Is $^*$ defined on the set $\{1, 2, 3, 4, 5\}$ by $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$ a binary operation? Justify your answer.

(D) binary operation $^*$ on a set $A$ is a function $^*: A \times A \to A$. This means that for all $a, b \in A$,the result $a \,^*\, b$ must also be an element of $A$.
Let $A = \{1, 2, 3, 4, 5\}$. The operation is defined as $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$.
For the operation to be a binary operation,$a \,^*\, b$ must belong to $A$ for all $a, b \in A$.
Let us test this with some elements from the set $A$:
Consider $a = 2$ and $b = 3$. Both $2, 3 \in A$.
$2 \,^*\, 3 = \text{L.C.M. of } 2 \text{ and } 3 = 6$.
Since $6 \notin \{1, 2, 3, 4, 5\}$,the result of the operation is not in the set $A$.
Similarly,$2 \,^*\, 5 = 10 \notin A$,$3 \,^*\, 4 = 12 \notin A$,etc.
Since there exist elements $a, b \in A$ such that $a \,^*\, b \notin A$,the operation $^*$ is not a binary operation on the set $\{1, 2, 3, 4, 5\}$.