Proving that the number of vertices of odd degree in any graph G is even

Let G be a graph with ′e′${}^{\prime }{e}^{\prime }$ edges and ′n′${}^{\prime }{n}^{\prime }$ vertices v1,v2,v3,...,vn$v_1,v_2,v_3,...,v_n$. Since each edge is incident on two vertices, it contributes 2$2$to the sum of degree of vertices in graph G$G$. Thus the sum of degrees of all vertices in G$G$ is twice the number of edges in G$G$. Hence,

∑i=1ndegree(vi)=2e.$$\sum _{i=1}^n\text{degree}(v_i)=2e.$$

Let the degrees of first r$r$ vertices be even and the remaining (n−r)$(n-r)$ vertices have odd degrees,then clearly,∑ri=1degree(vi)=even$\sum _{i=1}^r\text{degree}(v_i)=even$.Since,

∑i=1ndegree(vi)=∑i=1rdegree(vi)+∑i=r+1ndegree(vi)$$\sum _{i=1}^n\text{degree}(v_i)=\sum _{i=1}^r\text{degree}(v_i)+\sum _{i=r+1}^n\text{degree}(v_i)$$

⟹$⟹$ ∑ni=1degree(vi)−∑ri=1degree(vi)=∑ni=r+1degree(vi)$\sum _{i=1}^n\text{degree}(v_i)-\sum _{i=1}^r\text{degree}(v_i)=\sum _{i=r+1}^n\text{degree}(v_i)$

⟹$⟹$ ∑ni=r+1degree(vi)$\sum _{i=r+1}^n\text{degree}(v_i)$ is even.(WHY?$WHY?$)

But, the for i=r+1,r+2,...,n$i=r+1,r+2,...,n$ each d(vi)$d(v_i)$ is odd. So, the number of terms in ∑ni=r+1degree(vi)$\sum _{i=r+1}^n\text{degree}(v_i)$ must be even.

In lucid words,(n−r)$(n-r)$ is even.

Hence the result.