Question

The Ramsey number R(m, n), where m and n are positive integers greater than or equal to 2, denotes the minimum number of people at a party such that there are either m mutual friends or n mutual enemies, assuming that every pair of people at the party are friends or enemies.

Assume that in a group of six people, each pair of individuals consists of two friends or two enemies. Show that there are either three mutual friends or three mutual enemies in the group.

Answer 1

The Ramsey number R(3, 3) implies that in a group of six people, there are either three mutual friends or three mutual enemies. By selecting one person and applying the pigeonhole principle to their five connections, we conclude that it is inevitable to have at least one set of three mutual friends or enemies.

Step-by-step explanation:

Ramsey Number R(3, 3)

The question involves the concept of Ramsey numbers which is a part of combinatorial mathematics used to solve problems related to graph theory. In this specific question, we need to prove that in a group of six people (R(3, 3)=6), there will always be either three mutual friends or three mutual enemies. This is because the Ramsey number R(m, n) represents the smallest number of vertices (people in this context) needed to ensure that there is either a complete graph of size m (mutual friends) or its complement (mutual enemies) of size n within the graph.

Let's pick one person from the group and consider the five connections that person has. According to the pigeonhole principle, at least three of these connections must be of one kind - either friendships or enmities - since there are only two kinds of relationships possible. Suppose there are three friends (the other case can be handled analogously). These three friends form a triangle. If any of the three pairwise connections among these people is one of friendship, we have our triplet of mutual friends. If not, then they are all enemies of each other, and we have a triplet of mutual enemies.

The proof by contradiction also supports this argument. Assume that there are no three mutual friends or enemies. However, after picking any person and looking at their five connections, the assumption quickly becomes impossible to maintain due to the pigeonhole principle. Therefore, we must have at least one triplet of mutual friends or enemies.

To summarize, the six people at this party represent a complete graph of 6 vertices where each pair of vertices is connected by an edge representing either friendship or enmity. By considering the connections of one person and using the pigeonhole principle, it is inevitable to find either three mutual friends or three mutual enemies, thus confirming that R(3, 3) ≤ 6.