Let
*
be the binary operation on **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 **N**

(v) Which
elements of **N **are
invertible for the operation *?

The binary
operation *
on **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* &mnForE;
*a, b* ∈ **N**.

∴*a* *
*b* = *b ***
a*

Thus, the operation * is commutative.

(iii) For
*a, b*, *c *∈ **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* ∈ **N**

⇒ *a* *
1 = *a* = 1 *
*a* &mnForE; *a* ∈
**N**

Thus, 1 is the identity of *
in **N**.

(v) An element *a* in **N** is invertible with respect to the
operation *
if there exists an element *b* in **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 *.

**
**