Chapter 1Foundationsnotes.en.pdf:26; slides.en.pdf:59-60

Peano Axioms (P1–P5)

Five axioms nail down the naturals

Def 2.1.6 — Peano AxiomsDef 1.6 — Peano Axioms (slides)
Theorem 1.7-1.11 — Axioms P1-P5 as separate
Concept

The Peano axioms nail down what we mean by "the natural numbers".
There are five of them:

P1. 00 is a natural number.

0∈N0 \in \mathbb{N}

P2. Every natural number nn has a successor s(n)s(n).

∀n∈N. s(n)∈N\forall n \in \mathbb{N}.\ s(n) \in \mathbb{N}

P3. 00 is not the successor of anything.

∀n∈N. s(n)≠0\forall n \in \mathbb{N}.\ s(n) \neq 0

P4. The successor function is injective.

∀m,n∈N. s(m)=s(n)⇒m=n\forall m, n \in \mathbb{N}.\ s(m) = s(n) \Rightarrow m = n

P5. (Induction) If a set S⊆NS \subseteq \mathbb{N} contains 00 and is closed
under ss, then S=NS = \mathbb{N}.

∀S⊆N. (0∈S∧∀n.n∈S⇒s(n)∈S)⇒S=N\forall S \subseteq \mathbb{N}.\ (0 \in S \land \forall n. n \in S \Rightarrow s(n) \in S) \Rightarrow S = \mathbb{N}

These five axioms are enough to derive all of elementary arithmetic.

Animation — peano construction
Transcript — click a line to jump8 cues
  1. 0.0sPeano Arithmetic: Building the Natural Numbers
  2. 1.3sTwo rules: o-rule seeds 0 in N; s-rule lifts n to s(n).
  3. 3.1sA box on the right will collect every number we build.
  4. 3.9sStep one: 0 enters by the o-rule.
  5. 4.9sStep two: s(0) arrives via the s-rule.
  6. 6.3sStep three: s(s(0)) keeps applying s.
  7. 7.7sStep four: s(s(s(0))) by s-rule once more.
  8. 10.0sEach natural is built by finitely many o or s steps.
Worked example
Step 0 of 3
Practice — score 100% to advance
Multiple choice
Q1
Which axiom says the successor function is injective?
Q2
Which axiom is the induction principle?
Q3
What does P3 say?
Q4
Which axiom lets us prove 0=00 = 0?
Q5
Are the Peano axioms enough to derive all of elementary arithmetic?
Q6
Which is FALSE?
Match definitions
Match each concept on the left to its definition on the right.
Order the steps
Arrange these proof steps in the correct order using the arrows.
1
By P3, 00 is not the successor of anything, including nn.
2
By P4 (injectivity of ss), n=s(n)n = s(n).
3
But n=s(n)n = s(n) means nn is its own successor.
4
Assume for contradiction that s(n)=s(s(n))s(n) = s(s(n)).
5
Contradiction. So ∀n.s(n)≠s(s(n))\forall n. s(n) \neq s(s(n)). ∎
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