Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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FALSE: every two-colouring of N\mathbb N contains an infinite monochromatic arithmetic progression

Statement

Every two-colouring of N\mathbb N contains an infinite monochromatic arithmetic progression a,a+d,a+2d,a,a+d,a+2d,\ldots with d>0d>0.

Facts & Assumptions

Given: Natural numbers and their order as in The natural numbers N\mathbb{N} (von Neumann) and Order on the natural numbers.

[L1]

Every finite colouring of a sufficiently long initial interval has a monochromatic arithmetic progression whose common difference has the same colour (Van der Waerden's theorem, strengthened so the progression and its common difference have one colour).

Refutation

technique · constructive
1.1

Colour 00 red. For n1n\ge1, colour nn red when the unique mm with 2mn<2m+12^m\le n<2^{m+1} is even, and blue when mm is odd. Thus consecutive dyadic blocks alternate colours, with every power of two assigned to the block beginning there.

construct
2.1

Fix aNa\in\mathbb N and d>0d>0. For every sufficiently large mm, let qmq_m be the least qq with a+qd2ma+q d\ge2^m. Minimality gives a+qmd<2m+d<2m+1a+q_m d<2^m+d<2^{m+1}, so the progression meets the mmth dyadic block. It therefore meets infinitely many blocks of each parity and contains both colours.

step 1.1algebra
3.1

No infinite arithmetic progression is monochromatic in this colouring. This does not contradict [L1], which guarantees arbitrarily long finite progressions only. The displayed universal statement is false.

step 2.1L1discharge-construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 23 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources