Handshake Lemma

Theorem
Let $$G$$ be a graph, which may be a multigraph or a pseudograph, or both.

The number of vertices with odd degree in $$G$$ must be even.

Direct Proof
Let $$G = \left({V, E}\right)$$ be a simple graph, i.e. unweighted and undirected.

Consider the sum of the degrees of its vertices:
 * $$K = \sum_{v \in V} \deg_G \left({v}\right)$$.

From Sum of Graph Degrees is Twice the Size, we have that $$K = 2 \left|{E}\right|$$, which is an even integer.

Subtracting from $$K$$ the degrees of all vertices of even degree, we are left with the sum of all degrees of vertices in $$V$$ with odd degree.

That is:
 * $$\left({\sum_{v \in V} \deg_G \left({v}\right)}\right) - \left({\sum_{v \in V : \deg_G \left({v}\right) = 2 k} \deg_G \left({v}\right)}\right) = \left({\sum_{v \in V : \deg_G \left({v}\right) = 2 k + 1} \deg_G \left({v}\right)}\right)$$.

This must still be an even number, as it is equal to the difference of two even numbers.

Since this is a sum of exclusively odd terms, there must be an even number of such terms for the sum on the right side to be even.

Hence the result.