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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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2 Answers 2

This answer is not complete.

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).

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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)

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"It seems that if you order the lines according to the slope and starting from the line with the smallest slope (smallest bi) it is possible to choose di for each new line appropriately (based on the geometric representation above)". How would you treat this rigorously? –  Maaz-ul-Haq May 30 '13 at 22:29
    
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

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