Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.4, Set Language, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy,

1. PEDAGOGY OF
MATHEMATICS – PART II
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 - 1.4

12. Solution :
P = {1, 2, 5, 7, 9}
Q = {2, 3, 5, 9, 11}
R = {3, 4, 5, 7, 9}
S = {2, 3, 4, 5, 8}
(i) (P U Q) U R
(P U Q) = {1, 2, 3, 5, 7, 9, 11}
(P U Q) U R = {1, 2, 3, 4, 5, 7, 9, 11} -------(1)

13. (ii) (P n Q) n S
P n Q = {2, 5, 9}
(P n Q) n S = {2, 5} ------(2)
(iii) (Q n S) n R
Q n S = {2, 3, 5}
(Q n S) n R = {3, 5} ------(3)

15. Solution :
P = {3, 4, 5, 6}
Q = {2.01......, .............., 6.990,..............}
P U Q = Q U P
P n Q = Q n P

17. Solution :
Associative property for union :
A U (B U C) = (AU B) U C
B U C = {m, n, p, q, s, t}
A U (B U C) = {m, n, p, q, r, s, t} ------(1)
(AU B) = {m, n, p, q, r, s, t}
(AU B) U C = {m, n, p, q, r, s, t} ------(2)
(1) = (2)
Hence proved.

19. Solution :
Associative property for intersection :
A n (B n C) = (A n B) n C
B n C = {√3, √5}
A n (B n C) = { √5 } ------(1)
(A n B) = { √5 }
(A n B) n C = { √5 } ------(2)
(1) = (2)
Hence proved.

21. Solution :
A = {x : x = 2n, n ∈ W and n < 4}
A = {1, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
B = { 2, 4, 6, 8 }
C = {0, 1, 2, 5, 6}
Associative property for intersection :
A n (B n C) = (A n B) n C
(B n C) = {2, 6}
A n (B n C) = { } ---(1)
(A n B) = {4, 8}
(A n B) n C = { } ---(2)
Hence proved.