Let $^*$ be the binary operation on $N$ given by $a \, ^* \, b = \text{L.C.M. of } a \text{ and } b$. Is $^*$ commutative?
(A) The binary operation $^*$ is defined on the set of natural numbers $N$ as $a \, ^* \, b = \text{L.C.M. of } a \text{ and } b$.
For any $a, b \in N$,we know that the least common multiple of $a$ and $b$ is the same as the least common multiple of $b$ and $a$.
Therefore,$a \, ^* \, b = \text{L.C.M. of } a \text{ and } b = \text{L.C.M. of } b \text{ and } a = b \, ^* \, a$.
Since $a \, ^* \, b = b \, ^* \, a$ for all $a, b \in N$,the operation $^*$ is commutative.
For any $a, b \in N$,we know that the least common multiple of $a$ and $b$ is the same as the least common multiple of $b$ and $a$.
Therefore,$a \, ^* \, b = \text{L.C.M. of } a \text{ and } b = \text{L.C.M. of } b \text{ and } a = b \, ^* \, a$.
Since $a \, ^* \, b = b \, ^* \, a$ for all $a, b \in N$,the operation $^*$ is commutative.
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