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A simple question from Rudin's principles of mathematical analysis

  1. Jul 25, 2011 #1
    It's NOT a homework. I'm self-studying Rudin's Analysis and I came to this part that I can't follow the argument after a certain point:

    statement: if x,y [itex]\in[/itex] R, and x<y, then there exists a p [itex]\in[/itex] Q such that x<p<y.

    Proof: since x<y, we have y-x>0, and the Archemedean property furnishes a positive integer n such that:
    Apply the Archemedes' property again to obtain positive integers m1 and m2 such that m1>nx, m2>-nx. Then:

    [Here is where I get confused].
    Hence there is an integer m (with -m2<=m<=m1) such that: *
    m-1 <= nx < m. *
    If we combine these inequalities, we obtain
    nx < m <= 1 + nx < ny. *

    Since n>0, it follows that
    x < m/n <y.

    I don't understand the places that I've marked with *. I fail to follow the argument after the place that I've mentioned it. Can anyone clarify up these statements for me please?
  2. jcsd
  3. Jul 25, 2011 #2
    Hi AdrianZ! :smile:



    Then m will satisfy [itex]m-1\leq nx<m[/itex] (try to prove this). There is only one question: does the minimum exist? That is, does the set [itex]\{k~\vert~nx<k\}[/itex] nonempty and does it has a lower bound.

    What exactly don't you understand here?
    [itex]nx<m[/itex] by the choice of m
    [itex]m\leq 1+nx[/itex] since [itex]m-1\leq nx[/itex]
    [itex]1+nx<ny[/itex] since [itex]n(y-x)>1[/itex].

  4. Jul 25, 2011 #3
    well, since x is a positive real number and n is a natural number then nx>0. k>nx by definition, therefore k is a positive integer. to prove that the set is nonempty I should show that no matter how large nx is, there is always a k such that nx<k. well, suppose that It wasn't true. therefore for any positive integer k we had k<=nx which leads to a false statement that the set of positive integers has an upper bound. so there exists a positive integer k such that nx<k which means the set is non-empty. Now we know that the set [tex]A = \{k~\vert~nx<k\}[/tex] is non-empty and its members are positive integers. the well-ordering principle now guarantees the existence of the smallest k in the set which is the m we want. so the existence part of the theorem is now completed.
    Now to prove that [itex]m-1\leq nx<m[/itex] one need to realize that if [itext]m-1\leq nx[/itext] wasn't true, then we had nx<m-1. but m-1<m and m was the smallest positive number s.t nx<k, which is a contradiction. that proves the inequality part.

    Now I understand it completely.

    Are we done or there are still holes in the argument that we need to fill?
  5. Jul 25, 2011 #4
    OK, a few remarks: we don't know if x is positive or not. Second, it was easier if you would just use m1 and m2 in your argument. The existence of m
    provides a lower bound, and the existence of m1 says that the set is nonempty.

    It's alright, I think.
  6. Jul 25, 2011 #5
    Ah, right! well, thanks for the proof.
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