Let ${ }^{*}$ be the binary operation on $\mathbf{N}$ given by $a^{*} b=$ L.C.M. of $a$ and $b$. Find
(i) $5^{*} 7,20^{*} 16$ (ii) Is * commutative?
(iii) Is * associative? (iv) Find the identity of * in $\mathbf{N}$
(v) Which elements of $\mathbf{N}$ are invertible for the operation ${ }^{*}$ ?
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 *.
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