xy+x+y=0 What is the inverse Element?

Question

I have a task to show that $G:= \mathbb Q\setminus \{-1\}$ is a group to some binary operation *

(x is the "times" symbol) with $A*B:=A\times B + A + B$

I could show that it's associative, that the neutral Element is $0$, but I'm stuck finding the inverse Element...

Maybe I started in the wrong place and don't see the wood for the trees, but I just found a Element for $1$ with $A=1$ and $B=-1/2$

so that $A*B=[1 \times (-1/2)]+1 - 1/2 = 0$

but because of the one you can't derive a general rule from that

but is there a general inverse Element??

Answer 1

If the neutral element is $0$ and you want to find the inverse of $a \in G$, that means you want to find $B$ such that $0 = a*b$. This implies:

$0 = a*_{\scriptsize G} b = a\times_{\scriptsize \mathbb Q} b+_{\scriptsize \mathbb Q}a+_{\scriptsize \mathbb Q}b = (a+_{\scriptsize \mathbb Q}1)\times (b+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1$

$\iff 1 = (a+_{\scriptsize \mathbb Q}1)\times_{\scriptsize \mathbb Q}(b+_{\scriptsize\mathbb Q}1)$

$\iff 1/_{\scriptsize \mathbb Q}(a+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1 = b$

$ \qquad \qquad \qquad \qquad \qquad \:\:= -_{\scriptsize \mathbb Q}a/_{\scriptsize \mathbb Q}(a+_{\scriptsize\mathbb Q}1)$

Answer 2

Hint: What happens when you expand $(a+1)(b+1)$?