Every graph whose chromatic number is more than ____ is not planner.
My attempt:
The answer should be $4$ by four color theorem.
Somewhere, I read "Five color theorem"(See Theorem 6.3.8 at page no.-74). Somewhere, there is given "Six color theorem".
Can you give a planar graph such that chromatic number of that graph will $5$?
However, given question "Every graph whose chromatic number is more than ____ is not planner." does not seem correct. Since bipartite graph $K_{3, 3}$ has chromatic number is $2$, but it's not planar graph.
Can you explain in formal way, please?
You have to find for which $n$, the sentence "Every graph whose chromatic number is bigger than $n$ is not planar". This sentence is of the type $A\implies B$ where $A$ is "The chromatic number of $G$ is bigger than $n$" and $B$ is "$G$ is not planar" and your task is to find $n$ such that $A\implies B$ is true.
Of course, it won't tell you anything about graphs whose chromatic number is less or equal to $n$ but you don't really care about that.
Concerning the problem itself, $4$ is a right answer because of the four color theorem (every planar graph has chromatic number at most $4$). Since the four color theorem has been proved by a computer (they reduced all the planar graphs to just a bunch of different cases, about a million I think), most of the books show the proof of the five color theorem (which has a non-computer proof).
Note that $10^{42}$ is a right answer too, because they are not asking the smallest $n$ for which it is true. However, the expected answer is probably $4$, or at most $5$.