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Proving uniqueness of inverse by identity (Groups)

  1. Sep 13, 2013 #1
    1. Which of the following is a group?
    To find the identity element, which in these problems is an ordered pair (e1, e2) of real numbers, solve the equation (a,b)*(e1, e2)=(a,b) for e1 and e2.




    2. (a,b)*(c,d)=(ac-bd,ad+bc), on the set ℝxℝ with the origin deleted.



    3. The question also asks for you to find the inverse and I think implicitly for associativity and then for commutativity. I've got the other three down, but the identity axiom is giving me trouble.

    (a,b)*(a,b)-1=(aa-1-bb-1,ab-1+ba-1)
    Unless I'm really screwing up here, I think this implies that e1-e2 is the first value, which would be equivalent to e1. However, I don't know how to resolve the ab-1+ba-1) part.


    I think I've been dropped from this class for a lack of prerequisites, but I think I'm still going to try and finish the class nonetheless.
     
  2. jcsd
  3. Sep 13, 2013 #2

    pasmith

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    Homework Helper

    That certainly is a complex multiplication rule.

    That won't work; it's not the case here that [itex](a,b)^{-1} = (a^{-1},b^{-1})[/itex] (what inverses did you assign to [itex](1,0)[/itex] and [itex](0, 1)[/itex], both of which are members of this alleged group?).

    It is a consequence of the definitions that you cannot assign an inverse to an element until you have found the identity.

    As suggested, you need to set [itex](a,b)*(e_1,e_2) = (ae_1 - be_2, ae_2 + be_1) = (a,b)[/itex] and solve for [itex]e_1[/itex] and [itex]e_2[/itex].

    Then, to find [itex](a,b)^{-1} = (c,d)[/itex], set [itex](a,b)*(c,d) = (e_1,e_2)[/itex] and solve for [itex]c[/itex] and [itex]d[/itex].
     
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