Uncomputability of $a<b$ or $b<a$

Question

Is it possible to prove the existence of two real numbers $a, b$ that have the property that it is uncomputable whether or not $a<b$?

Answer 1

No, it cannot be proved, because no such numbers exist; that is, for any two real numbers $a$ and $b$, the value of the function $f_{<}: {\Bbb{R}}^2\to\{0,1\}$ such that $$f_{<}(a,b) = \begin{cases} & 1 \ \ \text{ if }\ a < b\\ & 0 \ \ \text{ otherwise } \end{cases}$$ can be computed for those particular $a$ and $b$. This is so because, regardless of how the two reals are supposed to be represented as input, the following two programs exist: program $P_0$ does nothing but output $0$ and halt, and program $P_1$ does nothing but output $1$ and halt. Consequently, one of these two programs correctly computes the number $f_{<}(a,b)$. (That we may not know which of these two programs is the correct one is not relevant to its existence.)

However, note that in reasonable models of real computation, the function $f_{=}: {\Bbb{R}}^2\to\{0,1\}$ such that $$f_{=}(a,b) = (1-f_{<}(a,b))(1-f_{<}(b,a)) = \begin{cases} & 1 \ \ \text{ if }\ a = b\\ & 0 \ \ \text{ otherwise } \end{cases}$$

is uncomputable (see a proof here), and hence the function $f_{<}$ is uncomputable -- in contrast to the computability of any single number such as $f_{<}(a,b)$.