Q. 165.0( 4 Votes )

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.[CBSE 2015, 2016]

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 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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