Question

Let G = (V, E) be a graph, and let v be a node in G with an odd number of neighbors. Prove that there is a node v' ≠ v in G such that v' also has an odd number of neighbors.

Answer 1

Let us prove this be contradiction method.

Assume there exists no v' such that number of neighbors of v' is odd. i.e. all other vertices except v have even number of neighbors(degree).

We know , any un-directed graph G with e number of edges, then sum of degrees of each vertex is 2e(a even number).

Now, our graph, has one odd degree vertex v and all other even degree vertices. degree sum of even degree vertices is even so adding this to degree of v will be a odd number.

i.e. degree sum of gigen graph is odd, which is not true.

Hence there must exist another vertex v', whose degree(number of neighbors) is odd, so that total sum of degrees would be even.