Proof that an undirected graph has an even number of vertices of odd degree.
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Let G be a graph with e edges and n vertices v1,v2,v3,...,vn.
Since each edge is incident on two vertices, it contributes 2 to the sum of degree of vertices in graph G. Thus the sum of degrees of all vertices in G is twice the number of edges in G:
Let the degrees of first r vertices be even and the remaining (n−r) vertices have odd degrees, then:
But, the for each is odd. So, the number of terms in
must be even. So, is even.