Euler Polyhedron Formula

Theorem
For any convex polyhedron with $V$ vertices, $E$ edges, and $F$ faces:


 * $V - E + F = 2$

Proof
Let $G$ be a planar graph with $V$ vertices and $E$ edges.

Let $V = 1$.

The number of faces $F$ is then given by:
 * $F = E + 1$

Hence:
 * $V - E + F = 1 - E + \left({E + 1}\right) = 2$

and the formula holds.

Otherwise, let $G$ be any (planar) graph.

Since contracting any edge decreases the number of vertices and edges each by one, the value of $V - E + F$ remains unchanged.

Hence by induction through contracting edges indefinitely, the value remains the same as if $G$ was the same as the one considered in the previous case.

Hence $V - E + F = 2$ for any (planar) graph.

From Polyhedra and Plane Graphs, any polyhedron's vertices, edges and faces may be represented by a (planar) graph.

Hence the formula applies to polyhedra.