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Homenatural number
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natural number
Given the ZermeloFraenkel axioms of set theory, one can prove that there exists an inductive set $X$ such that $\emptyset\in X$. 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:

$0:=\emptyset$

$1:=0^{{\prime}}=\{0\}=\{\emptyset\}$

$2:=1^{{\prime}}=\{0,1\}=\{\emptyset,\{\emptyset\}\}$

$3:=2^{{\prime}}=\{0,1,2\}=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$
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 $0,1,\dots,n1$, 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 $\mathbb{N}=\{1,2,3,\dots\}$. 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: $a\leq b$ if $a\subseteq b$.
Mathematics Subject Classification
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