For this problem I understand that a k-regular graph is a graph where all the vertices have even degrees within the graph. I understand that a k-regular graph has to have an even number of vertices, I just don't know how to proof the concepts this proof wants me to prove, I know definitions, but maybe I just don't entirely get the concept? Either way I really need help with this proof, I have no idea how to do it. Thanks so much in advance!!!!!
Here is the question in more detail:
We say a graph G=(V,E) is a k-regular if it's sheer sequence has the form (k,k,k,…k), I.e. if every vertex has degree k. Examples of 2-regular graphs are K_3 and the "square" graph( 4 vertices and 4 edges laid out as a square).
Prove that, when k is odd, a k-regular graph must have an even number of vertices.
Best Answer
The number of vertex-edge incidences is $2|E|$ and is also $k|V|$.