Size of Complete Graph

Theorem

Let $K_n$ denote the complete graph of order $n$ where $n \ge 0$.

The size of $K_n$ is given by:

$\size {K_n} = \dfrac {n \paren {n - 1} } 2$

Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\size {K_n} = \dfrac {n \paren {n - 1} } 2$

First we explore the degenerate case $\map P 0$:

\(\ds \size {K_0}\)

\(=\)

\(\ds 0\)

as $K_0$ is the null graph

\(\ds \)

\(=\)

\(\ds \dfrac {0 \paren {0 - 1} } 2\)

Thus $\map P 0$ is seen to hold.

Basis for the Induction

$\map P 1$ is the case:

\(\ds \size {K_1}\)

\(=\)

\(\ds 0\)

Complete Graph of Order 1 is Edgeless

\(\ds \)

\(=\)

\(\ds \dfrac {1 \paren {1 - 1} } 2\)

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:

$\size {K_k} = \dfrac {k \paren {k - 1} } 2$

from which it is to be shown that:

$\size {K_{k + 1} } = \dfrac {\paren {k + 1} k} 2$

Induction Step

This is the induction step:

Let $K_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to the complete graph $K_k$.

To do so, it is necessary to add an edge to join $v_{k + 1}$ to every vertex of $K_k$.

Thus there are a total of $k$ edges more in $K_{k + 1}$ than there are in $K_k$.

So:

\(\ds \size {K_{k + 1} }\)

\(=\)

\(\ds \size {K_k} + k\)

from the above analysis

\(\ds \)

\(=\)

\(\ds \dfrac {k \paren {k - 1} } 2 + k\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds \dfrac {k^2 - k + 2 k} 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {k^2 + k} 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {k + 1} k} 2\)

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

Therefore:

$\forall n \in \Z_{\ge 0}: \size {K_n} = \dfrac {n \paren {n - 1} } 2$

$\blacksquare$