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Math Help - Prove 1 is the least element in the positive integers

  1. #1
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    Prove 1 is the least element in the positive integers

    I did a proof, it sounds right to me, but I want to make sure.

    Prove:
    1 is the smallest element in Z+ (the positive integers)

    Suppose not, then there exists some integer a\in Z+ such that a<1

    Let S=[x\in Z+|x<1]. By our assumption that a<1, a\in S. Thus S is not the empty set and by the Well Ordering Principle there exists a smallest element, call it b. Since b<1 it follows that b-1<0.

    Thus, b(b-1)<0 which implies that b^2<b. It follows that b^2<1. Thus b^2\in S. But b^2<b, a contradiction since b is the smallest element in S. Thus 1 must be the smallest element in the positive integers.

    Is this right?
    Last edited by superevilcube; November 4th 2008 at 03:39 AM.
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  2. #2
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    Quote Originally Posted by superevilcube View Post
    I did a proof, it sounds right to me, but I want to make sure.

    Prove: 1 is the smallest element in Z+ (the positive integers)

    Suppose not, then there exists some integer a\in Z+ such that a<1

    Let S=[x\in Z+|x<1]. By our assumption that a<1, a\in S. Thus S is not the empty set and by the Well Ordering Principle there exists a smallest element, call it b. Since b<1 it follows that b-1<0.

    Thus, b(b-1)<0 which implies that b^2<b. It follows that b^2<1. Thus b^2\in S. But b^2<b, a contradiction since b is the smallest element in S. Thus 1 must be the smallest element in the positive integers.

    Is this right?


    Using the well ordering principal you can prove what you want to prove in the following way:


    The principal says that:

    For all sets S, S \subseteq N and S=/= Φ ( S being non empty) S has a least element.

    Since the above holds for all non empty subsets of N and since N is non empty and a subset of itself then N has a least member.


    Alternatively we can prove the above by induction in the following way:


    for n=1 we have .............1>=1 and the theorem is true for n=1


    suppose now the theorem true for n=k and k>=1 .And we need to prove that : k+1>= 1.


    But k>=1 ====> k+1>= 1+1 >1 +0, since 1>0 and so


    ................k+1>1 =====> k+1>=1 and the theorem is proved
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  3. #3
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    I think what poutsos is trying to say is that the well ordering principle is a consequence of 1 being the smallest element of \mathbb{Z}^+. He may or might not be right, depending on what axioms you are using. In the construction I learnt, 1 was defined as the smallest positive integer.

    Your proof is entirely sound if you are allowed the well ordering principle.
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