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Let A = N × 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.

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)
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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.

 

 

 

 

 

 

 

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