Let $r$ and $s$ be real numbers in the interval $[1, \infty)$ such that for all positive integers $a$ and $b$ with $a \mid b \implies \left\lfloor ar \right\rfloor$ divides $\left\lfloor bs \right\rfloor$. a) Prove that $\frac{s}{r}$ is a natural number. b) Show that both $r$ and $s$ are natural numbers. Here, $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x$.
Problem
Source: Romania National Olympiad 2023
Tags: floor function, number theory
Jjesus
08.04.2024 03:13
Any Solution?
MrOreoJuice
08.04.2024 09:06
For part (a) consider $a=b=n$ then \[\frac sr = \lim_{n \to \infty} \frac{s - \frac{\{ ns\}}{n}}{r - \frac{\{ nr\}}{n}} = \frac{\lfloor ns\rfloor}{\lfloor nr \rfloor} \in \mathbb Z\]For part (b) Let $s= t r$ where $t \in \mathbb Z$.
\[\frac{\lfloor bs \rfloor}{\lfloor ar \rfloor } = \frac{btr - \{btr\}}{ar - \{ar\}} \ge \frac{btr - \{btr\}}{ar} = \frac{bt}{a} - \frac{\{btr\}}{ar}\]and \[\frac{\lfloor bs \rfloor}{\lfloor ar \rfloor } \le \frac{btr}{ar - \{ar\}} = \frac{bt}{a} \times \frac{1}{1 - \frac{\{ar\}}{ar}} \approx \frac{bt}{a}\left(1 + \frac{\{ar\}}{ar}\right)\]for $a \to \infty$, giving equality everywhere, therefore $\{ar\} = 0$ for large enough $a$, meaning $r$ is rational but if $r$ is not integer then $a$ can be suitably selected (a prime for example) such that $\{ar\} \neq 0$ which would give strict inequalities. This gives a contradiction.