# numbers independent over $\mathbb{Q}$ but not BA? numbers that aren't a basis for a number field but are BA?

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 $$q\in \mathbb{Z}^k$$.

• There are multiple notions of "badly approximable"; what definition are you considering? – Greg Martin Feb 7 at 6:52
• @GregMartin Thanks I'ved edited the question – user135512 Feb 7 at 7:26
• 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. – Gro-Tsen Feb 7 at 8:02
• @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. – user135512 Feb 7 at 8:06
• @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 – user135512 Feb 7 at 8:12