# On the set $R -\{ -1\}$ a binary operation $\ast$ is defined by $a \ast b = a + b + ab$ for all $a,b \in R - \{-1\}$. Prove that $\ast$ is commutative as well as associative on $R -\{-1\}$. Find the identity element on $\ast$ if any.

$(A)\; 0$
$(B)\; 1$
$(C)\; 2$
$(D)\; 3$

(i)Given: $a \ast b = a + b+ ab -----(1)$
and $b \ast a = b + a + ab ---- (2)$
$(1) = (2)$
$\implies a \ast b = b \ast a$
Hence it is commutative on $R- \{-1\}$
\begin{align*}(ii)(a \ast b) \ast c & = (a + b + ab) \ast c \\ & = (a + b + ab) + c + (a + b + ab) c \\ & = a+ b + c + ab + bc + ac + abc ---(3)\end{align*}
\begin{align*}a \ast (b \ast c ) & = a + (b + c + ab) + a (b + c + bc ) \\ & = a + b + c + ab + bc + ac + abc ----(4) \end{align*}
Hence (3) = (4)
so$(a \ast b ) \ast c = a \ast (b \ast c)$
So it is associative on R-{-1}
(iii) Let e be the identity element.
$\therefore a \ast e = a = e \ast a$
$\implies a + e + ae = a$ and $e + a + ea = a$
$\implies e (1+a) = 0$
$\implies e = 0$
But $0 \in R-\{-1\}$
Hence 0 is the identity element.