Here a * b = H.C.F. of a and b, a, b ∈ N (i) H.C.F. (a, b) = H.C.F. (b, a)
∴ a * b = b * a
* Â Â Â is a commutative binary operation.
(ii) Let a, b, c ∈ N
∴ a * (b * c) = (a, H.C.F. (b, c))
= H.C.F. (H.C.F. (a,b),c)
= (a * b) * c ∴ a * (b * c) = (a * b) * c
∴ *    is associative binary operation.
(iii) If e is an identity element, then e * a = a * e = a for a ∈ N.
⇒ H.C.F. of a and e = a ∀ a ∈ N
⇒ a divides e ∀ a ∈ N.
Such a number e, which is divisible by every natural number, does not exist.
Let A = N x N and * be the binary operation on A defined by
(a, b) * (c, d) = (a + c, b + d)Â 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 ?
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) Â Â Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Â Â Â Is * commutative ?
(iii) Â Â Â Compute (2 * 3) * (4 * 5).