Prove that the language $\{a^ib^i | i\geq 0\}$ is not regular? (Do we just consider $a^nb^n$, where $n$ is the pumping length?
To apply the pumping lemma for regular languages:


You assume some pumping length $p &gt; 0$: Suppose a pumping length $p$.
You pick a string $s$ such that $|s| \geq n$: Indeed, $a^pb^p$ is a good one since $|a^pb^p| = 2p \geq p$.
Now, you have to consider all the partitions of $s$ as $xyz$, such that 


$|y|&gt;0$ and
$|xy| \leq p$.



In this case, the 2nd inequality restricts $y$ to contain only $a$'s. Therefore, all the possible partitions are of the form $y = a^k$, $x = a^{p-k-r}$ and $z = a^rb^p$ for $k$ and $r$ such that $k &gt; 0$ and $p \geq k + r$.


For every possible partition of the previous step, you have to show that there is some $i\in\mathbb{N}$ such that $xy^iz$ is not in the language.


Here, consider $i = 0$. Then $xy^0z = xz = a^{p-k-r}a^rb^p = a^{p-k}b^p$.
It is now a matter of showing that $p-k\neq p$, or, equivalently, that $k &gt; 0$ which is already true.