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.
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 ?
A = N x N and (a, b) * (c, d) = (ad + bc, bd)
(i) Let (a, b), (c, d), (e,f) be any three elements of A.
∴ {(a, b) * (c, d)} * (e,f) = (ad + bc, bd) * (e, f)
= ((ad + bc) f + (bd) e, (bd) f)
∴ {(a, b) * (c, d)} * (e,f) = (adf + btf + bde, bdf) ...(1)
Again (a, b) * {(c, d) * (e,f)} = (a, b) * (c f + de, d f)
= (a(d f) + b (c f + de), b (d f))
∴ (a, b) * {(c, d) * (e,f)} = (adf + bef + bde, bdf) ...(2)
From (1) and (2), we get,
(a, b) * {(c, d) * (e,f)} = {(a, b) * (c, d) * (e,f)} ∴ (A. *) is associative.
(ii) If possible, suppose that (x, y) is identity element in A.
∴ (a, b) * (x, y) = (a, b) ∀ (a, b) ∈ A
⇒ (ay + bx by) = (a, b) ∀ (a, b) ∈ A ⇒ ay + bx = a and by = b ∀ a, b ∈ N ⇒ x = 0, y = 1 ∀ a, b ∈ N This is not possible as 0 ∉ N ∴ our supposition is wrong ∴ (A, *) has no identity element.
(iii) Let (a, b), (c, d) be any two elements of A.
Now (a, b) *(c, d) = (ad + be , bd)
= (bc + ad,db)
= (cb + da , db)
∴ (a, b) * (c, d)) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈A ∴ (A, *) is 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).