1. The problem statement, all variables and given/known data Prove that, if G is a simple graph with no K5-minor and |V(G)| Does not = 0, then G has a vertex at most 5. 2. Relevant equations |E(G)| <= 3|V(G)| - 6 for |V(G)| >= 3 (Proved by earlier part of problem set) Handshake theorem (I don't believe we are allowed to use hadwiger's conjecture 3. The attempt at a solution Well I first used the handshake theorem to show that the sum of the degrees in G are equal to 2 times the edges in G. Sum(Deg (v)) for all v in G = 2 |E(G)| use V average and replace |E(G)| with 3|V(G)| - 6 so: Vaverage |V(G)| <= 2(3|V(G)| - 6) = 6|V(G)| - 12 divide by |V(G)| = 6 - 12/|V(G)| now for any V(G) less than 3, we can see it intuitively. However...I can't have |V(G)| > 12.....Help please! I've been a little frustrated. Thanks for your time!