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I am working with something on Diophantine approximation, and I found a high dimensional generation of Dirichlet approximation theorem which may be true; I will be very happy if this is true. The statement is following:

Dirichlet approximation theorem, high dimensional generation Let $\alpha_1,..,\alpha_n\in \mathbb R$ be n linear independent numbers on $\mathbb Q$. Then $\forall x\notin \mathbb Q(\alpha_1,...,\alpha_n)$, there are infinitely many tuples $c-(c_1,...,c_n)\in \mathbb Q^n$ such that, $$|x-\sum_{i=1}^n c_i\alpha_i|<\frac{1}{\|c\|^{n+1}_{hight}}$$ Where $\|c\|_{hight}:=\max_i \{\|c_i\|_{hight}\}$, $\|c_i\|_{hight}=\max {\{|p|,|q|, c_i=\frac{q}{p}, (p,q)=1\}}$.

It is not difficult to see this conjecture coincides with the classical Dirichlet problem when $n=1$.

Has this problem been studied? As the problem is so natural, I think it should be a classical result but I can not find it in classical books in Diophantine approximation.

I will appreciate for point out a reference or some advice, thanks!

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We can not get a proof more or less the same with Dirichlet approximation theorem.

Now let us consider a more nontrivial one, if we do some distortion on the linear structure, for example, is it true that:

conjecture Let $\alpha_1,..,\alpha_n\in \mathbb R$ be n linear independent numbers on $\mathbb Q$. Then $\forall x\notin \mathbb Q(\alpha_1,...,\alpha_n)$, there are infinitely many tuples $c-(c_1,...,c_n)\in \mathbb Q^n$ such that, $$min_{1\leq k\leq n}|x-\sigma_k(c_1\alpha_1,...,c_n\alpha_n)|<\frac{1}{\|c\|^{n+1}_{hight}}$$ Where $\|c\|_{hight}:=\max_i \{\|c_i\|_{hight}\}$, $\|c_i\|_{hight}=\max {\{|p|,|q|, c_i=\frac{q}{p}, (p,q)=1\}}$. $\sigma_k$ is the $k-$th symmetric sum, i.e. $$\sigma_k(\alpha_1,...,\alpha_n)=\sum_{1\leq i_1<...<i_k\leq n}\Pi_{j=1}^k\alpha_{i_j}$$

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  • $\begingroup$ Sorry, I posed this problem too fast, we can get a proof more or less the same with Dirichlet approximation theorem. The key point is the boxes principle and the linear structure. I have add a further problem, which seems to be nontrivial. $\endgroup$
    – Hu xiyu
    Commented Dec 29, 2017 at 6:37
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    $\begingroup$ You can delete the question, if you choose to. $\endgroup$
    – Hans
    Commented Dec 29, 2017 at 6:59
  • $\begingroup$ Sorry, the linear case have not been proved. My argument in mind is wrong. This two problem are both important to have an answer. $\endgroup$
    – Hu xiyu
    Commented Dec 29, 2017 at 7:12

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