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Math Help - Determine if this is a group

  1. #1
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    Determine if this is a group

    Problem:
    Determine if the set with the binary operation forms a group.
    Set S = All real numbers except -1.
    a*b = ab+a+b

    ****************

    I know I need to check three things: associativity, if there is an identity, and if there is an inverse for every element in S.

    I know how to do associativity since it's straightforward. For the identity, I just guess and checked and found that e=0 works. (Is there any way to find an identity without guessing/checking?)

    For the inverse, I am having difficulty.

    I know that this must be satisfied to have an inverse:

    a*a' = a'*a = e

    Since I know the identity element is 0,

    a*a' = a'*a = 0

    Expanding this out with the definition of the binary operation,

    aa'+a+a' = a'a+a'+a = 0

    I said that an inverse does not exist, because from the last equation, it is impossible to get aa'+a+a' equal to 0 for all a. And therefore, this is not a group.

    Am I correct?
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  2. #2
    MHF Contributor

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    Tejas
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    Re: Determine if this is a group

    let's try to find an identity in a more systematic way:

    we want a*x = a.

    (it should be obvious that a*b = b*a, so we just need to check for a one-sided identity).

    now a*x = ax + a + x. set this equal to a, and try to solve for x:

    ax + a + x = a
    ax + x = 0
    (a+1)x = 0. now, can we divide by a+1? yes we can, because a will not be -1.
    x = 0/(a+1) = 0.

    are you sure it is impossible to find an inverse?

    again, we want b such that a*b = 0. we want to solve for b in terms of a, if we can.

    ab + a + b = 0
    (a+1)b = -a.....i think you can finish. verify your answer.
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