Set of all rational numbers less than given computable real number is decidable
I would approach this problem as follows. Let $a$ be the computable real, and $A = \{x\in\mathbb{Q}\ |\ x&lt;a\}$.

If $a \in \mathbb{Q}$, then $A$ is decidable since $x&lt;a$ is a comparison between rationals, which is decidable.

Otherwise, we can assume $a$ is irrational. Our algorithm performs the following.
First, we take the candidate $x\in\mathbb{Q}$ as input.
We then query $f$ (the function which approximates $a$ as requested) repeatedly with $q=2^{-n}$ for every $n&gt;0$.
Eventually, for some $n$ we will obtain one of

$$
f(2^{-n}) + 2^{-n} &lt; x
\qquad \lor\qquad
x &lt; f(2^{-n}) - 2^{-n}
$$
Indeed, this will happen as soon as $2^{-n} &lt; |x-a| / 2$, which is a possibility because we know $x \neq a$.

In the first case, we return $x &gt; a$ ($x \notin A$), in the second we return $x&lt;a$ ($x \in A$).

The correctness comes from
$$
a \leq f(2^{-n}) + 2^{-n}
\qquad \land\qquad
f(2^{-n}) - 2^{-n} \leq a
$$
being true for all $n$.



Note that this solution is not uniform in $a$. That is, we only prove that there is an algorithm for any computable $a$, taking a candidate rational $x$ as input. We do not show that there is a single algorithm that, given both $a$ and $x$ as input is able to decide $&lt;$ (indeed, that would be undecidable).