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Math Help - binary operation

  1. #1
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    binary operation

    (a)The operation @ and * are defined real numbers as follows :

    a*b = a+b+ab
    a@b=ab+1

    For each operation , determine if @ is distributive on * .


    how can this be done ? Thanks for any help .
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  2. #2
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    Here's how to do the first one:

     x*(y+z) = x + (y+z) + x(y+z) = x + y + z + xy + xz
     x*y + x*z = (x + y + xy) + (x + z + xz) = 2x + y + z + xy + xz

    These are not the same (e.g. let x = 2), so the operation of * as defined here is not distributive.

    Take your time and understand this, and if you do you should be able to apply the same idea to the second.
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  3. #3
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    Re :

    For the second one :

    x*(y+z) = x(y+z)+1 = xy+xz+1

    x*y+x*z = xy+1+xz+1 = xy+xz+2

    Thus , it is not distributive .
    Any mistake ?
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  4. #4
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    Hello, mathaddict!

    The operation @ and * are defined real numbers as follows :

    . . a*b \:= \:a+b+ab .
    (Add the two and add their product)

    . . a\,@\,b \:=\:ab+1 .
    (Multiply the two and add 1)

    Determine if @ is distributive on *.

    If @ is distributive on *, then: . a\,@\,(b*c) \;=\;(a\,@\,b)*(a\,@\,c) . . . Is this true?


    On the left:

    . . a\,@\,(b*c) \;=\;a\,@\,(b+c+bc)

    . . . . . . . . . =\;a(b+c+bc) + 1

    . . . . . . . . . = \;ab + ac + abc + 1


    On the right:

    . . (a\,@\,b) * (a\,@\,c) \;=\;(ab+1) * (ac+1)

    . . . . . . . . . . . . = \;(a+b+1) + (a+c+1) + (ab+1)(ac+1)

    . . . . . . . . . . . . = \;a^2bc + ab + ac + 2a + b + c + 3


    They are not equal . . . @ is not distributive on *.

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