False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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FALSE: ordinal multiplication is commutative

Statement

FALSE. Ordinal multiplication (Ordinal multiplication $α \cdot β$ ) is commutative: $α \cdot β = β \cdot α$ for all ordinals $α$ and $β$ .

It fails at the smallest possible place: $2 \cdot ω = ω$ , while $ω \cdot 2 = ω + ω$ , which is strictly larger.

Facts & Assumptions

Given: The ordinals with the operations of Ordinal addition $α + β$ and Ordinal multiplication $α \cdot β$ , and $ω$ the least limit ordinal ( $ω$ is the least limit ordinal, Successor and limit ordinals).

[L1]

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

[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 $1 \cdot β = β$ : $1 \cdot μ = μ \cdot 1 = μ$ (claim (a)); $ν < θ$ implies $α + ν < α + θ$ (claim (b)); $μ \leq ν$ implies $μ γ \leq ν γ$ (claim (e)).

[L3]

For $m, n \in ω$ the ordinal $m \cdot n$ lies in $ω$ (claim (a) of On $ω$ the ordinal $+$ and $\cdot$ are the Peano operations: $ω$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product).

[L4]

$ω$ is a limit ordinal, so $⋃ ω = ω$ and $0 \in ω$ ( $ω$ is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, $μ \subseteq ν$ iff $μ \in ν$ or $μ = ν$ , and $μ \notin μ$ (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals).

Refutation

technique · direct

1.1L2L3L4

For every $n \in ω$ the ordinal $2 \cdot n$ lies in $ω$ by [L3], hence $2 \cdot n \subseteq ω$ by [L4]; and $n = 1 \cdot n \leq 2 \cdot n$ by [L2], since $1 \leq 2$ , hence $n \subseteq 2 \cdot n$ .

1.2L1L2

$ω \cdot 2 = ω \cdot 1^{+} = ω \cdot 1 + ω = ω + ω$ by [L1] and [L2].

2.1step 1.1L1L4

$2 \cdot ω = ⋃ {2 \cdot n : n \in ω}$ by [L1], and that union equals $ω$ : it is contained in $ω$ because each $2 \cdot n \subseteq ω$ by step 1.1, and it contains $ω$ because $ω = ⋃ ω = ⋃ {n : n \in ω}$ by [L4] and each $n \subseteq 2 \cdot n$ by step 1.1.

2.2step 1.2L1L2L4

$ω + ω \neq ω$ : since $0 \in ω$ , claim (b) of [L2] gives $ω = ω + 0 < ω + ω$ , and $μ \notin μ$ by [L4].

3.1step 2.1step 2.2step 1.2∎

Therefore $2 \cdot ω = ω$ while $ω \cdot 2 = ω + ω \neq ω$ , so $2 \cdot ω \neq ω \cdot 2$ and ordinal multiplication is not commutative.

Remarks

The picture. By $α \cdot β$ is the order type of $α \times β$ ordered by last differences, that is $β$ copies of $α$ , $2 \cdot ω$ is $ω$ copies of a two element set, laid end to end: that is a copy of $ω$ , since relabelling gives $0, 1, 2, \dots$ again. And $ω \cdot 2$ is two copies of $ω$ , one entirely above the other, which is $ω + ω$ and has no greatest element but does have an element with infinitely many predecessors. The convention that fixes which is which is stated in Ordinal multiplication $α \cdot β$ : the successor clause appends a copy of $α$ on the right, so $α \cdot β$ is $β$ copies of $α$ .

What survives. Multiplication is still associative and still distributes over addition on the left (Ordinal multiplication is associative, and $α \cdot (β + γ) = α \cdot β + α \cdot γ$ ), and it is still strictly increasing and cancellative in the right argument when the left factor is nonzero (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 $1 \cdot β = β$ ). Right distributivity is a separate casualty, refuted in FALSE: $(β + γ) \cdot α = β \cdot α + γ \cdot α$ for all ordinals.

Finite ordinals are not a counterexample to anything. On $ω$ the ordinal product is the Peano product (On $ω$ the ordinal $+$ and $\cdot$ are the Peano operations: $ω$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product), which is commutative. The failure is purely infinitary, and $2$ and $ω$ are the smallest pair that exhibits it.

FALSE: ordinal multiplication is commutative

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources