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It was proved here that if $a\in \mathbb{N}_{\geq3}$ then

$$\bigcap_{i = 1}^{a} \bigcup_{j = 0}^{i-1} \left[\frac{1+aj}{i},\frac{a(j+1)-1}{i}\right] = \varnothing \tag{1}$$

It may be conjectured that forcing $i\ne b$, where $1\leq b< a$, renders $(1)$ untrue, that is, the result is not an empty interval.

We look at the diagram here from the link above for $a=5$ to get a better picture:
 
Red is the interval, Yellow are the gaps between the intervals that cause the intersection to be a null set, white gaps do not effect the intersection. The conjecture here says that if we were to remove any one of the top $4$ strips, then there will form a region of intersection.

I tried analyzing the gaps but everything seems to meet up at a dead end.

What tools may one employ to handle such problems?

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  • $\begingroup$ It looks a bit Cantor-setish.. $\endgroup$
    – Felix Goldberg
    Commented May 26, 2013 at 10:46
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    $\begingroup$ Another way to state the original is that for every number $x\in [0,1]$ there is some $1 \le i \le a$ so that $i x$ is within $1/a$ of an integer $j$. Your conjecture is that for any $1 \le b \lt a$, there is some $x \in [0,1]$ so that the only one of $x, 2x, ..., ax$ which is within $1/a$ of an integer is $bx$. $\endgroup$
    – Douglas Zare
    Commented May 26, 2013 at 12:10
  • $\begingroup$ Which is just $x=\frac ca+\frac 1a^2$ when $b$ and $a$ are relatively prime with $c$ given by $bc\equiv -1\mod a$. So, when $a$ is prime, it is certainly true. $\endgroup$
    – fedja
    Commented May 26, 2013 at 16:12
  • $\begingroup$ Sorry, $\frac ca+\frac 1{a^2}$, of course... $\endgroup$
    – fedja
    Commented May 26, 2013 at 16:13
  • $\begingroup$ It's interesting that fedja's answer for $(a,b)=1$ and mine correspond to different endpoints of the interval of solutions for $b=2, a=5$. $\endgroup$
    – Douglas Zare
    Commented May 27, 2013 at 3:03

1 Answer 1

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As I commented, the original result can be restated as that for every $x \in [0,1]$ (or $x\in \mathbb R$) there is some $1\le i\le a$ so that $ix$ is within $1/a$ of an integer (and the proof is the pigeonhole principle -- two of the fractional parts of $0,x,2x,...,ax$ must be within $1/a$ of each other, so their difference is within $1/a$ of an integer). Your conjecture is that for any $1\le b \lt a$ there is some $x$ so that the only one of $x,2x,...,ax$ within $1/a$ of an integer is $bx$. For example, for $a=5$ and $b=2$, then if we take $x \in [\frac{44}{100},\frac{45}{100}] \cup [\frac{55}{100},\frac{56}{100}]$ then the only one of the first $5$ multiples within $1/5$ of an integer is the second.

If $2b \gt a$ then $x=1/b$ works. Since $b\lt a$, $i/b$ is within $1/a$ of an integer only when $i$ is a multiple of $b$.

For $2b \le a$ we can modify this to $x=1/b + 1/(2ab) = \frac{2a+1}{2ab}$. This is designed so that $bx = \frac{2ab + b}{2ab} = 1 + \frac{1}{2a}$ is within $1/a$ of $1$, but $2bx = 2 + \frac{1}{a}$ just barely misses being within $1/a$ of $2$. Larger multiples of $bx$ are also too large, $bix - i = \frac{i}{2a} \ge \frac{1}{a},$ while $(bi-1)x$ is too small to be within $1/a$ of $i$ when $bi-1 \le a$. $i - \frac{(bi-1)(2a+1)}{2ab} = \frac{2a - (bi-1)}{2ab} \ge \frac{a}{2ab} = \frac{1}{2b} \ge \frac{1}{a}.$

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