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Q is dense in R question about proof

  1. Aug 24, 2011 #1


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    In Rudin's Principles of Mathematical Analysis there's the following proof that Q is dense in R.

    Theorem: If [itex] x,y\in \mathbb{R}[/itex] and [itex]x < y [/itex] there exists a [itex]p \in \mathbb{Q} [/itex] such that [itex] x<p<y.[/itex]

    Proof: Since [itex]x<y[/itex], we have [itex]y-x>0[/itex]. It follow from the Archimedian property that there is a positive integer [itex]n[/itex] such that
    We again apply the Archimedian property to find positive integers [itex]m_1[/itex] a,d [itex]m_2[/itex] such that [itex]m_1>nx[/itex] and [itex]m_2>-nx[/itex]. Then
    Hence there is an integer [itex]m[/itex] (with [itex]-m_2\leq m\leq m_1[/itex]) such that
    [tex]m-1\leq nx < m.[/tex]

    We combine the inequalities to get
    [tex]nx < m \leq 1+nx < ny.[/tex]
    n is positive so
    [tex]x < \frac{m}{n} < y.[/tex]
    Which proves that [itex]\mathbb{Q}[/itex] is dense in [itex]\mathbb{R}[/itex].

    How one concludes that the m in the red bit exists is what's troubling me.

  2. jcsd
  3. Aug 24, 2011 #2
    Hmm, you're not the first one to ask that very question.

    Anyway, the key is to define

    [tex]m=\min\{k\in \mathbb{Z}~\vert~nx<k\}[/tex]

    However, there are some things we need to show:

    - Does the minimum exist?? I.e. is the set [itex]\{k\in \mathbb{Z}~\vert~nx<k\}[/itex] non-empty, does it have a lower bound? Why is the infimum a minimum?
    - does [itex]m-1\leq nx<m[/itex] hold.
  4. Aug 24, 2011 #3


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    i.e. are you aware of the well ordering principle?
  5. Aug 27, 2011 #4


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    Thanks for the replies, seems I don't know enough about integers yet, so I've been looking into them.
    I've found another book on analysis which does introduce the integers, it defines them as "those real numbers which are in every inductive set."

    But this isn't how the integers are usually introduced is it?

    What's a good book to learn enough about integers (and maybe a little more)?

  6. Aug 17, 2012 #5
    This is self evident .Any real number must lie between two successive integers . When You have the natural number system ,Irrationals fill the gaps between any 2 successive integers . It's very simple.
  7. Aug 17, 2012 #6
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