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Peano’s Arithmetic

As a reminder, here are Peano’s axioms, which concern the successor function s:\mathbb N\to \mathbb N, which intuitively “adds 1” to each natural integer.
These axioms allow us to reconstruct the whole arithmetic structure of the set \mathbb N of natural integers, and from naive set theory, all natural mathematical objects and structures.

Axiom 1

0 is not the successor of any natural number. In other words, there is no natural number n such that s(n)=0.

This axiom says in particular that the function s is not surjective, since the number 0 has no antecedent by s.

Axiom 2

If two natural numbers have the same successor, then they are equal. In other words, for all natural numbers m,n, if s(m)=s(n) then m=n.

We can reformulate this axiom by saying that the successor application is injective. It thus defines a bijection from \mathbb N onto s(\mathbb N). We recognise here the characterisation of an infinite set.

Proposition 1

The set \mathbb N of natural numbers is infinite.

Axiom 3 [Principle of induction (or recursion)]

If S is a subset of \mathbb N such that:

  1. 0\in S
  2. for all n\in S, s(n)\in S (“induction step”),

then we have S=\mathbb N.

This principle expresses intuitively that the set \mathbb N is entirely \og browsed” if we enumerate it starting from 0 and add each successive natural number, indefinitely.

We can deduce the determination of the image of s :

Proposition 2

If we admit the induction principle, then the image of s is \mathbb N^*, the set of non-zero natural numbers.


Let S be the set \{0\}\cup Im(s)=\{0\}\cup\{n\in\mathbb N: \exists m\in\mathbb N,\ n=s(m)\}. If we show that S=\mathbb N, then any non-zero natural number is in the image of s, so Im(s)=\mathbb N^*. By definition, we have 0\in S, and suppose that n is a natural number, and that n\in S. By definition, the natural integer s(n) is in S! By the induction principle, the set S is the whole set \mathbb N, and the proposition is proved. \square

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