# How to prove that a simple graph having 11 or more vertices or its complement is not planar?

It is an exercise on a book again. If a simple graph $$G$$ has 11 or more vertices,then either G or its complement $$\overline { G }$$ is not planar.

How to begin with this? Induction?

It follows from the Euler's formula that a simple planar graph $G$ with $m$ edges and $n\geq 3$ vertices must satisfy (see here) $$\tag{1}m\leq 3n-6.$$ For a graph $G$ with $m$ edges and $n$ vertices, its complement $\overline{G}$ has $\displaystyle\frac{n(n-1)}{2}-m$ edges. Therefore, if $\overline{G}$ is also planar, by $(1)$ we have $$\tag{2}\frac{n(n-1)}{2}-m\leq 3n-6.$$ Adding $(1)$ and $(2)$, we obtain $$\frac{n(n-1)}{2}\leq 6n-12,$$ which implies that $n\leq 10$.
• I've had this doubt, could it be that both $G \text{ and } \overline G$ are not planar? Sep 13, 2015 at 20:13
• @YoTengoUnLCD Of course. What this proved was that at least one of $G$ and $\overline{G}$ must be nonplanar. If you want specific constructions of graphs who are nonplanar whose complements are also nonplanar, see for example this question. May 25, 2016 at 17:46