Verification for a Proof By Contradiction concerning a 5-root being irrational

Question

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.

Answer 1

Well, you are on the right track, but you cannot consider when $p/q = n \in \mathbb Z$.

HINT

I guess you implicitly meant that $m$ is a non-negative integer. Then, we have:

$q^5 \cdot m = p^5$, where $m, q, p$ are non-negative integers, and $p$, $q$ are coprime with each other, or $p=1$ or $q=1$.

Can you continue?

Answer 2

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 $p$

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In fact, this argument can be generalized to $\sqrt[n]m$ where $m$ is not a perfect $n^\text{th}$ power