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Proving no upper bound in A, where A = {x in Q | x^2 < 2}

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that there is no upper bound in A, where A = {x in Q | x2 < 2}

    3. The attempt at a solution

    My attempt has been to assume that there is an upper bound p in A and then I have been trying to find a way to show that there is a number that is larger than p but still in A.

    So I have tried doing some inequality expressions and saying that I need to find some x such that when I take the sum of p and x and square it, it is less than 2. [(p + x)2 < 2]

    But in the end when I try to find this x using the just mentioned inequality statement I get that x is a number that is less than ((2 - p2) / (2p + x)). So, this number is less than some expression that contains itself. I am having some trouble figuring this out.

    So I need a q in Q, where q = p + x, and x is in Q+, and I need q2 < 2. That's about all I got so far.
     
    Last edited: Aug 10, 2011
  2. jcsd
  3. Aug 10, 2011 #2
    You need to show that [itex]\forall x \in A, \exists y \in A[/itex] where [itex]y > x[/itex].

    It seems easiest to do this by direct proof.

    Hint: try writing [itex]y[/itex] as [itex]x + \epsilon[/itex] for some [itex]\epsilon > 0[/itex]
     
  4. Aug 10, 2011 #3

    SammyS

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    Do you mean you need to prove there is no Least Upper Bound for A
     
    Last edited: Aug 10, 2011
  5. Aug 10, 2011 #4
    The question came up because I was watching these real analysis lectures and when the topic of least upper bound came up, the professor gave the example of the set A = {x in Q | x2 < 2} saying that this had no least upper bound.

    For some reason, I was thinking that there was a least upper bound and it is in A and it is just the rational number that is the closest that you could get to the square root of 2 but still being less than it and thus being the largest number in the set and then being a least upper bound. I was just trying to prove to myself that that notion is false.

    The professor did show a way to prove that there is no least upper bound for A but what I wanted to show was that for A, there is no upper bound that is in A.
     
  6. Aug 10, 2011 #5
    This is the crux of the issue. Does such a rational number exist? If so, name it.

    Or rather, to disprove it, turn the notion around. Show that, for any rational number less than [itex]\sqrt{2}[/itex], there exists another rational greater than it but less than [itex]\sqrt{2}[/itex].

    For instance, if [itex]x < \sqrt{2}[/itex], then there must be some number [itex]\alpha > 0[/itex] such that [itex]x + \alpha = \sqrt{2}[/itex].

    This is the start. The next step is to see if you can find an explicit expression which will give you [itex]y \in Q, x < y < \sqrt{2}[/itex].
     
    Last edited: Aug 10, 2011
  7. Aug 10, 2011 #6

    SammyS

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    The statement of the problem was just fine.

    There is no Least Upper Bound of A in all of Q, so of course, there is no Upper Bound of A in A.
     
  8. Aug 10, 2011 #7

    jbunniii

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    If [itex]x[/itex] and [itex]y[/itex] are real and [itex]x \neq y[/itex], then there is a nonzero distance between them: [itex]d = |x - y| > 0[/itex]. Thus the problem reduces to proving that any open interval of nonzero length contains a rational. (Choose [itex]y = \sqrt{2}[/itex] and [itex]x[/itex] to be any rational in [itex]A[/itex].)
     
  9. Aug 10, 2011 #8

    jbunniii

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    That's actually a good point, though. The statement that "A has no least upper bound in Q" is stronger than "A has no maximal element." Indeed, even [itex](0,1)\cap Q[/itex] has no maximal element despite having a least upper bound in Q. [itex](0,\sqrt{2}) \cap Q[/itex] lacks a maximal element for a more profound reason.

    So it seems more interesting (and not any harder) to prove the stronger statement "A has no least upper bound in Q" and answer the OP's question as a corollary.
     
    Last edited: Aug 10, 2011
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