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Math Help - Prove that there is no positive integer between 0 and 1

  1. #1
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    Prove that there is no positive integer between 0 and 1

    Theorem : Prove that there is no positive integer between 0 and 1


    <proof>
    Suppose there is a positive integer 'a' between 0 and 1. Let S = {n E Z+ | o < n <1}. Since 0 < a < 1, a E S, so S in nonempty.
    </proof>

    <me> So far fine from side</me>

    <proof>
    Therefore, by well ordering principle, S has a least element 'l', where o < l < 1
    </proof>

    <me> fine from my side</me>

    <proof>
    Then 0 < l^2 < l, so l^2 E S.
    </proof>

    <me>
    This is not fine. First of all from the set S is supposed to hold only +ve integers, given by the author as
    S = {n E Z+ | 0 < n 1}. And the square of a no is less than the original no, only if the no is less than 1. But when the author has assume that 'S' contains only +ve integers Z+, he can't assume that 'l' is a non-integer which he has done in this case by saying 0 < l^2 < l.

    Here is the rest of the proof anyways. If I am wrong I would greatly appreciate your help in understanding why I am wrong.
    </me>

    <proof>
    But l^2 < l, which contradicts our assumption that 'l' is a least element of S.Thus there are +ve integers
    </proof>
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  2. #2
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    Quote Originally Posted by scw_0611 View Post
    Therefore, by well ordering principle, S has a least element 'l', where o < l < 1
    Then 0 < l^2 < l, so l^2 E S.


    This is not fine. First of all from the set S is supposed to hold only +integers, given by the author as S = {n E Z+ | 0 < n 1}. And the square of a no is less than the original no, only if the no is less than 1.
    You are mistaken about that.
    Given any x \in (0,1) we have 0 < x^2  < x < 1. To see that simply multiply the inequality 0  < x < 1 through by x.
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  3. #3
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    Got it. I just went on to think that the inequality can't be true because an inequality a^2 < a only if 0 < a < 1, but I forgot that we are proving by contradiction.
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