i. We are given with the set R – {– 1}. 

A general binary operation is nothing but an association of any pair of elements a, b from an arbitrary set X to another element of X. This gives rise to a general definition as follows: 

A binary operation * on a set is a function * : A X A → A. We denote * (a, b) as a * b. 

Here the function *: R – {1}X R – {1} → R – {1} is given by a * b = a + b – ab 

For the ‘ * ’ to be commutative, a * b = b * a must be true for all a, b belong to R – {1}. Let’s check. 

1. a * b = a + b – ab2. b * a = b + a – ba = a + b – ab⇒ a * b = b * a (as shown by 1 and 2) 

Hence ‘ * ’ is commutative on R – {1} 

For the ‘ * ’ to be associative, a * (b * c) = (a * b) * c must hold for every a, b, c∈ R – {1}. 

3. a * (b * c) = a * (b + c – bc) 

= a + (b + c – bc) – a(b + c + bc) 

= a + b + c – ab – bc – ac + abc 

4. (a * b) * c = (a + b – ab) * c 

= a + b – ab + c – (a + b – ab)c 

= a + b + c – ab – bc – ac + abc⇒ 3. = 4. 

Hence ‘ * ’ is associative on R – {1} 

ii. Identity Element: Given a binary operation*: A X A → A, an element e ∈A, if it exists, is called an identity of the operation*, if a*e = a = e*a ∀ a ∈A.

Let e be the identity element of R – {1} and a be an element of R – {1}. 

Therefore, a * e = a⇒ a + e – ae = a⇒ e + ea = 0⇒ e(1 – a) = 0⇒ e = 0. 

(1 – a≠0 as the a cannot be equal to 1 as the operation is valid in R – {1}) 

iii. Given a binary operation * A X A → A with the identity element e in A, an element a ∈ A is said to be invertible with respect to the operation, if there exists an element b in A such that a * b = e = b * a and b is called the inverse of a and is denoted by a ^–1. 

Let us proceed with the solution. 

Let b R – {1} be the invertible element/s in R – {1} of a, here a ∈ R – {1}. 

∴a * b = e (We know the identity element from previous)

⇒ a + b – ab = 0⇒ b – ab = – a⇒ b(1 – a) = – a

⇒ b = -a/(1 -a) (Here a ≠ 1, b ≠ 1)