We consider $(\mathbb R\setminus \{-1\},\star)$, where \begin{equation}\label{2.1.1}a\star b:= ab+a+b,\quad a,b\in\mathbb R\setminus \{-1\}\end{equation} a. Show that $(\mathbb R\setminus \{-1\},\star)$ is an Abelian group. b. Solve $$3\star x \star x = 15$$ in the Abelian group $(\mathbb R\setminus \{-1\},\star)$, where $\star$ is defined in \eqref{2.1.1}.

Solution:

a. We check conditions in Definition 2.7 and the commutativity. Note that \begin{equation}\label{2.1.2}a\star b=(a+1)(b+1)-1.\end{equation} (1) Closure of $\mathbb R\setminus \{-1\}$ under $\star$: if $a,b\in\mathbb R\setminus \{-1\}$, then $(a+1)(b+1)\ne 0$. Therefore $ab+a+b\ne -1$, which shows $a\star b\in \mathbb R\setminus \{-1\}$.

(2) Associativity: if $a,b,c\in\mathbb R\setminus \{-1\}$, then it follows from \eqref{2.1.2} that \begin{align*}(a\star b)\star c = &\ ((a+1)(b+1)-1)\star c\\ =& \ (a+1)(b+1)(c+1)-1\end{align*}and \begin{align*}a\star(b\star c)=&\ a\star((b+1)(c+1)-1)\\ = &\ (a+1)(b+1)(c+1)-1.\end{align*}Thus $(a\star b)\star c=a\star(b\star c)$.

(3) Neural element: if $a\in\mathbb R\setminus \{-1\}$, then $$a\star 0= a\cdot 0+a+0=a$$and $$0\star a=0\cdot a +0+a=a.$$ (4) Inverse element: if $a\in\mathbb R\setminus \{-1\}$, then $a+1\ne 0$ and we have $a^{-1}=\dfrac{1}{a+1}-1$. This can be checked directly using \eqref{2.1.2}.

(5) Commutativity: if $a,b\in\mathbb R\setminus \{-1\}$, it is clear from \eqref{2.1.2} that \begin{align*}a\star b= &\ (a+1)(b+1)-1\\= &\ (b+1)(a+1)-1=b\star a.\end{align*} b. Recall the computation from part a.(2), we have $$3\star x \star x = (3+1)(x+1)(x+1)-1.$$Hence we have to it reduces to solve $$4(x+1)^2=16,$$which gives $x+1=\pm 2$. We have $x=1$ or $x=-3$.