# Ordinal Multiplication by One

## Theorem

Let $x$ be an ordinal.

Let $1$ denote the ordinal one.

 $\ds x \cdot 1$ $=$ $\ds x$ $\ds 1 \cdot x$ $=$ $\ds x$

## Proof

 $\ds x \cdot 1$ $=$ $\ds x \cdot \O^+$ Definition of One (Ordinal) $\ds$ $=$ $\ds \paren {x \cdot \O} + x$ Definition of Ordinal Multiplication $\ds$ $=$ $\ds \O + x$ Definition of Ordinal Multiplication $\ds$ $=$ $\ds x$ Ordinal Addition by Zero

$\Box$

The proof of the other equality shall proceed by Transfinite Induction.

### Basis for the Induction

 $\ds 1 \cdot \O$ $=$ $\ds \O$ Definition of Ordinal Multiplication

This proves the basis for the induction.

### Induction Step

 $\ds 1 \cdot x$ $=$ $\ds x$ Inductive Hypothesis $\ds \leadsto \ \$ $\ds \paren {1 \cdot x} + 1$ $=$ $\ds x^+$ Ordinal Addition by One $\ds 1 \cdot x^+$ $=$ $\ds \paren {1 \cdot x} + 1$ Definition of Ordinal Multiplication $\ds \leadsto \ \$ $\ds 1 \cdot x^+$ $=$ $\ds x^+$ Equality is Transitive

This proves the induction step.

### Limit Case

 $\ds \forall y \in x: \,$ $\ds 1 \cdot y$ $=$ $\ds y$ by hypothesis $\ds \leadsto \ \$ $\ds \bigcup_{y \mathop \in x} \paren {1 \cdot y}$ $=$ $\ds \bigcup_{y \mathop \in x} y$ Indexed Union Equality $\ds \leadsto \ \$ $\ds 1 \cdot x$ $=$ $\ds \bigcup_{y \mathop \in x} y$ Definition of Ordinal Multiplication $\ds \leadsto \ \$ $\ds 1 \cdot x$ $=$ $\ds x$ Union of Limit Ordinal

This proves the limit case.

$\blacksquare$