Proof that the integers have no zero divisors.

Hint

First prove by induction that for any $\langle a,b\rangle$ there is an $n\in\mathbb N$ so that $\langle a,b\rangle\sim \langle n,0\rangle$ or $\langle a,b\rangle\sim\langle 0,n\rangle$. Start with $\langle a,0\rangle$ and proceed by induction on $b$.

I presume you've proven that $\cdot_{\mathbb Z}$ is well-defined? That's actually a much harder proof, on some level.

This makes your cases far more clear.

Suppose $\langle a,b \rangle \cdot \langle c,d \rangle = 0_\mathbb{Z}$ and $\langle a,b\rangle \neq 0_\mathbb{Z}$. We prove that $\langle c , d \rangle = 0_\mathbb{Z}$.

So we have $ac + bd = ad + bc$ and $a \neq b $ and we want to show that $c=d$.

Since we are working on the Natural numbers, we can't use subtraction, which would make this easy. But we can work around this limitation by using the definition of the $<$ relation on $\mathbb{N}$.

Case 1: $a>b$

Then $a = b+k$ for some $k\in \mathbb{N}$, $k \neq 0$.

So $(b+k) c + bd = (b+k) d + bc$.

$\Rightarrow b(c+d) + kc = b(c+d) + kd$

$\Rightarrow kc = kd$

$\Rightarrow c=d$.

Here we used in the last two steps that terms and factors can be cancelled from both sides of an equation. You might have already proved that or you can prove it with induction.

Case 2: $a<b$

similar to case 1.