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Q. 11 Let $A = N \times N$ and ∗ be the binary operation on A defined by
$(a, b) * (c, d) = (a + c, b + d)$

Show that ∗ is commutative and associative. Find the identity element for ∗ on
A, if any.

Answers (1)

best_answer

$A = N \times N$ and ∗ be the binary operation on A defined by

$(a, b) * (c, d) = (a + c, b + d)$

Let $(a,b),(c,d) \in A$

Then, $a,b,c,d \in N$

We have

$(a, b) * (c, d) = (a + c, b + d)$

$(c,d)*(a,b) = (c+a,d+b)= (a+c,b+d)$

$\therefore \, \, \, \, \, (a,b)*(c,d)=(c,d)*(a,b)$

Thus it is commutative.

Let $(a,b),(c,d),(e,f) \in A$

Then, $a,b,c,d,e,f \in N$

$[(a, b) * (c, d)]*(e,f)= [(a + c, b + d)]*(e,f) =[(a+c+e),(b+d+f)]$

$(a, b) * [(c, d)*(e,f)]= (a,b)*[(c + e, d + f)] =[(a+c+e),(b+d+f)]$

$\therefore \, \, \, \, \, [(a,b)*(c,d)]*(e,f)=(a, b) * [(c, d)*(e,f)]$

Thus, it is associative.

Let $e= (e1,e2) \in A$ will be a element for operation * if $(a*e)=a=(e*a)$ for all $a= (a1,a2) \in A$ .

i.e. $(a1+e1,a2+e2)= (a1,a2)= (e1+a1,e2+a2)$

This is not possible for any element in A .

Hence, it does not have any identity.

Posted by

seema garhwal

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