Question

Let $\ast$ be the binary operation on N given by $a\ast b=LCM$ of a and b.
(i) Find $5\ast 7,20\ast 16$

(ii)Is $\ast$ commutative?

(iii)Is $\ast$ associative?

(iv) Find the identity of $\ast$ in N

(v)Which elements of N are invertible for the operation $\ast$ ?

Answer 1

The binary operation $\ast$ on N is defined as $a\ast b=LCM$ of a and b.
We have $5\ast 7=LCM$ of 5 and 7 =35
and $20\ast 16=LCM$ of 20 and 16=80

The binary operation $\ast$ on N is defined as $a\ast b=LCM$ of a and b.
It is known that
LCM of a and b =LCM of b and a for $a,b\in N.$
Therefore, $a\ast b=b\ast a$. Thus, the operation $\ast$ is commutative.

The binary operation $\ast$ on N is defined as $a\ast b=LCM$ of a and b.
For $a,b,c\in N$, we have
$(a\ast b)\ast c$ =(LCM of a and b) $\ast c$ =LCM of a,b, and c

$a\ast (b\ast c)=a\ast$(LCM of b and c) = LCM of a,b and c
$a\ast (b\ast c)=a\ast (b\ast c)$. Thus, the operation $\ast$ is associative.

The binary operation $\ast$ on N is defined as $a\ast b=LCM$ of a and b.
It is known that
LCM of a and 1 =a =LCM of 1 and a, $a\in N.$
$a\ast 1=a=1\ast a,a\in N$
Thus, 1 is the identity of $\ast$ in N.

The binary operation $\ast$ on N is defined as $\ast b=LCM$ of a and b.
An element a in N is invertible with respect to the operation $\ast$ if there exists an element b in N such that $a\ast b=e=b\ast a$.
Here, e=1, This means that
LCM of a and b =1 =LCM 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 $\ast .$