I was asked to prove that for a non negative $m$ number, that is not a 5-th power of an integer, its 5-root, $\sqrt[5]m$ is not a rational number.
Thanks for any feedback.
Proof By Contradiction:
Suppose Not,
Assume that for a non-negative $m$ number, that is not a 5-th power of an integer, its 5-root $\sqrt[5]m$ is a rational number.
Being a rational number we can say:$$\sqrt[5]m=\frac{p}{q}\quad \text{for} \;p,q\in\Bbb{Z}\;\text{and}\:q\neq0$$
Then by algebra,$$m=\left(\frac{p}{q}\right)^{5}\Rightarrow m=\frac{p^{5}}{q^{5}}$$
Consider when $\frac{p}{q}=n$, $n \in\Bbb Z$,
This is a contradiction with the fact that $m$ is not a 5-th power of an integer.
Hence, $\sqrt[5]{m}$ is not a rational number.
Best Answer
You may assume (wlog) that $q\gt 0$, so since $p^5/q^5=m\in\Bbb Z$, this implies $q^5\mid p^5$. You can furthermore assume (wlog) that $\gcd(p,q)=1$. This would imply that $q\mid p$, so $1=\gcd(p,q)=q$ but then $m=p^5$ leading to a contradiction that $m$ is a 5th power of an integer, namely $p$
In fact, this argument can be generalized to $\sqrt[n]m$ where $m$ is not a perfect $n^\text{th}$ power