Q. 253.7( 3 Votes )

Let S be the set of all rational numbers except 1 and * be defined on S by a*b = a + b – ab, for all a,bS.
Prove that:

i. * is a binary operation on S

ii. * is commutative as well as associative.

Answer :

Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except 1 i.e., R – {1} defined as a*b = a + b – ab


Let us assume a + b – ab = 1


a – ab + b – 1 = 0


a(1 – b) – 1(1 – b) = 0


(1 – a)(1 – b) = 0


a = 1 or b = 1


But according to the problem, it is given that a≠1 and b≠1 so,


a + b + ab≠1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’.


We know that commutative property is p*q = q*p, where * is a binary operation.


Let’s check the commutativity of given binary operation:


a*b = a + b – ab


b*a = b + a – ba = a + b – ab


b*a = a*b


Commutative property holds for given binary operation * on S.


We know that associative property is (p*q)*r = p*(q*r)


Let’s check the associativity of given binary operation:


(a*b)*c = (a + b – ab)*c


(a*b)*c = a + b – ab + c – ((a + b – ab)×c)


(a*b)*c = a + b + c – ab – ac – bc + abc ...... (1)


a*(b*c) = a*(b + c – bc)


a*(b*c) = a + b + c – bc – (a×(b + c + bc))


a*(b*c) = a + b + c – ab – bc – ac + abc ...... (2)


From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘S’.

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