Prove that the following language is not regular: $\{0^i1^j : i \neq j\}$
Let $L = \{0^i1^j : i \neq j\}$. You can prove that $L$ is not regular in many ways. Here are some examples.

Closure properties

If your language were regular then so would $0^*1^* \setminus L$ be, but that language is $\{0^n1^n : n \in \mathbb N\}$, which presumably you already know isn't regular.

Myhill&ndash;Nerode

The words $\{0^i : i \in \mathbb N\}$ are pairwise distinguishable modulo $L$: if $i \neq j$, then $0^i1^i \notin L$ but $0^j1^i \in L$. It follows that $L$ isn't regular.

Pumping lemma

If $L$ is regular then it satisfies the pumping lemma, say with constant $n$. Consider the word $w = 0^n 1^{n+n!} \in L$. According to the pumping lemma, there should be a decomposition $w = xyz$ such that $|xy| \leq n$, $|y| \geq 1$, and $xy^iz \in L$ for all $i \in \mathbb N$. Let $|y| = \ell$, so that $y = 0^\ell$. Pick $i = 1 + n!/\ell$. Then $xy^iz = 0^{n+n!} 1^{n+n!} \notin L$, contradicting the pumping lemma.