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Prove commutative and associative

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    a function 'd' is a closed binary operation on a set called 'T'.
    There is an identity element named j.
    for all elements a, b, and c in the set 'T', we have d(a, d(b,c)) = d((a,c), b)

    can anyone help me show that d is commutative and associative?



    2. Relevant equations

    f is commutative if function f(s,u) = f(u,s)
    f is associative if function f(x, f(y,z)) = f((x,y), z)
    q is an identity element if function f(q,a) = a and f(a,q) = a.

    3. The attempt at a solution
    The attempt began with proving commutative, but ended shortly after because there are three elements in the problem and as far as i know commutative only uses 2.
     
  2. jcsd
  3. Nov 15, 2011 #2

    Dick

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    Try putting one of your three variables to j.
     
  4. Nov 15, 2011 #3

    epenguin

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    May show up my unfamiliarity :shy: but have you quoted the associativity law right?

    Likewise in the first line d((a,c), b) ? You have got this (a,c) in the air with no operation on the pair or rule to do anything with them, if I am not mistaken. :shy:
     
  5. Nov 15, 2011 #4
    yes the law is correct. What was given to me in the problem is misleading regarding the law of associativity which is why i am so stumped. as for j, I am not sure what putting one of my variables in for j will do, but i will mess with it and see where it gets me. Thanks!
     
  6. Nov 15, 2011 #5

    epenguin

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    Then I am the one who needs help.

    Associativity is written x*(y*z) = (x*y)*z where * is an operation.

    Translating this into the f language, this is

    f(x, f(y,z)) = f((f(x,y),z) .

    Both are formulae saying combine two things then combine the result with this other thing can be done in this sequence, but if you do it in this other sequence the result is the same. But it seems to me that in your formula you also have these brackets e.g. (x,y) with no defined operation to combine them so they do not mean anything definite or defined?
     
  7. Nov 15, 2011 #6

    Dick

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    Yes, there's a 'd' missing in the given relation and an 'f' missing in the statement of associativity. I figured they were just typos.
     
  8. Nov 15, 2011 #7

    epenguin

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    I am relieved!

    That should make it easier. :biggrin:
     
  9. Nov 16, 2011 #8

    Deveno

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    this is NOT associativity, as stated. if we write d(a,b) as a*b, what he is given is:

    a*(b*c) = (a*c)*b (note the reversal of b and c).

    it is far easier to prove commutivity FIRST.

    this is what the hint on using j is meant to convey.
     
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