# Math Help - Inverse

1. ## 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

2. 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.

3. 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.