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Math Help - Prove or Disprove

  1. #31
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    I had to read this a couple of times myself as I thought it was an existence statement, and was starting to question my reading comprehension at the "for all".

    The devil is in the details.
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  2. #32
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    Well guys, I guess you are intent on confusing, rather than helping the OP, so I won't say any more!
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  3. #33
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    Quote Originally Posted by Archie Meade View Post
    Well guys, I guess you are intent on confusing, rather than helping the OP, so I won't say any more!
    We're just trying to get the statement right ... this kind of problem can get real confusing sometimes.
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  4. #34
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    Quote Originally Posted by Bacterius View Post
    That's what I thought too I was starting to doubt of myself

    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
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  5. #35
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    Quote Originally Posted by Archie Meade View Post
    Well guys, I guess you are intent on confusing, rather than helping the OP, so I won't say any more!

    This really doesn't help: either I (and perhaps some others) am wrong or you are. I'm not trying to impose my opinion by authority or, much less, by force. I gave some reasons, using logic symbols, and I think I made my point clear. If you think it is otherwise you must then deffend your stand with arguments, not by dismissing others' opinions saying that they're confusing the OP which, again imo, was precisely what you did.
    It wouldn't be the first time, and most probably neither the last one, I could be wrong, so I've no problem at all in being shown where my mistake is, in case it exists.

    Tonio
    Last edited by tonio; March 23rd 2010 at 05:11 AM.
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  6. #36
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    I came up with a new proof:

    Prove: there does not exist a real number x such that for all real numbers y, xy=1

    Contradiction:
    Suppose yes, that is supppose that there does exist a real number x such that for all real numbers y, xy=1.

    Let x=z for any real number z and let y=0.

    Then,
    xy=z*0 by substitution
    =0 by algebra

    This is a contradiction. We were trying to prove that for all real numbers y there is a real number x where xy=1. This is NOT the case when we use the real number 0 for y. Therefore, the original statement is true. There does NOT exist a real number x such that for all real numbers y, xy=1. **END OF PROOF**
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  7. #37
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    Quote Originally Posted by matthayzon89 View Post
    I came up with a new proof:

    Prove: there does not exist a real number x such that for all real numbers y, xy=1

    Contradiction:
    Suppose yes, that is supppose that there does exist a real number x such that for all real numbers y, xy=1.

    Let x=z for any real number z and let y=0.

    Then,
    xy=z*0 by substitution
    =0 by algebra

    This is a contradiction. We were trying to prove that for all real numbers y there is a real number x where xy=1. This is NOT the case when we use the real number 0 for y. Therefore, the original statement is true. There does NOT exist a real number x such that for all real numbers y, xy=1. **END OF PROOF**
    You have not proved it yet. You need to go back and read the logical expression by Bacterius, and simply take one counterexample from Archie Meade.
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  8. #38
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    can u elaborate and explain WHAT exactly is incorrect? why is my proof not good?
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  9. #39
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    Quote Originally Posted by novice View Post
    You have not proved it yet. You need to go back and read the logical expression by Bacterius, and simply take one counterexample from Archie Meade.
    Why?

    His proof is correct, although a bit redundant - there is no need to write "Let x=z for any real number z". Instead, you (matthayzon) should write: "Let x \in \mathbb{R}", or "Let x be a real number". Then let y=0 and procceed as before.

    I think tonio summed it up pretty well. There was some confusion as to what the poster was actually asked to prove. In my opinion it is rather apparent that what we need to show is that there is no constant x \in \mathbb{R} such that for ANY y \in \mathbb{R}, xy=1, which is exactly what tonio said.

    Quote Originally Posted by matthayzon89 View Post
    can u elaborate and explain WHAT exactly is incorrect? why is my proof not good?
    Your proof is correct, the only thing you need to fix is what I said in this post. It should look like this:

    Theorem: There does not exist a real number x \in \mathbb{R} such that for any y \in \mathbb{R}, xy=1.

    Proof: Assume, by contradiction, that there does exist such x, ie. that for any real number y, xy=1. Now, choose y=0. Since y (0) is a real number, by our assumption we get that xy=1. However, we know that x * 0 = 0 for any real number x. Therefore,  0=1, in contradiction.
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  10. #40
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    Quote Originally Posted by Defunkt View Post

    Theorem: There does not exist a real number x \in \mathbb{R} such that for any y \in \mathbb{R}, xy=1.

    Proof: Assume, by contradiction, that there does exist such x, ie. that for any real number y, xy=1. Now, choose y=0. Since y (0) is a real number, by our assumption we get that xy=1. However, we know that x * 0 = 0 for any real number x. Therefore,  0=1, in contradiction.
    Who said there exist no such real number x such that xy=1?

    When x=\frac{1}{y}, xy=1.

    At least one x does exist. Contradiction.
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  11. #41
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    Yah, you are both right. y=0 spoils it.
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  12. #42
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    y=0 contradicts it, but the claim that such x exists is false even if 0 is excluded.
    what he was asked to prove\disprove is that you can use the SAME x for ALL y. In your example, x depends on y.

    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.
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  13. #43
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    Quote Originally Posted by Defunkt View Post
    y=0 contradicts it, but the claim that such x exists is false even if 0 is excluded.
    what he was asked to prove\disprove is that you can use the SAME x for ALL y. In your example, x depends on y.

    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.
    This, I tihnk is what people have been trying to communicate - sometimes unsuccessfully (although some did put it in symbols and there was still debate). The positioning of the existence and for all statements has been muddled a bit, however it does appear from the OP's original statement, the the original order is there exists, followed by for-all.
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  14. #44
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    Quote Originally Posted by Defunkt View Post
    y=0 contradicts it, but the claim that such x exists is false even if 0 is excluded.
    what he was asked to prove\disprove is that you can use the SAME x for ALL y. In your example, x depends on y.

    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.
    I agree with this, this is how I understand the statement and how I would have (and have) proved it. But it looks like the OP is resolved to use his own proof ...
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  15. #45
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    Quote Originally Posted by Bacterius View Post
    I agree with this, this is how I understand the statement and how I would have (and have) proved it. But it looks like the OP is resolved to use his own proof ...
    Well, both proofs are correct, so I don't see any reason to prefer one over the other.
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