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Math Help - simple but puzzling proofs

  1. #1
    Member pberardi's Avatar
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    simple but puzzling proofs

    I am a bit puzzled on this. It is a bit embarrassing.

    Is the following statement true or false?
    For all y there exists an x s.t. x > y^2 + 1

    The problem says nothing else about x or y and this is confusing. I have a feeling it is true because I can always find a number that is bigger than another number. Am I thinking correctly? Or is it that I choose y to be zero and x to be zero and thus it is false by counterexample. I think that this is incorrect because it is for some x not for all x.
    Last edited by pberardi; May 5th 2009 at 05:35 PM. Reason: figured out the second one...thanks
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  2. #2
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    Quote Originally Posted by pberardi View Post
    I am a bit puzzled on this. It is a bit embarrassing.

    Is the following statement true or false?
    For all y there exists an x s.t. x > y^2 + 1
    Let y be any real number. Define x so that  x = y^2 + 2 .
    Then show that  x > y^2 + 1

    You might find the part I chapters 1-3 of my book Topology and the Language of Mathematics useful. There's a free download available through here: Bobo Strategy - Topology

    Hope that helps...
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  3. #3
    Member pberardi's Avatar
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    Thanks but could you please explain what gives me the right to define x = y^2 + 1 ?
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    Re:

    Quote Originally Posted by pberardi View Post
    Thanks but could you please explain what gives me the right to define x = y^2 + 1 ?
    You are supposed to show that for any y, you can come up with an x so that x is bigger than  y^2 + 1 . That's what we've done.

    For any y, we have shown there is an x that is bigger than  y^2 + 1 . And we've defined it as a function of the arbitrary y.

    Note you could also define x = 323 - it just wouldn't be useful as it wouldn't always be bigger than  y^2 + 1 (for example when y = 20).
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