Thanks for the feedback. I think you can construct all graphs and use it to prove the theorem if you prove that you can add an arbitrary number of additional edges and nodes to an arbitrary graph and keep the sum of the degrees of all nodes even, instead of just one additional node and one additional edge. I also see what you mean about this:
I can't quite tell if you actually rely on that construction.
I think the inductive hypothesis in the rest of that paragraph might be enough, and I just wrote down how I intuitively visualized the proof before that without realizing that it wasn't necessary (nor sufficient, I now know) for the argument to carry through.
If you have an idea of how you would write the proof, I'd be interested in seeing it. I looked at the book and the proof is actually even less formal there.
Lemma: sum of the degrees of the nodes is twice the number of edges.
Proof: We proceed by induction on the number of edges. If a graph has 0 edges, the the sum of degrees of edges is 0=2(0). Now, by way of induction, assume, for all graphs with n edges, the sum of the degrees of the nodes 2n; we wish to show that, for all graphs with n+1 edges, the sum of the degrees of the nodes is 2(n+1). But the sum of the degrees of the nodes is (2n)+2 = 2(n+1). ∎
The theorem follows as a corollary.
If you want practice proving things and haven't had much experience so ...
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