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Relations And Functions

Question
CBSEENMA12032338
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Let A = Q x Q. Let be a binary operation on A defined by (a, b) * (c, d) = (ac, ad + b).

Then

(i) Find the identify element of (A, *)
(ii) Find the invertible elements of (A, *)

Solution

(i) Let (x, y) be the identify element of (A, *).
∴ (a, b) * (x,y) = (a, b) ∀ a, b ∈ Q ⇒ (ax, ay + b) = (a, b)
⇒ ax = a, ay + b = b ⇒ ax = a, ay = 0 ∀ a ∈ Q ⇒ x = 1, y = 0
Now (a, b) * (1,0) = (a, b) ∀ a, b, ∈ Q Also (1,0)* (a, b) = (a, 1. b + 0) = (a, b)
∴ (1, 0) is the identity element of A.
(ii) Let (a, b) ∈ A be invertible ∴ there exists (c, d) ∈ A such that (a, b) * (c, d) = (1, 0) = (c, d) * (a, b)
∴ (ac, ad + b) = ( 1,0) ⇒ ac = 1, ad + b = 0
rightwards double arrow space space space space straight c equals 1 over straight a comma space space space straight d equals fraction numerator negative straight b over denominator straight a end fraction

Now a ≠ 0
∵ if a = 0, then (0, b) is not invertible as (0, b) invertible implies that here exists (c, d)∈ A such that (0, b) * (c, d) = (1, 0) or (0, b) = (1, 0), which is senseless.
therefore space space open parentheses 1 over straight a comma space fraction numerator negative straight b over denominator straight a end fraction close parentheses asterisk times left parenthesis straight a comma space straight b right parenthesis space equals space left parenthesis 1 comma space 0 right parenthesis rightwards double arrow space space left parenthesis straight a comma space straight b right parenthesis to the power of negative 1 end exponent space equals space open parentheses 1 over straight a comma space fraction numerator negative straight b over denominator straight a end fraction close parentheses space
therefore  invertible elements of A are (a, b). a not equal to0 and (a, b)-1 = open parentheses 1 over straight a comma space fraction numerator negative straight b over denominator straight a end fraction close parentheses.


Some More Questions From Relations and Functions Chapter

Give an example of a relation which is

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

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.

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