Recall that for a set $x$ its rank $\alpha$ is the least ordinal such that $x \in V_{\alpha+1}$. Or in other words: $x$ is built up out of $\alpha$ levels of braces and the empty set.

I think with the usual constructions of numbers (cartesian products, sets of equivalence classes, Dedekind cuts, etc.), we have

$rank(\mathbb{N})=\omega, rank(\mathbb{Z})=\omega+4, rank(\mathbb{Q})=\omega+8, rank(\mathbb{R})=\omega+10$

Now it is possible to find a bijection $\mathbb{Z} \cong \mathbb{N}$, so that there is a copy of $\mathbb{Z}$ of smaller rank, namely $\omega$. But this is, of course, nonsense. We should also consider the ring structure on $\mathbb{Z}$, which is given by two maps $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. Therefore we might ask the following:

Is there a ring $(R,+,\*)$, which is isomorphic to $(\mathbb{Z},+,\*)$, but $rank(R,+,\*) < rank(\mathbb{Z},+,\*)$? What about the other rings above?

Of course, this question is just out of curiosity. I doubt that anybody cares about these bounds of ranks (if not, please let me know).