Wikipedia defines a computable real number to be a number where there exists an algorithm that can approximate it to within any given precision. I believe this definition is standard.
The Wikipedia article then claims, without giving a specific source:
A real number is computable if and only if there is a computable Dedekind cut $D$ corresponding to it.
It is obvious that if the Dedekind cut for $\alpha$ is computable, then $\alpha$ is computable according to the first definition. The other direction sounds more iffy, though, and I've always been told it doesn't hold.
It is certainly not the case that given a Turing machine that approximates some $\alpha$ to within a given precision, we can effectively derive a Turing machine that decides its Dedekind cut. If we could do that, we could solve the halting problem by considering, for an arbitrary machine $M$, the number$$ \alpha_M = \begin{cases} 1/m & \text{if $M$ halts in $m$ steps} \\ 0 & \text{if $M$ diverges} \end{cases} $$It's easy to write down a concrete program that approximates $\alpha_M$ to any given precision. If we had an effective way to convert it to a Dedekind-cut decider, we could learn whether $M$ halts by asking the resulting machine whether $0 < \alpha_M$. (For this purpose I'm assuming Dedekind cuts where the lower set has no maximum; for the opposite convention consider $-\alpha_M$ instead).
However, this is not an example of a concrete number that has a machine of one type, but doesn't have one of the other type. No matter whether $M$ halts, $\alpha_M$ is even rational, so a machine that decides its Dedekind cut certainly exists, though we may never know which machine that is.
We can see, however, that even if Dedekind-cut deciding machines exist for all computable numbers, they are not something one can even begin to, ahem, compute with. It's easy to write programs that decide the cuts for $\sqrt2$ and $\alpha_M-\sqrt2$, so if we had an effective way to add cut-deciders, we could use it to discover whether $M$ halts.
This observation feels like a good reason not to use Dedekind cuts as a definition of computable reals. But it doesn't resolve the question of whether the pure (non-constructive) existence of a Dedekind-cut-deciding machine is equivalent to the (pure non-construtive) existence of an approximating machine.
Thus, my question: Do we know whether there exist computable real numbers whose Dedekind cuts are not computable?
