Prove that every $k$-chromatic graph has size $m\geq \binom k2$.
Here is what I know:
Let $G$ be a $k$-chromatic graph, that mean $\chi(G)=k$. Thus $G$ must have a subgraph of a complete $k$-partite graph, say $H$. Hence $m_G \geq m_H$.
Here are my questions
1) I'm kinda confused between $k$-coloring, $k$-colorable and $k$-chromatic.
I know that $G$ is $k$-colorable means $\chi (G) \leq k$ and $\chi(G)$ is the minimum number of colors needed for a legal coloring of $G$. Then the book said if $G$ is $k$-chromatic then there exists a $k$-coloring but no $(k-1)$-coloring.
I feel like the book is just kicking me around.
2) I know $\binom k2$ is the size of a complete graph of order $k$, but I can't see how a $k$-partite graph has that size.