# On Classical Ramsey Numbers

This paper has two parts. In the first part the question, “Why is it impossible to color the edges of kr(p,q) without forming either a red kp or a blue kq ?” is answered while in the second the question, “What is the smallest value of n for which kn[pic]kp,kq?” is changed to equivalent forms.

Introduction

A graph G is composed of a finite set V of elements called vertices and a set E of lines joining pairs of distinct vertices called edges. We denote the graph whose vertex set is V and whose edge set is E by G= (V,E). By a kn we mean a graph with n vertices and all edges joining these vertices with each other.

eg.

.

k1 k2

k3 k4

Fig.1

Ramsey’s theorem states that if p,q≥2 are integers, then there is a positive integer n such that if we color the edges of kn using two colors, red and blue, it is impossible to color the edges of the kn without forming either a red kp or a blue kq. In short we formulate it as kn[pic]kp,kq (read as kn arrows kp,kq). The smallest value of such n is denoted by r(p,q),known as the Ramsey number. A famous example for the two color Ramsey theorem is k6 which arrows k3,k3. To prove k6 [pic]k3,k3, let’s put 6 points on a plane and call one of them v. There are 5 edges joining v to the remaining 5 points. Let’s color them red or blue. At least 3 of them will have the same color, red, say. Consider the 3 vertices at the other ends of these 3 red edges:

v

Blue

Red

Fig.2

If any of the edges joining these 3 vertices with each other is red, then we have a red triangle. On the other hand if there is no red edge, we get a blue triangle. A complete subgraph with all its edges having the same color is called a monochromatic sub graph or a monochromatic clique. The word coloring refers to assigning either of the colors red or blue for the edges and is used in a different sense from edge coloring; i.e. adjacent edges can have the same color. A kn with one side missing is one which becomes a kn when the missing side is filled. Let’s denote it by kn-s. Clearly a k1 & a k2 with one side missing are vacuous sets.

A k3 -s A k4-s A k5 -s Fig.3

The main result

We have shown that k6[pic]k3,k3. But why is it impossible to color the edges of k6 with out forming a monochromatic k3 of either color? In a k6 there are 6C3=20 triangles. If these triangles were drawn separately, they would have 20[pic]3=60 edges. But a k6 has only 6C2= 15 edges. Hence an edge is shared by [pic] = 4 triangles in a k6. Now let’s try to form a k6 whose edges colored with red and blue but doesn’t consist of a monochromatic k3 of either color. What problem shall we encounter? Let’s start with a red edge, v1v2. As shown below, it is shared by 4 triangles. Let all triangles on v1v2 have blue sides, except v1v2.

v1 v1

v3 v5 v3 v5

v4 v6 v4 v6

v2...

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