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

Theorem
Let $$T$$ be a connected simple graph of order $$n$$.

Then $$T$$ is a tree iff the size of $$T$$ is $$n-1$$.

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

Sufficient Condition
Suppose $$T$$ is a tree with $$n$$ nodes. We need to show that $$T$$ has $$n-1$$ edges.

Proof by induction:

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.

Basis for the Induction
$$P(1)$$ 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 $$1$$. Such a node exists from Tree has Degree One Nodes‎.

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.

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

Remove any edge $$e$$ of $$T$$.

As $$T_{k+1}$$ has no circuits, $$e$$ must be a bridge, from Condition for an Edge to be a Bridge.

So removing $$e$$ disconnects $$T_{k+1}$$ into two trees $$T_1$$ and $$T_2$$, with $$k_1$$ and $$k_2$$ nodes, where $$k_1 + k_2 = k+1$$.

By the induction hypothesis, $$T_1$$ and $$T_2$$ have $$k_1 - 1$$ and $$k_2 - 1$$ edges.

Putting the edge $$e$$ back again, we see that $$T_{k+1}$$ has $$\left({k_1 - 1}\right) + \left({k_2 - 1}\right) + 1 = k$$ edges.

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

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

Necessary Condition
Suppose $$T$$ is a connected simple graph of order $$n$$ with $$n-1$$ edges.

We need to show that $$T$$ is a tree.

Suppose that $$T$$ is not a tree. Then it contains a circuit.

It follows from Condition for an Edge to be a Bridge that there is at least one edge in $$T$$ which is not a bridge.

So we can remove this edge and obtain a graph $$T'$$ which is connected and has $$n$$ nodes and $$n-1$$ edges.

However, it can easily be proved by induction that such a graph can not be connected.

Therefore all the edges in $$T$$ are bridges.

Hence $$T$$ can contain no circuits and so must be a tree.