Are some real numbers “uncomputable”?

Is there an algorithm to calculate any real number. I mean given $a \in \mathbb{R}$ is there an algorithm to calculate $a$ at any degree of accuracy ?

I read somewhere (I cannot find the paper) that the answer is no, because $\mathbb{R}$ is not a recursive set since it’s not a countable set.

But I’m missing something… can someone explain me in detail why so ?

Moreover it means that there exist real number that are uncomputable, and that you cannot neither write down nor express… it’s like they don’t exist.

Solutions Collecting From Web of "Are some real numbers “uncomputable”?"

As you have observed yourself, $\mathbb{R}$, the set of real numbers, is uncountable, but every recursively enumerable set is necessarily countable. This immediately implies that there exist uncomputable real numbers. The same argument shows that there are (formally) indescribable real numbers. Indeed, almost all real numbers are indescribable, and a fortiori, uncomputable. There is nothing wrong about this, though it may be disturbing.

Do uncomputable/indescribable real numbers ‘exist’? Well, that’s a philosophical question. A Platonist might say that they exist, even though we have no means of naming them specifically. A finitist might say they don’t exist, precisely because we have no algorithm to compute or even recognise such a number.

Does this impact the way we do mathematics? Not really. So what if the vast majority of real numbers are uncomputable? By and large we deal with generic real numbers, not specific ones. For example, the fact that every non-zero real number $x$ has an inverse does not rely on the computability properties of $x$.

Most real numbers are not computable. Informally speaking, a real number is computable if there is a machine or algorithm that computes its decimal expansion, one digit at a time, that is, you can ask for the $n$-th digit. Once you formalize machines and algorithms, which are finite animals, you see that there are only a countable number of them and so there are only a countable number of computable real numbers. Since the real numbers are not countable, most real numbers are not computable.

I think you are talking of computable numbers. You can read more about it on Wikipedia. Unfortunately, what I know regarding this is only a very little subset of what Wikipedia has. A relatively famous example for a non-computable number is the Chaitin’s constant.

Yes, you may calculate $a$ up to an arbitrarily small error.
But if you want to find the exact description, you will be able to do this only for a countable set of numbers. For instance, every description of a number contains a finite set of words, and thus the set of all such descriptions is countable. Since the set of all real numbers is not countable, many real numbers cannot be described, we only know that they exist.

Precisely, a real number $x$ is computable if there is a total computable function $\varphi_{e}$ such that $|x-q_{\varphi_{e}(n)}|<2^{-n}$, where $\langle q_i \rangle_{i \in \omega}$ is an effective numbering of the rationals. So $\varphi_{e}$ picks out a computably-fast converging sequence of rationals whose limit is $x$. Note that one can then compute the decimal expansion of $x$ to any desired precision (and know when that precision has been reached).

Since each $\varphi_{e}$ corresponds to at most one computable real and there are only countably many such $\varphi_{e}$’s, there can be at most countably many computable reals. Of course, each rational is computable, so there are a countable infinity of computable reals.

Regarding your comment: “…it’s like they don’t exists”. Here you have to understand what it means for something to exist in mathematics or—even better—in first-order logic. The point is that once you know the set of computable numbers is countable, Cantor’s diagonal argument can be carried out in ZFC to show that there is no surjection from the computable numbers onto all the reals; i.e. noncomputable numbers must “exist”.