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Dirchlet's Approximation Theorem states that for $k$ real numbers $b_1, b_2, \ldots, b_n$, and $N\in \mathbb{N}$ there exists a $q\in\mathbb{N}$ such that for all $1\le i\le k$, there exists integers $p_i$ such that $|b_iq-p_i|\le \frac{1}{N^{\frac{1}{k}}}$.

Note that in some sense these approximations can be both lower or upper bounds. For a particular application I need rational approximations of this form in only one direction. In particular, I was curious if the following result holds. For positive real numbers $b_1,\ldots,b_k$ and any $\epsilon>0$, there exists a positive integer $q$ such that $0\le (qb_i-\lfloor qb_i \rfloor)\le \epsilon$ for $1\le i \le k$.

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This is false in general, consider any irrational $b_1\in (0,1)$ and $b_2=1-b_1$. Then $qb_1-\lfloor qb_1 \rfloor+qb_2-\lfloor qb_2 \rfloor=1$ for any positive integer $q$, thus if $\epsilon<1/2$, your $q$ does not exist.

On the other hand, if $b_i$'s and 1 are rationally independent, we may say even more: for any intervals $\Delta_i\subset (0,1)$ there exist a positive integer $q$ such that fractional part of $qb_i$ lies in $\Delta_i$.

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  • $\begingroup$ Is it not a theorem that $q$ does exist (satisfying the positive relations in the problem) if there are no rational dependencies between the b_I? Gerhard "Not Remembering His Linear Algebra" Paseman, 2017.03.10. $\endgroup$
    – Gerhard Paseman
    Commented Mar 10, 2017 at 17:48
  • $\begingroup$ @GerhardPaseman if $b_i$'s and 1 are rationally independent, we may say even more: for any intervals $\Delta_i\subset (0,1)$ there exist a positive integer $q$ such that fractional part of $qb_i$ lies in $\Delta_i$. $\endgroup$
    – Fedor Petrov
    Commented Mar 10, 2017 at 18:09
  • $\begingroup$ Indeed. Good as your answers are, I think adding your comment to this answer would make it even better. Future readers might then discover the relevant approximation theorems sooner. Gerhard "Your Choice About Adding, However" Paseman, 2017.03.10. $\endgroup$
    – Gerhard Paseman
    Commented Mar 10, 2017 at 18:50
  • $\begingroup$ Out of curiosity how does one prove the second part of the statement? $\endgroup$
    – mssmath
    Commented Mar 12, 2017 at 22:34
  • $\begingroup$ @mssmath Actually the vectors $\{qb_i\},q=1,2,\dots$ are uniformly distributed on the torus $(\mathbb{R}/\mathbb{Z})^k$. This follows from the fact that for any function $f(x_1,\dots,x_k)=\exp(2\pi i(a_1x_1+\dots+a_kx_k))$, $a_i\in \mathbb{Z}$, not all $a_i$ are zeros, we have $\sum_{i=1}^N f(qb_1,\dots,qb_k)=o(N)$. $\endgroup$
    – Fedor Petrov
    Commented Mar 12, 2017 at 22:43

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