# How can different representations of the same integer be equivalent?

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).

• What properties do you want your positive integers to have? Can you check both your models have these and produce equivalent results? Essentially you can do this in two stages identifying $n$ with $(n,0)$ and then using $(n,0)\simeq(n+k,k)$ for natural numbers $n$ and $k$ with the positive integers being those with $n\ge 1$ Commented Feb 18 at 12:36

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).
• When you say that $M$ looks like $\mathbb{N}$, do you mean solely in terms of arithmetic operations? Also, if you don't mind, could you please give a concrete example of this, using some positive integer? Commented Feb 18 at 13:11
• For $k \ge 1$, let $P_k$ and $N_k$ be the be the equivalence classes of pairs of the form $(a+k,a)$ and $(a,a+k)$, respectively, and let $O$ be the class of pairs $(a,a)$. (In other words, the positive/negative integers and zero.) Then $$\mathbb{Z} = \{ \ldots,N_2,N_1,O,P_1,P_2,\ldots \} ,$$ and the map $$\mathbb{N} = \{ 0,1,2,\ldots \} \mapsto M = \{ O,P_1,P_2,\ldots \}$$ is a one-to-one correspondence which preserves addition and multiplication. For example, $P_2 + P_5$ is (by definition) the equivalence class containing $(a+2,a) + (b+5,b) = (a+b+7,a+b)$, i.e., $P_7$. Commented Feb 18 at 19:48