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Chapter 1 Class 12 Relation and Functions
Serial order wise

Example 43 - Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be relation

Example 43 - Chapter 1 Class 12 Relation and Functions - Part 2
Example 43 - Chapter 1 Class 12 Relation and Functions - Part 3 Example 43 - Chapter 1 Class 12 Relation and Functions - Part 4 Example 43 - Chapter 1 Class 12 Relation and Functions - Part 5 Example 43 - Chapter 1 Class 12 Relation and Functions - Part 6

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Example 43 (Method 1) Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R1 = R2. R1 = {(x, y) : x – y is divisible by 3} R2 = { (x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} } We will prove R1 = R2 by proving R1 ⊂ R2 and R2 ⊂ R1 i.e, all elements of R1 are in the set R2 and all elements of R2 are in the set R1 Proving R1 ⊂ R2 Let (x, y) ∈ R1 ⇒ x – y is a divisible 3 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ (x, y) ∈ R2. Hence, R1⊂ R2. Rough In {1, 4, 7} x – y = 1 – 4 = – 3 = 4 – 7 = –3 = 7 – 1 = 6 So, x – y is divisible by 3 Similarly, in {2, 5, 8} & {3, 6, 9} x – y is divisible by 3 Proving R2 ⊂ R1 Let (x, y) ∈ R2 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible 3 ⇒ (x, y) ∈ R1. Hence, R2 ⊂ R1. Hence, R1 ⊂ R2 & R2 ⊂ R1. ∴ R1 = R2 Hence shown Rough In {1, 4, 7} x – y = 1 – 4 = 3 = 4 – 7 = –3 = 7 – 1 = 6 So, x – y is divisible by 3 Similarly, in {2, 5, 8} & {3, 6, 9} x – y is divisible by 3 Example 43 (Method 2) Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R1 = R2. X = {1, 2, 3, 4, 5, 6, 7, 8, 9} R1 = {(x, y) : x – y is divisible by 3} Finding R1 R1 = { (1, 4), (1, 7) , (2, 5) , (2, 8), (3, 6) , (3, 9) , (4, 1), (4, 7), (5, 2), (5, 8) , (6, 3) , (6, 9), (7, 1) , (7, 4) , (8, 2), (8, 5), (9, 3), (9, 6) } Now, finding R2 R2 = { (x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} } R2 = { (1, 4), (4, 1) , (1, 7) ,(7, 1), (4, 7) , (7, 4) , (2, 5), (5, 2) , (2, 8) ,(8, 2), (5, 8) , (8, 5) , (3, 6), (6, 3), (3, 9) , (9, 3), (6, 9) , (9, 6) } ∴ R1 = R2 Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.