### No simple polyhedron has seven edges.

**Proof:** We show first that for any polyhedron we have *2E ≥ 3F* and *2E ≥ 3V*. The faces of the polyhedron are polygons, each bounded by a number of sides. Along each edge exactly two faces come together, so an edge corresponds to exactly two sides: the total number of sides is *2E*. We also notice that any face has *at least* 3 sides, so the total number of sides is
*at least* 3 times the number of faces. Thus we get:

The total number of sides = *2E*

and

The total number of sides ≥ *3F*.

Putting this together we get:

*2E ≥ 3F*,

proving our first inequality.

To prove the second inequality we count the total number of *ends* of edges. Each edge has two ends, so the total number of ends equals *2E*. At each vertex at least three edges come together, so the total number of ends of edges is at least 3 times the number of vertices. Putting this together we get:

*2E ≥ 3V*.

Now, if a polyhedron has 7 edges, then *3F ≤ 14* and *3V ≤ 14*. This means that both *F* and *V* cannot be bigger than 4. A little thought will convince you that every polyhedron has strictly more than three faces, so we must have *F=4*. Similarly we get that *V=4*. This gives

*V - E + F = 4 - 7 + 4 = 1 ≠ 2*.

This tells us that our hypothetical seven-edged polyhedron cannot exist, for if it did, then Euler's formula would hold and the above sum would have to be equal to 2 — QED!