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Math Help - For all a,b,c ϵ B | a+b = a+c -> b = c

  1. #1
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    For all a,b,c ϵ B | a+b = a+c -> b = c

    Can some tell me if this proof works and if not correct me please.

    a+b = a+c -> b=c
    ~aa+b = ~aa+c -> b=c
    b=c -> b=c

    Proved!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Kitizhi View Post
    Can some tell me if this proof works and if not correct me please.

    a+b = a+c -> b=c
    ~aa+b = ~aa+c -> b=c
    b=c -> b=c

    Proved!
    Your proof is not correct. You need to keep in mind that we are dealing in additive notation, so the inverse must be expressed in additive notation (i.e. a^{-1}\implies -a)

    It should be a+b=a+c\implies (-a)+a+b=(-a)+a+c\implies(-a+a)+b=(-a+a)+c\implies b=c

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Your proof is not correct. You need to keep in mind that we are dealing in additive notation, so the inverse must be expressed in additive notation (i.e. a^{-1}\implies -a)

    It should be a+b=a+c\implies (-a)+a+b=(-a)+a+c\implies(-a+a)+b=(-a+a)+c\implies b=c

    Does this make sense?
    Hmm kinda but i dont know of additive notation.
    The identity law states -a + a = 1
    so shouldnt it be 1+b = 1+c
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  4. #4
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    Hmm so this state is proven to be true right?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kitizhi View Post
    Hmm kinda but i dont know of additive notation.
    The identity law states -a + a = 1
    so shouldnt it be 1+b = 1+c
    what??

    no, -a + a = 0

    you are reading something wrong. please review your text

    Quote Originally Posted by Kitizhi View Post
    Hmm so this state is proven to be true right?
    what Chris did is correct. review it carefully
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  6. #6
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    oops for some reason i kept thinking multiplication.
    sorry my mistake
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