# Definition:Natural Numbers/Axiomatization

## Definition

The natural numbers $\N$ can be axiomatised in the following ways:

### Peano's Axioms

Peano's Axioms are intended to reflect the intuition behind $\N$, the mapping $s: \N \to \N: s \left({n}\right) = n + 1$ and $0$ as an element of $\N$.

Let there be given a set $P$, a mapping $s: P \to P$, and a distinguished element $0$.

Historically, the existence of $s$ and the existence of $0$ were considered the first two of Peano's Axioms. The other three are as follows:

 $(P3)$ $:$ $\displaystyle \forall m, n \in P:$ $\displaystyle \map s m = \map s n \implies m = n$ $s$ is injective $(P4)$ $:$ $\displaystyle \forall n \in P:$ $\displaystyle \map s n \ne 0$ $0$ is not in the image of $s$ $(P5)$ $:$ $\displaystyle \forall A \subseteq P:$ $\displaystyle \paren {0 \in A \land \paren {\forall z \in A: \map s z \in A} } \implies A = P$ Principle of Mathematical Induction: Any subset $A$ of $P$, containing $0$ and closed under $s$, is equal to $P$

### Naturally Ordered Semigroup

The concept of a naturally ordered semigroup is intended to capture the behaviour of the natural numbers $\N$, addition $+$ and the ordering $\le$ as they pertain to $\N$.

#### Naturally Ordered Semigroup Axioms

A naturally ordered semigroup is a (totally) ordered commutative semigroup $\left({S, \circ, \preceq}\right)$ satisfying:

 $(NO 1)$ $:$ $S$ is well-ordered by $\preceq$ $\displaystyle \forall T \subseteq S:$ $\displaystyle T = \varnothing \lor \exists m \in T: \forall n \in T: m \preceq n$ $(NO 2)$ $:$ $\circ$ is cancellable in $S$ $\displaystyle \forall m, n, p \in S:$ $\displaystyle m \circ p = n \circ p \implies m = n$ $\displaystyle p \circ m = p \circ n \implies m = n$ $(NO 3)$ $:$ Existence of product $\displaystyle \forall m, n \in S:$ $\displaystyle m \preceq n \implies \exists p \in S: m \circ p = n$ $(NO 4)$ $:$ $S$ has at least two distinct elements $\displaystyle \exists m, n \in S:$ $\displaystyle m \ne n$

### 1-Based Natural Numbers

The following axioms are intended to capture the behaviour of $\N_{>0}$, the element $1 \in \N_{>0}$, and the operations $+$ and $\times$ as they pertain to $\N_{>0}$:

 $(A)$ $:$ $\displaystyle \exists_1 1 \in \N_{> 0}:$ $\displaystyle a \times 1 = a = 1 \times a$ $(B)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a$ $(C)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle a + \paren {b + 1} = \paren {a + b} + 1$ $(D)$ $:$ $\displaystyle \forall a \in \N_{> 0}, a \ne 1:$ $\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1$ $(E)$ $:$ $\displaystyle \forall a, b \in \N_{> 0}:$ $\displaystyle$Exactly one of these three holds: $\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(F)$ $:$ $\displaystyle \forall A \subseteq \N_{> 0}:$ $\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$