Problem 1 - Regular Languages

Show that the following language is not regular.

$$\{ a^n b^{2n} \vert n \geq 0\} (\subseteq \{a, b\}^{*})$$

If $L$ has a regular grammar, then $L$ is a regular set. If $L$ is a regular set, then the pumping lemma holds:

$$(\exists N)(\forall z)((z \in L)(\wedge \vert z\vert \geq N)\righ... ..., \vert uv\vert \leq N, \vert v\vert \geq 1 \wedge (\forall i) (uv^iw \in L)))$$

Let's $N$ be the pumping lemma constant. We can choose $z=a^N b^{2N}\in L$ $(\vert z\vert\geq N)$. Following the above notations, we have $z=uvw$ with $\vert uv\vert\leq N$ and $\vert v\vert\geq 1$. It means that $v$ is composed of one or more $a$'s. By the pumping lemma, $uw\in L$ (case $i=0$). But since we deleted between 1 and $N$ $a$'s, it is not possible that the number of $a$'s be still half of the number of $b$'s, a contradiction. So, $L$ is not regular.