Answer :

Given that, * be the binary operation defined on Q.


A binary operation ‘*’ is commutative if a*b = b*a a, b Q


(i) a * b = a – b a, b Q


a * b = a – b = -b+a = -(b-a) = -(b*a)


a*b ≠ b*a


Hence, ‘*’ is not commutative on Q.


(ii) a * b = a2 + b2 a, b Q


a * b = a2 + b2


= b2 + a2 [ addition is commutative on Q a+b = b+a ]


= b*a


a*b = b*a


Hence, ‘*’ is commutative on Q.


(iii) a * b = a + ab a, b Q


a * b = a + ab and b*a = b + ab


a + ab ≠ b + ab


a*b ≠ b*a


Hence, ‘*’ is not commutative on Q.


(iv) a * b = (a – b)2 a, b Q


a * b = (a – b)2 = {-(b-a)}2


= (b-a)2


=b*a


a*b = b*a


Hence, ‘*’ is commutative on Q.


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