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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 130 / 1 / ii / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 130 1 ii
Created 2026-10-03 Updated 2026-10-05  0 By others on same topic  0 Discussions Create my own version
Pull the given finite coloring χ:N→[k] back to a finite coloring of unordered pairs by putting
χ​({a,b})=χ(ab2)(a<b).
(1)
The ordering a<b makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set X={x1​<x2​<⋯} has all its pairs of one color r under χ​. Consequently
χ(xi​xj2​)=r(i<j).​
(2)
Possible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

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