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Let \(\ast\) be the binary operation on \(N\) given by \(a\ast b=L.C.M\,.of\,a\,and\,b\). Find $\begin{array}{1 1}(i)\;\; 5 \ast7,\; 20 \ast 16 & (ii)\;\; Is\; \ast \;commutative?\\(iii)\;\; Is\; \ast \;associative? & (iv)\;\; Find\, the\,identity \, of \ast \, in N\\(v)\;\; Which \, elements\, of \, N\, are\, invertible\, for\, the\, operaation\,\ast ? & \;\end{array}$


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Useful Facts:-

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


$(i)$ $5 \ast 7 = L.C.M (5,7) = 35$ 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.

answered Dec 2, 2012 by sukhendu.math
edited Dec 3, 2012 by sukhendu.math
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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



answered Feb 27, 2013 by meena.p
edited Mar 20, 2013 by thagee.vedartham

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