# There exists a graph $G$ with chromatic number $\chi(G)\ge n$ but $G$ does not contain a triangle

Could you give an example or a proof for a graph that has no triangles but still has chromatic number $$\chi(G) \ge n$$, where $$n \gt 3$$ is the number of vertices? This is from coloring of a graph.

The only way I can think about it is in terms of complete graph $$K_n$$, but that has triangles and its chromatic number is $$\chi(K_n) \ge n$$. But can you give an example for a graph that is not complete and still has $$\chi(G) \ge n$$?

An upper bound for chromatic number is $$\chi(G) \le \Delta(G)+1$$ (see Greedy coloring). And we know that for any simple graph $$G$$, we have $$\Delta(G) \le n-1$$ (Note that if $$G$$ is not simple, we can consider its condensation by removing parallel edges as they don't affect chromatic number). Thus, we have $$\chi(G) \le n$$. So, it is not hard to see that $$\chi(G) = n$$ is possible only when we have $$G = K_n$$. But, as you noticed, this is not a triangle-free graph for $$n \ge 3$$.
I assume that by $$n$$ you mean the number of vertices. If a graph is not complete then there is a pair of vertices with no edge between them, color them in the same color and all of the other vertices in their own unique colors to obtain an $$(n-1)$$-coloring. So a graph that has chromatic number $$\ge n$$ is complete. A complete graph can still contain no triangle, as long as it has less than 3 vertices :). If it has 3 or more then take any 3 of them to form a triangle.