# Let an operation $\ast$ on the set of natural numbers N be defined by $a\ast b= a^{b}\cdot$ Find(i) wheteher $\ast$ is a binary or not, and (ii) if it is a binary, then is it commutative or not.

(i) As $a^{b}\, \epsilon N$ for all $a,b\: \epsilon N$
ie    $a\ast b\: \epsilon N$  $a,b\: \epsilon N$
so $\ast$ is binary
(ii) As $1^{2}\neq 2^{1}\; so\; 1\ast 2\neq 2\ast 1$    therefore $\ast$ is not commutative

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