# If $l>10$ and $\{a\}>\frac{1}{l}$ then there exists $k \in \{1,\dots, l\}$ such that $\{ka\} \in [1/10,2/10]$

It is probably a simple question, but I wasn't able to solve it.

Let $n=10$ and $l>n$ an integer. Let $a>0$ a real number such that $\{a\}>\frac{1}{l}$ (where $\{ \cdot \}$ denotes the fractional part).

I would like to prove (I think this is true) that

There exists an integer $k \in \{1,\dots, l\}$ such that $\{ka\} \in [1/n,2/n]$.

Since $l>n$, I know that at least two numbers from the set $X=\{ \{ka\} \mid k \in \{1,...,l\}\}$ belong to an interval of the form $[r/n,(r+1)/n]$. I don't know how to go further.

• This is not true. Try $l>10$ arbitrary and $a=\{a\}=\frac12>\frac1l$. Then $\{ka\}$ is either $0$ or $\frac12$, but never $\in[\frac1{10},\frac2{10}]$. – Hagen von Eitzen Feb 27 '16 at 21:04
Consider $l = 11$. Let $a = 2.5$. Clearly, $0.5 = \{2.5\} > \frac{1}{11}$.
Consider $k \in \{1,..., 11\}$. If $k$ is odd, then $\{2.5k\} = 0.5$. If $k$ is even, then $\{2.5k\} = 0$.
Since $0$ and $0.5$ are not contained in $[\frac{1}{10}, \frac{2}{10}] = [0.1,0.2]$, the statement is false.