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We were given yesterday couple of pigeonhole principle questions, i did solve most of them but these 3 i could not, i don't know even where to start.

1) 20 people are sitting around round table, on it there is a big pizza (20 slices) 10 of them with olives and 10 without, some of people like the pizza with olives while other don't, show that one can rotate the pizza such that at least half of the people(at least 10) are happy ?

2) Given a graph with 6 vertices prove that there is at least 3 vertices that have no edge between every two of them or that they all are connected ?

3) Given $x \in \mathbb{R}$ and $x \not \in \mathbb{Q}$ prove that for all $\epsilon >0$ there is $n>0$ such that $\{ n x \} < \epsilon$ ?

any help is appreciated, even if you know how solve one question, please leave a comment showing me how to prove it, or posting an answer,

Thank you.

Ahmad
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1 Answers1

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This problem can be retooled to be as such: Instead of considering no edge vs. having an edge, consider $K_6$ (the complete graph on 6 vertices) such that every edge has a color. A red edge corresponds to the edges from the graph you want to consider being connected, and a blue edge means those vertices have no edge. We're looking to show that there is at least one red or one blue triangle.

Consider an arbitrary vertex $v$. By generalized pigeonhole principle, vertex $v$ must have at least 3 edges the same color, WLOG say that it has 3 red edges, furthermore connect those edges to vertices $a,b$ and $c$. To avoid a monochromatic triangle, edges $\{a,b\}$ and $\{b,c\}$ must be blue.

But no matter the choice of color edge for $\{a,c\}$, you end up with a monochromatic triangle (a red triangle on $v,a,c$ or a blue triangle on $a,b,c$, leading to the situation described (three vertices all connected or three vertices with no edge between them).

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Joseph DeGaetani
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