Q. 35.0( 1 Vote )

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 :

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 + ab
2. 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 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
(Required invertible elements, a≠ – 1, b≠ – 1)


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