Question

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 *?

Answer 1

The given binary operation

∗ is defined as a∗b=L.C.M.( a,b ).

(i)

Apply the given binary operation on 5∗7.

Consider a∗b=5∗7.

a∗b=5∗7 =L.C.M.( 5,7 ) =35

Similarly, apply the given binary operation on 20∗16.

Consider a∗b=20∗16.

a∗b=20∗16 =L.C.M.( 20,16 ) =80

Thus, the value of 5∗7=35 and 20∗16=80.

(ii)

Consider any two random values of a and b. Let a=5 and b=7.

Apply the given binary operation on 5∗7.

a∗b=5∗7 =L.C.M.( 5,7 ) =35

Apply the given binary operation on 7∗5.

a∗b=7∗5 =L.C.M.( 7,5 ) =35

From the above calculation, it can be observed that L.C.M. of a and b=L.C.M. of b and a. So, it satisfies the commutative property.

Thus, the operation ∗ is commutative.

(iii)

Apply binary operation by taking three variables ( a∗b )∗c.

( a∗b )∗c=( L.C.M.( a,b ) )∗c =L.C.M.( a,b,c )

Apply binary operation by taking three variables a∗( b∗c ).

a∗( b∗c )=a∗( L.C.M.( b,c ) ) =L.C.M.( a,b,c )

From the above calculation, it can be observed that ( a∗b )∗c is equal to a∗( b∗c ). So, it satisfies associative property.

Thus, the operation ∗ is associative.

(iv)

Consider the value of b as 1for the identity function.

Determine the value of the binary operation.

a∗1=L.C.M.( a,1 ) =a 1∗a=L.C.M.( 1,a ) =a

From the above calculation, it is observed that a∗1=1∗a.

Thus, 1 is the identity of ∗ in N.

(v)

An element a in the domain N is said to be invertible with respect to the operation ∗, if there exist other element b in the same domain, such that a∗b=1 and b∗a=1.

So, the equation for the least common factor to be invertible is,

L.C.M.( a,b )=1 L.C.M.( b,a )=1

The above condition is only possible when a and b are equal to 1.

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