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Homework Help: Between any two distinct real numbers there is a rational number

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Let x and y be real numbers with x<y and write an inequality involving a rational
    number p/q capturing what we need to prove. Multiply everything in your inequality by q,
    then explain why this means you want q to be large enough so that q(y-x)>1 . Explain
    how you can rewrite this inequality and use the Archimedean property to find such a q.

    3. The attempt at a solution
    So, this is a question on a worksheet our teacher gave us to go along with the theorem in the book. Here is what I did so far:

    x < p/q < y
    Then multiply both sides by q as the question states:
    qx < p < yq
    0 < p-qx < yq-qx
    0 < p-qx < q(y-x)

    I am having trouble seeing why q(y-x)>1. It is obviously great than zero as the inequality states, but can someone help me see why it has to be >1??
     
  2. jcsd
  3. Feb 3, 2010 #2
    What you are trying to do is exhibit a rational number [tex]p/q[/tex] between [tex]x[/tex] and [tex]y[/tex], that is, to choose [tex]p[/tex] and [tex]q[/tex] so that [tex]x < p/q < y[/tex]. The argument you're being led down is a strategy for choosing [tex]q[/tex], then [tex]p[/tex], so that they satisfy the conditions you want.

    So, what you need to do is:
    1. prove that you can choose [tex]q[/tex] so that [tex]q(y - x) > 1[/tex], i.e., prove that such a [tex]q[/tex] exists; then
    2. figure out why that means that you can choose [tex]p[/tex] so that [tex]0 < p - qx < q(y - x)[/tex], i.e., give a proof that if you have [tex]q[/tex] satisfying part 1 then such a [tex]p[/tex] exists.
     
  4. Feb 3, 2010 #3
    So, we chose q>1/(y-x)?

    So then,
    0 < p-(x/(y-x)) < 1
    0 < p < (y/(y-x))

    I'm still lost on how we choose p then.
     
  5. Feb 4, 2010 #4
    You need to give one sentence (really, one phrase) of proof for why one can choose [tex]q > 1/(y - x)[/tex].

    Now you need [tex]p[/tex] to satisfy [tex]0 < p - qx < q(y - x)[/tex]. Rewrite this as [tex]qx < p < qy[/tex]. You need to find an integer [tex]p[/tex] satisfying this inequality. Why does your choice of [tex]q[/tex] guarantee that there is at least one such [tex]p[/tex]? (This is the only condition you need [tex]p[/tex] to satisfy!)
     
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