# 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,b∈S.Prove that:i. * is a binary operation on Sii. * is commutative as well as associative.

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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