# Is the language of words having same number of a's and b's context-free?

I'm trying to use the pumping lemma, to show that the language $$L = \{w \in \{a, b\}^+: na(w) = nb(w)\}$$ is not context free, where $$na(w)$$ is the number of $$a$$'s in $$w$$ and $$nb(w)$$ is the number of $$b$$'s in $$w$$.

I have this: By contradiction, if $$L$$ is context free, we use the pumping lemma, then let $$N$$, and $$w = a^{floor(N/2)}*b^{floor(N / 2)}$$, with $$|w| = N$$. Then if we divide $$w = uvxyz$$, with $$v \not = \epsilon$$ and $$x = \epsilon$$, we see that when repeating v there appear more a's than b's, then $$uv ^ kxy ^ kz$$ $$\notin$$ L. Contradiction.

Is this right or am I missing something?

This language L = { w | w has equal number of as and bs and w≠ ε (since you put a +) } is a CFL

A PDA would work as follows :

If the stack is empty or the top of the stack is a push a

If the top of stack is b pop b push nothing