(Cauchy's Equation) Find the functions $f: \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(x+y)=$ $f(x)+f(y)$ for all $x, y \in \mathbb{Q}$.
Let $f$ be a potential solution to the problem.

- For $x=y=0$, we get $f(0)=2 f(0)$, hence $f(0)=0$.
- By choosing $y=-x$, we observe that for any integer $x$, we have $f(x)+f(-x)=0$, and thus $f(-x)=-f(x)$. This ensures that it suffices to determine $f$ on the natural numbers to know $f$ on the entire $\mathbb{Z}$.
- We do not see an obvious relation to determine $f(1)$, so we set $a=f(1)$.
- By choosing $y=1$, we observe that $f(x+1)=f(x)+f(1)=f(x)+a$, for any integer $x$. It is then fairly easy to deduce that $f(2)=2a$, $f(3)=3a$, $f(4)=4a$. By induction, we deduce that $f(n)=na$ for any natural number $n$, and then for any integer $n$.
- If $q$ is a rational number, we can write $q=\frac{m}{n}$ where $m, n$ are integers such that $n>0$. Since we know $f$ well on the integers, the idea is to reduce it to this case. We observe that for $x=y$, we have $f(2x)=2f(x)$, and for $y=2x$ that $f(3x)=3f(x)$, and by induction, for any integer $k>0$ and any rational $x$, we have $f(kx)=kf(x)$. Therefore, we have $f(nq)=nf(q)$ and also $f(nq)=f(m)=ma$, so $nf(q)=ma$, hence $f(q)=\frac{m}{n}a=qa$. Thus, there are infinitely many solutions, which are the functions of the form $f: q \mapsto aq$ where $a$ is a fixed real number.