Size of Tree is One Less than Order/Necessary Condition

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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

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 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 graph.

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.

$\blacksquare$


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