Pedagogy of Mathematics (Part II) - Set language introduction and ex.1.2, Set Language, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy

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