Question
Check the commutativity and associativity of the following binary operations:
'*' on N defined by $a * b = 2^{ab}$​​​​​​​ for all $a, b \in N.$

Answer

Commutative: Let $\text{a, b}\in\text{N},$ Then $a * b = 2^{ab} = 2^{ba} = b * a$
 $\therefore$ a * b = b * a $\therefore$ * is commutative on $N.$
Associative: Let $\text{a, b, c}\in\text{N},$ Then$(\text{a}\ ^*\ \text{b}) *\ \text{c}=2^{\text{ab}}\ ^*\ \text{c}=2^{2{\text{ab}}.\text{c}}\ ...(\text{i})$
and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ 2^{\text{bc}}=2^{\text{a}.2^{\text{bc}}}\ ....(\text{ii})$ From (i) and (ii), we get$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
$\therefore$ * is not associative on $N.$

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