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Set of sequences of rationals.

  1. Jan 3, 2007 #1
    i need to prove that there are c sqequences of rational numbers.
    basically, i need to show that |Q^N|=c.
    here, are a few attempts from my behalf:
    i thought that Q^N is a subset of R^N, so |Q^N|<=c, but this doesnt help here, so i thought perhaps to find a bijection from {0,1}^N to Q^N.
    i know that each rational number can be represented in base 2 by the digits 0,1, but im having difficulty to formalise this idea.
     
  2. jcsd
  3. Jan 3, 2007 #2

    HallsofIvy

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    Any real number can be identified with a specific sequence of rational numbers. For example, the real number [itex]\pi[/itex] can be identified with the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, with each term (a rational number because it is a terminating decimal) being one more decimal place in the decimal expansion of [itex]\pi[/itex]. This gives a one-to-one function from the set set of real numbers to the set of all sequences of rational numbers and so shows that [itex]c\le |Q^N|[/itex].
     
  4. Jan 4, 2007 #3

    Hurkyl

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    Grr. If I remember correctly, it's a theorem that for infinite c:

    [tex]a^c = b^c[/tex]

    whenever [itex]a \leq b \leq c[/itex]. The proof isn't immediately leaping to mind, though. :frown:

    Oh, bother, I just checked Wikipedia, and it looks like you need the axiom of choice to prove that, so the proof won't be as easy as I had hoped.

    Anyways, LQG, remember that |Q| = |N|. It might be easier to try and prove |N||N| = 2|N| or |N||N| = |R| instead.

    But, we see HoI seems to have proven it directly, so don't listen to me. :wink:
     
    Last edited: Jan 4, 2007
  5. Jan 4, 2007 #4

    HallsofIvy

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    Oh, Hurkyl, I'm blushing.


    By the way, I had a friend who, on another forum, used the name "Hog on Ice" which was regularly abbreviated "HOI". So whenever I see "HOI" used for "HallsofIvy", I'm surprized!
     
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