Use the pigeonhole principle to prove that a graph of order n ≥2 always has two vertices of the

step 1 Hi, in this question we are asked to show that a simple graph with at least two vertices must co

Use the pigeonhole principle to prove that a graph of order...

Key Concepts

Pigeonhole Principle

The pigeonhole principle is a fundamental combinatorial concept that asserts if you place more objects than there are containers, at least one container will contain more than one object. In the context of graph theory, it is used to argue that when there are more vertices than possible distinct degree values, some vertices must share the same degree.

Vertex Degree in Graphs

The degree of a vertex in a graph is the number of edges incident to that vertex. In a simple graph (which excludes multiple edges and loops), the degree of any vertex must lie between 0 and n?1, where n is the total number of vertices.

Degree Constraints in Simple Graphs

For a simple graph with n vertices, the possible degree values range from 0 to n?1. However, due to the inherent restriction that a vertex cannot simultaneously have degree 0 (isolated) if another vertex has degree n?1 (adjacent to every other vertex), there are at most n?1 different degree values available. This limitation, combined with the fact that there are n vertices, guarantees that at least two vertices must share the same degree by the pigeonhole principle.

Simple Graphs versus Multigraphs

In a multigraph, multiple edges between the same pair of vertices are allowed, leading to a broader or less constrained set of possible degree counts. This relaxation means that the specific pigeonhole argument used for simple graphs might not apply to multigraphs, and it is possible for a multigraph to have vertices with all distinct degrees.

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