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Proof that the rationals are dense

  1. Jun 18, 2011 #1
    Is the following proof that the rationals are dense in the reals valid?

    Theorem: [tex]\forall x,y\in\mathbb{R}:x<y, \exists p\in\mathbb{Q}: x<p<y[/tex] Viewing x and y as Dedekind cuts (denoting the cuts as x* and y*), x* is a proper subset of y*, hence there exists a rational in x* but not in y*, i.e. there is a rational between x and y.
     
    Last edited: Jun 18, 2011
  2. jcsd
  3. Jun 18, 2011 #2
    To prove the rationals are dense in the reals, you need to prove for any real number, there exists a rational number which is arbitrarily close to the real number. I don't see this in your proof.
     
  4. Jun 18, 2011 #3

    HallsofIvy

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    Well, what do you mean by "arbitrarily close"? The usual definition of "there exist a rational number arbitrarily close to x" is "for any [itex]\epsilon> 0[/itex] there exist rational y such that [itex]|x- y|< \epsilon[/itex]" which is the same as [itex]-\epsilon< x- y< \epsilon[/itex] or, in turn, [itex]x-\epsilon< y< x+ \epsilon[/itex] which is true if and only if "between any two real numbers there exist a rational number".

    dalcde, when you say "x* is a proper subset of y*, hence there exists a rational in x* but not in y*" you have the inclusion wrong- there exist a rational in y* that is not in x*.

    Also, if you are defining real number as Dedekind cuts (set of rational numbers), how are you embedding the rationals in the reals? The rational you are getting is a member of the set y*, not a real number itself.

    (Yes, I know that you are associating the "rational cut", the set of all rational number less that r, with the rational number r. But you need to say that.)
     
  5. Jun 18, 2011 #4

    micromass

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    Hi dalcde! :smile:

    Your proof is not completed yet. You've proved that there is a rational in y*, but not in x*. So what will be the p* between x* and y* then? And why is p* not equal to x* and y*?
     
  6. Jun 18, 2011 #5
    x* is the set of all rationals "below" x and y* is the set of all rationals "below" y. Hence there is some rational below x and not below y.
     
  7. Jun 18, 2011 #6

    micromass

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    I don't quite follow the "hence" bit. What exactly is the rational. And why doesn't it equal x and y?
     
  8. Jun 18, 2011 #7
    According to you, x < y. Let x* = {a in Q : a < x} and y* = {b in Q : b < y}. If z is in x*, then z < x < y. So z is in y*. So, you are wrong: every rational in x* is also in y*. HallsofIvy had already corrected you.
     
  9. Jun 19, 2011 #8
    Sorry, I wanted to write x>y, and got it the wrong way round.
     
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