Prove that there is a constant $c$ for which $f(x) = cx$ for all $x \in \mathbb{Q}$

Question

Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Prove that there is a constant $c$ for which $f(x) = cx$ for all $x \in \mathbb{Q}$.

We know that $f(0) = 0$. Now set $x = n$ and $y=n$ to get $f(2n) = 2f(n)$ where $n$ is a rational number. But doesn't this condition imply all constant functions work? So we must get another condition.

Answer 1

Notice: $$ f\left(\frac{m}{n}\right) = m\cdot f\left(\frac{1}{n}\right) = m\cdot \left( \frac{f(1)}{n}\right) = f(1)\cdot \frac{m}{n} $$ where $f\left(\frac{1}{n}\right) = \frac{f(1)}{n}$ because $$ f\left(\frac{1}{n}\right) = f\left( \frac{n}{n} - \frac{n-1}{n}\right) = f(1) - (n-1)f\left(\frac{1}{n}\right) $$ by the linearity assumption.

Answer 2

You can demonstrate that for $n\in\mathbb{Z}$, $f(n)=nf(1)$ and $f(1)=nf(\frac1n)$ for $n\ne0$, so that for $x=\frac{p}{q}\in\mathbb{Q}$, $f(x)=\frac{p}{q}f(1)$, so $c=f(1)$.