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