The document provides a detailed exploration of various mathematical set concepts, including cardinality, finite and infinite sets, equivalent and unequal sets, and null and singleton sets. It includes specific examples and solutions related to cardinal numbers, properties of sets, and the relationships between them. The document serves as an educational resource for understanding and teaching the fundamental principles of set theory in mathematics.

1. Pedagogy of Mathematics – Part II
continuation
BY
DR. I. UMA MAHESWARI
PRINCIPAL
PENIEL RURAL COLLEGE OF EDUCATION, VEMPARALI, DINDIGUL
DISTRICT
IUMA_MAHESWARI@YAHOO.CO.IN

2. Std IX
Chapter 1 – Set Language
Ex - 1.2

15. Problem

26. Question 1 :
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
Solution :
Number of elements in the set is 6. Hence n(M) = 6.

27. (ii) P = {x : x = 3n + 2, n∈W and x < 15}
Solution :
Since n belongs to whole number, we have to start with 0.
By applying the values of n from 0 to 14, we get 15 different values
for x. Hence n(P) is 15.

28. (iii) Q = { y : y = 4/3n, n ∈ N and 2 < n ≤ 5}
Solution :
The values of n are 3, 4, 5. By applying the
above three values for n, we get different
values of y. Hence n(Q) is 3.

29. (iv) R = {x : x is an integers, x ∈ Z and –5 ≤ x < 5}
Solution :
The elements of R are
R = {-5,-4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R) = 10

30. (v) S = The set of all leap years between 1882 and 1906.
Solution :
The leap years 1884, 1888, 1892, 1896, 1900, 1904.
Hence n(S) = 6

32. Question 2 :
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
Solution :
Districts in Tamilnadu is countable. Hence it is finite set.

33. (ii) Y = The set of all straight lines passing
through a point.
Solution :
We may draw an infinite number of lines
through a point.
Hence it is infinite set.

34. (iii) A = { x : x ∈ Z and x < 5}
Solution :
Z means integers. The elements of A are 1, 2, 3, 4.
Hence set A is finite.

35. (iv) B = {x : x2–5x+6 = 0, x ∈N}
Solution :
x2–5x+6 = 0
(x - 2) (x - 3) = 0
x = 2 and x = 3
By solving the quadratic equation, we get two different values. Hence B is
finite set.

37. Question 3 :
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
Solution :
A = {a, e, i o, u}
B = {V, O, W, E, L}
n(A) = 5, n(B) = 5
Since n(A) = n(B), they are equivalent sets.

38. (ii) C = {2, 3, 4, 5}
D = { x : x ∈ W, 1 < x < 5}
Solution :
D = {2, 3, 4}
The number of elements are not same and set D contain 1 lesser element than
C. Hence they are unequal sets.

39. (iii) E = The set of A={ x : x is a letter in the word “LIFE”}
F = {F, I, L, E}
Solution :
E = { L, I, F, E }
F = {F, I, L, E}
In both sets we have same elements. Hence they are equal sets.

40. (iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution :
G = {5, 7, 11, 13, 17, 19} and n(G) = 6
H = {1, 2, 3, 6, 9, 18} and n(H) = 6
Since the number of elements of both sets are same. They are
equivalent sets.

42. Question 4 :
Identify the following sets as null set or singleton set.
(i) A = {x : x ∈N, 1 < x < 2}
Solution :
N is the natural number and it should be greater than 1 and
lesser than 2. Since it is not possible, the set A will not contain
any elements. It is null set.

43. (ii) B = The set of all even natural numbers which are not
divisible by 2
Solution :
Even number means, a number which is divisible by 2. But the set
B will contain the elements which are not divisible by 2. It is not
possible. It is null or empty set.

44. (iii) C = {0}.
Solution :
The set C contains only one element. Hence it is singleton set.
(iv) D = The set of all triangles having four sides.
Solution :
All triangles will have only three sides. Hence D is empty set.

46. Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n, f, h, s} = {f, a, s}
Since A ∩ B ≠ φ , A and B are overlapping sets.
(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = φ, C and D are disjoint sets.
(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ φ, E and F are overlapping sets.

48. Question 6 :
If S = {square, rectangle, circle, rhombus, triangle}, list the elements of
the following subset of S.
(i) The set of shapes which have 4 equal sides.
Solution :
Let A be the set, which contains the shape of the 4 equal sides.
A = {Square, Rhombus}

49. (ii) The set of shapes which have no sides.
Let B be the set, which contains the shape which have no sides.
B = { Circle }
(iii) The set of shapes in which the sum of all interior angles is 180o
Solution :
For any triangle the sum of all interior angles is 180, hence
C = {Triangle}

50. (iv) The set of shapes which have 5 sides.
Solution :
In the given set no shape will contain 5
sides. Hence it is a empty set.

52. Question 7 :
If A = {a, {a, b}}, write all the subsets of
A.
Solution :
Subset of A = {{ }, {a}, {a, b}, {a, {a, b}}}

54. Question 8 :
Write down the power set of the following sets.
(i) A = {a, b}
Solution :
Subset of A are
= { { }, {a}, {b}, {a, b} }

55. (ii) B = {1, 2, 3}
Solution :
Subset of B are
= { { }, {1}, {2}, {3}, {1, 2}, {2, 3} {3, 1} {1, 2, 3} }
(iii) D = {p, q, r, s}
Solution :
Subset of D are
= {{ }, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s} {r,
s}, {q, r, s} {p, q, r, s}}

56. (iv) E = ∅
Solution :
P(E) = { { } }

57. Question 9 :
Find the number of subsets and the number of proper subsets
of the following sets.
(i) W = {red, blue, yellow}

58. Solution :
The number of proper subsets of a set A is
n [P(A)] – 1 = 2m–1
n(A) = 3
n [P(A)] – 1 = 23–1
= 8 - 1
Number of proper subset :
n [P(A)] – 1 = 7

59. (ii) X = { x2 : x ∈ N, x2 ≤ 100}
X = {22, 32, 42, 52, 62, 72, 82, 92, 102}
n (X) = 10
n [P(A)] – 1 = 210 – 1
n [P(A)] = 1024
Number of proper subset :
Number of proper subset = 1024 - 1 = 1023

60. Question 10 :
(i) If n(A) = 4, find n [P(A)].
Solution :
n(A) = 4, find n [P(A)]
n [P(A)] = 24
= 16

61. (ii) If n(A) = 0, find n [P(A)].
Solution :
n [P(A)] = 2m
= 20
= 1

62. (iii) If n[P(A)] = 256, find n(A).
n [P(A)] = 256
2m = 256
2m = 28
m = 8