Prove that all vertices of a complete graph Kn have deg(v) = (n-1)
∑ deg(v) = 2|E|
|E| = ½(n)(n-1) for Kn
The Attempt at a Solution
I may have over thought this but this was my initial path at a formal proof.
Using the degree sum formula above and the formula for |E| for Kn,
∑ deg(v) = 2|E| = n(n-1)
Kn has n vertices so,
deg(v1) + deg(v2) + . . . . + deg(vn) (1/n) = (n-1)
∴ Each vertex , v, has degree (n-1)
EDIT: The degree sum formula was mistakenly labeled as the Handshaking lemma.