Exercise 4.1.23* (When $\lfloor n \alpha \rfloor, \lfloor m \beta \rfloor$ fill $\mathbb{N}^*$?)

Proof. $\text{∙}$ Suppose first that $\alpha$ and $\beta$ are positive irrational numbers such that $\frac{1}{\alpha }+\frac{1}{\beta }=1$. Then $1∕\alpha =1-1∕\beta <1$, thus $\alpha >1$. We can write $\alpha =1+\lambda$, where $\lambda$ is a positive irrational number.

Then $1∕\beta =1-1∕\alpha =(\alpha -1)∕\alpha =\lambda ∕(\lambda +1)$, so $\beta =1+{\lambda }^{-1}$. Therefore

where $\lambda$ is a positive irrational number.

By Problem 22, we know that $𝒮={\mathbb{N}}^{\text{∗}}$, and that every positive integer appears exactly once among the terms $\lfloor \mathit{\alpha x}\rfloor =\lfloor x+\mathit{\lambda x}\rfloor$ together with $\lfloor \mathit{\beta x}\rfloor =\lfloor x+{\lambda }^{-1}x\rfloor$, where $x\in {\mathbb{N}}^{\text{∗}}$.

$\text{∙}$ Conversely, suppose that $𝒮$ consists of every positive integer, each appearing exactly once.

If $\alpha <1$, then $\lfloor \alpha \cdot 1\rfloor =0\notin {\mathbb{N}}^{\text{∗}}$, in contradiction with the hypothesis. Therefore $\alpha >1$, and similarly $\beta >1$, so we can write $\alpha =1+\lambda$, $\beta =1+\mu$, where $\lambda >0,\mu >0$.

For all positive integers $n,m$, we define

As in part (a) of the solution of Problem 22, the sequences $u={(u_n)}_{n\in {\mathbb{N}}^{\text{∗}}},v={(v_m)}_{m\in {\mathbb{N}}^{\text{∗}}}$ are strictly increasing.

If $\lambda \ge 1$ and $\mu \ge 1$, then $u_1\ge 2$ and $v_1\ge 2$ thus the positive integer $1$ is not represented by any term of the sequences $u,v$. Therefore $\lambda <1$ or $\mu <1$.

If $\lambda <1$ and $\mu <1$, then $u_1=v_1=1$, so $1$ is not appearing exactly once. Since $\lambda$ and $\mu$ are irrational, $\lambda \ne 1$ and $\mu \ne 1$. Therefore the only two possibilities are $\lambda <1$ and $\mu >1$, or $\lambda >1$ and $\mu <1$. Since $\alpha$ and $\beta$ play a symmetric role, these two possibilities are treated in the same way, so we assume from now that

It remains to prove that $\mu ={\lambda }^{-1}$ (and then $1∕\alpha +1∕\beta =1$). As in part (b) of the solution of Problem 22, we prove that

Indeed,

Therefore $\lfloor (m+1)\mu \rfloor =\lfloor \mathit{m\mu }\rfloor$ or $\lfloor (m+1)\mu \rfloor =\lfloor \mathit{m\mu }\rfloor$, which gives (1):

Moreover, if $v_{m+1}-v_m=2$, then

thus $\mathit{m\mu }<\lfloor \mathit{m\mu }\rfloor +1=k$, so

so $m=\lfloor k{\mu }^{-1}\rfloor$.

Conversely, if $m=\lfloor k{\mu }^{-1}\rfloor$ for some $k\in {\mathbb{N}}^{\text{∗}}$, then, using $\mu \in ℝ\setminus \mathbb{Q}$, we obtain $m<k{\mu }^{-1}<m+1$ , thus $k{\mu }^{-1}<m+1<k{\mu }^{-1}+1$, so $k<(m+1)\mu <k+\mu <k+1$. Therefore $\lfloor (m+1)\mu \rfloor =k$. Moreover $m<k{\mu }^{-1}<m+1$ implies $\mathit{m\mu }<k<(m+1)\mu$, thus $k-1<k-\mu <\mathit{m\mu }<k$, so $k=1+\lfloor \mathit{m\mu }\rfloor$. This proves $\lfloor (m+1)\mu \rfloor =1+\lfloor \mathit{m\mu }\rfloor$. Thus

The equivalence (2) is proven.

These two items (1) and (2) mean that the only integers not represented by the sequence $v$ are the integers $v_{\lfloor k{\mu }^{-1}\rfloor }+1$, where $k\in {\mathbb{N}}^{\text{∗}}$. The least such integer is $v_{\lfloor {\mu }^{-1}\rfloor }+1$. This integer must be represented by the least term of the sequence $u$, so $u_1=v_{\lfloor {\mu }^{-1}\rfloor }+1$.

If we suppose that $u_k=v_{\lfloor k{\mu }^{-1}\rfloor }+1$ for all positive integers $k<n$, then $v_{\lfloor n{\mu }^{-1}\rfloor }+1$ must be represented by the least term of the sequence $u$ greater than $u_{n-1}$. Therefore $u_n=v_{\lfloor n{\mu }^{-1}\rfloor }+1$. This induction shows that, for all $n\in {\mathbb{N}}^{\text{∗}}$,

This is equivalent to

We know that $\lfloor \mathit{\lambda n}\rfloor \sim \mathit{\lambda n}$ in the neighborhood of $+\infty$, if $\lambda >0$. Therefore

Then $(1+\lambda )n\sim (1+{\mu }^{-1})n$ shows that $1+\lambda =1+{\mu }^{-1}$, thus $\lambda ={\mu }^{-1},\mu ={\lambda }^{-1}$. Since $\alpha =1+\lambda ,\beta =1+\mu$, we obtain

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