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Numbers which are not exactly calculable by any method.

  1. May 6, 2012 #1
    Is there any way to prove that a real number exists which is not calculable by any method?

    For example, you could have known irrational and/or transcendental numbers like e or π. You could have e^x where x is any calculable number, whether it be by infinite series with hyperbolic/normal trigonometric functions and an infinite number of random terms, and use that as the upper limit for an integral of whatever other type of function or combination of functions.

    Is there a possible way to prove that there exists any real number that is not equal to any combination of functions (apart from 0/0)?
  2. jcsd
  3. May 6, 2012 #2
  4. May 6, 2012 #3


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    It's easy to prove such number exists (the site micromass links to shows that the set of all "computable numbers" is countable while the set of all real numbers is uncountable. In a very real sense "almost all number are not computable".

    I don't believe, however, it is possible to give an example of such a number- in fact the very naming of such a number would probably be a description of how to compute it!
  5. May 6, 2012 #4
    Interesting. Just out of curiosity, is there a standard symbol for the "set of everything" or "set of all numbers"?
  6. May 6, 2012 #5
    The set of all real numbers is denoted [itex]\mathbb{R}[/itex]. The latex is \mathbb{R}.

    Of course that is not the set of "everything"; in standard set theory there is no set of everything. The set of real numbers is the set we're talking about when we talk about numbers in this thread. The computable numbers are a subset of the reals.

    Interestingly, there is a noncomputable real number that has a name and whose properties can be talked about. It's called Chaitin's constant.

  7. May 6, 2012 #6
    I actually began thinking about non-computable numbers, so I Googled it and found Ω. That is what led me to post here, to see if anyone knew of any other non-computable numbers with known properties.
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