Handshake Lemma/Corollary

Corollary to the Handshake Lemma
Let $G$ be a $\tuple {p, q}$-graph, which may be a multigraph or a loop-graph, or both.

The number of odd vertices in $G$ is even.

Proof
Let $G$ be a $\tuple {p, q}$-graph.

Consider the sum of the degrees of its vertices:
 * $\displaystyle K = \sum_{v \mathop \in V} \map {\deg_G} v$

From the Handshake Lemma:
 * $K = 2 \card E$

which is an even integer.

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

That is:
 * $\displaystyle \paren {\sum_{v \mathop \in V} \map {\deg_G} v} - \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k} \map {\deg_G} v} = \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k + 1} \map {\deg_G} v}$

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

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

Hence the number of odd vertices in $G$ must be even.