Ranks of $\mathbb{Z},\mathbb{Q}$ and $\mathbb{R}$ We have the following definitions of the rank of a set in the von Neumann hierarchy:
$$\mathrm{rank}(x)=\sup\{(\mathrm{rank}\,y)^+:y\in x\}.$$
I now want to find the ranks of $\mathbb{Z},\mathbb{Q}$ and $\mathbb{R}$.
$\mathbb{Z}$ can be realised as a subset of $2\times \omega$, each element having finite rank (regardless of implementation details like pairing method). Hence I get that the rank of $\mathbb{Z}$ is $\omega$ as well.
Similarly, $\mathbb{Q}$ can be thought of as a subset of $\mathbb{Z}\times \mathbb{Z}$, and again, regardless of implementation details, each element in this set has finite rank, hence I get that the rank of $\mathbb{Q}$ is $\omega$.
We construct $\mathbb{R}$ as subsets of $\mathbb{Q}$ (e.g. the lower part of Dedekind cuts), the relevant subsets having rank $\omega+1$. Hence the rank of $\mathbb{R}$ as defined like this is $\omega+2$.
My questions are:


*

*Some other answers seemed to suggest that the ranks of $\mathbb{Z}$ and $\mathbb{Q}$ depend on implementation details - however, with the argument above, I find this hard to see. So are the ranks both $\omega$ in most cases, or have I missed something?

*Is it true that, defined like this, the rank of $\mathbb{R}$ is $\omega+2$?


Other relevant links:
The real numbers and the Von Neumann Universe
Rank of $\mathbb Z$ and $\mathbb Q$
 A: "Regardless of pairing method"  What if we use a truly terrible pairing method, like $\langle a,b\rangle=\{\{\{a\},\{\omega_{17}\}\},\{\{a,b\},\{\omega_{17}\}\}\}$? Very few things are truly implementation-independent.

More seriously, at the second linked post a very different construction of $\mathbb{Z}$ (for example) is used: elements of $\mathbb{Z}$ are infinite sets of pairs of natural numbers, namely equivalence classes under the appropriate relation. Arguably (and I hold this opinion) this is actually more natural than your approach, since it captures the algebraic nature of the construction $\mathbb{N}\leadsto\mathbb{Z}$. And under this approach, each element of $\mathbb{Z}$ has rank $\omega$ and so $\mathbb{Z}$ itself has rank $\omega+1$. 
And the rank issue only gets worse as we go further. The algebraic approach says that elements of $\mathbb{Q}$ should be sets of pairs of elements of $\mathbb{Z}$, so by this approach an element of $\mathbb{Q}$ is an infinite set of infinite sets of natural numbers - and so has rank $\omega+1$, leading to $\mathbb{Q}$ itself having rank $\omega+2$. If we construe a real as a set of Cauchy sequences of rationals this then kicks $\mathbb{R}$ all the way up to rank $\omega+4$ - its elements are infinite sets of infinite sequences of infinite sets of infinite sets of natural numbers.
