Let $^*$ be the binary operation on $N$ defined by $a \,^*\, b = \text{H.C.F. of } a \text{ and } b$. Is $^*$ commutative? Is $^*$ associative? Does there exist an identity for this binary operation on $N$?

(N/A) The binary operation $^*$ on $N$ is defined as: $a \,^*\, b = \text{H.C.F. of } a \text{ and } b$.
$1$. Commutativity:
We know that the $\text{H.C.F.}$ of $a$ and $b$ is the same as the $\text{H.C.F.}$ of $b$ and $a$ for all $a, b \in N$.
Therefore,$a \,^*\, b = b \,^*\, a$.
Thus,the operation $^*$ is commutative.
$2$. Associativity:
For $a, b, c \in N$,we have:
$(a \,^*\, b) \,^*\, c = (\text{H.C.F. of } a \text{ and } b) \,^*\, c = \text{H.C.F. of } a, b, \text{ and } c$.
$a \,^*\, (b \,^*\, c) = a \,^*\, (\text{H.C.F. of } b \text{ and } c) = \text{H.C.F. of } a, b, \text{ and } c$.
Since $(a \,^*\, b) \,^*\, c = a \,^*\, (b \,^*\, c)$,the operation $^*$ is associative.
$3$. Identity Element:
An element $e \in N$ is the identity for $^*$ if $a \,^*\, e = a = e \,^*\, a$ for all $a \in N$.
This implies $\text{H.C.F.}(a, e) = a$,which means $a$ must be a divisor of $e$ for all $a \in N$. Since there is no such fixed element $e \in N$ that is a multiple of every natural number,there is no identity element for this operation.