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Relations and Functions

Short Answer Type

301. Is *defined on the set {1, 2, 3, 4, 5} by a * b = l.c.m. of a and b a binary operation ? Justify your answer.

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302. Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative ? Is * associative ? Does there exist identity for this binary operation on N ?

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Long Answer Type

303.

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.

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Short Answer Type

304.

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 ∧.

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305.

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.

194 Views

Long Answer Type

306.

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 ?

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307.

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).

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308. Consider the binary operations * : R x R → R and a : R x R → R defined as a * b = | a – b | and a o b = a ∀ a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a * (b o c ) = (a * b) o (a * b). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

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Short Answer Type

309. Given a non-empty set X, consider the binary operation * : P(X) x P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *.

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310. Given a non-empty set X, let * : P(X) x P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set ϕ is the identity for the operation * and all the elements A of P(X) are invertible with A^–1 = A.

Let A be any element of P(X).
(i) Now A * ϕ = (A – ϕ) ∪ (ϕ – A)
= (A ∪ ϕ’) ∪ (ϕ ∩ A')    [∵ A – B = A ∩ B']
= (A ∩ U) ∩ ϕ
= A u ϕ = A
∴ A * ϕ = A    ...(1)
Also    ϕ * A = (ϕ – A) ∪ (A – ϕ)
= (ϕ ∩ A') ∪ (A ∩ ϕ')
= ϕ ∪ (A ∩U)
= ϕ ∪ A = A
∴ ϕ * A = A    ...(2)
From (1) and (2). we get,
A * ϕ = A = ϕ * A ∀ A ∈ P(X)
∴ ϕ is the identity element of (P(A), *)
(ii)    A * A = (A – A) ∪ (A – A)
= ϕ ∪ ϕ = ϕ ∴ A * A = ϕ A is invertible and its inverse is A.
⇒ A is invertible ∀ A ∈ P(X) and A ^–1 = A.