Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a

Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.

RELATIONS AND FUNCTIONS

Class 12, NCERT Chapter 1, Example5

Solution

R is reflexive, as 2 divides (a – a) for all a ∈ Z.

Further, if (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric.

Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b - c) is even. So, (a-c) is divisible by 2. This shows that R is transitive.

Thus, R is an equivalence relation in Z.

Comments

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1, Exercise 1.2, Q7 (i) Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1 = 3 – 4x 2 ⇒ – 4x 1 = –4x 2 ⇒ x 1 = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1 = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1 = 2 and x 2 = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2) 2 = 5 and f (x 2 ) = f(–2) = 1 + (–2) 2 = 5 ∴ f (x 1 ) = f(x 2 ) but x 1 ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2 = –2 ⇒ x 2 = –3 = ± √–3 ∈R So, f is not onto function.

Let f : R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1, Exercise 1.2, Q11 f : R → R is defined as f (x) = x 4 Let x , y ∈ R such that f ( x ) = f ( y ). ∴ does not imply that x 1 =x 2 . For instance, f (1) = f (-1) = 1 ∴ f is not one-one. Consider an element 2 in co-domain R . It is clear that there does not exist any x in domain R such that f ( x ) = 2. ∴ f is not onto. Hence, function f is neither one-one nor onto. The correct answer is D

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.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1 Example2 Example3 ←you are here

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