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

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(B) On the set $Q$,the operation $^*$ is defined by $a * b = (a - b)^2$. For commutativity,we check if $a * b = b * a$ for all $a, b \in Q$: $a * b =

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Commutative but not Associative

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(B) On the set $Q$,the operation $^*$ is defined by $a * b = (a - b)^2$. For commutativity,we check if $a * b = b * a$ for all $a, b \in Q$: $a * b = (a - b)^2$ $b * a = (b - a)^2 = [-(a - b)]^2 = (a - b)^2$ Since $a * b = b * a$,the operation $^*$ is commutative. For associativity,we check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in Q$: Consider $a = 1, b = 2, c = 3$: $(1 * 2) * 3 = (1 - 2)^2 * 3 = (-1)^2 * 3 = 1 * 3 = (1 - 3)^2 = (-2)^2 = 4$ $1 * (2 * 3) = 1 * (2 - 3)^2 = 1 * (-1)^2 = 1 * 1 = (1 - 1)^2 = 0$ Since $(1 * 2) * 3 
eq 1 * (2 * 3)$,the operation $^*$ is not associative.

$^*$	$1$	$2$	$3$	$4$	$5$
$1$	$1$	$1$	$1$	$1$	$1$
$2$	$1$	$2$	$2$	$2$	$2$
$3$	$1$	$2$	$3$	$3$	$3$
$4$	$1$	$2$	$3$	$4$	$4$
$5$	$1$	$2$	$3$	$4$	$5$

Let $^$ be a binary operation on the set $Q$ of rational numbers defined as $a b = (a - b)^2$. Determine whether the operation is commutative and associative.

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Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a * b = (a - b)^2$. Determine whether the operation is commutative and associative.