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I am stumped. I will give the problem and what I have so far.

For a positive real number α, define $S(α)$ = {$\lfloor nα$$\rfloor$ : n = 1,2,3,...}. Prove that {1,2,3,...} cannot be expressed as the disjoint union of three sets $S(α)$,$S(β)$ and $S(γ)$.

What I have so far is since $\alpha$ is a real number, one could write $\alpha$ as $a.$$\rho$, where $a$ is an integer and $\rho$ is real. Thus we have that $\lfloor nα$$\rfloor$=$na$. Thus $S(\alpha$) is the set of all multiples of n. In other words, $S(a)$=$a\mathbb N$. So we (assume by way of contradiction) that $\mathbb N$=$a\mathbb N$ $\cup$ $b\mathbb N$ $\cup$ $y\mathbb N$, for $a,b,y$ $\in$ $\mathbb Z$.

That is all I have so far. From here I am done. I was thinking that if you could split up $\mathbb N$ into a union of multiples of $\mathbb N$, you would miss some integers that are neither multiples of $a,b$ or $c$. So maybe one could do something like this, but I cannot formalize it. Maybe I am wrong entirely.

Mike Pierce
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combinatorialist46Carey2
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  • Your first statement is true, since we can write $a=1, \rho = \alpha$, but it should be apparent the sequence $\lfloor n\alpha\rfloor \neq n$ except when $\alpha = 1$. We stop there (none of the rest can be recovered) – Brian Moehring Feb 20 '21 at 19:03
  • $a$ need not be $1$ though. Right? And I think I got something. We may assume without loss of generality that $gcd(a,b,y)$=$r$$geq 1$. $r$ must be an integer. So we can all write $a,b,y$ as multiples of $r$ and maybe show that $r$ is not an integer for some values of $x$ $\in$ $mathbb Z$. – combinatorialist46Carey2 Feb 20 '21 at 19:19
  • Yeah that could work! For instance, consider the integer $\frac{r^2n}{r^2}$, where $n$ is not a multiple of $r$. None of the $a,b,y$ will ever "reach" or "touch" this integer. But this integer was arbitrary. Does this work? – combinatorialist46Carey2 Feb 20 '21 at 19:23
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    The proof. See also this wiki page. – WhatsUp Feb 20 '21 at 21:56
  • My point is you have not defined $a$, so your question about whether it must be $1$ or not is sort of the point. Until we resolve that, none of the rest of your argument is worth talking about. For a concrete example, suppose $\alpha = 4/3$: what is $a$? And once you have that value, is $\lfloor n\alpha \rfloor = na$? (No matter which $a$ you choose, these two sequences cannot equal) – Brian Moehring Feb 20 '21 at 22:45
  • But isn't $\lfloor \frac{4n}{n} \rfloor$=$na$? Oh wait, choose $n=3$. Ok thank you. The idea of taking fractional parts keep popping in my head. – combinatorialist46Carey2 Feb 20 '21 at 23:53
  • There is an earlier thread where this problem was discussed. I gave a few references and outlined an approach. My answer is not very strong, so I should not be the first to cast a vote to close as a duplicate. – Jyrki Lahtonen Feb 28 '21 at 20:25
  • Thank you very much sir! – combinatorialist46Carey2 Feb 28 '21 at 20:48

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