Romanian Mathematical Olympiad - District Round

The function $f$ is injective if and only if $\frac{a}{b} \in \mathbb{Q}$. Indeed, suppose $f$ is injective and $\frac{a}{b} \in \mathbb{R} \setminus \mathbb{Q}$. Then $f\left(\frac{a}{b}\right) = a = f(1)$, a contradiction. Suppose now that $\frac{a}{b} \in \mathbb{Q}$ and let $x_1, x_2$ such that $f(x_1) = f(x_2)$. If $x_1, x_2 \in \mathbb{R} \setminus \mathbb{Q}$ or $x_1, x_2 \in \mathbb{Q}$, then $x_1 = x_2$. If $x_1 \in \mathbb{R} \setminus \mathbb{Q}$ and $x_2 \in \mathbb{Q}$, then $a x_2 = b x_1$ and consequently $\frac{a}{b} = \frac{x_1}{x_2} \in \mathbb{R} \setminus \mathbb{Q}$, a contradiction. Hence $f$ is injective.

The function $f$ is surjective if and only if $\frac{a}{b} \in \mathbb{Q}$. To this end, suppose $f$ is surjective and let $x$ with $f(x) = b$. If $x \in \mathbb{R} \setminus \mathbb{Q}$ then $x = 1$, a contradiction. Therefore $x$ is rational and so $a x = b$, implying $\frac{a}{b} = \frac{1}{x} \in \mathbb{Q}$. Conversely, suppose $\frac{a}{b} \in \mathbb{Q}$ and let $y \in \mathbb{R}$. Since $\frac{y}{a} \in \mathbb{R} \setminus \mathbb{Q}$ and $\frac{y}{b} \in \mathbb{Q}$ can not occur simultaneously, we have either $f\left(\frac{y}{a}\right) = y$ or $f\left(\frac{y}{b}\right) = y$. Hence $f$ is surjective.

The claim now follows.