show that if $a,K$ are positive integers and $a≠m^k$ then $a^\frac{1}{K}$ is irrational.
I assumed $a=m^K$ and $a^\frac{1}{K}∈ℝ\backslash ℚ$,it's clear that $a$ is a power of $m$, means $a=m×...×m$($K$ copies of $m$), then if $K$ is even and $m$ is a negative integer we have $a^\frac{1}{K}=\left|m\right|$ otherwise $a^\frac{1}{K}=m$ , in both cases $a^\frac{1}{K}$ is equail to an element which is not irrational, a contradiction,implies $a≠m^k⇒a^\frac{1}{K}∈ℝ\backslash ℚ$ as desired.
I DON'T KNOW MY PROOF IS CORRECT OR NOT...
Hint: Assume $a \neq m^k$ and that there exist $p,q \in \mathbb{Z}$ coprime s.t. $a^{\frac{1}{k}} = \frac{p}{q}$. Try then to derive a contradiction.
Let $a,k \in \mathbb{N}$ s.t. for any $m \in \mathbb{Z}: a \neq m^k$ and assume there exist $p,q \in \mathbb{Z}$ coprime s.t. $a^{\frac{1}{k}} = \frac{p}{q}$. Then $a = \frac{p^k}{q^k}$, hence $a q^{k-1} = \frac{p^k}{q}$. The left hand side is an integer, so $\frac{p^k}{q} \in \mathbb{Z}$. But $p,q$ are coprime, so $q = 1$, and thus $a = p^k$. A contradiction to the assumption since $p \in \mathbb{Z}$. Thus $a^{\frac{1}{k}}$ is irrational.