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Math Help - Group Structure

  1. #1
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    Group Structure

    Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

    not that G=R\{-1} means that G is all real numbers excluding -1.
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  2. #2
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    Quote Originally Posted by TexasGirl
    Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

    not that G=R\{-1} means that G is all real numbers excluding -1.
    Binary Operation is Closed: The only case where it would not be closed is when, a,b \not = -1
    a*b=-1
    Thus,
    a+b+ab=-1
    Thus,
    b(a+1)=-a-1=-(a+1)
    Thus,
    b=-1
    Impossible, thus the binary operation is closed.

    Associative: We have,
    (a*b)*c=(a+b+ab)*c=a+b+ab+c+ac+bc+abc
    And,
    a*(b*c)=a*(b+c+bc)=a+b+c+bc+ab+ac+abc
    Thus, a*(b*c)=(a*b)*c

    Identity: We have, that,
    0*a=0+a+a0=a and,
    a*0=a+0+00=a.

    Inverse: Notice that,
    a*-\frac{a}{a+1}=-\frac{a}{a+1}*a=0
    Finally, we need to show if, a\not = -1 then, the element -\frac{a}{a+1}\not = -1
    Assume, it were then,
    -\frac{a}{a+1}=-1
    Thus,
    \frac{a}{a+1}=1
    Thus,
    a=a+1
    Thus,
    1=0
    Impossible, thus, -\frac{a}{a+1}\in G
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