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Example 19 - R = {(a, b): a, b Q and a – b Z}. Show (a, a)

Example 19 - Chapter 2 Class 11 Relations and Functions - Part 2

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Example 19 - Chapter 2 Class 11 Relations and Functions - Part 3 Example 19 - Chapter 2 Class 11 Relations and Functions - Part 4 Example 19 - Chapter 2 Class 11 Relations and Functions - Part 5

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Example 19 - Chapter 2 Class 11 Relations and Functions - Part 6 Example 19 - Chapter 2 Class 11 Relations and Functions - Part 7 Example 19 - Chapter 2 Class 11 Relations and Functions - Part 8

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Example 19 Let R be a relation from Q to Q defined by R = {(a, b): a, b ∈ Q and a – b ∈ Z}. Show that (i) (a, a) ∈ R for all a ∈ Q Given R = {(a, b): a, b ∈ Q and a – b ∈ Z} Hence we can say that (a, b) is in relation R if a, b ∈ Q i.e. both a & b are in set Q a – b ∈ Z i.e. difference of a & b is an integer We need to prove both these conditions for (a, a) a, a ∈ Q , i.e. a is in set Q Example 19 Let R be a relation from Q to Q defined by R = {(a, b): a, b ∈ Q and a – b ∈ Z}. Show that (i) (a, a) ∈ R for all a ∈ Q Given R = {(a, b): a, b ∈ Q and a – b ∈ Z} Hence we can say that (a, b) is in relation R if a, b ∈ Q i.e. both a & b are in set Q a – b ∈ Z i.e. difference of a & b is an integer We need to prove both these conditions for (a, a) a, a ∈ Q , i.e. a is in set Q Also, a – a = 0 & 0 is an integer, ∴ a – a ∈ Z Since both conditions are satisfied, (a, a) ∈ R Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and a – b ∈ Z}. Show that (ii) (a, b) ∈ R implies that (b, a) ∈ R Given R = {(a, b): a, b ∈ Q and a – b ∈ Z} Given (a, b) ∈ R , i.e. (a, b) is in relation R . So, the following conditions are true a, b ∈ Q i.e. both a & b are in set Q a – b ∈ Z i.e. difference of a & b is an integer We need to prove both these conditions for (b, a) 1. a, b ∈ Q, hence b, a ∈ Q 2. b – a ∈ Z i.e. difference of b & a is an integer First condition is satisfied as a, b ∈ Q is given For second condition Since negative of integer is also an integer, i.e. –(a – b) is also integer – a + b is integer, b – a is integer b – a is integer ∴ b – a ∈ Z Since both conditions are satisfied, (b, a) ∈ R Example 19 Let R be a relation from Q to Q defined by R = {(a,b): a,b ∈ Q and a – b ∈ Z}. Show that (iii) (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R Given R = {(a, b): a, b ∈ Q and a – b ∈ Z} Given (a, b) ∈ R , i.e. (a, b) is in relation R . So, following conditions are true a, b ∈ Q i.e. both a & b are in set Q a – b ∈ Z i.e. difference of a & b is an integer Given (b, c) ∈ R , i.e. (b, c) is in relation R . So, following conditions are true b, c ∈ Q i.e. both b & c are in set Q b – c ∈ Z i.e. difference of b & c is an integer We need to prove both these conditions for (a, c) 1. Given a, b & b, c ∈ Q, Hence, a, c ∈ Q 2. Given (a – b) & (b – c) are integers, Sum of integers is also an integer So, (a – b) + (b – c) is also integer a – b + b – c is integer, a – c is integer ∴ a – c ∈ Z Since both conditions are satisfied, ∴ (a, c) ∈ R 2. Given (a – b) & (b – c) is an integer, Sum of integers is also an integer So, (a – b) + (b – c) is also integer a – b + b – c is integer, a – c is integer ∴ a – c ∈ Z Since both conditions are satisfied, (a, c) ∈ R

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.