Prove that for $\alpha\not\in\mathbb{Q},\alpha>0$, $([n+n\alpha])_{n\in\mathbb{N}}\sqcup([n+n\alpha^{-1}])_{n\in\mathbb{N}}=\mathbb{N} $

Question

My Quesion: Prove that for $\alpha\not\in\mathbb{Q},\alpha>0$, $([n+n\alpha])_{n\in\mathbb{N}}\bigcup([n+n\alpha^{-1}])_{n\in\mathbb{N}}=\mathbb{N}$, where $[k]$ means the integral part of $k\in\mathbb{Z}$, and $\sqcup$ means disjoint union.

My Idea: I just realized that $$\frac{1}{1+\alpha}+\frac{1}{1+\alpha^{-1}}=1, \quad\text{where } 1+\alpha, 1+\alpha^{-1} \text{ are positive irrational numbers}$$ So the interested sequences are Beatty's sequences. And here is a proof (page 11-13) for Beatty sequences.

You already provided a complete solution to your problem – Maximilian Janisch Oct 16 '19 at 14:26 — Maximilian Janisch, Oct 16 '19 at 14:26

score 0 · Accepted Answer · answered Oct 16 '19 at 20:43

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"You already provided a complete solution to your problem" Yes, I just proved it.

answered Oct 16 '19 at 20:43

WLOG

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If you propose a Q and answer it and accept your own answer do you gain reputation points for the acceptance? I don't really care. I never thought about it. – DanielWainfleet Oct 17 '19 at 07:21
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@DanielWainfleet No, I can't get any reputation points by answering my own question. Not wanting to draw any further attention to this solved question, I posted my answer to it. – WLOG Oct 18 '19 at 17:50

Prove that for $\alpha\not\in\mathbb{Q},\alpha>0$, $([n+n\alpha])_{n\in\mathbb{N}}\sqcup([n+n\alpha^{-1}])_{n\in\mathbb{N}}=\mathbb{N} $

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