Math Problem Statement

During a lecture, each of 26 mathematicians fell asleep exactly once. Prove that there are either 6 mathematicians such that no two were asleep at the same time, or 6 mathematicians such that there was some point in time during which all 6 were asleep.

Solution

We are tasked with proving that among 26 mathematicians who each fell asleep exactly once during a lecture, there are either:

  1. A group of 6 mathematicians such that no two were asleep at the same time.
  2. A group of 6 mathematicians such that there was some point in time during which all 6 were asleep.

This problem can be tackled using Ramsey theory, which focuses on finding order in seemingly random or chaotic situations. Specifically, we will use Ramsey's theorem for two-color edge graphs, denoted by R(k,l)R(k, l), which tells us the smallest number of vertices NN such that any graph of size NN contains either a complete subgraph KkK_k (all edges between the kk vertices are present) or an independent set of size ll (no edges between the ll vertices are present).

Approach

Let’s interpret the problem in the following way:

  1. Each mathematician is represented as a vertex in a graph.
  2. Two vertices are connected by an edge if the corresponding mathematicians were asleep at the same time. Thus, the edge means they overlapped in time while sleeping.

We are looking for either:

  • A group of 6 mathematicians, such that no two were asleep at the same time (this forms an independent set), or
  • A group of 6 mathematicians, such that there is a point in time when all of them were asleep (this forms a complete subgraph).

Ramsey's Theorem Application

Ramsey's theorem tells us that R(6,6)=31R(6, 6) = 31. This means that in any graph with at least 31 vertices, no matter how the edges are colored (i.e., whether two vertices are connected by an edge or not), there will always be either a complete subgraph of size 6 or an independent set of size 6.

However, we are working with only 26 mathematicians, not 31. But this still aligns with Ramsey’s theorem since we are interested in a weaker condition. Specifically, we need to prove that among the 26 vertices (mathematicians), there exists either a complete subgraph of size 6 (where 6 mathematicians were asleep at the same time) or an independent set of size 6 (where no two of the 6 were asleep at the same time).

Conclusion

Since R(6,6)=31R(6, 6) = 31, and we have fewer than 31 vertices (26 mathematicians), we cannot guarantee that both a complete subgraph of size 6 and an independent set of size 6 are absent. Therefore, there must be one or the other. Hence, there are either:

  1. A group of 6 mathematicians such that no two were asleep at the same time (an independent set), or
  2. A group of 6 mathematicians such that all were asleep at the same time (a complete subgraph).

This completes the proof.


Do you have any questions, or would you like further details on this proof?

Here are 5 related questions to explore:

  1. What is Ramsey's theorem, and how is it applied to graph theory problems?
  2. Can we generalize this problem to larger numbers of mathematicians?
  3. How do we calculate the Ramsey number R(k,l)R(k, l) for other values of kk and ll?
  4. What are other real-world scenarios where Ramsey theory can be applied?
  5. How does the probabilistic method relate to problems like this one?

Tip: Ramsey theory is a powerful tool for finding structure in complex systems and is widely applicable in computer science, combinatorics, and beyond.

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Math Problem Analysis

Mathematical Concepts

Graph Theory
Ramsey Theory
Combinatorics

Formulas

R(k, l) - Ramsey number formula

Theorems

Ramsey's Theorem

Suitable Grade Level

Grades 11-12, University