# Principle of Mathematical Induction/One-Based

## Theorem

Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.

Suppose that:

- $(1): \quad \map P 1$ is true

- $(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$

Then:

- $\map P n$ is true for all $n \in \N_{>0}$.

## Proof 1

Let $S$ be the set defined as:

- $S := \set {n \in \N_{>0}: \map P n \text { is false} }$

Aiming for a contradiction, suppose $S \ne \O$.

From the Well-Ordering Principle it follows that $S$ has a minimal element $m$.

From $(1)$ we have that $\map P 1$ holds.

Hence $1 \notin S$.

Therefore $m \ne 1$.

Therefore $m - 1 \in \N_{>0}$.

But $m$ is the minimal element of $S$.

So $m - 1 \notin S$.

Therefore $\map P {m - 1}$ is true.

Hence by $(2)$ it follows that $\map P m$.

But then $m \notin S$.

This contradicts our supposition that $m \in S$.

Hence there can be no such $m \in S$.

So $S = \O$ and the result follows.

$\blacksquare$

## Proof 2

Let $M$ be the set of all $n \in \N_{>0}$ for which $\map P n$ holds.

By $(1)$ we have that $1 \in M$.

By $(2)$ we have that if $k \in M$ then $k + 1 \in M$.

From the Axiomatization of $1$-Based Natural Numbers, Axiom $(F)$, it follows that $M = \N_{>0}$.

$\blacksquare$

## Proof 3

We have that Natural Numbers are Non-Negative Integers.

Then we have that Integers form Well-Ordered Integral Domain.

The result follows from Induction on Well-Ordered Integral Domain.

$\blacksquare$

## Also see

- Results about
**Proofs by Induction**can be found here.

## Sources

- 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf{I}$ - 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.4.2$: Mathematical induction: Definition $1.29$

- 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): Appendix $\text{A}$: Set Theory: Induction - 2012: M. Ben-Ari:
*Mathematical Logic for Computer Science*(3rd ed.) ... (previous): Appendix $\text{A}.6$