Finite Connected Graph is Tree iff Size is One Less than Order

Theorem
Let $$T$$ be a tree of order $$n$$.

Then the size of $$T$$ is $$n-1$$.

Proof
Proof by induction:

By definition, the order of a tree is how many nodes it has, and its size is how many edges it has.

Let $$T_n$$ be a tree with $$n$$ nodes.

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition that a tree with $$n$$ nodes has $$n-1$$ edges.


 * $$P(1)$$ says that a tree with $$1$$ vertex has no edges.

It is clear that $$T_1$$ is $$N_1$$, the null graph, which has $$1$$ node and no edges.

So $$P(1)$$ is (trivially) true.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * Any tree with $$k$$ nodes has $$k-1$$ edges.

Then we need to show:
 * Any tree with $$k+1$$ nodes has $$k$$ edges.

Induction Step
Let $$T_{k+1}$$ be any tree with $$k+1$$ nodes.

Take any node $$v$$ of $$T_{k+1}$$ of degree $$1$$. There has to be at least one, otherwise $$T_{k+1}$$ would have a circuit.

Let us consider $$T_k$$, the subgraph of $$T_{k+1}$$ created by removing $$v$$ and the edge connecting it to the rest of the graph.

By Subgraph of Tree, $$T_k$$ is itself a tree.

The order of $$T_k$$ is $$k$$, and it has one less edge than $$T_{k+1}$$ by definition.

By the induction hypothesis, $$T_k$$ has $$k-1$$ edges.

So $$T_{k+1}$$ must have $$k$$ edges.

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore tree with $$n$$ nodes has $$n-1$$ edges.