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Math Help - Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x = 0

  1. #1
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    Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x = 0

    Would someone please give me some help with my case 2 below?

    The problem: Let x  \ge 0, x  \in \Re such that for any \varepsilon > 0, x  \le \varepsilon . Prove that x = 0

    Solution:

    Case 1: Let x > 0 and \varepsilon = \frac{x}{2}

    then

    \varepsilon = \frac{x}{2} > 0 because x > 0

    since \varepsilon > 0 by given, x  \le \varepsilon

    Case 2: Let x = 0 and \varepsilon = \frac{x}{2}

    then

    \varepsilon = 0/2 = 0

    since \varepsilon = 0 = x, I cannot conclude x  \le \varepsilon

    unless I let \varepsilon = x + y (where y  \in {\Re ^ + }). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.

    Thanks,
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  2. #2
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    Re: Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x

    Quote Originally Posted by Mathlv View Post
    Case 2: Let x = 0 and \varepsilon = \frac{x}{2}

    then

    \varepsilon = 0/2 = 0

    since \varepsilon = 0 = x, I cannot conclude x  \le \varepsilon

    unless I let \varepsilon = x + y (where y  \in {\Re ^ + }). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.

    Thanks,
    Why do you have a case 2? There are two possibilities. Either x=0 or x>0. First, you showed that if x>0, you arrive at a logical contradiction because there exists an \varepsilon>0 such that x>\varepsilon>0. Hence x is not greater than 0. That only leaves one possibility. If x=0 and \varepsilon>0, then x\le \varepsilon is true.
    Last edited by SlipEternal; October 9th 2013 at 06:30 PM.
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  3. #3
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    Re: Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x

    SlipEternal,

    Thank you so much for your help. I am understood now. It was a logical contradiction for the case of x > 0; therefore, that case was false. That left x = 0 is true for x <= espilon.
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