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The rational numbers are clearly countable. The irrational numbers are uncountable, which means that for every rational number there are infinitely many irratio

• btilly 9y ago
  But for every integer there are also an infinite number of rationals. And since we can put the rationals into 1-1 correspondence with integers, that means that for every rational there are an infinite number of rationals. Infinity is funny like that. The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over. Accepting that philosophy also leads to the conclusion that not only do numbers which can in no way ever be represented exist, but there are more of them than numbers which we can explicitly name. Now I ask you, in what sense do they REALLY exist?
  • umanwizard 9y ago
    > The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over Can you elaborate? The proof that there are more irrationals than rationals is very straightforward, from my perspective.
    • btilly 9y ago
      The easiest way to understand it is to look at the question from a philosophical framework where it makes no sense to claim that there are "more" irrationals than rationals. And then untangle why it came to a different answer. In Constructivism, all statements have 3 possible values, not 2. They are true, false, and not proven. All possible objects must have a construction. So instead of talking about a vague "Cauchy sequence", we might instead have a computer program that given n will return a rational within 1/n of the answer. The first thing to notice is that all possible things that could exist is contained within a countable set of all possible constructions. There can't be "more" irrationals than rationals. But what about diagonalization? That proof still works. You still can't enumerate the reals. But why not? The answer is because determining whether a given program represents a real is a decision problem that cannot in general be solved by any algorithm. You are running into the same category of challenges that lie behind Gödel's theorem and The Halting Problem. It is not that there are "more" irrationals than rationals. It is that there are specific programs which you can't decide whether they represent reals. From a constructivist's eyes, the traditional proof that there are more irrationals falls apart because you're reasoning about unprovable statements about arbitrary sequences whose logic could have been constructed with the same sort of logic that you're applying to them. Is it any wonder that you wind up concluding the "existence" of things that clearly don't actually exist? Now stepping back from BOTH philosophies, the differences between them lie in different attitudes about existence and truth. Attitudes that underly the axioms which we use, and cannot possibly be proven one way or another. (Gödel actually proved that. Any contradiction in Constructivism is immediately a contradiction in classical mathematics. But conversely there is a purely mechanical transformation of any proof in classical mathematics which resulted in a contradiction, into a constructive proof that also results in a contradiction.)
      • umanwizard 9y ago
        Excellent explanation, thanks! But I feel compelled to point out that only a small minority of working mathematicians are constructivists.
        • btilly 9y ago
          This is true. However there is no logical argument that can disprove the constructivist view, in which there aren't "more" irrationals than rationals. And therefore the logical argument that there is must have some hidden implicit assumptions.
• enedil 9y ago
  For every rational, there are infinitely many rational numbers too, since finite cartesian product of countable sets is countable.
• umanwizard 9y ago
  I wasn't talking about irrational numbers.
  • deleted 9y ago
    [deleted]
  • pcvarmint 9y ago
    But all non-computable numbers are irrational.
    • TuringTest 9y ago
      No, there are also non-computable numbers that are imaginary, complex, or transfinite.
      • pcvarmint 9y ago
        All rational numbers are real. Therefore all non-real numbers are irrational.
        • TuringTest 9y ago
          Uh, no, "irrational" is defined as a subset of real numbers.
          • pcvarmint 9y ago
            Non-rational, then. This is about as consequential as debating whether 1 is a prime number.
            • TuringTest 9y ago
              This thread was about "all the numbers that it's theoretically impossible to compute". If you think the difference between "irrational" and "non-rational" in this context is irrelevant, you have a weak grasp of number theory. Yes, all are non-computable, but in different ways.