Linear transformation from irrationals to rationals with a constraint

Question

Can we, $\forall\space x\in\mathbb{R}$, find a linear transformation of the form $ax+b$ where $a,b\in\mathbb{R}$ such that if $x\in\mathbb{R}$ \ {$\mathbb{Q}$}, then $(ax+b) \in\mathbb{Q}$ and if $x\in\mathbb{Q}$ then $(ax+b)\in\mathbb{Q}$.

So for example, find $a,b\in\mathbb{R}$ such that $(a\pi+b)\in\mathbb{Q}$ and $(an+b)\in\mathbb{Q}$ where $n\in\mathbb{Q}$.

Answer 1

The answer is no:

If $aq+b\in\mathbb Q$ for any $q\in\mathbb Q$, then $a=2a+b-(a+b)\in\mathbb Q$. So $b=(a+b)-a\in\mathbb Q$. This implies that if $x$ is irrational, then $ax+b$ cannot be rational, otherwise $x=\frac{ax+b-b}{a}$ would be rational.

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Hope this helps.