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We have an algebraic number $a$ and a real number $b$. Can the following inequality have infinitely many solutions for $n \in \mathbb{N}$?

$$ \{an\} \in [b - \frac{1}{2^n}, b + \frac{1}{2^n}] $$

Here $\{x\}$ denotes the fractional part of $x$, $\{x\} = x - [x]$.

Background. I encountered this problem when I was dealing with some kronecker approximations, I wanted to show that for a given point $(b_1t \mod 2\pi, b_2t \mod 2\pi)$ where $b_1,b_2$ are algebraic numbers who are linearly independent over rational numbers, one can not get exponentially close to a given fixed point $(\phi_1,\phi_2) \in [-1,1]^2$. Some special cases of this problem could be solved by Baker but I was unable to solve it in the general case.

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  • $\begingroup$ crossposted in MSE: math.stackexchange.com/questions/4545287/… $\endgroup$
    – Franz Lemmermeyer
    Oct 5, 2022 at 11:10
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    $\begingroup$ Yes, even for $a = \sqrt{2}$. By the Baire category theorem (BCT), the intersection $\cap_{N=1}^\infty \cup_{n = N}^\infty \left(\{an\}-\frac{1}{2^n},\{an\}+\frac{1}{2^n}\right)$ is nonempty, since each $\cup_{n = N}^\infty \left(\{an\}-\frac{1}{2^n},\{an\}+\frac{1}{2^n}\right)$ is open and dense. (One of course doesn't need to appeal to BCT and can just keep recursively choosing nested intervals, the intersection of which $b$ will lie; but that's basically the proof of BCT anyways). $\endgroup$
    – mathworker21
    Oct 6, 2022 at 19:01
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    $\begingroup$ @IlyaBogdanov They are dense because the sequence $(\{\alpha n\})_{n=1}^\infty$ is dense. The Lebesgue measure tending to $0$ is irrelevant; e.g. putting a small ball around each rational will result in a dense set with the measure being as small as you wish. Your answer is the same as my comment. $\endgroup$
    – mathworker21
    Oct 7, 2022 at 11:52
  • $\begingroup$ Ah, yes, sorry, something mixed up in my head... $\endgroup$
    – Ilya Bogdanov
    Oct 7, 2022 at 13:15

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Such $b$ exists for every real $a$.

Define an increasiung sequence of positive integers $n_1,n_2,\dots$ as follows. Picj $n_1$ so that $\{an_1\}<1/2$. If $n_i$ has been already defined, choose $n_{i+1}>n_i$ such that $$ \{an_{i+1}\}\in\left[\{an_i\},\{an_i\}+\frac1{2^{n_i+1}}\right]; $$ such $n_{i+1}$ clearly exists (notice here that $\{an_i\}\leq 1-1/2^{n_i}$). Then the sequence of segments $$ \left[\{an_i\},\{an_i\}+\frac1{2^{n_i}}\right] $$ is nested, hence its intersection is a desired point $b$.

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