Introduction to graph theory/Proof of Theorem 5

Statement
A tree of order $$n$$ has $$n-1$$ edges.

Corollary of
Theorem 4:For a graph $$G$$, the following are equivalent:
 * $$G$$ is a tree
 * $$G$$ is a minimal connected graph, in the sense that the removal of any edge will leave a disconnected graph.
 * $$G$$ is a maximal acyclic graph, in the sense that the addition of any missing edge will create a cycle.

Proof
For $$n=1$$, all graphs have $$0=n-1$$ edges. For $$n=2$$, there are two graphs, one with an edge and one without. The graph without an edge is disconnected, and therefore is not a tree, whereas the one with an edge is connected and has no cycles and therefore is a tree.

For $$n=3$$, there are 4 graphs, one each with 0, 1, 2 or 3 edges. The graphs with 0 and 1 edges are disconnected, while the graph with 3 edges has a cycle. The graph with 2 edges is connected and has no cycle, so the theorem is proved for $$n\le 3$$.

Now suppose we have proved this theorem for all $$n<k$$, and let $$T$$ be a tree of order $$k$$. Remove an edge $$e$$ from $$T$$ to form a new graph $$T-e$$. By Theorem 4, $$T$$ is a minimal connected graph, and hence $$T-e$$ is disconnected. Any connected component of $$T-e$$ (by the connectedness of $$T$$) must have one of the vertices of $$e$$ in it, and hence $$T-e$$ has exactly two components.

The two components are connected, and have no cycles, and are therefore trees. Let there be $$n_1, n_2$$ vertices in the two components of $$T-e$$. Then $$n_1+n_2=k$$. Further, there are $$n_1-1, n_2-1$$ edges in the components, so $$T-e$$ has $$k-2$$ edges in total. Therefore, putting edge $$e$$ back into the graph, $$T$$ has $$k-1$$ edges.