Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have not been able to find a proof to the statement that if a graph $G$ has $\chi(g)=k$, then it must have at least $\binom{k}{2}$ edges. Would you be able to show me a simple proof?

share|cite|improve this question
up vote 10 down vote accepted

Suppose we have a k-coloring of our graph, with $\chi$(g)=k. Then for any two colors, call them red and blue, there must be some edge that connects them; if there weren't, we could paint every red edge blue, and we would have a (k-1)-coloring of our graph. There are $\binom{k}{2}$ pairs of colors, and any given edge cannot connect more than 2 colors, so there must be at least $\binom{k}{2}$ edges in our graph.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.