# Size of Tree is One Less than Order

## Theorem

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

Then $T$ is a tree if and only if the size of $T$ is $n-1$.

## Proof

By definition:

and:

### Necessary 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 $n \in \N_{>0}$, 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:

Then we need to show:

#### Induction Step: Proof 1

Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.

Take any node $v$ of $T_{k+1}$ of degree $1$.

Such a node exists from Finite Tree has Leaf Nodes.

Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.

By Connected Subgraph of Tree is 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 hypothesis, $T_k$ has $k - 1$ edges.

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

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

$\Box$

#### Induction Step: Proof 2

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

Remove any edge $e$ of $T_{k + 1}$.

By definition of tree $T_{k + 1}$ has no circuits

Therefore from Condition for Edge to be Bridge it follows that $e$ must 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 hypothesis, $T_1$ and $T_2$ have $k_1 - 1$ and $k_2 - 1$ edges.

Putting the edge $e$ back again, it can be seen that $T_{k + 1}$ has $\paren {k_1 - 1} + \paren {k_2 - 1} + 1 = k$ edges.

Therefore a tree of order $k+1$ is of size $k$.

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

$\Box$

### Sufficient Condition

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

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

Aiming for a contradiction, suppose $T$ is not a tree.

Then it contains a circuit.

It follows from Condition for Edge to be 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 $n - 2$ edges.

Let us try and construct a connected graph with $n$ nodes and $n - 2$ edges.

We start with the edgeless graph $N_n$, and add edges till the graph is connected.

We pick any two vertices of $N_n$, label them $u_1$ and $u_2$ for convenience, and use one edge to connect them, labelling that edge $e_1$.

We pick any other vertex, label it $u_3$, and use one edge to connect it to either $u_1$ or $u_2$, labelling that edge $e_2$.

We pick any other vertex, label it $u_4$, and use one edge to connect it to either $u_1, u_2$ or $u_3$, labelling that edge $e_3$.

We continue in this way, until we pick a vertex, label it $u_{n - 1}$, and use one edge to connect it to either $u_1, u_2, \ldots, u_{n - 2}$, labelling that edge $e_{n - 2}$.

That was the last of our edges, and the last vertex still has not been connected.

Therefore a graph with $n$ vertices and $n-2$ edges that such a graph *cannot* be connected.

Therefore we cannot remove any edge from $T$ without leaving it disconnected.

Therefore all the edges in $T$ are bridges.

Hence $T$ can contain no circuits.

Hence, by Proof by Contradiction, $T$ must be a tree.

$\blacksquare$

## Beware

Just because a simple graph $G$ has an order one greater than its size does not necessarily make it a tree.

$G$ still has to be connected.

Take this simple graph, for instance: