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.