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Math Help - quantifier problem

  1. #1
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    quantifier problem

    \exists t \geq 0 \forall s \geq 0, we have s \geq t

    t = 0 works.

    \exists s \geq 0 \forall t \geq 0, we have s \geq t

    Why does s = t+1 not work? The book said
    \exists s \geq 0 \forall t \geq 0, we have s \geq t is false since its negation is true. But I would like to know why could s = t+1 not be used?

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    Last edited by mr fantastic; June 13th 2009 at 05:16 AM. Reason: The question may still be useful for others
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  2. #2
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    Quote Originally Posted by armeros View Post
    [tex]
    \exists s \geq 0 \forall t \geq 0, we have s \geq t
    Why does s = t+1 not work? The book said
    \exists s \geq 0 \forall t \geq 0, we have s \geq t is false since its negation is true. But I would like to know why could s = t+1 not be used?
    The answer may appear to be too subtle: it is a matter of order.
    Let’s translate the sentence into standard English.
    It would be: “Some non-negative number is greater than or equal to every non-negative number.”
    Surely put that way the expression is clearly false.

    Do you see why order makes a different?
    In a word, you are saying that ‘s is fixed’, it comes first, and it has a certain relation to every other t.
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