Let A = N ´ N and * be a binary operation on A defined by (a, b) * (c

1.	Let A = N ´ N and * be a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Also, find the identity element for * on A, if any.
Answer» Given A = N x N * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d) (i) Commutativity: Let (a, b), (c, d) ∈ N x N Then (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) (∵ a, b, c, d ∈ N, a + c = c + a and b + d = d + c) = (c, d) * b Hence, (a, b) * (c, d) = (c, d) * (a, b) ∴ * is commutative. (ii) Associativity: let (a, b), (b, c), (c, d) Then [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) = {a + (c + e), b + (d + f)] (∵ set N is associative) = (a, b) * (c + e, d + f) = (a, b) * {(c, d) * (e, f)} Hence, [(a, b) * (c, d)] * (e, f) = (a, b) * {(c, d) * (e, f)} ∴ * is associative. (iii) Let (x, y) be identity element for ∀ on A, Then (a, b) * (x, y) = (a, b) ⇒ (a + x, b + y) = (a, b) ⇒ a + x = a, b + y = b ⇒ x = 0, y = 0 But (0, 0) ∉ A ∴ For *, there is no identity element.

Let A = N ´ N and * be a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Also, find the identity element for * on A, if any.

Answer»

Given A = N x N

* is a binary operation on A defined by

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

(i) Commutativity: Let (a, b), (c, d) ∈ N x N

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

(∵ a, b, c, d ∈ N, a + c = c + a and b + d = d + c)

= (c, d) * b

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

∴ * is commutative.

(ii) Associativity: let (a, b), (b, c), (c, d)

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

= {a + (c + e), b + (d + f)] (∵ set N is associative)

= (a, b) * (c + e, d + f) = (a, b) * {(c, d) * (e, f)}

Hence, [(a, b) * (c, d)] * (e, f) = (a, b) * {(c, d) * (e, f)}

∴ * is associative.

(iii) Let (x, y) be identity element for ∀ on A,

Then (a, b) * (x, y) = (a, b)

⇒ (a + x, b + y) = (a, b)

⇒ a + x = a, b + y = b

⇒ x = 0, y = 0

But (0, 0) ∉ A

∴ For *, there is no identity element.