Draw a non-planar graph whose complement is a non-planar graph I have been teaching myself graph theory. I am stuck at solving this problem on my own.
Please provide an example of such a graph. 
What approach would you take to draw such a graph? 
 A: One way to think of non-planar graphs is graphs with too many edges. You can use Euler's polyhedron formula $V - E + F = 2$ to quantify this based on the vertices $V$, edges $E$, and faces $F$ of a graph, and you get that if a graph is planar then $E \leq 3V-6$.
Since a graph and its complement add to the complete graph with $\frac{n(n-1)}{2}$ edges, if you pick a non-planar graph with fewer than $\frac{n(n-1)}{2} - (3n-6)$ edges, the complement has too many edges and they are both non-planar.
So, for example, take a graph on $10$ vertices and connect $5$ of them into $K_5$, while leaving the others disconnected. In the complement, the disconnected vertices form another $K_5$, so they are both non-planar.
A: Non-planar graphs include (in a certain sense) either the complete graph $K_5$ or the complete bipartite graph $K_{33}$. To me the latter seems easier to work with. Now, if we just take $K_{33}$, it's complement is not non-planar. But if we take $K_{33}$ and add a second component consisting of three vertices, then when we take the complement this component will form a $K_{33}$ with each of the existing components, so the resulting graph will also be non-planar.
We can connect the three new vertices to each other however we want without affecting the result, and if we want a connected graph we can connect them to one of the existing parts of the $K_{33}$--in the complement graph they will still form a $K_{33}$ with the other component.
Here's an example of such a graph:

A: I drew UG and added six more isolated vertices. Now, in G' K6 is a subgraph and therefore G' is nonplanar.
