Proving Gaussian Integers are countable

Question

I know that Gaussian Integers are a subset of complex numbers. They are numbers in the form

$G = \{a + ib \,\vert\, a,b \in \mathbb{Z}\}$

So to prove that a set is countable, I need to find a function $G \rightarrow \mathbb{N}$ such that every $n$ has finitely many preimages (from tricki.org).

How should I go about proving that it is countable? I don't really understand what preimages mean.

Answer 1

$G$ has the same cardinality as $\mathbb{Z} \times \mathbb{Z}$. In general, the finite cartesian product of countable sets is countable.

If you want an explicit bijection for this problem, here's a hint: $$0,1,i,-1,-i,1+i,-1+i,1-i,-1-i,2,2i,-2,-2i,\ldots$$