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Understanding Sets, Relations, and Functions in Mathematics: Session 2 Objectives & Exercises

This session covers essential concepts in set theory, relations, and functions. We explore the Cartesian products of given sets, identifying elements common to both A × B and B × A. Exercises include defining relations, determining domains and ranges, as well as testing properties of equivalence relations. Students will engage in various class exercises aimed at reinforcing these concepts through practical examples, fostering a solid understanding of mathematical relationships.

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Understanding Sets, Relations, and Functions in Mathematics: Session 2 Objectives & Exercises

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  1. Mathematics

  2. Session Set, Relation & Function Session - 2

  3. Session Objectives

  4. Class Test

  5. If A = {1, 2, 3} and B = {3, 8}, thenis • {(3, 1), (3, 2), (3, 3), (3, 8)} • {(1, 3), (2, 3), (3, 3), (8, 3)} • {(1, 2), (2, 2), (3, 3), (8, 8)} • {(8, 3), (8, 2), (8, 1), (8, 8)} Class Exercise - 1

  6. Solution Hence, answer is (b).

  7. LetA and B be two non-empty setssuch that . Then provethat A × B and B × A have n2elementsin common. Class Exercise - 2

  8. have n2 elements is common. Solution

  9. R be a relation on set of natural numbers N defined as . Find the following. (i) R, R–1 as sets of ordered pairs (ii) Domain of R and R–1(iii) Range of R and R–1(iv) R–1 oR (v) R–1 in set-builder form Class Exercise - 3

  10. (i) (v)R–1 = {(a, b) | a + 3b = 12, a, } Solution = {(1, 9), (2, 6), (3, 3)} R–1 = {(9, 1), (6, 2), (3, 3)} (ii) Domain (R) = {1, 2, 3}Domain (R–1) = {9, 6, 3} (iii)Range (R) = {9, 6, 3}Domain (R–1) = {1, 2, 3} (iv) {(9, 1), (6, 2), (3, 3)} o {(1, 9), (2, 6), (3, 3)}= {(1, 1), (2, 2), (3, 3)}

  11. Let R be a relation from A = {2, 3, 4, 5, 6}to B = {3, 6, 8, 9, 12} defined asExpress R asset of ordered pairs, find the domain andthe range of R, and also find R–1 inset-builder form (where x | y means xdivides y). Class Exercise - 4

  12. = {(2, 6), (2, 8), (2, 12), (3, 3), (3, 6), (3, 9), (3, 12), (4, 8), (4, 12), (6, 6), (6, 12)} and R–1 x is divisible by y} Solution Domain (R) = {2, 3, 4, 6} Range (R) = {6, 8, 12, 3, 9} R–1 = {(6, 2), (8, 2), (12, 2), (3, 3), (6, 3), (9, 3),(12, 3), (8, 4), (12, 4), (6, 6), (12, 6)}

  13. Let R be a relation on Z defined asExpressR and R–1 as set of ordered pairs.Hence, find the domain of R and R–1. Class Exercise - 5

  14. Domain (R) = Domain (R–1) = {0, 3, 4, 5, –3, –4, –5} Solution R = {(0, 5), (0, –5), (3, 4), (3, –4), (4, 3), (4, –3),(5, 0), (–3, 4), (–3, –4), (–4, 3), (–4, –3), (–5, 0)} R–1 = {(5, 0), (–5, 0), (4, 3), (–4, 3), (3, 4), (–3, 4),(0, 5), (4, –3), (–4, –3), (3, –4), (–3, –4), (0, –5)} = R

  15. Let S be the set of all the straight lines on a plane, R be a relation on S defined as. Then check R for reflexivity, symmetry and transitivity. Class Exercise - 6

  16. Reflexive: as a line cannot beperpendicular to itself. Symmetric: Let R is symmetric Transitive: Let , i.e. Solution

  17. Let f = ‘n/m’ means that n is factorof m or n divides m, where .Then the relation ‘f’ is (a) reflexive and symmetric(b) transitive and symmetric(c) reflexive, transitive and symmetric(d) reflexive, transitive and not symmetric Class Exercise - 7

  18. Reflexive: a/a As a is factor of a Reflexive Symmetric: Let i.e. a/b or a is a factor of b Solution

  19. Transitive: Let , i.e. a is factor of c Solution contd.. Hence, answer is (d).

  20. Let where R is set of realsdefinedas Check S for reflexive, symmetric andtransitive. Class Exercise - 8

  21. Reflexive: Let , i.e. Hence, only for two values of R not Not reflexive Symmetric: If , i.e. a2 + b2 = 1 Solution

  22. Transitive: If i.e. a2 + b2 = 1 and b2 + c2 = 1 may not be 1 Not transitive Solution contd..

  23. Class Exercise - 9 Let A be a set of all the points in space. Let R be a relation on A such that a1 Ra2 if distance between the points a1 and a2 is less than one unit. Then which of the following is false? (a) R is reflexive(b) R is symmetric(c) R is transitive(d) R is not an equivalence relation

  24. Hence, Symmetric: If Distance between isless than 1 unit. Distance between a2 and a1 is less than 1 unit. a2Ra1Symmetric Transitive: If Solution Reflexive: Distance between a1 and a1is 0 less than one unit.

  25. Distance between a1 and a2 is lessthan 1 unit and distance between a2 anda3 is less than 1 unit Distance between a1 and a3 is less than 1 unit. For example, Distance between a1 and a3 = 1.8 > 1 Not transitive Solution contd.. Hence, R is not an equivalence relation. Hence, answer is (c).

  26. Let N denote the set of all naturalnumbers and R be the relation onN × N defined by Show that R isan equivalence relation. Class Exercise - 10

  27. Solution Reflexive: (a, b) R (a, b) As ab(b + a) = ba(a + b) Symmetric: If (a, b) R (c, d)

  28. af(b + e) = be(a + f) (a, b) R (e, f) Solution contd.. Transitive: If (a, b) R (c, d)and(c, d) R (e, f) Hence, R is an equivalence relation.

  29. Thank you

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