Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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Erdős's finite counting bound R(k,k)>2k/2 for every k≥3

Statement

Facts & Assumptions

Given: A natural k≥3 and N:=⌊2k/2⌋; binomial coefficients are as in The set [A]k of k-element subsets and the binomial coefficient (nk):=∣[n]k∣.

[L1]

If A and B are finite, then AB is finite and ∣AB∣=∣A∣∣B∣ (The set AB of functions B→A between finite sets is finite, with ∣AB∣=∣A∣∣B∣).

[L2]

If n,k∈N and k≤n, then (nk)⋅k!⋅(n−k)!=n!, equivalently (nk) k!=nk‾ ((nk) k! (n−k)!=n! for k≤n; hence (nk) k!=nk‾, the quotient n!/(k!(n−k)!) is a natural number, and (nk)=(nn−k)).

[L3]

Every real x has a unique integer ⌊x⌋ with ⌊x⌋≤x<⌊x⌋+1 (Integer part: for every real x there is exactly one integer m with m≤x<m+1).

Proof

technique · direct
1.1

There are (N2) edges in KN, and [L1] therefore counts exactly 2(N2) red-blue edge colourings.

L1
2.1

For a fixed k-vertex set, exactly 2⋅2(N2)−(k2) colourings make all its edges monochromatic. Summing these finite bad sets over the (Nk) choices, with overlaps allowed, shows that a colouring with no monochromatic k-set exists whenever 2(Nk)2−(k2)<1.

step 1.1L1
3.1

If N<k, then (Nk)=0 by the definition of the binomial coefficient. If k≤N, [L2] gives (Nk) k!=Nk‾≤Nk, so again (Nk)≤Nk/k!. Since N≤2k/2, the left side in step 2.1 is therefore at most 21+k/2/k! in either case. At k=3 this is 25/2/6<1; thereafter the ratio of the bound for k+1 to that for k is 2/(k+1)<1. Hence the strict inequality holds for every k≥3.

step 2.1L2algebra
4.1

Step 2.1 supplies a colouring on N vertices with no monochromatic k-set, so R(k,k)>N. As R(k,k) is an integer and N=⌊2k/2⌋, [L3] implies R(k,k)≥N+1>2k/2.

step 3.1L3∎

Depends on

Used by

Dependency tree · two levels

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Sources