Q:

Showing that if $\chi(G) = n$ then $G$ must be $K_n$

How should I go about trying to proof that if $\chi(G) = n$ then $G$ must be $K_n$? Hints anyone?

A:

Suppose that there is no edge between vertices $u$ and $v$ in $G$.  Can you color them the same color?  What does that make $\chi(G)$?

