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 and ∗ be the binary operation on A defined by
(a, b) ∗ (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
now , (a, b) * (c , d) = (a + c, b + d)
(c, d) * (a , b) = (c + a , d + b ) = (a + c , b + d)
here, (a , b) * (c , d) = (c ,d) * (a , b)
therefore * is commutative.

Now, (a, b), (c, d), (e, f) ∈ A
Then, a, b, c, d, e, f ∈ N
Now, 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)
here, {(a, b) * (c , d)} * (e, f) = (a, b) * {(c, d) * (e, f)}
therefore , * is associative.

Now, an element e = (e₁, e₂) ϵ A will be an identity for the operation *
if a * e = a = e * a$\forall$ a = (a₁, a₂) ϵ A,
(a₁ + e₁, a₂ + e₂) = (a₁, a₂) = (e₁ + a₁, e₁ + a₂)
which is not true for any element in A.
Therefore, the operation * does not have any identity element