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Math Help - What kind of relation is that?

  1. #1
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    What kind of relation is that?

    (exist z) x=yz

    What kind of relation is that?

    ***************
    I did it this way in my exam:

    For z = 1 the equation is correct, because x=y*1 is x=y

    The relation is symmetric
    ***************
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aRTx View Post
    (exist z) x=yz

    What kind of relation is that?

    ***************
    I did it this way in my exam:

    For z = 1 the equation is correct, because x=y*1 is x=y

    The relation is symmetric
    ***************
    We appear to be missing some information. For example, how did you know that x = y?

    -Dan
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  3. #3
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by topsquark View Post
    We appear to be missing some information. For example, how did you know that x = y?

    -Dan
    If he says that x=yz and z =1 than x=y*1 --> x=y

    though that's how I interpreted it...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shinhidora View Post
    If he says that x=yz and z =1 than x=y*1 --> x=y

    though that's how I interpreted it...
    Hmmm... It looks more to me that the problem is written
    \exists z \text{ such that }x = yz
    So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking x, y, z \in \mathbb{R}. Possibly this would be some sort of "maximal" relation? (My ignorance here is probably going to make a fool out of me, so I'm likely done trying to help out on this one.)

    -Dan
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  5. #5
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by topsquark View Post
    Hmmm... It looks more to me that the problem is written
    \exists z \text{ such that }x = yz
    So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking x, y, z \in \mathbb{R}. Possibly this would be some sort of "maximal" relation? (My ignorance here is probably going to make a fool out of me, so I'm likely done trying to help out on this one.)

    -Dan
    I thought he meant:

    He knows for sure that:

    x=yz and z = 1 and then it's obvious that x=y

    Edit: I would believe what you, Topsquark and Isomorphism say, If I'd understand it I just don't, so I stick with the stuff I understand xD
    Last edited by shinhidora; June 18th 2008 at 06:45 AM.
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Hmmm... It looks more to me that the problem is written
    \exists z \text{ such that }x = yz
    So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking x, y, z \in \mathbb{R}.
    -Dan
    I, too, believe this. Isnt it an equivalence relation?

    \text{xRy if}\, \exists z \text{ such that }x = yz

    A) Reflexive: xRx by choosing z=1.

    B) Symmetric: xRy implies there exists z such that x = yz. But x=yz => y = x(1/z) . Thus yRx.

    A) Transitive: xRy , yRw => there exists a and b such that x =ay and y = bw. But this means x = ay = a(bw) = (ab)w. Thus xRw.
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  7. #7
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    We really do need to know the domain of this relation.
    If the domain contains 0 then it is not symmetric.
    For example: \left( {0,2} \right) \in R \quad  \mbox{but} \quad \left( {2,0} \right) \notin R .
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