Short Answer Type
Long Answer TypeLet A = N x 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.
A = N x N and (a, b) * (c, d) = (a + c, b + d)
(i) Let (a, b), (c, d), (e, f) be any three elements of A.
∴ (a, b)* {(c, d) * (a, f)} = (a, b) * (a + e, d + f)
= (a + (c + e), b + (d + f))
= ((a + c) + e, (b + d) + f)
= (a + c, b + d) * (e, f)
∴ (a, b) * {(c. d) * (e,f)} = {(a, b) * (c, d) } * (e, f) ∀ (a, b), (c, d), (e, f) ∈ A ∴ (A, *) is associative.
(ii) Let (a, b), (c, d) be any two elements of A.
∴ (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b)
∴ (a, b) * (c, d) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈ A ∴ (A, *) is commutative
(iii) if possible , suppose that (x, y) the is identity element in A.
∴ (a, b) * (x + y) = (a, b) ∀ (a, b), ∈ A
⇒ (a + x, b + y) = (a, b) ∀ (a, b)∈ A ⇒ a + x = a, b + y = b ∀ a, b ∈ N ⇒ x = 0, y = 0 ∀ a, b ∈ N This is impossible as 0 ∉ N ∴ our supposition is wrong.
∴ (A, *) does not have any identity element.
Short Answer TypeConsider the binary operation A on the set {1, 2, 3, 4, 5} defined by a ∧ b = min. {a, b}. Write the operation table of the operation ∧.
Let A = N x N and let ‘*’ be a binary operation on A defined by (a, b) * (c, d) = (a c, b d). Show that
(i) (A, *) is associative (ii) (A, *) is commutative.
Long Answer TypeLet A = N x N and let ‘*’ be a binary operation on A defined by
(a, b) * (c, d) = (ad + bc, bd). Show that
(i) (A, *) is associative, (ii) (A, *) has no identity element,
(iii) Is (A, *) commutative ?
Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
|
∧ |
1 |
2 |
3 |
4 |
5 |
|
1 |
1 |
1 |
1 |
1 |
1 |
|
2 |
1 |
2 |
1 |
2 |
1 |
|
3 |
1 |
1 |
3 |
1 |
1 |
|
4 |
1 |
2 |
1 |
4 |
1 |
|
5 |
1 |
1 |
1 |
1 |
5 |
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative ?
(iii) Compute (2 * 3) * (4 * 5).
Short Answer Type
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