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?
Thanks for your help!
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Maybe using Kuratowki's theorem. – Josué Tonelli-Cueto Apr 06 '12 at 10:54
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4Or maybe Euler's formula. What is the maximum number of edges that a simple planar graph with 11 vertices can have? – Peter Shor Apr 06 '12 at 10:55
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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 $n\leq 10$.
Martin Sleziak
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Paul
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Oh,the book I am reading now doesn't have this formula m≤3n−6 ... Now it seems not to be a good book. – tamlok Apr 07 '12 at 01:49
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1I've had this doubt, could it be that both $G \text{ and } \overline G$ are not planar? – YoTengoUnLCD Sep 13 '15 at 20:13
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@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. – JMoravitz May 25 '16 at 17:46