Proof

Again, we proceed by contradiction. Suppose \(K_{3,3}\) were planar. Then by Euler's formula there will be 5 faces, since \(v = 6\text{,}\) \(e = 9\text{,}\) and \(6 - 9 + f = 2\text{.}\)

How many boundaries surround these 5 faces? Let \(B\) be this number. Since each edge is used as a boundary twice, we have \(B = 2e\text{.}\) Also, \(B \ge 4f\) since each face is surrounded by 4 or more boundaries. We know this is true because \(K_{3,3}\) is bipartite, so does not contain any 3-edge cycles. Thus

But this would say that \(20 \le 18\text{,}\) which is clearly false. Thus \(K_{3,3}\) is not planar.