Simple Graph of Maximum Size is Complete Graph

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Theorem

Let $G$ be a simple graph of order $n$ such that $n \ge 1$.

Let $G$ have the largest size of all simple graphs of order $n$.


Then:

$G$ is the complete graph $K_n$
its size is $\dfrac {n \paren {n - 1} } 2$.


Proof

By definition, $K_n$ is the simple graph of order $n$ such that every vertex of $K_n$ is adjacent to all other vertices.

So, let $G$ have the largest size of all simple graphs of order $n$.

Then by definition of largest size, it is not possible for another edge to be added to $G$.

The only way that could be is if all the vertices of $G$ are already adjacent to all other vertices.

Hence $G = K_n$ by definition.


The size of $G$ then follows from Size of Complete Graph.

$\blacksquare$


Examples

Maximum Size Simple Graph of Order 3

Let $G$ be the simple graph of order $3$ whose edge set $E$ is as large as possible.

Then the size of $G$ is given by:

$\size E = 3$


Maximum Size Simple Graph of Order 4

Let $G$ be the simple graph of order $4$ whose edge set $E$ is as large as possible.

Then the size of $G$ is given by:

$\size E = 6$


Maximum Size Simple Graph of Order 5

Let $G$ be the simple graph of maximum size whose vertex set is:

$V = \set {v_1, v_2, v_3, v_4, v_5}$

Then the edge set $E$ such that $\size E$ is as large as possible is:

$E = \set {v_1 v_2, v_1 v_3, v_1 v_4, v_1 v_5, v_2 v_3, v_2 v_4, v_2 v_5, v_3 v_4, v_3 v_5, v_4 v_5}$

Thus:

$\size E = 10$

and it is seen by inspection that $G = K_5$, that is, the complete graph of order $5$.


Also see


Sources

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