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.
Best Answer
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
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.