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Mathematics

Mathematics. Session. Set, Relation & Function Session - 2. Session Objectives. Class Test. If A = {1, 2, 3} and B = {3, 8}, then is. {(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.

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Mathematics

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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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