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Let $x_1,x_2,\ldots, x_k \in [0,1]$ be irrational numbers.

I'm interested in what, if anything, can be said about the values $\{nx_i \bmod 1: n\in \mathbb{N}\}$. Specifically, I'm interested if there are constraints that we can place on the $x_i$'s such that for any set of $y_i$'s and any $\varepsilon \in (0,1/2)$, there exists $n\in \mathbb{N}$ simultaneously satisfying

$$nx_i- y_i \bmod 1 \equiv \varepsilon_i$$ for all $i\in \{1,2,\ldots, k\}$, where all $\varepsilon_i \in (-\varepsilon, \varepsilon)$.

I know that this is impossible in general. For example, if $x_1=\sqrt{2},x_2=2\sqrt{2}$ then we could not have $x_1 n\bmod 1\approx 1/3$ and $x_2 n \bmod 1 \approx 1/3$ simultaneously. However, I maybe if we place some restrictions on $x_i$, e.g., requiring them all to be pairwise linearly independent over $\mathbb{Q}$, then we could say something of this form?

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Your guess is correct. For $\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_d) \in \mathbb{R}^d$, the values of $m \alpha \bmod 1$ are equidistributed (and in particular dense) in $(\mathbb{R}/\mathbb{Z})^d$ if and only if, for all nonzero $k \in \mathbb{Z}^d$, we have $k \cdot \alpha \not\in \mathbb{Z}$. See, for example, Exercise 5 in Terry Tao's notes.

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  • $\begingroup$ Thanks! This is exactly what I was looking for. If I understand correctly my guess was not quite correct, I guessed that this only required pairwise linear independence, i.e., $x_i \neq k x_j$ for any $k\in \mathbb{Q}$ for all $i\neq j$, but it looks like it requires $\{x_1,x_2,\ldots, x_n\}$ to be linearly independent over $\mathbb{Q}$. $\endgroup$
    – Alek Westover
    Commented Jul 29, 2023 at 21:04
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    $\begingroup$ @AlekWestover The correct condition is that $1,x_1,x_2,\dotsc,x_n$ are linearly independent over $\mathbb{Q}$. Density is due to Kronecker, equidistribution is due to Weyl. For a short proof of density, see my post at mathoverflow.net/questions/106819/… . $\endgroup$
    – GH from MO
    Commented Jul 29, 2023 at 21:21
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    $\begingroup$ @GHfromMO Thanks for the correction. You are right, we need $1$, $x_1$, $x_2$, ..., $x_n$ lin indep, not just $x_1$, $x_2$, ..., $x_n$ lin indep. That looks like a typo in Tao's notes. I've fixed my post now. $\endgroup$
    – David E Speyer
    Commented Jul 29, 2023 at 22:22
  • $\begingroup$ @DavidESpeyer Perhaps Tao meant modulo $1$, in which language $k\cdot\alpha\not\in\mathbb{Z}$ reads $k\cdot\alpha\neq 0$. I meant to correct Alek Westover, not you :-) $\endgroup$
    – GH from MO
    Commented Jul 29, 2023 at 22:26

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