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Thread: Test

  1. #1
    Junior Member
    Joined
    Jan 2010
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    41

    Test

    The Problem: Let S = (0,1)

    Prove for each $\displaystyle \varepsilon $ > 0 there exist an x $\displaystyle \in $ S such that x < $\displaystyle \varepsilon $.

    Solution:

    For every $\displaystyle \varepsilon $ > 0, 0 < $\displaystyle \frac{\varepsilon }{2}$ < $\displaystyle \varepsilon $

    Consider $\displaystyle \frac{\varepsilon }{2}$ and x = $\displaystyle \frac{1}{2}$
    then

    x < $\displaystyle \varepsilon $

    $\displaystyle \Leftrightarrow $ $\displaystyle \frac{1}{2}$ < $\displaystyle \frac{\varepsilon }{2}$

    $\displaystyle \Leftrightarrow $ 1 < $\displaystyle \varepsilon $

    Thus

    $\displaystyle \varepsilon $ > 0 and $\displaystyle \varepsilon $ = 1 > x = $\displaystyle \frac{1}{2}$

    Therefore x <
    $\displaystyle \varepsilon $

    Q.E.D
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  2. #2
    MHF Contributor
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    Re: Test

    See my post in the other forum.
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