Question

Let * be a binary operation on Q − {−1} defined by
a * b = a + b + ab for all a, b ∈ Q − {−1}
Then,
(i) Show that '*' is both commutative and associative on Q − {−1}.
(ii) Find the identity element in Q − {−1}
(iii) Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.

Answer 1

(i) Commutativity:
$$\begin{gathered} \text{Let}a,b\in Q-\left\{-1\right\}.\mathrm{Then}, \\ a*b=a+b+ab \\ =b+a+ba \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in Q-\left\{-1\right\} \end{gathered}$$
Thus, * is commutative on Q $-${$-$1}.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in Q-\left\{-1\right\}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+c+bc\right) \\ =a+b+c+bc+a\left(b+c+bc\right) \\ =a+b+c+bc+ab+ac+abc \\ \left(a*b\right)*c=\left(a+b+ab\right)*c \\ =a+b+ab+c+\left(a+b+ab\right)c \\ =a+b+c+ab+ac+bc+abc \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in Q-\left\{-1\right\}. \end{gathered}$$
Thus, * is associative on Q $-$ {$-$1}.

(ii) Let e be the identity element in Q$-$ {$-$1} with respect to * such that
$$\begin{gathered} a*e=a=e*a,\forall a\in Q-\left\{-1\right\} \\ \Rightarrow a*e=a\text{and}e*a=a,\forall a\in Q-\left\{-1\right\} \\ \Rightarrow a+e+ae=a\text{and}e+a+ea=a,\forall a\in Q-\left\{-1\right\} \\ \Rightarrow e\left(1+a\right)=0,\forall a\in Q-\left\{-1\right\} \\ \Rightarrow e=0,\forall a\in Q-\left\{-1\right\}\left[\text{∵}\mathit{\text{a}}\text{≠-1}\right] \end{gathered}$$
Thus, 0 is the identity element in Q $-$ {$-$1} with respect to *.

$$\begin{gathered} \left(\mathrm{iii}\right)\text{Let}a\in Q-\left\{-1\right\}\text{and}b\in Q-\left\{-1\right\}\text{be the inverse of}a.\mathrm{Then}, \\ a*b=e=b*a \\ \Rightarrow a*b=e\text{and}b*a=e \\ \Rightarrow a+b+ab=0\text{and}b+a+ba=0 \\ \Rightarrow b\left(1+a\right)=-a\in Q-\left\{-1\right\} \\ \Rightarrow b=\frac{-a}{1+a}\in Q-\left\{-1\right\}\left[\text{∵}a\ne -1\right] \\ \text{Thus,}\frac{-a}{1+a}\text{is the inverse of}a\in Q-\left\{-1\right\}. \end{gathered}$$