1

Show that if $ G $ is a graph with $ n $ vertices such that $ | E (G) | > \frac {n^2}{4} $ then $ G $ contains a triangle.

My try:

Suppose $ G $ does not contain any triangle, then there is no clique of size $ k \geq 3 $. By the Turán theorem $$ | E (G) | \leq \left(1- \frac{1}{3-1} \right) \frac{n^2}{2}= \frac{n^2}{4} $$ Therefore, if $ G $ is a graph with $ n $ vertices such that $ | E (G) | > \frac {n^2}{4} $ then $ G $ contains a triangle.

Sofía Contreras
  • 1,494
  • 2
    Suggestions for what? Your proof looks fine to me (besides you mean no clique of size $\geq 3$ don’t you?) – Jonas Linssen Jun 02 '20 at 05:38
  • 2
    I am guessing tat the point of this exercise is to not invoke Turan's theorem and give a self-contained proof. – caffeinemachine Jun 02 '20 at 07:26
  • I got the same feeling. Because seems to easy with the Turan´s theorem. – Sofía Contreras Jun 02 '20 at 13:22
  • There are certainly elementary proofs of this fact that do not appeal to Turán's theorem. For instance, if $G$ is a graph with more than $\frac{n^2}{4}$ edges, then $G$ contains a vertex whose deletion leaves a graph with more than $\frac{(n-1)^2}{4}$ edges (this is proven by simple counting and considering even/odd cases). Using this fact, the proof of your statement would follow by induction immediately. – Paralyzed_by_Time Jun 03 '20 at 00:33
  • Linked. – Alex Ravsky Jun 07 '20 at 10:22

0 Answers0