natural number

Given the Zermelo-Fraenkel axioms of set theory, one can prove that there exists an inductive set $X$ such that $\mathrm{\varnothing }\in X$. The natural numbers $\mathbb{N}$ are then defined to be the intersection of all subsets of $X$ which are inductive sets and contain the empty set as an element.

The first few natural numbers are:

• $0:=\mathrm{\varnothing }$

• $1:=0^{\prime }=\{0\}=\{\mathrm{\varnothing }\}$

• $2:=1^{\prime }=\{0,1\}=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$

• $3:=2^{\prime }=\{0,1,2\}=\{\mathrm{\varnothing },\{\mathrm{\varnothing }\},\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}\}$

Note that the set $0$ has zero elements, the set $1$ has one element, the set $2$ has two elements, etc. Informally, the set $n$ is the set consisting of the $n$ elements $0,1,\mathrm{\dots },n-1$, and $n$ is both a subset of $\mathbb{N}$ and an element of $\mathbb{N}$.

In some contexts (most notably, in number theory), it is more convenient to exclude $0$ from the set of natural numbers, so that $\mathbb{N}=\{1,2,3,\mathrm{\dots }\}$. When it is not explicitly specified, one must determine from context whether $0$ is being considered a natural number or not.

Addition of natural numbers is defined inductively as follows:

• $a+0:=a$ for all $a\in \mathbb{N}$

• $a+b^{\prime }:={(a+b)}^{\prime }$ for all $a,b\in \mathbb{N}$

Multiplication of natural numbers is defined inductively as follows:

• $a\cdot 0:=0$ for all $a\in \mathbb{N}$

• $a\cdot b^{\prime }:=(a\cdot b)+a$ for all $a,b\in \mathbb{N}$

The natural numbers form a monoid under either addition or multiplication. There is an ordering relation on the natural numbers, defined by: $a\le b$ if $a\subseteq b$.