Vertex degree sum in a graph

Proof

Lets prove this by induction on the number $\vert E \vert$ within a graph.

Suppose a graph has no edges. Therefore $\deg(v) = 0$ for all $v \in V$ as there are no edges to be incident to $v$. Thus

$$\sum_{v \in V} \mbox{deg}(v) = 0 = 2 \vert E \vert.$$

Suppose we have shown the statement true for all graphs where $\vert E \vert < k$ and suppose we have a graph with $\vert E \vert = k$. Pick any edge $e = (x,y)$ and remove it from the graph to get $G^{\ast} = (V, E^{\ast})$ . From the induction hypothesis we have

$$ \sum_{v \in V} \mbox{deg}_{G^{\ast}}(v) = 2 (\vert E \vert - 1)$$

Where $\mbox{deg}_{G^{\ast}}$ is the degree in $G^{\ast}$. Note that $\mbox{deg}_{G}(v) = \mbox{deg}_{G^{\ast}}(v)$ for all $v \in V \backslash \{x, y\}$. Whereas, $\mbox{deg}_{G^{\ast}}(v) + 1 = \mbox{deg}_{G}(v)$ for $v \in \{x,y\}$ (as it is incident to $e = (x,y)$ as well as all the edges in $E^{\ast}$). Therefore

$$\begin{align*} \sum_{v \in V} \mbox{deg}_{G}(v) & = \sum_{v \in V \backslash \{x,y\}} \mbox{deg}_{G^{\ast}}(v) + \sum_{v \in \{x,y\}} (\mbox{deg}_{G}(v) + 1)\\ & = 2 + \sum_{v \in V} \mbox{deg}_{G^{\ast}}(v) \\ & = 2 + 2 (\vert E \vert - 1)\\ & = 2 \vert E \vert. \end{align*}$$

This shows the inductive case and proves the statement.