1e. Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.5

PEDAGOGY OF
MATHEMATICS – PART II
By
Dr. I. Uma Maheswari
Principal
Peniel Rural College of Education,Vemparali, Dindigul District
iuma_maheswari@yahoo.co.in

STD IX
CHAPTER 1 – SET LANGUAGE
Ex - 1.5

Solution :
(i) A - B = {4, 6, 3}
(ii) B - C = {-1, 5, 7}
(iii) A' U B'
A' = {-3, 0, 1, 2, 5, 7, 8}
B' = {-3, 0, 1, 2, 3, 4, 6}
A' U B' = {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}

(iv) A' n B'
A' n B' = {-3, 0, 1, 2}
(v) (B U C)'
B U C = {-3, -,2, -1, 0, 3, 5, 7, 8}
(B U C)' = {4, 6, 1, 2}

(vi) A − (B U C)
A = {-2, -1, 3, 4, 6} and B U C = {-3, -,2, -1, 0, 3, 5, 7, 8}
A − (B U C) = {4, 6}
(vii) A − (B n C)
B n C = {-2, 8}
A = {-2, -1, 3, 4, 6}
A − (B n C) = {-1, 3, 4, 6}

Solution :
(i) K U (L n M)
(L n M) = { b, c, d }
K U (L n M) = {a, b, c, d, e, f}
(ii) K n (L U M)
(L U M) = {a, b, c, d, g, h}
K n (L U M) = {a, b, d}

(iii) (K U L) n (K U M)
(K U L) = {a, b, c, d, e, f, g}
(K U M) = {a, b, c , d, e, f, h}
(K U L) n (K U M) = {a, b , c, d , e, f}
(iv) (K n L) U (K n M)
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(K n L) = {b, d}
(K n M) = {a, b, d}
(K n L) U (K n M) = {a, b, d}

Solution :
A = {x : x ∊ Z,−2 < x ≤ 4}
A = {-1, 0, 1, 2, 3, 4}
B = {x : x ∈ W, x ≤ 5}
B = {0, 1, 2, 3, 4, 5}
C = {−4,−1, 0, 2, 3, 4}
A U (B n C) = (A U B) n (A U C)

(B n C) = {0, 2, 3, 4}
A U (B n C) = {-1, 0, 1, 2, 3, 4} ----(1)
(A U B) = {-1, 0, 1, 2, 3, 4, 5}
(A U C) = {-4, -1, 0, 1, 2, 3, 4}
(A U B) n (A U C) = {-1, 0, 1, 2, 3, 4}----(2)
(1) = (2)

Solution :
(B n C) = {a, d, g}
A - (B n C) = {b, c, e, h} ----(1)
(A − B) = {b, e, h}
(A − C) = {b, c}
(A − B) U (A−C) = {b, c, e, h} ----(2)
(1) = (2)
Hence proved

Solution :
A = {x : x = 6n, n ∈ W and n < 6}
A = {0, 6, 12, 18, 24, 30}
B = {x : x = 2n, n ∈ N and 2 < n ≤ 9}
B = {6, 8, 10, 12, 14, 16, 18}
and C = {x : x = 3n, n ∈ N and 4 ≤ n < 10}
C = {12, 15, 18, 21, 24, 27}

(B n C) = {12, 18}
A−(B n C) = {0, 6, 24, 30} ----(1)
(A − B) = {0, 24, 30}
(A − C) = {0, 6, 30}
(A − B) U (A − C) = {0, 6, 24, 30} ----(2)

Solution :
(B U C) = {-1, 0, 2, 5, 6, 7}
A − (B U C) = {-2, 1, 3} ----(1)
(A − B) = {-2, 1, 3}
(A − C) = {-2, 0, 1, 3}
(A − B) n (A − C) = {-2, 1, 3} ----(2)
(1) = (2)
Hence proved.

Solution :
A = {y : y = (a + 1)/2, a ∈ W and a ≤ 5}
A = {1/2, 3/2, 2, 5/2, 3}
B = {y : y = (2n - 1)/2, n ∈ W and n < 5}
B = {-1/2, 1/2, 3/2, 5/2, 7/2}
C = {-1, -1/2, 1, 3/2, 2}
A - (BUC) = (A - B) n (A - C)

B U C = {-1/2, -1, 1/2, 1, 3/2, 2, 5/2, 7/2}
A - (BU C) = {3} -----(1)
(A - B) = {2, 3}
(A - C) = {1/2, 5/2, 3}
(A - B) n (A - C) = {3} -----(2)
(1) = (2)
Hence proved.

Solution :
De morgan's laws for complementation :
(i) (A U B)' = A' n B'
(ii) (A n B)' = A' U B'
A = {7, 8, 11, 12} and B = {4, 8, 12, 15},
U = {4, 7, 8, 10, 11, 12, 15, 16}

(i) (A U B) = {4, 7, 8, 11, 12, 15}
(A U B)' = {10, 16} ----(1)
A' = {4, 10, 15, 16}
B' = {7, 10, 11, 16}
A' n B' = {10, 16} ----(2)
Hence proved.

(ii) (A n B) = {8, 12}
(A n B)' = {4, 7, 10, 11, 15, 16} ----(1)
A' = {4, 10, 15, 16}
B' = {7, 10, 11, 16}
A' U B' = {4, 7, 10, 11, 15, 16} ----(2)
Hence proved.

1e. Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.5

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