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Has anyone discovered a vector of algebraic real numbers $(a_1,...,a_k)$ such that $1,a_1,...,a_k$ are linearly independent over $\mathbb{Q}$ and such that $(a_1,...,a_k)$ is not "badly approximable"?

Has anyone discovered a vector of algebraic real numbers $(a_1,...,a_k)$ such that $$[\mathbb{Q}(a_1,...,a_k):\mathbb{Q}]>k+1$$ but such that $(a_1,...,a_k)$ is "badly approximable"?

knowing whether either of these is open would be a great help.

By badly approximable I mean that there exists $c>0$ such that the $\ell_\infty$ distance from $(qa_1,...,qa_k)$ to $\mathbb{Z}^k$ is greater than $cq^{-1/k}$ for all large $q\in \mathbb{N}$.

Or the equivalent notion that there exists $c>0$ such that the distance from $\sum_i q_i a_i$ to $\mathbb{Z}$ is greater than $c\|q\|_{\ell_\infty}^{-k}$ for all large $q\in \mathbb{Z}^k$.

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  • $\begingroup$ There are multiple notions of "badly approximable"; what definition are you considering? $\endgroup$ – Greg Martin Feb 7 at 6:52
  • $\begingroup$ @GregMartin Thanks I'ved edited the question $\endgroup$ – user135512 Feb 7 at 7:26
  • $\begingroup$ It would help in understanding the question if you could clarify what it says when $k=1$ and how it relates to Roth's theorem on approximation of algebraics. $\endgroup$ – Gro-Tsen Feb 7 at 8:02
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    $\begingroup$ @Gro-Tsen Yes, I should have mentioned that when k=1, the first question seems to be open: mathoverflow.net/questions/177481/…. So if I had to guess, I would assume that the k>1 case is open too, but I wasn't sure. $\endgroup$ – user135512 Feb 7 at 8:06
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    $\begingroup$ @Gro-Tsen Roth's theorem says that algebraic numbers are not "very well approximable". Being badly approximable is stronger than being not very well approximable $\endgroup$ – user135512 Feb 7 at 8:12

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