# Existence of a solution to a system of Diophantine Inequalities

Does the solution to the following system of inequalities exist?

$$a-1\geq a\left( b_ic -d_i\right)\geq 1$$

where $a\in \mathbb{N}_{\geq3}$, $c\in \mathbb{R}$ and $b_i,d_i \in \mathbb{N}$. Moreover, $0< c < 2\pi$ and $0\leq d_i < b_i$ and $i$ runs from $1$ to $a-1$ because of which we have a system of $a-1$ inequalities. The unknowns are $c$ and $d_i$.

Update: With some consideration, the bounds can be improved. Instead of $0 < c < 2\pi$ and $0 \leq d_i < b_i$, it is sufficient enough to consider $0 < c \leq \frac{1}{2}$ and $0 \\leq d_i < \lfloor \frac{b_i-1}{2} \rfloor$.

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From your inequalities we get $$\frac{d_i}{b_i}+\frac{a-1}{ab_i}\ge c\ge \frac{d_i}{b_i}+\frac{1}{ab_i} \ \ \ (1)$$ (apropos, this implies $c< 2\pi$). So it is enough to solve the system $$\frac{d_j}{b_j}+\frac{a-1}{ab_j}\ge \frac{d_i}{b_i}+\frac{1}{ab_i}$$ for all $1\le i,j\le a-1$, i.e. $$d_jb_i-d_ib_j\ge \frac{b_j-(a-1)b_i}{a}$$ and then to choose $c$ from (1).
If you divide on $a$ and let $a$ to be big you have basically a set of linear inequalities of the form $1-o(1) \geq b_ic-d_i \geq o(1)$. If you look at $b_ic$ geometrically, it is a beam of lines emanating from zero with slopes $b_1, ..., b_{a-1}$. And you want to find a point $c$ such that the line $x = c$ intersects each line in a strip $1+d_i-o(1) \geq y \geq d_i+o(1)$. It seems that if you order the lines according to the slope and starting from the line with the smallest slope (smallest $b_i$) it is possible to choose $d_i$ for each new line appropriately (based on the geometric representation above)
Basically, for each $b_i$, you take $d_i$ such that $d_i+1-o(1) > b_ic > d_i+o(1)$. To be able to find such $d_i$ you just need $$\|b_ic\| > 2*o(1)$$ which is possible provided $a$ is sufficiently large. –  DmitryZ May 30 '13 at 22:35
Ah, so if $a$ is given the last inequality can be satisfied if $a > 4$ or so. –  DmitryZ May 30 '13 at 22:39
So you say a solution always exists if $\|b_ic\| > 2*o(1)$ which implies $a > 4$? Can you elaborate this transition a bit? –  Maaz-ul-Haq May 30 '13 at 23:51
the inequality is $|b_ic mod 1| > 2/a$. so if a is less or equal to 4 it cannot be satisfied. otherwise i believe it something similar to the Kronecker approximation theorem, but i didn't write it down rigourously. –  DmitryZ May 31 '13 at 4:42