I can't seem to figure out this proof. Any help would be greatly appreciated!
Prove that if G is a graph of order p and minimum degree of G (d(G)) >= p/2, then the edge connectivity of G (k_1(G))= minimum degree of G.
Relevant statements: I'm not sure that this is relevant.. but d(G)>=(|V(G)|-1)/2 implies connectivity of G. Also the left side of the proof also implies that G is Hamiltonian (possible to connect all vertices of G in a cycle) which could be relevant to its edge connectivity..
My understanding so far: I've been considering the vertex v, with deg(v)=d(G). Clearly, the edges whose removal will disconnect G will be connected to v (or some other vertex with minimum degree). I can't seem to grasp the implications of the minimum degree being larger than half the order of G...
Sorry about my notation (first post)! And Thanks in advance!