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Math Help - Binary operations?

  1. #1
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    Binary operations?

    hey guy could somebody help me to solve this :
    is the binary operation * defined for real numbers x,y by x*y=x^2y^2
    is it commutative? is it associative?

    thank you
    Last edited by boomshine57th; October 15th 2008 at 01:38 PM. Reason: for got to say please and thank you
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  2. #2
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    Hello, boomshine57th!

    Is the binary operation * defined for real numbers x,y by: . x*y \:=\:x^2y^2 ?

    If x,y are any real numbers, then: . x*y \:=\:x^2y^2 is a real number.

    The operation is defined for real numbers.




    Is it commutative?

    If the operation is commutative, then: . x*y \:=\:y*x

    We have: . \begin{array}{cccc}x*y &=& x^2y^2 \\ y*x &=&y^2x^2 \end{array}


    Since x^2y^2 = y^2x^2, the operation is commutative.




    Is it associative?

    If the operation is associative, then: . x*(y*z) \:=\:(x*y)*z

    We have: . \begin{array}{ccccccc}<br />
x*(y*z) &=& x*(y^2z^2) &=& x^2(y^2z^2)^2 &=& x^2y^4z^4 \\<br />
(x*y)*z &=& (x^2y^2)*z &=& (x^2y^2)^2 z^2 &=& x^4y^4z^2\end{array}\quad\hdots not equal

    The operation is not associative.



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  3. #3
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    thank you

    thank you soooo much! you are a life saver
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