For any $F\subseteq {\mathbb R}$, say that a map $f:(0,\infty)\to (0,\infty)$ is $F$-affine when $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$ whenever $t\in F$, $x,y,tx+(1-t)y \gt 0$.
My question : can we construct a map $(0,\infty)\to (0,\infty)$ that's ${\mathbb Q}$-affine but not ${\mathbb R}$-affine ?
My thoughts : $f$ cannot be continuous, obviously. If $\alpha_1,\alpha_2,\ldots,\alpha_r$ are $\mathbb Q$-linearly independent positive real numbers and $C$ is the cone generated by them (so that $C$ is the set of all $\sum_{k=1}^{r} r_k\alpha_k $ where the $r_k$ are nonnegative rationals), then the map $f : C \to C$ defined by
$$ f\bigg(\sum_{k=1}^{r} r_k\alpha_k\bigg)=\beta_0+\sum_{k=1}^{r} r_k\beta_k $$
(where $\beta_0,\beta_1,\ldots,\beta_r$ are positive constants), is $\mathbb Q$-affine. Unfortunately, although there are Hamel bases of $\mathbb R$ over $\mathbb Q$ and we can certainly choose one to contain only positive numbers (replacing each negative base element by its opposite), some positive real numbers will inevitably have some negative coordinates in this base so this seems to be a dead end.
Also, if $f$ is a solution, $f$ can uniquely be extended to a $\mathbb Q$-affine map ${\mathbb R}\to {\mathbb R}$.
Suppose $f:(0,\infty)\to(0,\infty)$ is $\mathbb{Q}$-affine. Then I claim $f$ is order-preserving. Indeed, suppose $a<b$ but $f(a)>f(b)$. Then $qb+(1-q)a>0$ for all $q>0$, so $f(qb+(1-q)a)=qf(b)+(1-q)f(a)$ for all positive rationals $q$. But when $q$ is sufficiently large, $qf(b)+(1-q)f(a)$ is negative since $f(a)>f(b)$, so this is a contradiction.
Now I claim $f$ is continuous and hence $\mathbb{R}$-affine. Since $f$ is order-preserving, if $f$ were not continuous, there would be $a<b$ such that the image of $f$ omits an interval contained in $[f(a),f(b)]$. But now note that for any rational $q\in [0,1]$, $f(qa+(1-q)b)=qf(a)+(1-q)f(b)$, and the set of such values is dense in $[f(a),f(b)]$. Thus the image of $f$ intersects every interval contained in $[f(a),f(b)]$, so $f$ must be continuous.