Question

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.

Answer 1

A = N × **N **

• is a binary operation on A and is defined by:
  (a, b) * (c, d) = (a + c, b + d)
  Let (a, b), (c, d) ∈ A
  Then, a, b, c, d ∈ **N **
  We have:

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

[Addition is commutative in the set of natural numbers]
∴ (a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈A
Then, a, b, c, d, e, f ∈ N We have:

((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)
∴ ((a,b) * c,d)) * (e,f)= (a,b) * ((c,d) * (e,f))

Therefore, the operation * is associative.
An element e=(e1,e2)∈A will be an identity element for the operation * if ae=a=ea∀ a=(a1,a2)∈A,i.e,(a1+e1,a2+e2)=(a1,a2)=(e1+a1,e2+a2),which is not true for any element in A.

Therefore, the operation * does not have any identity element.