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How do i show if this is associative or not?

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    (x*y)=x+2y+4

    2. Relevant equations


    3. The attempt at a solution
    first i did this but i'm not sure if it is correct
    (x*y)*z=x+2y+4*z=x+2y+4+z+1
    x*(y*z)=x*y+2z+4=x+y+2z+4+1
     
    Last edited: Nov 19, 2015
  2. jcsd
  3. Nov 19, 2015 #2

    Svein

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    No, it is not correct. Study the definition closely. Remember that x does not stand for "x", but for "any expression to the left of "*". Vice versa for y.
     
  4. Nov 19, 2015 #3
    I don't understand what you mean?
     
  5. Nov 19, 2015 #4

    SteamKing

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    I think Svein means that instead of using the notation (x*y) = x + 2y + 4, you should use something like (x⊗y) = x + 2y + 4.

    Using '*' as an operator confuses, as we are used to thinking of '*' as indicating multiplication of the quantities x and y.

    By using the special ⊗ symbol as an operator, it becomes less confusing.

    For example, if we wrote (6⊗y), that would mean 6 + 2y + 4;
    likewise (x⊗6) would mean x + 2*6 + 4;
    (g⊗x) = g + 2x + 4
     
  6. Nov 19, 2015 #5
    Oh ok then did I write it out correctly
     
  7. Nov 19, 2015 #6

    SteamKing

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    It's not clear what you mean here.

    To take one of the items from the OP:
    (x ⊗ y) ⊗ z would mean finding out what (x ⊗ y) was first and then combining that quantity with z.

    You could re-write the original expression as

    (x ⊗ y) ⊗ z = (x ⊗ y) + 2z + 4 and then expand (x ⊗ y) according to the definition.
     
  8. Nov 20, 2015 #7

    SammyS

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    I don't see any problem in using the asterisk (*) for the operation defined here. Just be sure not to confuse it with traditional multiplication.

    When you write:
    (x*y)*z = x+2y+4*z ,​
    you really should use parentheses,
    (x*y)*z = ( x+2y+4 )*z​

    What you write after that is incorrect. It should be equal to ( x+2y+4 ) + 2z + 4 , etc.
     
  9. Nov 21, 2015 #8

    symbolipoint

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    Not showing all my steps, NOT associative.
    The two sides give x+2y+2z+8 and x+2y+4z+12, so these are unequal.
     
  10. Nov 21, 2015 #9
    so if (x*y)=x+2y-xy Then
    x*(y*z)=x+2(y+2z-yz)-x(y+2z-yz) and
    (x*y)*z=(x+2y-xy)+2z-(x+2y-xy)z
     
  11. Nov 21, 2015 #10

    SammyS

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    Yes.

    But it's not clear, looking at that, whether those two expressions are equivalent.
     
  12. Nov 21, 2015 #11
    they are not the same
     
  13. Nov 21, 2015 #12

    SammyS

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    i know that, but it's not all that clear from those two expressions.
     
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