# Principle of Mathematical Induction/Peano Structure

## Theorem

Let $\struct {P, s, 0}$ be a Peano structure.

Let $\map Q n$ be a propositional function depending on $n \in P$.

Suppose that:

- $(1): \quad \map Q 0$ is true

- $(2): \quad \forall n \in P: \map Q n \implies \map Q {\map s n}$

Then:

- $\forall n \in P: \map Q n$

### Principle of Finite Induction

Let $\struct {P, s, 0}$ be a Peano structure.

Let $S \subseteq P$.

Suppose that:

- $(1): \quad 0 \in S$

- $(2): \quad \forall n: n \in S \implies \map s n \in S$

Then:

- $S = P$

## Proof

Let $A \subseteq P$ be defined by:

- $A := \set {n \in P: \map Q n}$

From $(1)$, $0 \in A$.

From $(2)$:

- $\forall n \in P: n \in A \implies \map s n \in A$

As this holds for all $n \in P$, it holds a fortiori for all $n \in A$.

Thus the condition:

- $n \in A \implies \map s n \in A$

is satisfied.

So by Axiom $(\text P 5)$ of the Peano Axioms:

- $A = P$

That is:

- $\forall n \in P: \map Q n$

$\blacksquare$

## Also defined as

Some treatments of Peano's axioms define the non-successor element (or **primal element**) to be $1$ and not $0$.

The treatments are similar, but the $1$-based system results in an algebraic structure which has no identity element for addition, and so no zero for multiplication.

## Also see

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts*... (previous) ... (next): Introduction $\S 4$: The natural numbers - 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 1$: Introduction

- For a video presentation of the contents of this page, visit the Khan Academy.