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Reading an article, I see: $$\mathbb{Z}=\mathbb{N}\times\mathbb{N}/\sim\\(a,b)\sim(c,d)\iff a+d=b+c$$

In this formulation of the integers, what would a number actually look like? I’m familiar with the set theory treatment of the naturals, that is $0=\emptyset,1=\{0\},2=\{0,1\},3=\{0,1,2\}\cdots$ but here would the equivalence class $(0,1)\sim(1,2)\sim(2,3)\sim\cdots$ be the equivalence class representing the integer $1$? And would $-1$ be represented by $(1,0)\sim(2,1)\sim\cdots$? And how could one express (I assume with a basis in ZF theory) this article’s formulation of $1$ in terms of purely sets? This ordered pairs notion means what, exactly? $1\in\Bbb Z=\{\{0,1\},\{1,2\},\{2,3\},\cdots\}$? Wouldn’t this contradict the notion $1\in\Bbb N=\{\emptyset\}$?

Additionally, this equivalence relation assumes commutativity of natural arithmetic. Building up from scratch, axiomatically, this article seems to have a flimsy construction of the integers - what is the canonical way?

FShrike
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  • Pretty much all of the questions on https://math.stackexchange.com/questions/linked/1553537 are essentially the same as this; as are https://math.stackexchange.com/questions/1055501/why-are-integers-subset-of-reals and the ones linked to it (see on the right-hand side menu under "Linked"); and more. – Asaf Karagila Aug 07 '21 at 20:48

2 Answers2

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That is the canonical way of defining the integers. With that definition, $1$ is$$\{(1,0),(2,1),(3,2),\ldots\}$$and $-1$ is$$\{(0,1),(1,2),(2,3),\ldots\}.$$The idea is that the pair $(a,b)$ represents the integer $a-b$.

And, yes, with this definition $1\ne\{\emptyset\}$. But the integer number $1$ is not the natural number $1$. What happens is that there is a natural injective map from $\Bbb N$ into $\Bbb Z$ which maps $1(\in\Bbb N)$ into $1(\in\Bbb Z)$. It also maps $2$ into $2$, $3$ into $3$ and so on …

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This is indeed the canonical description of the integers. Let $\pi : \mathbb{N}^2 \to \mathbb{Z} = \mathbb{N}^2 / \sim$ be the quotient projection map - that is, $\pi(a, b) = \{(c, d) \in \mathbb{N}^2 | (c, d) \sim (a, b)\}$.

Then the "positive injection map" is given by $i_+ : \mathbb{N} \to \mathbb{Z}$, $i_+(n) = \pi(0, n)$. It can be shown that $i_+$ is an injection.

So numbers like $1, 2, 3$ are represented as $\pi(0, 1), \pi(0, 2), \pi(0, 3)$.

The "negative injection map" is given by $i_- : \mathbb{N} \to \mathbb{Z}$, $i_-(n) = \pi(n, 0)$. So negative numbers like $-1, -2, -3$ are represented as $\pi(1, 0)$, $\pi(2, 0)$, $\pi(3, 0)$.

Yes, the formulation does indeed rely on the properties of the natural numbers. There are other ways to more succinctly describe $\mathbb{Z}$ such as "the free group on 1 element", but they involve understanding even more theory.

There are many different instances of numbers like 1. There is the natural number 1, the integer 1, the rational 1, the real 1, the complex 1, etc. The key is not to get bogged down in "what a number really is" and instead look at how numbers relate to other things (using principles like induction and tools from abstract algebra).

Mark Saving
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