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Math Help - Integer Corollary Proof

  1. #1
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    Integer Corollary Proof

    I would appreciate some help with this proof!

    Let n be an element of the integers. There exists no integer x such that n < x < n+1.

    Thanks!
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    Quote Originally Posted by jstarks44444 View Post
    I would appreciate some help with this proof!

    Let n be an element of the integers. There exists no integer x such that n < x < n+1.

    Thanks!
    Here is a proof that there is no integer between 0 and 1 as an example for you.

    S=\{n\in\mathbb{Z} | 0<n<1\}

    Now, suppose a\in (0,1) and a is an integer. Now, since 0 < a < 1 is in S, then S is non-empty. By the well-ordering principle, S has a least element, l, so 0 < l < 1.

    Multiple the inequality by l.

    0*l < l*l < 1*l\Rightarrow 0<l^2<l. We have reached a contradiction since l^2<l
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    How would I go about doing this for n and n+1 though? using this method i arrive at

    ln < l^2 < ln + l

    but can any contradiction be made there?
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    Quote Originally Posted by jstarks44444 View Post
    I would appreciate some help with this proof! Let n be an element of the integers. There exists no integer x such that n < x < n+1.
    Quote Originally Posted by jstarks44444 View Post
    How would I go about doing this for n and n+1 though? using this method i arrive at
    ln < l^2 < ln + l but can any contradiction be made there?
    You have us at a clear disadvantage. We have no way to know what text material you are using. In the title of this post is the word ‘corollary’. That means it is in addition to some theorem. What theorem? You just have let in on the what basis you are proving things.
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    Well the theories that come before this corollary are,

    *For all k in the Naturals, k >= 1
    *There exists no integer x such that 0<x<1
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    Quote Originally Posted by jstarks44444 View Post
    Well the theories that come before this corollary are,
    *For all k in the Naturals, k >= 1
    *There exists no integer x such that 0<x<1
    I suspect that you also have the following definition:
    m>n if and only if m-n\in \mathbb{N}.
    That means that if n\in \mathbb{N} then n-0=n\in \mathbb{N} so that n>0
    If the theorem to which the current question is a corollary is n\ge1 for all n\in \mathbb{N} the how could there be an integer between 0~\&~1~?
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  7. #7
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    But n is an element of the integers, not of the naturals....

    where is the inductive step in this, or do you not need one? contradiction?
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  8. #8
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    Quote Originally Posted by jstarks44444 View Post
    But n is an element of the integers, not of the naturals....
    If you continue to refuse to post a complete list of axioms, definitions, and theorems you will not receive help.
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  9. #9
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    Sorry..axioms for the natural numbers:

    If m,n are in the naturals then m+n is in the naturals
    If m,n are in the naturals then mn is in the naturals
    0 is not in the naturals
    For every m in the integers, we have m in the naturals or m=0 or -m in the naturals
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  10. #10
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    Quote Originally Posted by jstarks44444 View Post
    Sorry..axioms for the natural numbers:

    If m,n are in the naturals then m+n is in the naturals
    If m,n are in the naturals then mn is in the naturals
    0 is not in the naturals
    For every m in the integers, we have m in the naturals or m=0 or -m in the naturals
    Thank you for that.
    BUT what is the definition for m>n where \{m,n\}\subset\mathbb{Z}.

    In case you do no know this, your text material is non-standard.
    We need to know the axiom set for the integers.
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