Let's prove that $K_{3,10}$ is not planar:
First, how many vertices and how many edges does $K_{3,10}$ have?
$v = \text{ and } e =$
Suppose, for the sake of contradiction, that $K_{3,10}$ is planar. Then how many faces would it have?
$f =$
However, since every face is bounded by at least edges, and every edge borders exactly faces, we can get a bound on the number of faces. What is the largest number of faces possible based on this line of reasoning?
$f \le$
This is a contradiction, so $K_{3,10}$ is not planar. QED.
