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Math Help - Another Analysis statement

  1. #1
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    Another Analysis statement

    Here's a new one, all values are assumed to be real:

     (\forall x) (\forall y), (\exists z) \ni xz=y

    To me, this says: For all x and y values, there's some z where x times z equals y.

    This is false, but I answered true. The reason why is because if z=1, any x would be equal to any possible y. I thought this meant it was true, but apparently it's not.

    Wait, just now I was thinking:

    The statement

    (\forall x)(\forall y), x=y

    is false, isn't it? Does this mean that any possible values of x and y have to be equal to each other?
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  2. #2
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    Quote Originally Posted by Skinner View Post
    Here's a new one, all values are assumed to be real:

     (\forall x) (\forall y), (\exists z) \ni xz=y

    To me, this says: For all x and y values, there's some z where x times z equals y.

    This is false, but I answered true. The reason why is because if z=1, any x would be equal to any possible y. I thought this meant it was true, but apparently it's not.

    Wait, just now I was thinking:

    The statement

    (\forall x)(\forall y), x=y

    is false, isn't it? Does this mean that any possible values of x and y have to be equal to each other?
    The answer for the first part is false, however, your reasoning is not quite correct. The statement is that given any x and y, there exist z such that xz=y is true. We do not need to consider what z is. This fails when x=0 and y\neq 0. i.e. if y=5 there no z such that 0*z = 5.

    And second statement is false like you have stated.
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  3. #3
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    Okay, Let me make sure that I understand.

    This is false because if  x=0 , Then no  z would exist to make  xz=y true for all real values  y .

    We know that this is false because it must work for every possible x and y value, and since it doesn't for this one case, we can be sure that it is false.

    Is that sound about right?
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  4. #4
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    Quote Originally Posted by Skinner View Post
    Okay, Let me make sure that I understand.

    This is false because if  x=0 , Then no  z would exist to make  xz=y true for all real values  y .

    We know that this is false because it must work for every possible x and y value, and since it doesn't for this one case, we can be sure that it is false.

    Is that sound about right?
    Correct
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