Using the above, solve the Cauchy equation on the set $X$ of numbers of the form $a+b \sqrt{2}$, with $a$ and $b$ in $\mathbb{Z}$.
Since $\sqrt{2}$ is irrational, the representation of a given real number in the form $a+b \sqrt{2}$, if it exists, is unique. From the above, if we set $x=f(1)$ and $y=f(\sqrt{2})$, then $f(a+b \sqrt{2})=a x+b y$. All functions of this form are solutions.

Continuous solutions of the Cauchy equation The following exercise requires knowledge of the concept of continuity. One can use the fact that every real number is the limit of a sequence of rational numbers (in other words: $\mathbb{Q}$ is dense in $\mathbb{R}$), and use this to reduce the problem to what was discussed above.