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Numbers which are not ultimately functions of integers

  1. Mar 14, 2012 #1
    Do there exist numbers which are not ultimately functions of the integers?

    Would they necessarily be transcendental numbers or otherwise uncountable?
  2. jcsd
  3. Mar 14, 2012 #2
    A rational is defined from two integers, and a real from an infinite sequence of rationals. In some sense, thus, all reals are defined from an infinite sequence of integers. But I wouldn't use the word 'function' loosely; I'd want a precise statement of how can a function be defined with an infinite cartesian product (is there such a thing?) as the domain.
  4. Mar 14, 2012 #3
    I would guess that most of the physical constants are not "ultimately functions of the integers" although they can be expressed by an infinite decimal string.
  5. Mar 14, 2012 #4
    Looks like Loren is levelling us here, and in this strict forum I expect some moderators to intervene shortly.
  6. Mar 14, 2012 #5

    I believe my questions to be legitimate. I am not trying to "level" anyone here.

    By "functions" I mean "well-defined functions."

    I am attempting to define certain numbers which are a subset of irrational numbers.

    Please refer to my original post as well.
  7. Mar 15, 2012 #6
    Then I must apologize for my misplaced remark. I did btw take your previuos posts into consideration, which are all hard to make sense of.

    Regarding your question about numbers not being ultimately well defined functions of the integers, you may be thinking of some version of undefinable numbers (wiki, google) which is an uncountable set. You may also be thinking of the numbers that will result from applying operations to the integers, where the allowed operations are: [here, unfortunately, you must specify the list]. You were probably not thinking like me, that all numbers are in fact functions of integers, in the most trivial sense: given any number a, define f(x)=a.

    It seems like you are working on some bigger underlying problem involving densities and randomness, and despite appearances to the contrary, I am interested in hearing more about your project.
  8. Mar 17, 2012 #7
    Hypothesis: Uncountable combinations of rational terms as infinite sequences may define an uncountable set of irrationals.

    Given: The set of rationals is countable, the set of irrationals is uncountable.

    Conclusion: A set of sequential functions includes uncountable sequences of countable terms, postulating that transforms of countable rationals to uncountable irrationals exist -- so such a set of functions must itself be uncountable.
  9. Mar 17, 2012 #8
    All right, this is reminiscent of a popular project from computer class.

    For starters: What is meant by "sequential functions"? What is the logic of the first sentence? Is it "For any set of..." or "There exists a set of..." or perhaps something else? What is the word "postulating" doing in your conclusion?

    If you imagine yourself being someone else, knowing all of college math, but new to this particular issue, would you have understood the meaning of your writings above? If not, please try to make your text pass that test.
  10. Mar 17, 2012 #9
    The real numbers can be defined as limits of (countable) sequences of rationals. Is that what you're saying?

    I don't know what you mean by "uncountable combinations of rational terms." Can you explain what you mean? Typically a sequence is defined as a function whose domain is the natural numbers. So a sequence by definition is a countable set of terms, ordered by indexing each term with a natural number.

    One could in theory define uncountable sequences via indexing by an uncountable ordinal, but that's not the usual meaning of the term; nor would it give us any more real numbers than we can already get by considering countable sequences.
  11. Mar 17, 2012 #10
    By "sequential functions" I mean any set of well-behaved functions, with repeating terms of similarly derived countable members (terms).

    One property of rational and irrational numbers is that, although the former is countable and the latter uncountable, a set of combinations (infinitely countable sequences of terms) of rationals maps into an existing set of uncountable rationals.

    I rushed with the word 'postulating." Rather, it should be more like "assuming."



    I hope I have addressed some of your concerns below as well.


    Given: The set of rationals is countable and the set of irrationals is uncountable.

    Hypothesis: There exist uncountable functions which map the countable set of rationals onto an uncountable subset of irrationals.

    Observation: Like that which calculates pi or 2^(1/2), there exist sequences (sequential functions) with rational terms. Uncountable sets of such sequences, each of which number countable terms, produce an uncountable subset of irrationals.

    Conclusion: The set of such mapping functions, of which sequences are the most outstanding, convert the countable set of rationals into a uncountable subset of irrationals. The set of sequential functions are thus uncountable, but function only from a rational domain to an irrational range.
  12. Mar 17, 2012 #11
    The set of sequences of rationals is uncountable. And the reals can be defined as certain equivalence classes of sequences of rationals. Is that what you're saying?
    Last edited: Mar 17, 2012
  13. Mar 18, 2012 #12
    Yes, SteveL27, with pith. Is this conclusion of any significance?
    Last edited by a moderator: Mar 18, 2012
  14. Mar 18, 2012 #13
    It's definitely significant, but it's well-known. The construction of the reals from sequences (or sometimes subsets) of rationals is done in real analysis; and the proof that the collection of subsets of the rationals is uncountable is done in real analysis or basic set theory.
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