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Math Help - Prove/disprove the following statement

  1. #1
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    Prove/disprove the following statement

    For all x, there exists a y such that, Prove/disprove the following statement-codecogseqn.gif.

    My attempt at the solution:

    Assuming x=/= y,
    x|z can be written as xk=z
    y|z can be written as yk=z
    xk=yk
    x=y
    Therefore, the statement is false.


    I have been told this solution is wrong. Can someone please tell me what I did wrong or give me a suggestion how to go about proving this?

    Thanks
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  2. #2
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    Re: Prove/disprove the following statement

    Quote Originally Posted by nubshat View Post
    For all x, there exists a y such that, Click image for larger version. 

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ID:	30956.
    My attempt at the solution:
    Assuming x=/= y,
    x|z can be written as xk=z
    y|z can be written as yk=z
    xk=yk
    x=y
    Therefore, the statement is false.
    Your mistake is in using the variable $k$ twice.
    Moreover, we must assume that all variables are positive integers?

    You are correct to say that the statement is false.
    The only way to show a statement is false is to provide a counter-example.

    Think about $x=2$
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  3. #3
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    Re: Prove/disprove the following statement

    Quote Originally Posted by Plato View Post
    Your mistake is in using the variable $k$ twice.
    Moreover, we must assume that all variables are positive integers?

    You are correct to say that the statement is false.
    The only way to show a statement is false is to provide a counter-example.

    Think about $x=2$
    Sorry I forgot to mention the universe of discourse is Z\{0}.

    Why would x=2 not work?

    If x=2, when z is not divisible by x (example: z=7), there will always exist a y that is also not divisible by 7(example: y=25).
    Also, when z is divisible by x (example: z=8), there will always be a y that is a multiple of z (example: y=1).
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  4. #4
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    Re: Prove/disprove the following statement

    Quote Originally Posted by nubshat View Post
    Why would x=2 not work?
    If x=2, when z is not divisible by x (example: z=7), there will always exist a y that is also not divisible by 7(example: y=25).
    Also, when z is divisible by x (example: z=8), there will always be a y that is a multiple of z (example: y=1).
    You do not understand how to read the statement. We are in $\mathbb{Z}\setminus\{0\}$ so now the statement is true.
    .It says "For every number $x$ there is a number $y\ne x$ such that for every number $z$ it is the case that $x|z~{\color{blue}\iff}~y|z$

    That is an if an if and only if which means if $2|6$ then $y|6$ must be true.
    Then what is $y$? $y\ne 3$ because $2|8\text{ but}3\not | 8$.

    So what is $y$ it has to work for very even $z$ ? Can you find one?
    Last edited by Plato; May 18th 2014 at 03:31 PM.
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