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Math Help - Inverse

  1. #1
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    Exclamation Inverse

    Define an operation * on Z by

    a*b = ab - a - b +2 for a, b in Z. We know that * is commutative and associative and that * has an identity element, namely 2. Determine which elements in Z have inverses relative to *. Justify your answer.


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  2. #2
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    Quote Originally Posted by suedenation
    Define an operation * on Z by

    a*b = ab - a - b +2 for a, b in Z. We know that * is commutative and associative and that * has an identity element, namely 2. Determine which elements in Z have inverses relative to *. Justify your answer.


    Thanks very much
    Given x\in \mathbb{Z} we attempt to find a y\in \mathbb{Z} such as,
    x*y=2 because 2 is the identity element.
    Thus,
    xy-x-y+2=2 by definition on *.
    Thus,
    xy-x=y or more elegantly xy=x+y
    Thus,
    x(y-1)=y
    Now we examine 2 cases:
    Case 1 y=1: Thus, 0=1. Impossible.
    Case 2 y\not =1: Thus, y-1\not =0. Thus if y=0 then x=0 Checking this we find that 0(1-0)=0 which is true. Thus, x=0 has an inverse. Now if y\not = 0 also then,
    x(y-1)=y with \gcd(y-1,y)=1 because they are right next to each other. Thus, the left hand side must be divisible by y, thus, y|x this is Euclid's Lemma. Thus, x=ky. Thus, ky(y-1)=y now since y\not =0 thus, k(y-1)=1 happens only when y-1=1 \mbox{ and } k=1. Thus,
    y=2 \mbox{ and } x=(1)(2)=2 Thus, x=2 is the other inverse. Thus, 0,2 are the only possible inverses.
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  3. #3
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    It may help to note that (a+1)*(b+1) = (a+1)(b+1)-(a+1)-(b+1)+2 = ab+1. Thus the map (Z,*) -> (Z,.) given by a -> a+1 is an isomorphism of semigroups.
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