Let ^* be a binary operation on the set Q of | vedclass

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(B) On $Q$,the operation $^*$ is defined as $a \,^* \,b = a^{2} + b^{2}$.For $a, b \in Q$,we have:$a \,^* \,b = a^{2} + b^{2} = b^{2} + a^{2} = b \,^*

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The operation $^*$ is commutative but not associative.

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(B) On $Q$,the operation $^*$ is defined as $a \,^* \,b = a^{2} + b^{2}$.For $a, b \in Q$,we have:$a \,^* \,b = a^{2} + b^{2} = b^{2} + a^{2} = b \,^* \,a$.Therefore,$a \,^* \,b = b \,^* \,a$,which means the operation $^*$ is commutative.Now,check for associativity:$(1 \,^* \,2) \,^* \,3 = (1^{2} + 2^{2}) \,^* \,3 = (1 + 4) \,^* \,3 = 5 \,^* \,3 = 5^{2} + 3^{2} = 25 + 9 = 34$.$1 \,^* \,(2 \,^* \,3) = 1 \,^* \,(2^{2} + 3^{2}) = 1 \,^* \,(4 + 9) = 1 \,^* \,13 = 1^{2} + 13^{2} = 1 + 169 = 170$.Since $(1 \,^* \,2) \,^* \,3 
eq 1 \,^* \,(2 \,^* \,3)$,the operation $^*$ is not associative.Thus,the operation $^*$ is commutative but not associative.

Let $^$ be a binary operation on the set $Q$ of rational numbers defined as $a \,^ \,b = a^{2} + b^{2}$. Which of the following is true?

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Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a \,^* \,b = a^{2} + b^{2}$. Which of the following is true?