Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step

Theorem

Let the following hold:

For all $j \le k$, a tree of order $j$ is of size $j - 1$.

Then this holds:

A tree of order $k + 1$ is of size $k$.

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

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

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

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

We have by hypothesis that $T_k$ has $k - 1$ edges.

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

Let $T_{k + 1}$ be an arbitrary 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 $k_1 + k_2 = k + 1$.

Then, 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 $k$.

1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.2$