Answer :

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

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 = (a^{2} + b^{2} – 2ab – c)^{2} ...... (1)

⇒ a*(b*c) = a*((b – c)^{2})

⇒

⇒ a*(b*c) = (a^{2} – b^{2} – c^{2} + 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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