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Math Help - Defining an operaor

  1. #1
    Member Jones's Avatar
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    Defining an operaor

    Hai,

    I have some toubles with this question.

    Define an operator * on R by

    x*y = 2xy -x -y

    a) is * commutative?
    b) is * associative?

    I can easily see that * is commutative, but how do i test for associativity?

    The rule states that (x*y)*z = x*(y*z)

    But what is z ?
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hai,

    I have some toubles with this question.

    Define an operator * on R by

    x*y = 2xy -x -y

    a) is * commutative?
    b) is * associative?

    I can easily see that * is commutative, but how do i test for associativity?

    The rule states that (x*y)*z = x*(y*z)

    But what is z ?
    * is associative if \forall x,y,z \in \mathbb{R} : (x*y)*z = x*(y*z)

    (x*y)*z = (2xy - x - y)*z
    x*(y*z) = x*(2yz - y - z)

    Can you finish?

    Spoiler:
    (x*y)*z = (2xy - x - y)*z = 2(2xy-x-y)z - (2xy - x - y) - z = 4xyz -2xz -2yz -2xy +x +y-z
    x*(y*z) = x*(2yz - y - z) = 2x(2yz - y - z) - x - (2yz - y - z) = 4xyz - 2xy - 2xz -x -2yz + y + z
     4xyz - 2xy - 2xz -2yz -x + y + z \neq 4xyz -2xy -2xz -2yz +x +y-z
    Last edited by Defunkt; September 27th 2009 at 09:11 AM.
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  3. #3
    Member Jones's Avatar
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    Ah, thank you. I think i understand now. So the conclusion must be that the function is indeed associative because the two results are different?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Jones View Post
    Ah, thank you. I think i understand now. So the conclusion must be that the function is indeed associative because the two results are different?
    No - a function is associative if the two results are equal, not different.

    The function is not associative (if Defunkt did his working right).
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  5. #5
    Member Jones's Avatar
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    Quote Originally Posted by Defunkt View Post
    * is associative if \forall x,y,z \in \mathbb{R} : (x*y)*z = x*(y*z)

    (x*y)*z = (2xy - x - y)*z
    x*(y*z) = x*(2yz - y - z)

    Can you finish?

    Spoiler:
    (x*y)*z = (2xy - x - y)*z = 2(2xy-x-y)z - (2xy - x - y) - z = 4xyz -2xz -2yz -2xy +x +y-z
    x*(y*z) = x*(2yz - y - z) = 2x(2yz - y - z) - x - (2yz - y - z) = 4xyz - 2xy - 2xz -x -2yz + y + z
     4xyz - 2xy - 2xz -2yz -x + y + z \neq 4xyz -2xy -2xz -2yz +x +y-z
    Im sorry, some of the steps i do not understand.
    (x*y)*z = (2xy - x - y)*z
    = 2 (2xy-x-y)z - (2xy - x - y) - z

    Where did you get 2 from?
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