Handshake Lemma

Theorem
The number of vertices with odd degree in a graph must be even.

Direct Proof
Let $$G = \left({V, E}\right)$$ be an unweighted, undirected graph.

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

Each edge $$e \in E$$ will be counted exactly twice by this sum, once for each vertex to which it is incident.

So this sum must be equal to the twice the number of edges in $$G$$.

That is, $$K = 2 \left|{E}\right|$$, 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.