Vector-space almost linear function Find a function $f:V \rightarrow W$, where $V$ and $W$ are vector spaces (and V is defined on $\mathbb{K}$), such that $$f(x+y) = f(x) + f(y), \forall x,y \in V$$ but $$\exists a \in 
\mathbb{K}: f(ax)\neq af(x).$$
I have so long proved with some algebra that this is impossible for functions on $\mathbb{Q}$, which can be some nice exercise as well for who might want to do this exercise.
 A: Hint: When working in $\mathbb{Z_p}, (x+y)^p = x^p + y^p$ (Where $p$ is a prime.)
A: Let $K=\mathbb Q(\sqrt 2)$ be our ground field, $V=W=K$ onedimensional and define $f\colon V\to W$, $a+b\sqrt 2\mapsto a-b\sqrt 2$. Or, take $V=W=K=\mathbb C$ and $f(z)=\bar z$, which is even continuous.
If the ground field is $\mathbb Q$, you are right: Additivity implies $f(n\cdot x)=n\cdot f(x)$ and hence also $f(\frac xn)=\frac1nf(x)$ and ultimately for arbitrary fractions.
If the ground field is $\mathbb R$, it is a field extension of $K=\mathbb Q(\sqrt 2)$ above and the $f$ defined above can be extended (using the Axiom of Choice, admittedly) accordingly to give an example.
A: For $f:\mathbb{R}\longrightarrow\mathbb{R}$ satisfying Cauchy's functional equation $f(x+y)=f(x)+f(y)$, we have
$$
f(\alpha x)=\alpha f(x) \quad \forall \alpha,x\in \mathbb{R}
$$
if and only $f$ is continuous.
There are lots of discontinuous solutions to this equation. They are known as Cauchy-Hamel functions. You can find here a way to construct them.
Note Hamel proved that their graph is dense in $\mathbb{R}^2$.
