Prove that if G is a graph of order n such that delta(G)>=(n-1)/2, then lambda(G)=delta(G).
Well, because delta(G)>=(n-1)/2, we know G is connected from a previous result. Also, this bound was proven to be sharp. We also know by a theorem that delta (G)<=lambda(G) for every graph G. So if we could just show that it is not the case that delta (G) < lambda(G), we would be done.
This is where I am stumped. I am pretty sure it involves the inequality with delta(G), but I don't know what to do with it....
Something isn't right here, but maybe it is just me. However, you said that "We also know by a theorem that for every graph ." By definition should remain connected after we fewer than egdes. But then, just remove the edges that are attached to the vertex of minimum degree to get a disconnected graph...so at best you have equality.
Do you perhaps mean ?
Something isn't right here, but maybe it is just me. However, you said that "We also know by a theorem that for every graph ." But then by definition should remain connected after we fewer than egdes. But then, just remove the edges that are attached to the vertex of minimum degree to get a disconnected graph...so at best you have equality.
Do you perhaps mean ?
EDIT: For example, two triangles connected by a single line. Here, ...