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Math Help - proof by contradiction

  1. #1
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    proof by contradiction

    Hi,

    I have to prove the following by contradiction

    the set S = [all natural numbers n such that n is a multiple of 13] has no greatest element

    I have started by saying let greatest value = x, where x is a member of S and x is greater than or equal to n

    then let n=13k for some natural number k

    x is greater than or equal to 13k

    not sure where to go from here??

    thanks
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  2. #2
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    Re: proof by contradiction

    Quote Originally Posted by AU11 View Post
    I have to prove the following by contradiction
    the set S = [all natural numbers n such that n is a multiple of 13] has no greatest element
    We know that (\forall k\in\mathbb{N})[k<k+1]~.

    If n=\max(S) is it true n=13k~\&~n=13k<13(k+1)=n+13~?
    Last edited by Plato; October 31st 2012 at 02:03 PM.
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  3. #3
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    Re: proof by contradiction

    Hi,

    I don't get why you have written k=13k?

    plus do I not talk about x (the greatest element) anymore?
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  4. #4
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    Re: proof by contradiction

    Quote Originally Posted by AU11 View Post
    Hi,

    I don't get why you have written k=13k?

    plus do I not talk about x (the greatest element) anymore?
    See my edit.

    If x is a real number x+1 is a real number and x<x+1 so there cannot a greatest real number.

    If 0<x then 0<\frac{x}{2}<x so there can be no smallest positive number.
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  5. #5
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    Re: proof by contradiction

    ohh right, I understand it now!

    thanks for the help
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