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*Useful Facts:-*

See the definitions from **Q7. ,Q11.** and **Q13.**

*@Ans:-*

$(i)$ $5 \ast 7 = L.C.M (5,7) = 35$ and $20 \ast 16 = L.C.M(20,16) = 80$.

$(ii)$ Yes, $\ast$ is **commutative**. As we know that $L.C.M(a,b) = L.C.M(b,a)$ for any two natural number in $N$.

$\implies \, a \ast b = b \ast a \, \forall a,b \in N$.

$(iii)$ Yes, $\ast$ is **associative**. As we know that $\,\,\,\,L.C.M(a, L.C.M(b, c)) = L.C.M(L.C.M(a,b), c)$

$\implies a \ast (b \ast c) = (a \ast b) \ast c \, \forall a,b,c \in N$.

$(iv)$ The element $e = 1$ in $N$ is the **identity element**, because $a \ast 1 = 1 \ast a = a, \, \forall a \in N$.(as we know that $L.C.M(a,1) = a$).

$(v)$ No element other then $identity \, element \, 1$ is invertible.

Proof: If possible let $a' \in N$ be the inverse of $a \in N$. Then $a \ast a' = a' \ast a = 1\, \implies \, L.C.M(a,a') = 1$ which is not possible for any element other then $a = a' = 1$. So, $a \neq 1 $ in $N$ is not invertible.

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- $a*b$ is commutative if $a_{ij}=a_{ji}$ for all $i,j \in \{1,2,3,4,5\}$ or $a*b=b*a \; \forall a,b \in N$
- * operation is associative if $(a*b)*c=a*(b \times c). \qquad a,b,c \in \{1,2,3,4,5\}$
- An element $a \in N$ is invertible if then exists an element b in N such that $a*b=e$ where e is identify element $[e \in N]$

Step 1.consider two elements 5,7

$5*7=LCM \;of\; 5\; and\; 7=35$

$ 20 * 16=LCM\;of\;20*16=80$

Step2. To check if * is commutative

LCM of 5 and 7=LCM of 7 and 5

This is true for all $a,b \in N$

LCM of a and b =LCM of b and a for $a,b \in N$

Hence * is commutative

Step3. To check if * operation is associative

For $a,b,c \in N$ we have

$(a*b)*c=(LCM \;of\; a\; and\; b)*c$

=LCM of a and b and c

$a*(b*c)=a*$ (LCM of b and c)

=LCM of a,b,c

Therefore $ (a*b)*c=a*(b*c)$

$*$ operation is associative

Step 4. To find the identity element

We see that If e is the identity element

$a*e=a=e*a$. This is possible only if e=1 ,hence identity element is 1

Step 5. To find Invertible element

a will be an invertible element if we can find another element b such that a*b=1

=> LCM of a and b should be 1.Only element possible is 1.

Hence 1 is the only element which is invertible with respect to * operation defined by$a*b=LCM $ of a and b

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