Maximum Number of Arcs in Digraph

Theorem
Let $D_n$ be a digraph of order $n$ such that $n \ge 1$.

Let $D_n$ have the greatest number of arcs of all digraphs of order $n$.

The number of arcs in $D$ is given by:
 * $\size {D_n} = n \paren {n - 1}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\size {D_n} = n \paren {n - 1}$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\size {D_k} = k \paren {k - 1}$

from which it is to be shown that:
 * $\size {D_{k + 1} } = \paren {k + 1} k$

Induction Step
This is the induction step:

Let $D_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to $D_k$.

To do so, it is necessary to add $2$ arcs to join $v_{k + 1}$ to every vertex of $D_k$.

Thus there are a total of $2 k$ arcs more in $D_{k + 1}$ than there are in $D_k$.

So:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \size {D_n} = n \paren {n - 1}$