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Math Help - Ring and Groups

  1. #1
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    Ring and Groups

    This was a challange question and I am not sure how to do this.. help would be appreciated thx
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  2. #2
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    \mbox{a) }\, (a\, +\, b)^2\, =\, (a\, +\, b)(a\, +\, b)\, =\, (a\, +\, b)a\, +\, (a\, +\, b)b

    =\, a^2\, +\, ba\, +\, ab\, + \, b^2\, =\, a\, +\, ba\, +\, ab\, +\, b\, =\, (a\, +\, b)

    Subtract (a\, +\, b) from either side. See if that leads anywhere helpful...?

    \mbox{b) }\, (ab)^2\, =\, (ab)(ab)\, =\, abab\, =\, a^2 b^2

    a^{-1}ababb^{-1}\, =\, a^{-1}a^2 b^2 b^{-1}

    ebae\, =\, eabe

    ...where e is the group's identity element. The result follows immediately.
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  3. #3
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    Quote Originally Posted by stapel View Post
    \mbox{a) }\, (a\, +\, b)^2\, =\, (a\, +\, b)(a\, +\, b)\, =\, (a\, +\, b)a\, +\, (a\, +\, b)b

    =\, a^2\, +\, ba\, +\, ab\, + \, b^2\, =\, a\, +\, ba\, +\, ab\, +\, b\, =\, (a\, +\, b)

    Subtract (a\, +\, b) from either side. See if that leads anywhere helpful...?

    \mbox{b) }\, (ab)^2\, =\, (ab)(ab)\, =\, abab\, =\, a^2 b^2

    a^{-1}ababb^{-1}\, =\, a^{-1}a^2 b^2 b^{-1}

    ebae\, =\, eabe

    ...where e is the group's identity element. The result follows immediately.
    thx for the reply and I can understand the second part and for the first part if I takeaway a+b from both sides then I will get ab = -ba but is it ok to say that it is commutative just from that step or do I have to go further any more steps to prove it

    I think for the part (ii) I have to do like as follows:

    b(a+a) = 0;
    therefore a+a =0 is it right
    thx once again for ur help
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