Size of Tree is One Less than Order/Necessary Condition

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

Then 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.

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 $n \in \N_{>0}$, let $\map P n$ be the proposition that a tree with $n$ nodes has $n-1$ edges.

Basis for the Induction
$\map P 1$ says that a tree with $1$ vertex has no edges.

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

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

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ 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: Proof 1
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Induction Step: Proof 2
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.