Let 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.
Consider 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.
A = N x N and (a, b) * (c, d) = (ac, bd)
(i) Let (a, b), (c, d), (e, f) be any three elements of A Now (a, b) * {(c, d) * (e, f)} = (a, b) * (ce, d f)
= (a(ce)), 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 Now (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.
Let 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).