Ramsey Theorem: $R(3,4)\le 10$ Proof: Why is the number of friends be at-least 6 instead of 5(by pigeon rule, $\lceil 9/2\rceil=5$)? In this webpage, https://plus.maths.org/content/friends-and-strangers under the section, FINDING $R(3,4)$,
The author assumes that 10 people/points are necessary and takes out one point, say A. This A is connected to 9 other points. Now by pigeon hole rule, at least $\left\lceil \frac{9}{2} \right\rceil$. So at least 5 must be connected via blue. But the author writes, "There must either be at least 6 red ones, or at least 4 blue ones (otherwise there won't be 9 in all!) " I am trying hard to understand why it is so, can anyone help?
 A: If there are fewer than $6$ red edges and fewer than $4$ blue edges, then the total number of edges incident to $A$ is strictly less than $5+3=8$, which is a contradiction.
We could prove similar claims by the same reasoning ("there must be at least $5$ red edges or at least $5$ blues edges", or "there must be at least $8$ red edges or at least $2$ blues edges", etc.), but the author has chosen these numbers in particular for the purpose of finding a Ramsay structure of the necessary form.
A: If I am understanding the problem based on the very quick skim I gave it, it's about connected graphs with coloring. It doesn't have to be at least 5 blue. For example, all 9 could be red. We know, however that there has to be at least 5 of one of the colours, be it blue or red. Whatever configuration we have, it is either the case that at least 6 or red, or less than six are red (meaning at least 4 are blue). As one of these is true, we can consider both cases, and show that in each case, you will find the shape you need.
