**Question:**

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 ${ }^{*}$ ?

**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 *.*

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.