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Homework Help: Prove in any interval there exists an irrational z

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that in any interval there exists an irrational z.

    2. Relevant equations

    3. The attempt at a solution
    My professor wrote this for me when trying to explain how to prove this:

    [itex]a \notin Q[/itex], [itex]\epsilon[/itex] rational

    [itex][r, s]\in a[/itex]

    [itex]l([r, s])<\frac{\epsilon}{2}[/itex]





    I don't see why you use [itex][r, s]\in a[/itex] or how my professor went from [itex]s-\frac{\epsilon}{2}+\epsilon<a+\epsilon[/itex] to [itex]s<s+\epsilon<a+\epsilon[/itex]

    I also don't know where to go from here to show what needs to be proved.
    I'd really appreciate any help.
  2. jcsd
  3. Dec 18, 2009 #2
    As people were saying in the previous thread, the problem appears to be that much of what you have written down is nonsense ([tex][r, s] \in a[/tex], which is backward, and the dubious inequality deduction you mention).

    What might be going on here is an attempt to make the following argument:
    Given that there is a rational number in every interval, and the existence of one single irrational number [tex]a[/tex], prove that there is an irrational number in every interval.

    ([tex]a[/tex] might be [tex]\sqrt{2}[/tex], using the standard proof.)

    Here is how this argument would go. Let [tex](r,s)[/tex] be the interval in which we wish to find an irrational number. We try to do this by adding a rational number [tex]q[/tex] to [tex]a[/tex], because the sum of a rational and an irrational is irrational. So we need to find [tex]q[/tex] so that [tex]r < q + a < s[/tex]. Clearly it suffices to have [tex]r - a < q < s - a[/tex], i.e., choose [tex]q[/tex] to be any rational number in [tex](r - a, s - a)[/tex]. We can do this by hypothesis, so the proof is done.

    Does this make sense?
  4. Dec 18, 2009 #3
    I think so. Is this proof correct:
    Let [itex]a\in[/itex] Q, [itex]\epsilon[/itex] rational, where [itex]\epsilon\in[r-a, s-a][/itex].
    Then [itex]r-a<\epsilon<s-a[/itex], so [itex]r<\epsilon+a<s[/itex]
    So [itex]\epsilon+a\in[r, s][/itex] and [itex]\epsilon+a[/itex] is irrational.
    Therefore in any interval, there exists an irrational number.
  5. Dec 18, 2009 #4


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    But, since [itex]a \in \mathbb{Q}[/itex] and [itex]\epsilon \in \mathbb{Q}[/itex] clearly [itex]a + \epsilon \in \mathbb{Q}[/itex]. Thus [itex]a + \epsilon[/itex] cannot be irrational.
  6. Dec 18, 2009 #5
    Oops, meant to write [itex]a\notin[/itex] Q
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