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I am asked to find the rank of $\mathbb Z$ as a set. I understand that $\mathbb Z$ may be constructed as the result of quotienting out $\mathbb N \times \mathbb N $ by the equivalence relation $(a,b) \sim (c,d) \Leftrightarrow a + d = c + b$.

If I understand correctly, this act of quotienting out leaves use with a set of equivalence classes, each with a representative. For simplicity, we can choose our representative to have a 1 either in the first or last position of the ordered pair, depending on if our integer is positive or negative.

My question however, is how do I take this rank of this set of equivalence classes? More specifically, what is the rank of the equivalence class? Is it just the rank of the representative ordered pair? Or should it be the rank of the set of ordered pairs that is the equivalence class?

Another question related to this is, how is $\mathbb Q$ defined? Most places I've seen formulate it as the equivalence class again but I just want to know is there a restriction on which one of the ordered pair can be negative?

For example if have the equivalence relation $(a,b) \sim (c,d) \Leftrightarrow ad = bc$ then is $(-1,-2)$ in the equivalent relation corresponding to $\frac{1}{2}$? It seems natural to me that we would restrict it to only allow for negative terms in the first component, but at the same time one could argue that it is very unnatural to do such a thing.

user366818
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    What is the rank of a set? – Bernard Apr 24 '18 at 21:09
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    @Bernard: you could have googled for that. Roughly speaking, the rank counts the maximum depth of nested curly brackets you'd need to write down an element of the set. See https://en.wikipedia.org/wiki/Von_Neumann_universe – Rob Arthan Apr 24 '18 at 21:11
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    As you have said, the integers may be constructed like that, but I'm sure there are other sensible descriptions you could use. The question as stands is not well-posed: it depends on the construction you choose. – tomasz Apr 24 '18 at 21:30

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You have the right idea - to find the rank of $\mathbb{Z}$ or $\mathbb{Q}$ you need to know exactly which set is taken to be $\mathbb{Z}$ or $\mathbb{Q}$. There are many conventions for that, so we can't directly answer it from the information in the question.

One reasonable definition is that $\mathbb{Z}$ is the set of equivalence classes of $\mathbb{N}\times\mathbb{N}$ under the relation $\equiv_1$: $$ (a,b)\equiv_1 (c,d) \leftrightarrow d + a = b + c. $$

Similarly, one reasonable definition of $\mathbb{Q}$ is the set of equivalence classes of $\mathbb{Z} \times (\mathbb{Z}\setminus \{0\})$ under the equivalence relation $\equiv_2$: $$ (p,q) \equiv_2 (r,s) \leftrightarrow ps = rq. $$

Once you make a definition of $\mathbb{Z}$ or $\mathbb{Q}$ it is not too hard to compute the rank, knowing that the rank of $\mathbb{N}$ is $\omega$.

You will also need to know which pairing function you use to define $\mathbb{N} \times \mathbb{N}$ and $\mathbb{Z} \times \mathbb{Z}$, because different pairing functions have different effects on the rank. Several possibilities are discussed under ordered pair on Wikipedia.

Carl Mummert
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