Proof

The proof is by contradiction. So assume that \(K_5\) is planar. Then the graph must satisfy Euler's formula for planar graphs. \(K_5\) has 5 vertices and 10 edges, so we get

\begin{equation*} 5 - 10 + f = 2\text{,} \end{equation*}

which says that if the graph is drawn without any edges crossing, there would be \(f = 7\) faces.

Now consider how many edges surround each face. Each face must be surrounded by at least 3 edges. Let \(B\) be the total number of boundaries around all the faces in the graph. Thus we have that \(3f \le B\text{.}\) But also \(B = 2e\text{,}\) since each edge is used as a boundary exactly twice. Putting this together we get

\begin{equation*} 3f \le 2e\text{.} \end{equation*}

But this is impossible, since we have already determined that \(f = 7\) and \(e = 10\text{,}\) and \(21 \not\le 20\text{.}\) This is a contradiction so in fact \(K_5\) is not planar.