Ilan Bernstein

I like numbers. I really like numbers. One of those weird numbers is $-1$ . That’s a pretty weird number huh? Let’s make it from scratch! This post assumes you have some understanding of set theory (I have almost none, so it should be pretty easy to understand).

Creating the Naturals

If you want a more formal definition, check these out, but for now I’ll shortcut it.

$\begin{aligned} 0 & = \{\} &\;=\;& \emptyset \\ 1 & = \{0\} &\;=\;& \{ \emptyset \} \\ 2 & = \{0, 1\} &\;=\;& \{ \emptyset, \{\emptyset\} \}, \\ 3 & = \{0, 1, 2\}&\;=\;& \{ \emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\} \} \end{aligned}$

Let’s make a function to find the next natural number (this is called a successor function):

$S(n)=n\cup\{n\}$

Now we can say $3=S(2)$ ; $4=S(3)$ ; et cetera!

$\mathbb{N}$ can now be defined as the smallest set containing the entire domain of $S(n)$ and containing $0$ .

Awesome! Let’s keep going.

Creating Ordered Pairs

We’re going to use Kuratowski’s ordered pairs. Here’s an example: $(x, y)\coloneqq\{x,\{x,y\}\}$

For example, let’s create the ordered pair $(0,1)$ :
$(0,1)=\{0,\{0,1\}\}$

Cartesian Product of Naturals

Here’s what we want to define:

Now we only have to define it for two ordered pairs: $(a,b)$ and $(c,d)$ , so let’s do that: $\mathbb{N}\times\mathbb{N}\coloneqq\{(a,b):a\in\mathbb{N},b\in\mathbb{N}\}$

Now you might be questioning: why can’t we just define $(\mathbb{N},\mathbb{N})$ ? Well think of it this way. Let’s define two sets:

$Suits = \{♥,♦,♣,♠\}$

$Ranks = \{A,2,3,\dots,K\}$

Well if we say $(Suits,Ranks)$ we get the set $(\{♥,♦,♣,♠\},\{A,2,3,\dots,K\})$

Now if we say $Suits\times Ranks$ we get the set $\{(♥,A),(♥,2),\dots,(♦,A),(♦,2),\dots,(♠,K)\}$

Alright now we can continue.

The Equivalence Relation

Defining Addition

Let’s start by defining addition using the successor function we defined all the way at the top:

$\begin{aligned} & a+0 = a & (1) \\ & a+S(b) = S(a+b) & (2) \end{aligned}$

The Relation

$(a,b)\sim(c,d)\iff a-b=c-d$

This relation defines that for any tuple $(a,b)$ , the difference between $a$ and $b$ must be equal to the difference between $c$ and $d$ . So for example, if $a=0$ and $b=1$ , then some possible values for $c$ and $d$ might be $c=5$ and $d=6$ ; another possible combination might be $c=2$ and $d=3$ .

But we can’t use subtraction since we haven’t defined it, so let’s reorder our terms to only use addition.

$(a,b)\sim(c,d)\iff a+d=b+c$

Defining -1

$-1\coloneqq [0,1]=\{(a,b)\in \mathbb{N}\times \mathbb{N}:a+1=b\}$

Here’s a quick explanation: $[0,1]$ contains a ‘representative’ of the above relation. This means that $[0,1]$ is one possible value for $-1$ under that relation. We then formalize that on the next line using $a+1=b$ . We did it! We defined $-1$ . Wasn’t that fun?

Have a good one, (ba-dum crash)
Ilan Bernstein