Question
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $a * b = ab^2$ for all $a, b \in Q.$
'*' on Q defined by $a * b = ab^2$ for all $a, b \in Q.$
✓
Answer
Commutativity: Let $\text{a, b}\in\text{Q}.$ Then,$a * b = ab^2$
$b * a = ba^2$
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (bc^2)$
$= a(bc^2)^2$
$= ab^2c^4$
$(a * b) * c = (ab^2) * c$
$= ab^2c^2$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
$b * a = ba^2$
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (bc^2)$
$= a(bc^2)^2$
$= ab^2c^4$
$(a * b) * c = (ab^2) * c$
$= ab^2c^2$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
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