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Math Help - Prove or Disprove

  1. #46
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Defunkt View Post
    Well, both proofs are correct, so I don't see any reason to prefer one over the other.
    Ah, I didn't see his proof actually changed. Yes it is correct (qlthough a bit risky to take zero as an example). These proofs are all alike, I must be tired
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  2. #47
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    I plotted for all real number y as a diagonal line on x-y axis and then plotted x= \frac{1}{y} . By plotting x= \frac{1}{y} , I have made sure none of the real number y is excluded. I found that all \forall y \in \mathbb{R}-\{0\} is possible on the diagonal line, but for x, I found it a very different story, namely, x=\frac{1}{y} approaches neither of the axes--neither x nor y, which means that if xy=1, neither x nor y can be a zero.

    Quote Originally Posted by Defunkt View Post

    For example, could have taken y=1 at first, to get x*1=1 -> x=1, and then take y=2 and get 1*2 = 1, which is obviously a contradiction.
    Since the statement says \sim \exists x \forall y:xy=1, we choose any y then find x; in other words, by choosing any x then find y would makes no sense.

    If you take y=2, you need not stick to x=1 because it did not say for all x. Since \exists x means at least one, I do have one, namely x=1/2.

    At any rate, it's very nice of you to help me. I enjoy reading many of you posts. You are quite good at math.
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  3. #48
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    Quote Originally Posted by novice View Post
    If you take y=2, you need not stick to x=1 because it did not say for all x. Since \exists x means at least one, I do have one, namely x=1/2.
    You are still confusing the order of the quantifiers :P
    \exists x \forall y:xy=1 - this means that there is a single x, such that for every y that you pick, xy=1. So, logically, how do you disprove such a statement? you simply find one such y, for which xy \neq 1.

    This is the essence of the proof - you assume by contradiction that there exists such an x. That means that for every y that you pick, xy=1 should be true.

    Specifically, it should be true for y=1. That means 1*x = 1 -> x = 1 (remember, one x for all y).
    So you have that x*y = 1*y = 1 for all real y. To prove that is not true, simply take any other y - for example y2 = 2. Your assumption implies that x*y2 = 1*y2 = 1, but 1*y2 = 1*2 = 2. So you get 2=1, which is obviously a contradiction.
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  4. #49
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    Emakarove, where are you? We need to exterminate this flea about the order of quantifiers.
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  5. #50
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    Quote Originally Posted by novice View Post
    Emakarove, where are you? We need to exterminate this flea about the order of quantifiers.

    There is no flea, imo: I already explained it, Defunkt already explained it (read his last post: cyrstal clear)...

    Tonio
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  6. #51
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    Quote Originally Posted by tonio View Post
    Writing with logic symbols may help sometimes, though in this case it comes really close: the statement \neg\exists x\,\forall y P(x,y) is logically equivalent to the

    statement \forall x\,\exists y\,\neg P(x,y) , so the difference, imo, between what it is and what was being understood by some is only the negation of P(x,y) in the last formula...

    Tonio
    Quite clear. Grazie tanto!
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  7. #52
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    So why is setting y=0 so bad when trying to prove the statement is TRUE using contradiction?
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  8. #53
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    Quote Originally Posted by matthayzon89 View Post
    So why is setting y=0 so bad when trying to prove the statement is TRUE using contradiction?
    It is not bad; your proof is valid.
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