Let * be the binary operation on N defined by a * b = H.C.F. of a and b.

The binary operation * on N is defined as:

a * b = H.C.F. of a and b

It is known that:

H.C.F. of $a$ and $b=$ H.C.F. of $b$ and $a \& m n F o r E ; a, b \in \mathbf{N}$.

$\therefore a^{*} b=b^{*} a$

Thus, the operation * is commutative.

For $a, b, c \in \mathbf{N}$, we have:

$\left(a^{*} b\right)^{*} c=(\text { H.C.F. of } a \text { and } b)^{*} c=$ H.C.F. of $a, b$, and $c$

$a^{*}\left(b^{*} c\right)=a^{*}($ H.C.F. of $b$ and $c)=$ H.C.F. of $a, b$, and $c$

Thus, the operation * is associative.

Now, an element $e \in \mathbf{N}$ will be the identity for the operation ${ }^{*}$ if $a^{*} e=a=e^{*} a \forall a \in \mathbf{N}$.

But this relation is not true for any $a \in \mathbf{N}$.

Thus, the operation * does not have any identity in $\mathbf{N}$.