# Division Theorem for Ordinals

## Theorem

Let $x$ and $y$ be ordinals.

Let $0$ denote the zero ordinal.

Suppose $y \ne 0$.

Then there exist unique ordinals $z$ and $w$ such that:

- $x = \paren {y \times z} + w$ and $w < y$.

## Proof

### Existence of $z$ and $w$

First, it must be proven that such $z$ and $w$ exist.

By Relation between Two Ordinals, it follows that:

- $x < y$ or $y \le x$

The proof shall proceed by cases.

#### Case 1

If $x < y$, then set $z = 0$ and $w = x$.

Since $w = x$, then $w < y$.

Furthermore, by Ordinal Multiplication by Zero:

- $x = \paren {y \times z} + w$

#### Case 2

If $y \le x$, then set $z = \bigcup \set {v : \paren {y \times v} \le x}$.

Set $w$ equal to the unique $u$ such that $\paren {y \times z} + u = x$.

By Ordinal Multiplication by One, it follows that $1 \le z$.

By Union of Ordinals is Least Upper Bound, it follows that $z$ is the least upper bound of $\set {v : \paren {y \times v} \le x}$.

Take any $u \in z$.

Then $u \in v$ for some $\paren {y \times v} \le x$

But then, by Membership is Left Compatible with Ordinal Multiplication, it follows that $\paren {y \times u} \le x$.

So $z$ is a transitive subset of $\On$ and therefore, it is an ordinal.

Since $z \ne 0$, then by the definition of limit ordinal, then $z = u^+$ for some $u$ or $z$ is a limit ordinal.

Suppose $z = u^+$.

Then $u < z$ and $y \times u \le x$.

Moreover, $u < v$ for some $y \times v \le x$.

This means that $v \not < z$ by No Natural Number between Number and Successor.

By Ordinal Membership is Trichotomy, $z \le v$.

Therefore:

- $y \times z \le x$

Suppose $z \in K_{II}$.

Note that $\forall u \in z: y \times u \le x$ so:

\(\ds y \times z\) | \(=\) | \(\ds \bigcup_{u \mathop \in z} \paren {y \times u}\) | Definition of Ordinal Multiplication | |||||||||||

\(\ds \) | \(\le\) | \(\ds \bigcup_{u \mathop \in z} x\) | Indexed Union Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds x\) | Set Union is Idempotent |

In either case, $y \times z \le x$.

Moreover, suppose that $y \times z^+ \le x$.

Then $z$ is not an upper bound of the set $\set {v : \paren {y \times v} \le x}$, which is a contradiction.

So $\paren {y \times z} + w = x$ for some $w \in y$ by Ordinal Subtraction when Possible is Unique.

### Uniqueness of $z$ and $w$

Suppose that $\paren {y \times z_1} + w_1 = x$ for some $w_1 \in y$.

Suppose that $\paren {y \times z_2} + w_2 = x$ for some $w_2 \in y$.

Then for both $z = z_1$ and $z = z_2$, the following inequality is established:

\(\ds y \times z\) | \(\le\) | \(\ds x\) | Ordinal is Less than Sum | |||||||||||

\(\ds \) | \(<\) | \(\ds \left({ y \times z }\right) + y\) | Membership is Left Compatible with Ordinal Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds y \times z^+\) | Definition of Ordinal Multiplication |

It follows that $y \times z_1 < y \times z_2^+$ and $y \times z_2 < y \times z_1^+$.

By Membership is Left Compatible with Ordinal Multiplication, $z_1 < z_2^+$ and $z_2 < z_1^+$.

\(\ds z_1 < z_2^+\) | \(\leadsto\) | \(\ds z_1 < z_2 \lor z_1 = z_2\) | Definition of Successor Set | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds z_1 \le z_2\) | ||||||||||||

\(\ds z_2 < z_1^+\) | \(\leadsto\) | \(\ds z_2 \le z_1\) | Definition of Successor Set | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds z_1 = z_2\) | Definition 2 of Set Equality |

Therefore $z$ is unique.

Since $z$ is unique, then $w$ is unique by Ordinal Subtraction when Possible is Unique.

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 8.27$