The Origin of Numbers

We will now use sets and functions to define the primary star of mathematics—the natural numbers.

Definition 1 (Natural Numbers). The natural numbers, denoted $\mathbb N$ , is defined as follows:

$0 := \emptyset \in \mathbb N$ .
For any $n \in \mathbb N$ , $n \cup \{n\} \in \mathbb N$ .

We assert that this set exists by the axiom of infinity. Using these two rules, we obtain:

$0 = \emptyset \in \mathbb N$ ,
$1 := 0 \cup \{0\} = \emptyset \cup \{0\} = \{0\} \in \mathbb N$ ,
$2 := 1 \cup \{1\} = \{0\} \cup \{1\} = \{0,1\} \in \mathbb N$ ,
$3 := 2 \cup \{2\} = \{0,1\} \cup \{2\} = \{0,1,2\} \in \mathbb N$ ,

so on and so forth. In this formulation of $\mathbb N$ , we will assume that $0 \in \mathbb N$ for convenience.

Define the successor function $s : \mathbb N \to \mathbb N$ by $s(n) = n \cup \{n\}$ . The natural numbers satisfies five crucial properties known as Peano’s axioms.

Theorem 1 (Peano’s Axioms). The natural numbers satisfies Peano’s axioms:

$0 \in \mathbb N$ .
For any $n \in \mathbb N$ , $s(n) \in \mathbb N$ .
For any $n \in \mathbb N$ , $s(n) \neq 0$ .
$s$ is injective.

Furthermore, for any subset $K \subseteq \mathbb N$ ,

$0 \in K \wedge (\forall n \in \mathbb N\ (n \in K \Rightarrow s(n) \in K)) \quad \Rightarrow \quad K = \mathbb N.$

Proof. The proofs of the first three properties are obvious, the fourth property is nontrivial, and we will prove the fifth property here. Observe that for the fifth property,

$\mathbb N \subseteq K \cap \mathbb N \subseteq K.$

In particular, if $K \subseteq \mathbb N$ , then $K = \mathbb N$ , as required.

For completeness, let’s prove the fourth property as well. Fix $m, n \in \mathbb N$ and suppose $s(m) = s(n)$ . Then $m \cup \{m\} = n \cup \{n\}$ . If $m \neq n$ , then $m \in n$ and symmetrically, $n \in m$ . Since $m \in n$ , we have $m \subseteq n$ , so that $n \in n$ , a contradiction to the axiom of regularity.

The fifth property is precisely what creates the proof technique known as proof by mathematical induction.

Theorem 2 (Mathematical Induction). Let $\phi(n)$ be a predicate on $\mathbb N$ . Suppose $\phi(0)$ and for any $n \in \mathbb N$ , $\phi(n) \Rightarrow \phi(s(n))$ . Then $\forall n \in \mathbb N\ \phi(n)$ .

Proof. Define $K := \{n \in \mathbb N : \phi(n)\} \subseteq \mathbb N$ .

Since $\phi(0)$ , $0 \in K$ .
For any $n \in \mathbb N$ , $n \in K \Rightarrow \phi(n) \Rightarrow \phi(s(n)) \Rightarrow s(n) \in K$ .

By the fifth of Peano’s axioms, $K = \mathbb N$ . In particular, fix $n \in \mathbb N$ . Then $n \in K$ . This implies $\phi(n)$ , as required.

It is induction that allows us to formally define $+$ and $\times$ in terms of $+$ , and show that they satisfy the properties that we expect they do.

Definition 2 (Addition). Define the addition map $+ : \mathbb N \times \mathbb N \to \mathbb N$ as follows:

For any $m \in \mathbb N$ , $+(m, 0) := m$
For any $m,n \in \mathbb N$ , $+(m, s(n)) := s(+(m,n))$ .

For readability purposes, we denote $+(m, n) = m + n$ , so that $+$ is defined inductively by the following:

For any $m \in \mathbb N$ , $m + 0 := m$
For any $m,n \in \mathbb N$ , $m + s(n) := s(m+n)$ .

Theorem 3. $1 + 1 = 2$ .

Proof. Using $s(0) = 0 \cup \{0\} = 1$ , we have

$s(n) = s(n+0) = n+s(0) = n+1,$

so that setting $n = 1$ we have the comically rigorous proof

$1 + 1 = s(1) = 1 \cup \{1\} = 2.$

Many of the properties of addition follow from applying induction on the desired results. Here, we will prove a simple result.

Theorem 4. For any $k,m, n \in \mathbb N$ , $(k+m)+n = k+(m+n)$ .

Proof. Fix $k, m \in \mathbb N$ . Define the predicate $\phi(n)$ on $\mathbb N$ by

$\phi(n) = (\forall n \in \mathbb N\quad (k+m)+n = k+(m+n)).$

We claim that $\forall n \in \mathbb N\ \phi(n)$ by induction.

For the base case,

$(k+m) + 0 = k+m = k+(m+0)\quad \Rightarrow \quad \phi(0).$

For the induction step, fix $n \in \mathbb N$ and suppose $\phi(n)$ . Then

$\begin{aligned}(k+m) + s(n) &= s((k+m) + n) \\ &= s(k+(m+n)) \\ &= k + s(m+n) \\ &= k + (m + s(n)).\end{aligned}$

Therefore, $\phi(s(n))$ , as required.

Furthermore, since $s(n) = n+1$ , we can write induction in terms of its usual formulation.

Corollary 1. Let $\phi(n)$ be a predicate on $\mathbb N$ . Suppose $\phi(0)$ and for any $n \in \mathbb N$ , $\phi(n) \Rightarrow \phi(n+1)$ . Then $\forall n \in \mathbb N\ \phi(n)$ .

Multiplication also ought to make intuitive sense, once we define it formally, and satisfy the properties we expect it to through various proofs by induction.

Definition 3 (Multiplication). Define the multiplication map $+ : \mathbb N \times \mathbb N \to \mathbb N$ as follows:

For any $m \in \mathbb N$ , $\times(m, 0) := 0$
For any $m,n \in \mathbb N$ , $\times(m, s(n)) := \times(m,n) + m$ .

For readability purposes, we denote $\times(m, n) = m \times n \equiv mn$ .

We can also order the natural numbers in the expected manner, formulated in the language of relations.

Definition 4 (Ordering). Define the not-more-than relation $\leq$ by

$\leq \quad:= \quad\{(m,n) \in \mathbb N \times \mathbb N : (\exists k \in \mathbb N : n = m + k)\} \quad \subseteq \quad \mathbb N \times \mathbb N .$

We write $a \leq b$ to mean $(a, b) \in\ \leq$ for readability.

With this, we have finally defined properly the natural numbers $\mathbb N$ . With a few more discrete mathematical tools, we can define the integers $\mathbb Z$ and the rational numbers $\mathbb Q$ .

Assuming the existence of the real numbers $\mathbb R$ , we can define the complex numbers $\mathbb C$ as well, giving us the chain of commonly used number systems: $\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$ , each number system an extension of the previously discussed one.

—Joel Kindiak, 4 Dec, 0410H

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The Origin of Numbers

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