Q. 65.0( 1 Vote )

# On R – {1}, a binary operation * is defined by a * b = a + b – ab. Prove that * is commutative and associative. Find the identity element for * on R – {1}. Also, prove that every element of R – {1} is invertible.

Answer :

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 – ab

2. 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 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

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