I am working on practicing proofs by induction. Can you please take a look at the proof below and tell me if I proved it correctly? I am particularly worried about the inductive step.
Definition 1. A tournament graph is a directed graph G = (V, E) without loops, in which for every two vertices u and v, exactly one of (u, v) or (v, u) are in E.
Prove using induction that for every tournament graph G = (V, E), it holds that E = [|V|(|V|-1)]/2.
Let the statement be P(n). For the base of a graph with 1 vertex, it holds that P(1) which is: E = 0 = 1(1-1)/2 = 2. We can now assume that for some arbitrary positive integer K that: E = k(k-1)/2. To complete the inductive step, we must use the inductive hypothesis to prove k+1 holds for: E = (k+1)((k+1)-1) / 2 = (k+1)k/2 By the inductive hypothesis, we know that this equals: k(k-1)/2 + k = k(k-1) +2k/2 = k^2 - k + 2k/2 = k(k - 1 + 2)/2 = k(k+1)/2 So by the inductive hypothesis, we know that P(k+1) holds for some arbitrary k. Since k is arbitrary, P(n) holds for any positive integer.