For language $L = \{a^m\, b^n\: |\: m \:\%\: n = 0 \}$, that is, $m$ is a multiple of $n$, I'm trying to find a proof that it isn't a context free. I know it isn't regular, but it also doesn't seem to be context-free to me. However, I am not sure how I would go about proving that it isn't context-free. My quick attempt follows:First, assume to contrary that $L$ is context-free. Let $p$ be the guaranteed pumping length. Choose string $s\,=\,a^{\,p+1}\,b^{\,p+1}$, where $s\,\in\,L$. By the pumping lemma, we can write $s$ as $s\,=\,uxwvy$. If $vx$ contains either $a's$ or $b's$, then $uv^2wx^2y\:\not\in\:L$. If $vx$ contains both $a's$ and $b's$, then $uv^2wx^2y\:\not\in\:L$. Therefore, we have a contradiction and $L$ cannot be context-free. However, I don't believe I have the proof completely right, and I would like an elaboration how this proof could be better presented. EDIT: Is this language context-free? Why or why not? 
Proving that $L = \{a^m b^n | m \% n = 0 \}$ is not context-free
formal languages;context free;pumping lemma