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(A) On $Q$,the operation $^*$ is defined by $a ^* b = ab + 1$.For commutativity:We know that $ab = ba$ for all $a, b \in Q$.Therefore,$ab + 1 = ba + 1

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$^*$ is commutative but not associative.

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(A) On $Q$,the operation $^*$ is defined by $a ^* b = ab + 1$.For commutativity:We know that $ab = ba$ for all $a, b \in Q$.Therefore,$ab + 1 = ba + 1$ for all $a, b \in Q$.This implies $a ^* b = b ^* a$ for all $a, b \in Q$.Thus,the operation $^*$ is commutative.For associativity:We check if $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Q$.Consider $(1 ^* 2) ^* 3 = (1 	imes 2 + 1) ^* 3 = 3 ^* 3 = 3 	imes 3 + 1 = 10$.Consider $1 ^* (2 ^* 3) = 1 ^* (2 	imes 3 + 1) = 1 ^* 7 = 1 	imes 7 + 1 = 8$.Since $(1 ^* 2) ^* 3 
eq 1 ^* (2 ^* 3)$,the operation $^*$ is not associative.

For each binary operation $^$ defined below,determine whether $^$ is commutative or associative. On $Q$,define $a ^* b = ab + 1$.

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For each binary operation $^*$ defined below,determine whether $^*$ is commutative or associative. On $Q$,define $a ^* b = ab + 1$.