# Let $G$ be a connected graph. Prove that if $E(G)$ is greater than or equal to $V(G)$ then $G$ contains at least three edges that are not bridges.

I was wondering how to solve this problem.

The idea is something along these lines if there are more edges than vertices (or equal), you have a cycle.

Since this is only possible with Graphs of order $$3$$ or more, we have a cycle of $$3$$ or more.

This means that you could take any of those edges out, and the graph is connected.

Making it not a bridge. Am I right?

• Yes, that's correct. Sep 19, 2020 at 12:04
• Try to consider some cycle $C$ in graph $G$. What can you say about the graph obtained from $G$ by deleting one of the edges of $C$? Sep 19, 2020 at 18:08