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# Check the commuta

Given that * is a binary operation on Q defined by a*b = (a – b)2 for all a,bQ.

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)2

b*a = (b – a)2 = (a – b)2

b*a = a*b

Commutative property holds for given binary operation * on Q.

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)2)*c

(a*b)*c = (a2 + b2 – 2ab – c)2 ...... (1)

a*(b*c) = a*((b – c)2)

a*(b*c) = (a2 – b2 – c2 + 2bc)2 ...... (2)

From (1) and (2) we can clearly say that associativity doesn’t hold for the binary operation ‘*’ on ‘Q’.

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