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For $α > 0$ every ordinal $β$ is $α \cdot ξ + ρ$ with $ρ < α$ , in exactly one way

Statement

Let $α$ and $β$ be ordinals (Ordinal (von Neumann)) with $α > 0$ . Then there are unique ordinals $ξ$ and $ρ$ with

$β = α \cdot ξ + ρ and ρ < α .$

$ξ$ is the quotient and $ρ$ the remainder of $β$ on division by $α$ ; concretely, $ξ$ is the largest ordinal with $α \cdot ξ \leq β$ , and $ρ$ is what For $α \leq β$ there is exactly one ordinal $γ$ with $α + γ = β$ returns from $α \cdot ξ \leq β$ .

No choice principle is used.

Facts & Assumptions

Given: Ordinals $α > 0$ and $β$ , with $+$ and $\cdot$ as in Ordinal addition $α + β$ and Ordinal multiplication $α \cdot β$ . For a set $A$ of ordinals, $\sup A = ⋃ A$ is its least upper bound.

[L1]

$α \cdot 0 = 0$ , $α \cdot δ^{+} = α \cdot δ + α$ , and $α \cdot λ = \sup {α \cdot ζ : ζ \in λ}$ for limit $λ$ (Ordinal multiplication $α \cdot β$ ).

[L2]

From Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + β = β$ and $1 \cdot β = β$ : $1 \cdot μ = μ$ and $μ + 0 = μ$ (claim (a)); $μ < ν$ implies $α + μ < α + ν$ , left cancellation for $+$ , and $α \leq α + μ$ (claim (b)); for $α > 0$ , $μ < ν$ implies $α μ < α ν$ , hence $μ \leq ν$ implies $α μ \leq α ν$ (claim (d)); and $μ \leq ν$ implies $μ γ \leq ν γ$ (claim (e)).

[L3]

If $μ \leq ν$ there is exactly one $γ$ with $μ + γ = ν$ (For $α \leq β$ there is exactly one ordinal $γ$ with $α + γ = β$ ).

[L4]

Every nonempty set of ordinals has an $\in$ -least element, and exactly one of $μ \in ν$ , $μ = ν$ , $ν \in μ$ holds (Trichotomy and well-ordering of the ordinals).

[L5]

$μ^{+}$ is an ordinal, $μ \subseteq ν$ if and only if $μ \in ν$ or $μ = ν$ , and $μ \notin μ$ (claims (b), (c), (f) of Basic closure properties of ordinals); consequently $μ < ν$ if and only if $μ^{+} \leq ν$ .

[L6]

Every ordinal is exactly one of $0$ , a successor, or a limit (Successor and limit ordinals).

Proof

technique · direct

1.1

$β < α \cdot β^{+}$ : since $α > 0$ gives $1 \leq α$ , claim (e) of [L2] gives $β^{+} = 1 \cdot β^{+} \leq α \cdot β^{+}$ , and $β \in β^{+}$ .

L1L2L5

1.2

Uniqueness: suppose $α ξ_{1} + ρ_{1} = α ξ_{2} + ρ_{2} = β$ with $ρ_{1}, ρ_{2} < α$ ; if $ξ_{1} < ξ_{2}$ then $ξ_{1}^{+} \leq ξ_{2}$ by [L5], so $α ξ_{1} + α = α ξ_{1}^{+} \leq α ξ_{2} \leq α ξ_{2} + ρ_{2} = β = α ξ_{1} + ρ_{1} < α ξ_{1} + α$ by [L1] and [L2], which [L5] forbids; by symmetry $ξ_{2} < ξ_{1}$ is impossible too, so $ξ_{1} = ξ_{2}$ by [L4] and then $ρ_{1} = ρ_{2}$ by left cancellation.

L1L2L4L5

2.1

The collection $C = {η \in (β^{+})^{+} : β \in α \cdot η}$ is a set of ordinals by Separation, and it is nonempty, because $β^{+} \in (β^{+})^{+}$ and $β \in α \cdot β^{+}$ by step 1.1.

step 1.1L5

3.1

Let $η_{0}$ be the $\in$ -least element of $C$ , which exists by [L4].

step 2.1L4

4.1

$η_{0}$ is a successor: it is not $0$ , since $α \cdot 0 = 0$ and $β \notin 0$ ; and it is not a limit $λ$ , for then every $ζ \in λ$ would lie in $(β^{+})^{+}$ by transitivity and outside $C$ by minimality, so $α ζ \leq β$ by [L4], making $β$ an upper bound of ${α ζ : ζ \in λ}$ and hence $α \cdot λ \leq β$ by [L1], contradicting $β \in α \cdot λ$ ; so $η_{0} = ξ^{+}$ for a unique ordinal $ξ$ by [L6].

step 3.1L1L4L5L6

5.1

With that $ξ$ : $ξ \in η_{0} \subseteq (β^{+})^{+}$ and $ξ \notin C$ by minimality of $η_{0}$ , so $α ξ \leq β$ by [L4]; and $β \in α \cdot ξ^{+} = α ξ + α$ because $η_{0} = ξ^{+} \in C$ .

step 4.1step 3.1L1L4L5

6.1

By [L3] applied to $α ξ \leq β$ there is exactly one $ρ$ with $α ξ + ρ = β$ , and $α ξ + ρ = β < α ξ + α$ forces $ρ < α$ , since $α \leq ρ$ would give $α ξ + α \leq α ξ + ρ$ by [L2].

step 5.1L2L3L4

7.1

Existence is step 6.1 and uniqueness is step 1.2, so $β = α \cdot ξ + ρ$ with $ρ < α$ in exactly one way.

step 6.1step 1.2∎

Remarks

Why the least $η$ with $β < α \cdot η$ has to be a successor. Because $η \mapsto α \cdot η$ is continuous at limits: at a limit stage its value is the supremum of the earlier values, so it cannot overtake $β$ for the first time there. That is claim (f) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + β = β$ and $1 \cdot β = β$ in the form used at step 4.1, and it is the only place the limit clause of Ordinal multiplication $α \cdot β$ is used.

The bound $(β^{+})^{+}$ is a Separation device. "The least $η$ with $β < α η$ " quantifies over all ordinals, which is not a set; step 1.1 supplies a specific witness inside $(β^{+})^{+}$ , so the collection can be cut out of a set. Nothing depends on the particular bound.

Uniqueness is proved before existence, and independently of it. Step 1.2 uses only the monotonicity laws, so it applies to any two representations whatever their origin. This is the order used again in Cantor normal form: every nonzero ordinal is $ω^{β_{0}} \cdot c_{0} + \dots + ω^{β_{k - 1}} \cdot c_{k - 1}$ with $β_{0} > \dots > β_{k - 1}$ and each $c_{i}$ a nonzero natural number, in exactly one way, where uniqueness is what licenses the definite article in "the Cantor normal form".

The remainder can be $0$ and the quotient can be $0$ . If $β < α$ then $ξ = 0$ and $ρ = β$ ; if $α$ divides $β$ exactly then $ρ = 0$ . Neither case is excluded, and neither needs separate treatment.

Depends on

Used by

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20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ordinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 2 (Ordinal numbers) (standard reference, not scraped)

For α>0α>0 every ordinal ββ is α⋅ξ+ρα⋅ξ+ρ with ρ<αρ<α, in exactly one way