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Prove if a < b, there is an irrational inbetween them

  1. Apr 30, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    No giving up :biggrin:!

    The question : http://gyazo.com/08a3726f30e4fb34901dece9755216f3

    2. Relevant equations

    A lemma and a theorem :

    http://gyazo.com/f3b61a9368cca5a7ed78a928a162427f
    http://gyazo.com/ca912b6fa01ea6c163c951e03571cecf

    The fact ##\sqrt{2}## and ##2^{-1/2}## are irrational.

    3. The attempt at a solution

    Suppose 0 < b - a. We must show that ##\exists x \in ℝ - \mathbb{Q} \space | \space a < x < b##

    Since 0 < b - a, we know we can apply theorem 1.3 to find ##r \in \mathbb{Q} \space | \frac{a}{\sqrt{2}} < r < \frac{b}{\sqrt{2}}## because of the denseness of ##\mathbb{Q}##.

    We know : ##ℝ \setminus \mathbb{Q}## is the set of irrationals.

    Also, since ##r \in \mathbb{Q}##, we can take ##r = \frac{p}{q}## for some ##p, q \in \mathbb{Z}##

    This yields ##a < \frac{p \sqrt{2}}{q} < b##.

    Therefore, because ## \frac{p \sqrt{2}}{q} \in ℝ - \mathbb{Q}## the claim is proven true.

    EDIT : Fixed a small error.
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2

    micromass

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    The proofs is ok.

    But, I really don't get this line:

     
  4. Apr 30, 2013 #3

    Zondrina

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    I was just trying to highlight that it was the set of irrationals.

    Do you know how to write backslashes in latex? I was trying to write R\Q, but it treats '\' like an escape character.
     
  5. Apr 30, 2013 #4

    micromass

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    OK, but the right-hand side you wrote down really isn't equal to the set of irrationals...

    \setminus
     
  6. Apr 30, 2013 #5

    LeonhardEuler

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    Don't forget that numbers like [itex]\pi[/itex] and e are also irrational (among many others).
     
  7. Apr 30, 2013 #6

    Zondrina

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    ##ℝ \setminus \mathbb{Q}## Yay :)!

    Hmm would general elements look like ##c + d \sqrt{e}## instead then? Trying to get a grasp on what the set elements look like.
     
  8. Apr 30, 2013 #7

    micromass

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    There is no real good description of general elements of the irrational numbers. Elements like ##c+d\sqrt{e}## are still very special.

    The reality is that the set of irrationals is huge. Most elements of the irrational numbers can't be explicitely described.
     
  9. Apr 30, 2013 #8
    There is no general way of writing the elements of the irrationals in the way you are trying to do. The transcendental numbers are a subset of the irrationals and they cannot, by definition, be expressed as the roots of polynomials with rational coefficients. We only know the general form of a few families of transcendental numbers. However, we know that almost all real (and complex) numbers are transcendental, so you've only captured a very tiny subset with your definition.
     
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