Consider a binary operation on

(i) For a binary operation $*$, e identity element exists if $a * e=e^{*} a=a .$ As $a * b=a+b-a b$

$a * e=a+e-a e(1)$

$e^{*} a=e+a-e a(2)$

using $a^{*} e=a$

$a+e-a e=a$

$e-a e=0$

$e(1-a)=0$

either $e=0$ or $a=1$ as operation is on $Q$ excluding 1 so $a \neq 1$, hence $e=0$.

So identity element $e=0$.

(ii) for a binary operation $*$ if e is identity element then it is invertible with respect to $*$ if for an element $b, a^{*} b=e=b^{*} a$ where $b$ is called inverse of $*$ and denoted by $a^{-1}$.

$a * b=0$

$a+b-a b=0$

$b(1-a)=-a$

$\mathrm{b}=\frac{-\mathrm{a}}{(1-\mathrm{a})} \Rightarrow \frac{\mathrm{a}}{(\mathrm{a}-1)}$

$a^{-1}=\frac{a}{(a-1)}$