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Math Help - proof by contradiction

  1. #1
    Banned
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    proof by contradiction

    given a set with the symbols :
    " + " for addition
    " - " for inverse
    the constants :
    0
    AND the axioms:
    for all a,b,c : a+(b+c) = (a+b) + c
    for all a : a+0 = a
    for all a : a +(-a) =0
    for all a,b : a+b = b+a
    PROVE using CONTRADICTION
    1) THE uniqness of zero
    2) THE uniqness of the inverse
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  2. #2
    Super Member malaygoel's Avatar
    Joined
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    India
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    Quote Originally Posted by alexandros View Post
    given a set with the symbols :
    " + " for addition
    " - " for inverse
    the constants :
    0
    AND the axioms:
    for all a,b,c : a+(b+c) = (a+b) + c
    for all a : a+0 = a
    for all a : a +(-a) =0
    for all a,b : a+b = b+a
    PROVE using CONTRADICTION
    1) THE uniqness of zero
    2) THE uniqness of the inverse
    1) Let 0^, be a number other than zero such that for all a,

    a+0^,=a

    then, since zero is number

    0+0^,=0

    also by our fourth axiom,  0+0^,=0^,+0 which is 0^, by our second axiom.

    which implies 0^, is same as 0....hence, zero is unique.
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  3. #3
    Junior Member
    Joined
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    2) The inverse of a is -a. Assume that there exists another inverse element denoted as a^{-1} such that a^{-1}\neq-a

    Now, by the definition of inverse element, we have:

    a+a^{-1}=0

    But from the third axiom, we have:

    a+(-a)=0

    Ergo:

    a+a^{-1}=a+(-a)

    a^{-1}=-a

    Which contradicts our original assumption a^{-1}\neq-a, so the inverse element must be unique.
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