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2. Use induction on v(G) to prove that every loop less graph G with at least one edge contains a bipartite subgraph that is spanning (i.e. contains every vertex) and has more than e(G)/2 edges. Proof: Let P(n) be the statement that every graph on n-vertices has a bipartite subgraph with at least two edges. We need to show that P(n) implies P(n+1). Let G be a graph with n+1 vertices and is a bipartite subgraph with at least 2 edges? Pick a vertex, v, of G and let H (a new graph) be the graph obtained from G by deleting v and all edges of G incident at v. This means that e(H) = e(G)-d. By induction, H has a bipartite subgraph B with at least 2 edges? Let v1 and v2 be the vertex of this subgraph. Now, we assume that the union of v1 and v2 = v(H). This means that B keeps the same vertices of H since B is a subgraph of H and now we can see that we can contain the bipartite subgraph B and it has more than two edges.

          2. Use induction on v(G) to prove that every loop less graph G with at least one edge contains a bipartite subgraph that is spanning (i.e. contains every vertex) and has more than e(G)/2 edges.

Proof: Let P(n) be the statement that every graph on n-vertices has a bipartite subgraph with at least two edges. We need to show that P(n) implies P(n+1). Let G be a graph with n+1 vertices and is a bipartite subgraph with at least 2 edges? Pick a vertex, v, of G and let H (a new graph) be the graph obtained from G by deleting v and all edges of G incident at v. This means that e(H) = e(G)-d. By induction, H has a bipartite subgraph B with at least 2 edges? Let v1 and v2 be the vertex of this subgraph. Now, we assume that the union of v1 and v2 = v(H). This means that B keeps the same vertices of H since B is a subgraph of H and now we can see that we can contain the bipartite subgraph B and it has more than two edges.

2. Use induction on v(G) to prove that every loop less graph G with at least one edge contains a bipartite subgraph that is spanning (i.e. contains every vertex) and has more than e(G)/2 edges.

Proof: Let P(n) be the statement that every graph on n-vertices has a bipartite subgraph with at least two edges. We need to show that P(n) implies P(n+1). Let G be a graph with n+1 vertices and is a bipartite subgraph with at least 2 edges? Pick a vertex, v, of G and let H (a new graph) be the graph obtained from G by deleting v and all edges of G incident at v. This means that e(H) = e(G)-d. By induction, H has a bipartite subgraph B with at least 2 edges? Let v1 and v2 be the vertex of this subgraph. Now, we assume that the union of v1 and v2 = v(H). This means that B keeps the same vertices of H since B is a subgraph of H and now we can see that we can contain the bipartite subgraph B and it has more than two edges.

Added by Shaun N.

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