I am looking for a proof of the theorem:
If a planar graph, $G$, has $v$ vertices ($v \geq 3$) and no cycles of length 3 then, $e \leq 2v-4$.
I remember doing this in a graph theory course and I think we used Euler's Formula $v-e+f =2$ but I forget how exactly.
Any reference or solution would be appreciated.
By Euler's formula, $v-e+f=2.$
None of the faces are triangles (so each face has at least four edges) and every edge is in exactly two faces, so $2e \geq 4f.$
Combining these results we get $v-e+\frac{e}{2} \geq 2,$ which can be rearranged to give the desired result.