$K_{3,3}$ an exception to Planar Graph formula

Question

So I have learned that for a graph to be considered Planar, if it has at least 3 vertices, you can apply the following formula to test for planarity:

number of edges ≤ 3(number of vertices) - 6

also known as $m \leq 3n-6$, where $m$ is the number of edges (the size), $n$ is the number of vertices (the order).

Why is the complete graph known as $K_{3,3}$ (which I have linked for convenience) not considered planar, even though it follows this rule? Is it an exception?

Answer 1

It's not really a test; it's a fact that falls out of a graph if it is planar. So if a graph is planar, then we have $m \leq 3n -6$ ($n$ vertices, $m$ edges). But not all graphs with this property will be planar.

In other words:

$G$ planar $\Rightarrow m \leq 3n-6$

but

$m \leq 3n-6 \nRightarrow G$ is planar.

Answer 2

Your misunderstanding is to do with logic, not with graph theory. The test says:

if the graph is planar, then $e\le 3v-6$.

This is the same as saying

if $e>3v-6$, then the graph is not planar.

But it is not the same as saying

if $e\le 3v-6$, then the graph is planar.

In fact, if $e\le 3v-6$, then the graph may be planar and may be non-planar. In terms of your comment on Alfred Yerger's answer, the formula does not "determine whether a graph is planar or not". It may give you the answer "non-planar" or "don't know", but can never give you the answer "planar".