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. We are given with the set Q – {– 1}. 

A general binary operation is nothing but 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 *: Q – {– 1}X Q – {– 1} → Q – {– 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 Q – {– 1}. Let’s check. 

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

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

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

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

(1 + a ≠ 0 as the a is not equal to 1 as given in the question) 

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 ∈ Q – {– 1} be the invertible element/s in Q – {– 1} of a, here a ∈ Q – {– 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) (Required invertible elements, a≠ – 1, b≠ – 1)