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This question is inspired by my own answer to this question. For a real number $x > 0$, define $$ S(x) = \{\lfloor kx \rfloor \mid k \in \mathbb N\}. $$ Are there positive real numbers $x, y$ such that $S(x) \cap S(y)$ is finite?

If $x, y$ are integers then the answer is clearly no: every of their common multiples is in there. This rules out rationals, too: when multiplied by the least common multiple of their denominators they reduce to the integral case.

This leaves irrational numbers, and intuitively it seems obvious to me that these would be "even worse": something about how the fractional parts of $(kx)_{k \in \mathbb N}$ for $x$ irrational are dense in the unit interval, maybe? Something about irrationality measures? But I don't know how to turn this into an argument.

Mees de Vries
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    In a somewhat tantilizing experimental attempt, I tried to take $x = 10$ and choose a $y$ so that $S(y) = \Bbb N\setminus S(x)$. Turns out, $y = \frac{10}9$ clearly doesn't work (as $9\cdot \frac{10}{9}\in S(x)$) and for $\epsilon>0$, $\frac{10}{9} - \epsilon$ will always be just a tiny bit too small, no matter how small you make $\epsilon$. – Arthur Jun 08 '18 at 08:48

1 Answers1

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For any positive irrational $r$, the Beatty sequence is the sequence of numbers

$$\mathcal{B}_r = \lfloor r \rfloor, \lfloor 2r \rfloor, \lfloor 3r \rfloor, \ldots$$

Rayleigh theorem (also known as Beatty's theorem) states that

Given any irrational $r > 1$, there exists irrational $s > 1$ so that $\mathcal{B}_r$ and $\mathcal{B}_s$ partition the set of positive integers. i.e. each positive integer belongs to exactly one of the two sequences.

For this to happen, the condition is $$\frac1r + \frac1s = 1 \iff s = \frac{r}{r-1}$$ Translate this to the problem at hand. For irrational $x > 1$, just take $y = \frac{x}{x-1}$ and we will have

$$S(x) \cup S(y) = \mathbb{N}\quad\text{ and }\quad S(x) \cap S(y) = \{ 0 \}$$

For more details, see the wiki entry for Beatty's sequences and refs there.

achille hui
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    The classic examples are $x= \sqrt{2}$ and $y=\sqrt{2}+2$ , and $x= \varphi = \frac{1+\sqrt{5}}{2}$ and $y= \varphi+1 = \frac{3+\sqrt{5}}{2}$ – Henry Jun 08 '18 at 14:02