Is the sign of a real number decidable?

Question

I'm working on the following problem in a class on provability.

Consider how $\mathbb{R}$ might be presented. Is the property of being positive decidable?

How could the reals possibly be presented in such a way that this isn't decidable? Presumably the most natural presentation would be using a number's decimal expansion, but this doesn't get around the fact that we need to put in the $+$ or $-$ sign somehow. The algorithm "check whether the number is preceded by the $+$ sign" is certainly effective...

Can someone explain how I'm misinterpreting the question?

Answer 1

I don't quite know if I've understood the question, but I think that for all possible presentations of $\mathbb{R}$, it's not computable. Indeed, there are two ways to interpret your question, and neither is computable.

• Suppose we could compute whether $x \geq 0$. Then we could compute whether $x = 0$ by determining whether both $x \geq 0$ and $-x \geq 0$. But that's uncomputable. So we can't compute whether $x \geq 0$.
• Suppose we could determine whether $x > 0$. Then we could compute whether $x = 0$ by determining whether $x > 0$ and $-x > 0$ are both false.

For a specific example, take a presentation of machines which spit out first a sign, then a digit every time-step, with exactly one decimal point at some time. The problem you have here is that you can't convert zero to a canonical form of $+0$. That is, if you've got a Turing machine which outputs $-0.00000\dots$, you can't tell whether that machine is representing $0$ or not.