Q: Prove that a graph in which triangular regions are permitted is planar if and only if $e \le 3v-6$.
This is one of the exercise questions from a discrete mathematics textbook. I feel this questions is wrong. Basically, we can prove that a graph in which triangular regions are permitted is planar implies $e \le 3v-6$. But we cannot prove the converse.
If all the regions are triangular, then we have $3r \le 2e$. So $e+2 =r+v \le \frac23e+v$ after simplification, we get $e\le 3v-6$.
A graph in which triangular regions are permitted. $\Rightarrow e\le 3v-6$
The textbook has another exercise as follows:
Q: Prove that a bipartite graph can only be planar if $e\le 2v-4$.
Basically, this question is equivalent to prove if a bipartite graph is planar, then $e\le 2v-4$. Am I right to say that?