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Let $\alpha>0$ and define $S(\alpha)=\{\lfloor n \alpha \rfloor: n\in\Bbb Z^+ \}$. (Here $\lfloor x\rfloor$ is the integer part of $x$ and $\mathbb Z^+$ the set of positive integers.)

Is there any known characterization for the pairs of positive irrationals $\alpha$ and $\beta$ for which $S(\alpha)\cap S(\beta)=\varnothing$?

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See also… – Gerry Myerson Dec 11 '13 at 22:19

3 Answers 3

Not a complete answer: these sequences are called Beatty sequences. It's a nice exercise to show that $S(\alpha)$ and $S(\beta)$ partition the positive integers when $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, hence for $\alpha, \beta$ satisfying this condition we can take $S(n \alpha), S(m \beta)$ where $n, m \in \mathbb{N}$.

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It seems to me that $S(\alpha)\cap S(\beta)=\varnothing$ if and only if $m/\alpha+n/\beta=1$, for some $m,n$ positive integers. – smyrlis Dec 11 '13 at 20:59

You are right about $S(\alpha)\cap S(\beta)=\varnothing\iff \dfrac{n}{\alpha}+\dfrac{m}{\beta}=1$ for some $n,m\in\mathbb Z^+$ (see Theorem 8 of the cited paper).

The implication $\Longleftarrow$ is easy so let's assume that $\frac{n}{\alpha}+\frac{m}{\beta}\neq 1$ for all $n,m\in\mathbb Z^+$. Then one of the following is true.

  1. The numbers $1,\dfrac{1}{\alpha},\dfrac{1}{\beta}$ are linearly independent over $\mathbb Q$,
  2. there exist some $n,m,k\in\mathbb Z^+$ such that $\left|\dfrac{n}{\alpha}-\dfrac{m}{\beta}\right|=k$,
  3. there exist some $n,m,k\in\mathbb Z^+$ such that $\dfrac{n}{\alpha}+\dfrac{m}{\beta}=k$ with $k>1$, and $\gcd(n,m,k)=1$.

It is enough to show that any one of 1., 2. or 3. implies that $S(\alpha)\cap S(\beta)=\emptyset$. Elementary proofs of these can be found in this paper (Theorem 5, Theorem 6 and Theorem 7 respectively).

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I suggest looking in Joe Roberts amazing book I thought I remembered a partition into 3 such sequences but actually it is $\lfloor \tau \lfloor \tau n \rfloor \rfloor$,$\lfloor \tau \lfloor \tau^2 n \rfloor \rfloor$,$\lfloor \tau^2 n \rfloor $ So not exactly what was requested. Maybe that is not so deep.

I put in this answer partly to get the link to the book in.

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It was a problem on the 1995 Putnam exam that there is no partition of the natural numbers into three sequences $\lfloor n\alpha\rfloor$, $\lfloor n\beta \rfloor$, and $\lfloor n\gamma\rfloor$. – Lucia Dec 12 '13 at 1:21
Indeed, it is possible to prove that three such sequences are never pairwise disjoint. – Rodrigo Dec 12 '13 at 2:29
Ha. OP has posted the 3-sequence question to… – Gerry Myerson Dec 12 '13 at 5:23
The Putnam exam version has easier proof, because the statement is weaker. – i707107 Dec 13 '13 at 18:59

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