An additive map that is not a linear transformation over $mathbb{R}$, when $mathbb{R}$ is considered as a $mathbb{Q}$-vector space

I am looking for an example of an additive map that is not a linear transformation over $\mathbb{R}$, when $\mathbb{R}$ is considered as a $\mathbb{Q}$-vector space. I mean, I want to find an example of a map $T:\mathbb{R}\rightarrow\mathbb{R}$ such that $T(u+v)=T(u)+T(v)$ for all $u,v\in \mathbb{R}$, but $T(\alpha v)=\alpha T(u)$ is not true for all $\alpha \in\mathbb{R}$.

Thanks for your kindly help.

Answer

Let $\{r_\alpha\}$ be a Hamel basis of $\Bbb R$ over $\Bbb Q$. Let $\phi$ map $x$ to $c_{\alpha_1}+\cdots+c_{\alpha_k}$, where the (unique) basis representation of $x$ is $c_{\alpha_1}r_{\alpha_1}+\cdots+c_{\alpha_k}r_{\alpha_k}$. Then $\phi(x+y)=\phi(x)+\phi(y)$, but takes on only rational values.

If $\phi(\alpha v)=\alpha\phi(v)$ for all $\alpha$, $v$ in $\Bbb R$, then $\phi$ would be onto.
As this isn't the case, $\phi$ is not $\Bbb R$-linear.

It is $\Bbb Q$-linear, though. In fact, any additive map would automatically be $\Bbb Q$-linear.

As far as I know, you need the axiom of choice to construct a function of this type (?).