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Relations And Functions

Question
CBSEENMA12032329
Wired Faculty App

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 ?

Solution

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 .

Some More Questions From Relations and Functions Chapter

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by R = {(a,b) : a ≤ b3} is refleive, symmetric or transitive.

If R and R’ arc reflexive relations on a set then so are R ∪ R’ and R ∩ R’. 

If R and R’ are symmetric relations on a set A, then R ∩ R’ is also a sysmetric relation on A.

Show that the union of two symmetric relations on a set is again a symmetric relation on that set.

Let A = {1. 2. 3}. Then show that the number of relations containing (1,2) and (2. 3) which are reflexive and transitive but not symmetric is four.

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