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  = (e_1, e_2)` will be an identity element for the operation * if

`a"*"e  = a = e"*"a "∀" a = (a_1,a_2) in A`   

, i.e.,`(a_1 + e_1, a_2 + e_2) = (a_1, a_2) = (e_1 + a_1, e_2 + a_2)` which is not true for any element in A.

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

which is not true for any element in A.

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