid q$.
Since
$(a, b) = 1$,
$p \not\mid b$.
Then
$b^np^{m}
= a^n
= (p^k q)^n
= p^{kn} q^n
$.
Since $p$ does not divide
either of
$b$ and $q$,
$m = kn$,
so that
$n \mid m$.
Since
$(m, n) = 1$,
this implies that
$n = 1$,
which is the
only way that
$p^{m/n}$
is rational.
And we are done.
A: First a comment: the problem should have the hypothesis $m\ne0$ and $n>1$, otherwise the statement is false.
We can assume $m<n$; indeed, with the division algorithm, write $m=nq+r$, with $0\le r<n$. Then
$$
p^{m/n}=p^q\,p^{r/n}
$$
and this is rational if and only if $p^{r/n}$ is rational. It can't be $r=0$, because otherwise $\gcd(m,n)=n>1$.
So we can assume $0<m<n$.
Suppose
$$
p^{m/n}=\frac{a}{b}
$$
for $0<m<n$. Then
$$
b^np^m=a^n
$$
and so $p\mid a^n$, which implies $p\mid a$. Therefore $a=pc$ and
$$
b^np^m=p^nc^n
$$
Since $0<m<n$ we can write this as
$$
b^n=p^{n-m}c^n
$$
and so $p\mid b^n$. Hence $p\mid b$: a contradiction.
A: 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.
