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Math Help - Binary operations on groups

  1. #1
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    Binary operations on groups

    Hey guys,

    I am trying to prove that a set with the specified binary operation below is a group? Here is the problem

    on R* x R, (a, b) * (c,d) = (ac, bc + d)

    I know that I have to prove that it is has
    1.) an identity: a * x = a
    2.) an inverse: a * x = e
    3.) associative

    The thing that throws me off is the (ordered pair). How do I start it off?

    This is how I started off trying to prove the identity
    a*x = a
    (a,b) * (x1, x2) = (a,b) ?? DOn't know if I can use x1,x2 ??
    (a*x1, b*x1 + x2 ) = (a,b)
    then I get two equations:

    a *x1 = a --> x1 = 1
    b*x1 + x2 = b --> b*1 + x2 = b --> x2 = 0

    NOw I plug x1 and x2 back into original eqn:

    (a *1, b*1 + 0 ) = (a, b)
    (a,b) = (a,b)

    haha, I started off not knowing how to go about and the mere fact of coming to this site and trying to ask a question got me going in the right direction and I ended up solving my own problem, . My thanks for this website!
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  2. #2
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    Ok guys, so I just thought I would go ahead and post if anybody else needed the help. I can do the rest, so thanks for watching out!
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  3. #3
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    Quote Originally Posted by spearfish View Post
    I am trying to prove that a set with the specified binary operation below is a group? Here is the problem
    on R* x R, (a, b) * (c,d) = (ac, bc + d)
    I know that I have to prove that it is has
    1.) an identity: a * x = a
    2.) an inverse: a * x = e
    3.) associative
    The identity is (1,0).
    But you have to show (x,y)*(1,0)=(1,0)*(x,y)=(x,y).

    The inverse of (a,b) is \left(\frac{1}{a},\frac{-b}{a}\right).
    Again you have to show that inverse properties hold.
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  4. #4
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    Thanks for the clarification Plato! It's kind of funny, but before I went to bed yesterday, it just popped into my head that I still needed to do the (x,y) * (1,0) = (1,0) * (x,y) = (x,y). I ll work on that and the rest of the problem today. Again, thanks!
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