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$W(3,2)=9$ by an explicit colouring of $\{0,\ldots,7\}$ and an exhaustive symmetry-reduced proof for $\{0,\ldots,8\}$

Example

With the zero-based natural-number convention of The natural numbers $\mathbb{N}$ (von Neumann) and Order on the natural numbers, the van der Waerden number of The van der Waerden number $W(k,c)$ as the least interval length forcing a monochromatic $k$ -term arithmetic progression is $W(3,2)=9$ . Translation identifies the intervals $\{0,\ldots,N-1\}$ and $\{1,\ldots,N\}$ , so this is the same convention used by Van der Waerden's theorem, strengthened so the progression and its common difference have one colour.

Facts & Assumptions

Given: Two colours, red and blue, and three-term progressions $a,a+d,a+2d$ with $d>0$ .

[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).

Verification

technique · direct

1.1

On $\{0,\ldots,7\}$ colour $0,1,4,5$ blue and $2,3,6,7$ red. Checking the possible differences $d=1,2,3$ shows that every three-term progression meets both two-point colour blocks. Thus $W(3,2)>8$ .

construct

1.2

Suppose $\{0,\ldots,8\}$ has an avoiding colouring. Exchange colour names to make $4$ red. The progression $0,4,8$ has a blue endpoint; reflect the interval if necessary to make $0$ blue. If $2$ is red, the progressions $2,3,4$ and $2,4,6$ force $3,6$ blue, making $0,3,6$ blue, a contradiction. Hence $2$ is blue, and $0,1,2$ forces $1$ red.

2.1

If $3$ is red, then $1,3,5$ forces $5$ blue, $1,4,7$ forces $7$ blue, $2,5,8$ forces $8$ red, and $4,6,8$ forces $6$ blue; now $5,6,7$ is blue. If $3$ is blue, then $0,3,6$ forces $6$ red, $1,4,7$ forces $7$ blue, and $3,5,7$ forces $5$ red; now $4,5,6$ is red. Both cases contradict avoidance, so every colouring of nine consecutive integers has a monochromatic three-term progression.

step 1.2

3.1

Steps 1.1 and 2.1 give the lower and upper bounds, hence $W(3,2)=9$ .

step 1.1step 2.1∎

Depends on

Used by

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Sources

MIT OpenCourseWare 18.310, Chapter 3 (standard reference, not scraped)

W(3,2)=9W(3,2)=9W(3,2)=9 by an explicit colouring of {0,…,7}\{0,\ldots,7\}{0,…,7} and an exhaustive symmetry-reduced proof for {0,…,8}\{0,\ldots,8\}{0,…,8}