I did the following proof which seems correct to me but does not match the approach of the answer provided by my professor, and seems pretty different from the question here in terms of notation and style. If I could get a verification that I'm correctly using induction on the number of edges of a graph, that would be great. Any tips on the clarity of my mathematical writing are also welcome. Thanks!
Question: prove the handshake lemma for simple graphs using induction on the number of edges. $$G=(V,E), \; \sum_{u\in V} \deg(u)=2|E|$$
Proof:
Base Case:
$|E|=1$. $\sum_{u\in V} \deg(u) = 2|E| = 2(1) =2$.
Inductive Step: Assume that all graphs w/ $|E| = n\geq 1, \: n \in \mathbb{N}$, fufill equality.
Let $G=(V,E)$ w/ $|E| = n+1$. Remove any edge from $G$, making $G' = (V, E')$, where now $|E'|=n$. Therefore, inductive assumption applies and the identity holds. $$\sum_{u\in V} \deg(u) = 2|E|$$
Now, add the removed edge back to $G'$. Because this edge is indecent on two vertices, we add two to the previous sum.
\begin{align}
\sum_{u\in V} \deg(u) +2 &= 2|E| + 2 \tag{by Inductive Assumption}\\
&=2(n) + 2\\
&=2(n+1)
\end{align}
As required.
Best Answer
No, I wouldn't say your proof is substantially different to the linked one; they're both correct.
The only advice I'd give is about your base case. Primarily, I think it needs some words. It's not clear from the one line of equalities, why a graph with one edge must have $2$ as the sum of its vertices' degrees. You can explain more clearly that given $m$ vertices, having one edge between them means that two of the vertices have degree $1$, and the rest have degree $0$, making the sum of the degrees $2$.
Or, even simpler and more generally, start with $n = 0$. If there are no edges in a graph, then the degree of every vertex is $0$, so the sum of degrees is also $0$.