What is a Negative Number?

What is $-1$ ? Effectively, subtracting is conceptualised as reverse-addition with the following property:

$a-b := -(b-a),\quad a<b.$

But that still begs the question on what the right-hand side refers to. We need to construct $\mathbb Z$ , and thankfully, we have $\mathbb N$ to help us with the process.

Intuitively, we want to say that

$-1 = 0-1 = 1-2 = 2-3 = \cdots.$

This suggests that we need to consider the pairs $(k, k+1)$ as equal in some meaningful way. We will informally denote such an equivalence using the symbol $\equiv$ . In a sense, then,

$(0,1) \equiv (1,2) \equiv (2,3) \equiv (3,4) \equiv \cdots.$

There’s something interesting here: observe that $1 + 4 = 2 + 3$ . This suggests that we can “link” the pairs up using this identity.

Definition 1. Denote $\mathbb N^2 = \mathbb N \times \mathbb N$ . For ordered pairs $(x_1,y_1), (x_2,y_2) \in \mathbb N^2$ , denote

$\begin{aligned} (x_1,y_1) \equiv (x_2,y_2) \quad &\Leftrightarrow \quad ( (x_1,y_1), (x_2,y_2)) \in\ \equiv \\ &\Leftrightarrow \quad x_1 + y_2 = x_2 + y_1. \end{aligned}$

In fact, this leads us to the study of equivalence relations, which are generalisations of $=$ in surprisingly useful ways.

Theorem 1. The relation $\equiv\ \subseteq \mathbb N^2 \times \mathbb N^2$ satisfies the following properties:

(Reflexivity) For any $(x,y) \in \mathbb N^2$ , $(x,y) \equiv (x,y)$ .

(Symmetry) For any $(x_1,y_1),(x_2,y_2) \in \mathbb N^2$ ,

$(x_1,y_1) \equiv (x_2,y_2) \quad \Rightarrow \quad (x_2,y_2) \equiv (x_1,y_1).$

(Transitivity) For any $(x_1,y_1),(x_2,y_2),(x_3,y_3) \in \mathbb N^2$ ,

$(x_1,y_1) \equiv (x_2,y_2) \wedge (x_2,y_2) \equiv (x_3,y_3) \quad \Rightarrow \quad (x_1,y_1) \equiv (x_3,y_3).$

Proof. Reflexivity is obvious since $x+y = x+y$ . Symmetry holds since

$\begin{aligned} (x_1,y_1) \equiv (x_2,y_2)\quad &\Leftrightarrow \quad x_1 + y_2 = x_2 + y_1 \\ &\Leftrightarrow \quad x_2 + y_1 = x_1 + y_2 \\ &\Leftrightarrow \quad (x_2,y_2) \equiv (x_1,y_1).\end{aligned}$

For transitivity, we first expand the definitions to obtain

$\begin{aligned}(x_1,y_1) \equiv (x_2,y_2) \quad &\Leftrightarrow \quad x_1 + y_2 = x_2 + y_1,\\(x_2,y_2) \equiv (x_3,y_3) \quad &\Leftrightarrow \quad x_2 + y_3 = x_3 + y_2. \end{aligned}$

Adding the equations yields

$\begin{aligned} (x_1 + y_3) + (x_2 + y_2) = (x_3 + y_1) + (x_2 + y_2).\end{aligned}$

By the cancellation law for natural number addition,

$x_1 + y_3 = x_3 + y_1 \quad \Leftrightarrow \quad (x_1,y_1) \equiv (x_3,y_3),$

as required.

This construction can be generalised to any general equivalence relation.

Definition 2. Let $K$ be a set. The relation $\sim\ \subseteq K \times K$ on $K$ is an equivalence relation if it satisfies the following three properties:

(Reflexivity) For any $x \in K$ , $x \sim x$ .
(Symmetry) For any $x_1,x_2 \in K$ , $x_1 \sim x_2 \Rightarrow x_2 \sim x_1$ .
(Transitivity) For any $x_1,x_2,x_3 \in K$ , $x_1 \sim x_2\wedge x_2 \sim x_3 \Rightarrow x_1 \sim x_3$ .

Note that we write $a \sim b\Leftrightarrow (a, b) \in\ \sim$ for readability.

We now have

$(0,1) \equiv (1,2) \equiv (2,3) \equiv (3,4) \equiv \cdots,$

but this seems to not define a single object, namely, $-1$ . To do that, we need to contain them all in a single box. More is true.

Theorem 2. Let $\sim$ be an equivalence relation on a set $K$ . For any $x \in K$ , define the set

$[x] := \{y \in K : x \sim y\} \subseteq K.$

Then the following properties hold:

For any $x \in K$ , $[x] \neq \emptyset$ .
For any $x,y \in K$ , $[x] \cap [y] \neq \emptyset \Leftrightarrow [x] = [y]$ .
$\bigcup_{x \in K} [x] = K$ .

In this case, define the quotient set of $K$ by

$K/{\sim} := \{[x] : x \in K\}.$

We say that $K/{\sim}$ forms a partition of $K$ .

Proof. The first and third properties, as well as the $(\Leftarrow)$ direction of the second property, are immediate by the reflexivity of $\sim$ , which gives $x \sim x \Rightarrow x \in [x]$ .

For the $(\Rightarrow)$ direction of the second property, assume that $[x] \cap [y] \neq \emptyset$ . Then there exists $z \in K$ such that $z \in [x]$ and $z \in [y]$ . This implies $x \sim z$ and $y \sim z$ . By symmetry, $z \sim y$ . By transitivity, $x \sim y$ .

We now aim to prove that $[x] = [y]$ . For the direction $(\subseteq)$ , fix $w \in [x]$ . This means that $x \sim w$ . By symmetry, $w \sim x$ . By transitivity, $w \sim y$ . By definition, $w \in [y]$ . Therefore, $[x] \subseteq [y]$ .

For the direction $(\supseteq)$ , we don’t need to repeat our hard work. We have shown that $x \sim y \Rightarrow [x] \subseteq [y]$ . By symmetry, $y \sim x$ . Therefore, $[y] \subseteq [x]$ , as required.

Since $\equiv$ is an equivalence relation on $\mathbb N^2$ , we have

$[(0,1)] := \{(x,y) \in \mathbb N^2 : (0,1) \equiv (x,y)\}.$

Since

$(0,1) \equiv (1,2) \equiv (2,3) \equiv (3,4) \equiv \cdots,$

they all belong to $[(0,1)]$ by transitivity. This is how we will define $\mathbb Z$ .

Definition 3 (Integers). The set of integers is defined to be $\mathbb Z := \mathbb N^2/{\equiv}$ . For any natural number $n \in \mathbb N$ , denote $-n := [(0,n)]$ . This is how we define a negative number.

To finally answer the question we started with: $-1 = [(0,1)]$ .

But how do we know that this agrees with the negative numbers that we were indoctrinated to believe in? We answer this question by completing the needful bookkeeping in the next post.

—Joel Kindiak, 5 Dec 2024, 0007H

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