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Abstract Algebra - showing commutativity and associativity

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to determine if a*b=ab+1 is commutative and associative.
    * is any arbitrary operation

    2. Relevant equations



    3. The attempt at a solution
    a*b=ab+1 is commutative, but not associative.
    I'm getting stuck in showing why.
    b*a=ba+1
    a*(b*c)=a(b+1)
    ?????
     
  2. jcsd
  3. Feb 11, 2009 #2

    cristo

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    What are you definitions of associativity and commutivity?
     
  4. Feb 11, 2009 #3
    commutative is iff a*b=b*a for all a and b.
    associative is iff (a*b)*c=a*(b*c)
     
  5. Feb 11, 2009 #4

    cristo

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    Ok, so have you tried to use your operation for * in these expressions? If so, what did you get?
     
  6. Feb 11, 2009 #5
    That's where I get confused in using the operation.
    a*b=ab+1
    so, b*a=ba+1

    a*b=ab+1
    so (a*b)*c=a*(b*c)
    ab+1*c=a*(bc+1)
    (ab+1)c+1=a(bc+1)+1
    abc+c+1=abc+a+1
    So, not associative.

    Is this the right idea?
     
  7. Feb 11, 2009 #6

    cristo

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    Yes, but why? You should state "because multiplication of real numbers in commutative..."

    This is a very sloppy way of doing things, since the equals signs don't hold: i.e. the thing on the left is not equal to the thing on the right.

    Better would be to set it up something like this:

    * is associative if(f) (a*b)*c=a*(b*c). Then,

    LHS: (a*b)*c=(ab+1)*c=(ab+1)c+1=abc+c+1 ..(#)
    RHS: a*(b*c)=a*(bc+1)=a(bc+1)+1=abc+a+1 ..(§)

    Since (#)[itex]\neq[/itex](§), then * is not associative.
     
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