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Math Help - Proving 1 isn't in my group

  1. #1
    Junior Member beebe's Avatar
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    Aug 2011
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    Proving 1 isn't in my group

    Let G=\mathbb{Q}-\{1\}
    a*b=a+b-ab \forall a,b \in G

    I need to show my group is closed under *, and I'm happy with my proof that all of the possible outcomes are rational numbers. So now I need to show that 1 is not a possible outcome. Is my proof of this acceptable?

    Assume a+b-ab=1

    a-ab=1-b

    a(1-b)=1-b

    a=\frac{1-b}{1-b}

    a=1

    1 \notin G

    Therefore, a+b-ab \neq 1
    Last edited by beebe; December 13th 2011 at 04:17 PM. Reason: Prettying up the formatting
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  2. #2
    MHF Contributor Amer's Avatar
    Joined
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    Jordan
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    Re: Proving 1 isn't in my group

    Quote Originally Posted by beebe View Post
    Let G=\mathbb{Q}-\{1\}
    a*b=a+b-ab \forall a,b \in G

    I need to show my group is closed under *, and I'm happy with my proof that all of the possible outcomes are rational numbers. So now I need to show that 1 is not a possible outcome. Is my proof of this acceptable?

    Assume a+b-ab=1

    a-ab=1-b

    a(1-b)=1-b

    a=\frac{1-b}{1-b}

    a=1

    1 \notin G

    Therefore, a+b-ab \neq 1
    it is correct but I think you just need to modified it like this want to prove that for all a,b\;\in \mathbb{Q} - \{1\}
    a*b \ne 1
    your proof is by contradiction so we will say
    suppose in the contrary that there exist a,b \in \mathbb{Q} - \{1\} such that a*b = 1

    a+b - ab = 1
    ... in the end we face a contradiction a \in G and a = 1 this gives 1 \in G contradiction
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