Given the Zermelo-Fraenkel 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:

• $0 := \emptyset$
• $1 := 0' = \{0\} = \{ \emptyset \}$
• $2 := 1' = \{0,1\} = \{\emptyset, \{ \emptyset \} \}$
• $3 := 2' = \{0,1,2\} = \{\emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}$

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 $0, 1, \dots, n-1$ , 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 $\mathbb{N} = \{1,2,3,\dots\}$ . 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 $a \in \mathbb{N}$
• $a + b' := (a+b)'$ for all $a,b \in \mathbb{N}$

Multiplication of natural numbers is defined inductively as follows:

• $a \cdot 0 := 0$ for all $a \in \mathbb{N}$
• $a \cdot b' := (a\cdot b) + a$ for all $a,b \in \mathbb{N}$