Show that the binary operation on

Show that the binary operation $\ast$ on $A= \mathbb{R}-\left \{ -1 \right \}$ defined as $a\ast b= a+b+ab$ for all $a,b\, \epsilon\, A$ is commutative and associative on A, Also find the identity element of $\ast$ in A and prove that every element of A is invertible.

Answers (1)

$a,b\, \varepsilon\, R\Rightarrow a+b\, \varepsilon \, R\, and\, ab\, \varepsilon \, R$
$\Rightarrow a+b+ab\, \varepsilon \, R$
Hence $\ast$ is a binary operation on R
commutativity for all $a,b\, \varepsilon \, R$
$a\ast b= a+b+ab$
          $= a+b-ba$ [community of addition and multiplication]
         $= b\ast a$
$a\ast b= b\ast a$ is a commutative on R
Associativity: $\left ( a\ast b \right )\ast c= \left ( a+b+ab \right )\ast c$
                               $= a+b+ab+c+\left ( a+b+ab \right )c$
                                $= a+b+ab+c+ac+bc+abc$
Similiraly you can solve for $a\ast \left ( b\ast c \right )$ we get:
$\left ( a\ast b \right )\ast c= a\ast \left ( b\ast c \right )$ $\ast$ is a associative on R Identity element: Let e be the identity element doenot of R and b be the innverse of a
$a\ast b= 0= b\ast a$
$a\ast b= 0$
$a+b+ab= 0$
$b\left ( 1+a \right )= -a\Rightarrow b= \frac{-a}{1+a}$
Innverse exist only if $a\, \varepsilon \, R-\left \{ -1 \right \}$
so, every element of R is innvertible expt -1 every element $a\left ( \neq -1 \right )\, \varepsilon \, R$ has its innverse $\frac{-a}{1+a}$