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Set of irrational numbers between 9 and 10 are countable

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data
    The set of irrational numbers between 9 and 10 is countable.


    2. Relevant equations



    3. The attempt at a solution
    My belief is that I can prove by contradiction.

    first, i must prove by contradiction using diagonalization that the real numbers between 9 and 10 are uncountable. (1)

    second, i take set of rational numbers Q is countable, hence a subset Q (9,10) is also countable. (2)

    third i can prove by contradiction stating I (9, 10) is countable
    R(9,10) =
    Q(9,10) [Countable per item (2) ] U I(9,10) [ Countable per statement]


    This would imply that by closure properties R(9,10) is countable. which is a condtradiction of what we found in 1.

    Is this logic sound? Can I prove (1) using the same diagonalization method used for (0,1)?
     
  2. jcsd
  3. Aug 27, 2010 #2

    jgens

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    Gold Member

    Your argument should work to prove that the set of irrational numbers between 9 and 10 is uncountable. However, if you know that the set of irrational numbers between 0 and 1 is uncountable already, why not just construct a bijection between the irrational numbers in (0,1) and the irrational numbers in (9,10)? This would show that they have the same cardinality and the set in question is thus uncountable.
     
  4. Aug 27, 2010 #3
    Well, I agree, I should be able to do a bijection as visually and logically we would expect the numbers between 9 and 10 to have the same cardinality as between 0 and 1, but I am not sure on how I would prove this.... in class we taught the diagonaliation proof for reals between 0 and 1, and the union proof for irrationals being uncountable using R = I U Q, but we haven't attacked something such as mapping all the numbers in 0,1 to 9,10

    I mean, i could prove bijection I *think* as follows:

    for a function
    F: R [tex]\rightarrow[/tex] R +9 : f(n) = n + 9

    Prove it is a bijection
    one-to-one:
    If f(x_1) = f(x_2) [tex]\rightarrow[/tex] x_1 = x_2
    f(x_1) = n_1 + 9 AND f(x_2) = n_2 + 9
    If n_1 + 9 = n_2 + 9 [tex]\rightarrow[/tex] n_1 = n_2 [tex]\rightarrow[/tex] x_1=x_2

    ONTO:
    For all y [tex]\in[/tex] R + 9 [tex]\exists[/tex] x [tex]\in[/tex] R : f(x) = y
    For all y = n + 9 [tex]\in[/tex] R + 9 [tex]\exists[/tex] x = n [tex]\in[/tex] R : f(n) = n + 9

    Hence I proved any real number can be mapped to a real number +9, so I can map R(0,1) to R(9,10)

    Is that right? Or I could do that for irrationals, i presume... but i figured getting the reals over would make my life easier...
     
  5. Aug 27, 2010 #4

    jgens

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    Gold Member

    There's nothing wrong with your initial argument, so if you're more comfortable with it, then by all means, stick with it; I just thought that constructing a bijection from the irrationals of (0,1) to the irrationals of (9,10) would be simpler. For example, you could have proven this result with the following argument:

    Let [itex]A[/itex] be the set of irrational numbers in the open interval [itex](0,1)[/itex] and [itex]B[/itex] the set of irrational numbers in the open interval [itex](9,10)[/itex]. Consider the function [itex]f:A \to B[/itex] defined by [itex]f(x) = x+9[/itex]. Since [itex]0 < x < 1[/itex], it follows that [itex]9 < x+9 < 10[/itex]; moreover, because [itex]x[/itex] is irrational, [itex]x+9[/itex] is also irrational. Therefore, every member of [itex]A[/itex] maps to a unique member of [itex]B[/itex]. This proves that [itex]|A| \leq |B|[/itex], and thus [itex]B[/itex] is uncountable.
     
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