I'd be very surprised if it were true in general. But there certainly has to be some non-computational component to the reals being a complete totally ordered A

I'd be very surprised if it were true in general. But there certainly has to be some non-computational component to the reals being a complete totally ordered Archimedean field, because the question of deciding the equality of two reals is uncomputable, so there is no procedure to decide the total order.