Given the following clues, is this number irrational?

Question

Given $ a, b \in \mathbb{N} $ and $ c \in \mathbb{Q}_{+}$, if $ a^c $ and $ b^c $ are irrational, with $ a $ and $ b $ coprime, is $ \frac{a^c}{b^c} $ irrational too?

Answer 1

Adapting Anas Ibrahim's answer:

We have (from Anas) $c=m/n$ in its lowest terms (so $m$ and $n$ have no common prime factor). Also \begin{align*} qa^c&=pb^c\\ \implies q^n a^m &= p^n b^m \end{align*} $a$ and $b$ have no common prime factor. Nor do $p$ and $q$. So, by unique factorization, the only way the LHS and RHS can be the same integer is if $a^m=p^n$ and $b^m=q^n$. But $m$ and $n$ have no common prime factor either. So the only way that $a^m$ and $b^m$ can both be $n$th powers is if $a$ and $b$ are themselves $n$th powers. But in that case $a^c$ and $b^c$ are rational, which contradicts the premiss of the question.

Answer 2

Yes. Because suppose for the sake of contradiction that$$\frac{a^c}{b^c}=\frac{p}{q} \iff qa^c=pb^c$$ Now let $c=\frac{m}{n}$ for coprime $m$ and $n$, we have $$q\sqrt[n]{a^m}=p\sqrt[n]{b^m}$$ But since $a$ and $b$ are coprime, then the right handside and the left handside are two different irrational numbers, meaning $q\sqrt[n]{a^m} \ne p\sqrt[n]{b^m}$, a contradiction.