For a positive real number $\alpha$, define \[S(\alpha)=\{ \lfloor n\alpha\rfloor \; \vert \; n=1,2,3,\cdots \}.\] Prove that $\mathbb{N}$ cannot be expressed as the disjoint union of three sets $S(\alpha)$, $S(\beta)$, and $S(\gamma)$.
Problem
Source:
Tags: floor function, Irrational numbers
Peter
30.08.2007 01:51
Solution by jmerry: http://www.mathlinks.ro/viewtopic.php?t=127810
H.HAFEZI2000
04.04.2020 12:27
we assume that $\alpha<\beta<\gamma$ and if $\alpha<1 \implies S(\alpha)=N$ now assume that three number exist such as that $1<\alpha<\beta<\gamma$ and they satisfy the problem, so for every natural number $n$ if $n$ isn't divisible by $a,b,c$ , the number of members of $S(\alpha)$ that are less than $n$ is equal to $\lfloor\alpha^{-1}n\rfloor$ and respectively we get $\lfloor\beta^{-1}n\rfloor$ for $S(\beta)$ and $\lfloor\gamma^{-1}n\rfloor$ for $S(\gamma)$. from now on suppose that $n$ isn't divisible by $a,b,c$ and because they are infinitive natural numbers with this property this wouldn't be a problem. because the three sets are disjoint and all numbers $1$ to $n-1$ belong exactly to one of these sets then we must have $\forall n>1 \in N :\lfloor\alpha^{-1}n\rfloor+\lfloor\beta^{-1}n\rfloor+\lfloor\gamma^{-1}n\rfloor=n-1$ . I didn't make any progross from here!