Acyclic Undirected Graph

Let $G=(V,E)$ be an undirected graph.

Prove or disprove: If $|E|\le |V| - 1$ then $G$ is acyclic.

I am unsure about if this is even true or not in the first place. I know that trees have $n-1$ edges, but I do not know if this quality can deem that $G$ is acyclic (a tree).

Take a triangle and add an isolated vertex, you have a graph with $|E|\leq |V|-1$ and a cycle.
If you add the hypothesis $G$ is connected however, then it is true.
• Doesn't the fact that it is not specified that $G$ must be connected as @Gamamal stated below make this not true for ANY undirected graph $G$ though? – Groups n' Stuff Apr 29 '15 at 4:10
• Why do you say my graph does not satisfy the formula? $3\leq 4-1$ – Yorch Apr 29 '15 at 4:17