Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find

Solution:

The binary operation * on $\mathbf{N}$ is defined as $a^{*} b=$ L.C.M. of $a$ and $b$.

(i) $5^{*} 7=$ L.C.M. of 5 and $7=35$

$20 * 16=$ L.C.M of 20 and $16=80$

(ii) It is known that:

L.C.M of $a$ and $b=$ L.C.M of $b$ and $a \& m n$ ForE; $a, b \in \mathbf{N}$.

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

Thus, the operation * is commutative.

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

(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of $a$ and $1=a=$ L.C.M. 1 and $a$ \&mnForE; $a \in \mathbb{N}$

⇒ a * 1 = a = 1 * a &mnForE; a ∈ N

Thus, 1 is the identity of * in $\mathbf{N}$.

(v) An element $a$ in $\mathbf{N}$ is invertible with respect to the operation ${ }^{*}$ if there exists an element $b$ in $\mathbf{N}$, such that $a^{*} b=e=b^{*} a$.

Here, e = 1

This means that:

L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.