Question
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by a * b = (a - b)2 for all a, b ∈ Q.
'*' on Q defined by a * b = (a - b)2 for all a, b ∈ Q.
a * b = (a - b)2
= (b - a)2
= b * a
Therefore,
a * b = b * a,
$\forall\ \text{a, b}\in\text{Q}$Thus, * is commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$
Then,a * (b * c) = a * (b - c)2
= a * (b2 + c2 - 2bc)
= (a - b2 - c2 + 2bc)2
(a * b) * c = (a - b)2 * c
= (a2 + b2 - 2ab) * c
= (a2 + b2 - 2ab - c)2
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
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$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$