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Thread: 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.


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