# Is there computable function to compute each computable real number?

Recall that a computable real number is one which can be calculated to any precision, like $\pi$ or $e$. It does not matter that these numbers are irrational, computability is about being able to compute rational approximations.

Is there a computable function which outputs an (infinite) sequence of Turing machines, each of which computes a real number, and such that all real numbers are so included?

• I don't think this question is suitable here, it seems to be mainly confusion about basic definitions. A real number is often represented by a function (e.g. a function over natural numbers). The computability of infinite objects like real numbers are usually defined using such representations of real numbers. A computable real number is also given by a finite algorithm. $\sqrt{2}$ and $\sqrt{x}|_{x=2}$ are the same object. Aug 9, 2013 at 7:35
• I have edited the question so that it makes sense. If this is not what @Dims is asking, then he should try harder. Aug 9, 2013 at 10:49
• "an (infinite) sequence of Turing machines, each of which computes a real number, and such that all real numbers are so included" Note that this is not possible simply because any sequence is only countably infinite, while the set of reals is uncountable. (As far as I can tell, OP is not asking "such that all computable real numbers are included"...) Feb 2 at 17:49

A real number $x$ is said to be computable if there a Turing machine, which upon input $n \in \mathbb{N}$ outputs a pair of numbers $a, b \in \mathbb{N}$ such that $|x - a/b| < 2^{-n}$. That is, there is a machine that computes arbitrarily precise rational approximations of $x$.
• perhaps in summary: the set of computable real numbers is countable, but not recursive enumerable (reduction from $A_{HALT}$ the problem of deciding if a TM halts on all inputs). Aug 9, 2013 at 14:51
• Not quite, because recursive enumeration only applies to natural numbers. So slightly more precise would be: there is no recursively enumerable set of codes of computable reals such that each computable real has a code in it. It is better to introduce the notion of a numbered set and then: "There is no surjective computable map from $\mathbb{N}$ onto the numbered set of computable reals." Aug 9, 2013 at 18:05