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Oct 24, 2016 at 8:10 comment added Martin Sleziak In your second comment you ask why the authors started the proof in this way. Short answer is: because it works. I do not have sufficient insight into this area to be able what might be a reasonable proof strategy for some result. Maybe for somebody who has seen several proofs of this type, it is almost immediate that this is a reasonable start.
Oct 24, 2016 at 8:08 comment added Martin Sleziak About your first comment: If you have a look into the version of the paper which was actually published (assuming you have access to it), the claim about $R(Q_n,Q_n)$ is omitted in Theorem 1 and it contains the above inequality instead. (Which is proved in the paper, however I did not look at the details of the proof.)
Oct 24, 2016 at 2:45 comment added Verse why do we divide the set with $n^2+n=n(n+1)$ elements into $(n+1)$ sets with $n$ elements each ?
Oct 24, 2016 at 1:35 comment added Verse Thank you, it's very helpful. but the paper did not mention the proof of the $$n+m \le R(Q_n,Q_m) \le mn+n+m.$$
Oct 23, 2016 at 1:56 history answered Martin Sleziak CC BY-SA 3.0