Planarity found in inducing $K_5$

Question

I was interested in studying whether or not if when we remove an arbitrary two edges from $K_5$, we get a planar graph. I understand that a planar graph has at most $3v-6$ edges, where $v$ is the number of vertices in the graph. Hence, given that $K_5$ has $\binom{5}{2} = 10$ and because it is apparent that $10 > 9 = 3(5)-6$, $K_5$ is obviously not planar. However, given the fact that this is not necessarily an "if-and-only-if" assumption on the bounds of the edges of a planar graph, I wasn't entirely sure if removing edges on a graph such that $E(G) \leq 9$ was going to be a simple solution. Is there an induced subgraph of $K_5$ that has $8$ edges and is not planar?

Answer 1

$K_5-e$ is planar for any edge. By Kuratowski's theorem, a graph is planar iff it contains no subgraph that is a subdivision of $K_5$ or $K_{3,3}$. Since $K_5-e$ cannot contain a subdivision of $K_5$, and doesn't have enough vertices to contain a subdivision of $K_{3,3}$, it is planar. The same can be said if you delete two edges.

Answer 2

We may just take $K_5$, remove an edge (any edge has the same role by transitivity) and show that the resulting graph is planar. How? This way: