How to show that if $G$ is a graph with $\delta (G) \geq 2$ then $G$ contains a cycle? I know a cycle is a closed trail with no repeated vertices except the first and the last (starts and ends at the same vertex). But I'm not sure how to go about proving this
 A: (Assuming $\delta(G)$ refers to the minimum degree of the graph...)


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*Pick your favorite node.

*Walk from that node to another node along your favorite edge, deleting the edge when you're done.

*Repeat until you can't anymore.  There must be no more edges attached to your final node.  That means you've deleted at least two edges attached to this node, so you've been here before!  The path you have traveled since your last visit to this node must be a cycle.

A: Assume that $G$ does not contain a cycle. Then $G$ is a tree. However, since $\delta(G) \geq2$ this implies that $G$ contains no end-vertices and so $G$ is not a tree. Thus a contradiction has been found.
A: While all trees do have leaves, you can get into trouble assuming it if it is not a theorem in the book or something you've already proven. And assuming a tree also assumes connectivity. The connectivity assumption turns out to be okay, since it gives us an upper bound on the number of edges.
So I'd use a handshake lemma argument here. If $G$ is a tree or a forest, $\sum_{v \in V(G)} d(v) \leq 2V - 2$, as a tree has $V - 1$ edges. So if $\delta(G) \geq 2$, $\sum_{v \in V(G)} d(v) = 2V > 2V - 2$. Thus, $G$ cannot be a tree or a forest.
