17.11. The numbers $[a],[2 a], \ldots,[N a]$ are distinct from each other, and the numbers $[1 / a],[2 / a], \ldots,[M / a]$ are also distinct from each other. Find all such $a$.

## 17.3. Approximations of irrational numbers by rational numbers
17.11. Answer: $\frac{N-1}{N} \leqslant|a| \leqslant \frac{M}{M-1}$.

The numbers $[x]$ and $[y]$ are different if and only if the numbers $[-x]$ and $[-y]$ are different. Therefore, it is sufficient to consider the case when $a>0$. If $a<\frac{N-1}{N}$, then among the numbers $[a],[2 a], \ldots,[N a]$ there are coinciding ones, since these $N$ numbers are contained among the $N-1$ numbers $0,1, \ldots, N-2$. Therefore, $a \geqslant \frac{N-1}{N}$. The same reasoning for the number $1 / a$ shows that $\frac{1}{a} \geqslant \frac{M-1}{M}$, i.e., $a \leqslant \frac{M}{M-1}$.

We will show that if $\frac{N-1}{N} \leqslant a \leqslant \frac{M}{M-1}$, then all the numbers $[a],[2 a], \ldots, [N a]$ are different. Indeed, if $a \geqslant 1$, then all the numbers $[a],[2 a],[3 a], \ldots$ are different, and if $a<1$, then $1-1 / N \leqslant a<1$, $2-2 / N \leqslant a<2, \ldots, N-1 \leqslant a<N$, so $[k a]=k-1$ for $k=1,2, \ldots, N$. The reasoning for the numbers $[1 / a],[2 / a], \ldots, M / a]$ is analogous.