# Tree components of a graph

If a graph $$G$$ has $$n$$ vertices and $$n-2$$ edges and no isolated vertices (no connected component has only one vertex) we want to show that at least $$2$$ of its connected components are trees with at least $$2$$ vertices each.

I tried contradiction by supposing we have at most $$1$$ tree with at least $$2$$ vertices, which means either we had such a tree or not. In case we don't have such a tree, since the component is connected, I'm trying to get a contradiction by supposing that each of the components has at least $$1$$ cycle, but I haven't got further.

a connected component on $$c$$ vertices has at least $$c-1$$ edges, with equality only if the component is a tree with at least $$2$$ vertices or is an isolated vertex. Therefore adding over all components we get the number of edges in $$G$$ is larger than or equal to $$n-t$$ where $$t$$ is the number of components that are trees with at least $$2$$ vertices.
The Euler characteristic $$\chi(G)=V-E$$ of a connected finite graph satisfies $$\chi(G) \le 1$$ with equality if and only if it is a tree.
For a disconnected finite graph $$G$$ with components $$G=G_1 \cup \cdots \cup G_K$$, we have $$\chi(G) = \chi(G_1) + \cdots + \chi(G_K)$$ Your graph $$G$$ has $$\chi(G) = n - (n-2)=2 > 1$$, so it is disconnected. In order for the above sum to be equal to $$2$$, at least two terms have to be $$\ge 1$$. But each term is $$\le 1$$. So at least two terms are $$=1$$. Thus, at least two of the components of $$G$$ are trees.