I recently read about a way to define the set of integers as the set of all equivalence classes for some equivalence relation $\simeq$ satisfying $(a,b)\simeq(c,d)$ for $(a, b),\;(c,d)\in\mathbb{N}\times\mathbb{N}$ iff $a+d=b+c$:
$\mathbb{Z}=\mathbb{N}\times\mathbb{N}/ \simeq$
I think I get the general idea behind this definition. The reason we define every integer as an equivalence class instead of a specific pair of tuples is that infinitely many pairs of tuples satisfy the relation. I am trying to get some intuition for this representation of integers. Particularly, I am curious as to how this representation of the positive integers corresponds to their definition as von Neumann ordinals.
Consider the number $1$. As a von Neumann ordinal, it's defined as:
$1=\{\emptyset\}$
If we use our new definition of the integers, another representation of $1$ would be:
$[1, 0]=\{ (\{ \emptyset \},\emptyset ),(\{ \emptyset,\{ \emptyset \} \},\{ \emptyset\})... \}$
I have a hard time understanding how these are equivalent. I realize these are both definitions, but ideally, definitions of the same thing would coincide with one another. (I am sure they do in fact coincide, I just fail to see how).
If you define $\mathbb{Z}$ like that, then strictly speaking $\mathbb{N}$ is not a subset of $\mathbb{Z}$. But $\mathbb{Z}$ contains a subset $M$ which looks exactly like $\mathbb{N}$, including how addition and multiplication works, namely the set of equivalence classes $[(a,b)]$ where $a \ge b$ (or $a>b$, depending on whether you count zero as a natural number or not). So it's customary to identify $\mathbb{N}$ with $M$, so that $\mathbb{N}$ can be considered as a subset of $\mathbb{Z}$ (as we are used to thinking about it).