Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the comment thread of an answer, I said:

The computable numbers are based on the intuitionistic continuum, and are not finitary.

To which T.. replied:

Computable numbers are not based on the intuitionistic continuum.

This disagreement contains, I think, a good example of a philosophical question: are the computable reals within the scope of finitistic mathematics?


  1. Bendegem, 2010, Finitism in Geometry
  2. Edalat, 2009, A computable approach to measure and integration theory
  3. Zach, 2001, Hilbert's Finitism
  4. Zach, 2003, Hilbert's Program
share|cite|improve this question

A real number $r$ is defined to be computable if there is an algorithm that, given any natural number $n$, will produce a rational number within distance $1/n$ of $r$. Thus each individual computable number is defined using a finite amount of information (the algorithm). On the other hand, there is a well-known intuitive sense in which an arbitrary real number can contain an infinite amount of information. In this sense, computable real numbers are finitary in a way that arbitrary real numbers are not.

The definition of a computable real number can be stated in either classical mathematics or in intuitionistic mathematics. Nothing in the concept of a computable real number ties you to one logic or the other.

share|cite|improve this answer
The core of the disagreement is about finitism. In the comment of mine I cited, I argued that finitism can't make use of the intuitionistic continuum, which is needed to support the computable reals. – Charles Stewart Aug 9 '10 at 18:18
We can define (codes for) computable reals and prove basic properties about them in weak systems of first-order arithmetic. I don't understand your argument why any continuum is required to study just the computable reals, or why the intuitionistic continuum in particular is necessary. – Carl Mummert Aug 9 '10 at 18:31

The following are unrelated concepts:

  1. Finitism

  2. The "intuitionistic continuum" (does that mean Weyl's book? the real numbers as defined in Bishop's book on constructive analysis?)

  3. Computable real numbers

For example, you can read the definition of computable real numbers under a classical or an "intuitionistic" interpretation of the words, and they will differ as to whether "1 if Goldbach conjecture is true and 2 if it's false" is a computable real number. But the theory of computable reals will make sense in both environments.

share|cite|improve this answer

Depends on what you exactly mean by finitary. A computable real has a finite description (a Turing Machine), so it is a finite object.

But many properties of computable real numbers are not finitary.

We can develope an interesting amount of analysis in very weak arithmetic theories (see this), theories which are much weaker PRA (which is often associated with Hilbert's finitism).

share|cite|improve this answer

I'm not so sure about the "finitary" definition, but I'm fairly certain that the class of partially computable functions (those for which algorithms can be devised) is countable. I'd thus conjecture that the computable reals are, at the very least, countable. Can anybody confirm or deny this, and/or relate what I have said to this concept of "finitary"?

share|cite|improve this answer
The set of computable numbers is indeed countable. – Alex Becker Dec 25 '11 at 4:40
This is not an answer to the question; it sounds like a comment or a separate question. If you have another question, please use the "Ask Question" link at the upper right of the page. – robjohn Feb 1 '13 at 8:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.