Given a strongly connected graph, for every vertex in the graph the indegree+outdegree is even, I need to prove that in this case indegree is equal to outdegree (if im wrong about that assumption please correct me) for all vertex of the graph.
I'm not sure how to show that.
Your help is appreciated.
Best Answer
Let's orient a $K_5$ to have arcs 12, 23, 34, 45, 51 (which I think is enough to make it strongly connected), 13, 14, 24, 25, 35. Each vertex has total degree $4$, which is even, but vertex $1$ has indegree $1$ and outdegree $3$.