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Math Help - well ordering principle

  1. #1
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    well ordering principle

    Use the WOP of N to prove that there is no  n \in N such that 0<n<1
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  2. #2
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    Well if there were then since n\in\mathbb{N} and n>0, it follows that (n-1)\in\mathbb{N}. Now you have that (n-1)\neq{0}, so continue in this way to get a contradiction.
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  3. #3
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    Quote Originally Posted by flower3 View Post
    Use the WOP of N to prove that there is no  n \in N such that 0<n<1
    I am a bit puzzled by the wording of this question.
    I my experience the following is a standard theorem: \left( {\forall x \in \mathbb{R}} \right)\left( {\exists !n \in \mathbb{Z}} \right)\left[ {n \leqslant x < n + 1} \right].
    The unique integer is of course the floor of x.
    I see what putnam120 is trying to do. But I an not at all sure where W.O. be used.

    The theorem I quoted above with x=0 give the result at once.

    PS. Of course, this is all mute if your course includes a rigorous construction of the natural numbers.
    Last edited by Plato; August 18th 2009 at 02:00 PM.
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  4. #4
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    W.O. states that every set of non-negative integers has a least element. Thus since \mathbb{N} only contains non-negative integers it to must have a minimal element. So using the construction I started you show that the set doesn't have a minimal element.

    I usually see \mathbb{N} defined as follows:

    1) 0\in\mathbb{N}

    2) (n\neq{0})\in\mathbb{N}\Longrightarrow (n\pm{1})\in\mathbb{N}

    3) (n=0)\in\mathbb{N}\Longrightarrow (n+1)\in\mathbb{N}
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