Prove that a smiple graph with at least two vertices would have two or more vertices of the same degree.
thanks.
Realize that if a simple graph has n vertices then the maximum degree of any vertex is n-1. Thus suppose that we have a simple graph G with a degree sequence of <0,1,2,…,n-1>, that is all the n-vertices are of different degree. Now remove the vertex of degree 0. We now have a new graph G’ with (n-1) vertices and the same number of edges. But it has a vertex of degree (n-1). But that is impossible; so the original graph is impossible. Thus the graph G would have to have at least two vertices of the same degree.
Actually the graph theory proof above does use the pigeonhole principle implicitly. Think of 15 pigeonholes numbered 0,1,2,…,14. Assign each person to the hole that matches the number of acquaintances at the meeting that he/she has. If the hole #14 is empty then the fifteen will be assigned to 14 holes so there is a hole 0-13 with at least two. If hole #14 is not empty then someone there knows each other person there. Hence hole #0 must be empty(WHY?) Thus the fifteen will be assigned to 14 holes so there is a hole 1-14 with at least two.