19. Write the following intervals in set-builder form:
(i) (–3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [–23, 5)
Answer:
(i) (–3, 0) = {x: x ∈ R, –3 < x < 0}
(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] ={x: x ∈ R, 6 < x ≤ 12}
(iv) [–23, 5) = {x: x ∈ R, –23 ≤ x < 5}
20. What universal set (s) would you propose for each of the following?
(i) The set of right triangles
(ii) The set of isosceles triangles
Answer:
(i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.
(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.
21. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universals set (s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) Φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Answer:
(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6}
B ⊂ {0, 1, 2, 3, 4, 5, 6}
However, C ⊄ {0, 1, 2, 3, 4, 5, 6}
Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.
(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
Therefore, Φ cannot be the universal set for the sets A, B, and C.
(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.
(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.
22. Find the union of each of the following pairs of sets:
(i) X = {1, 3, 5}; Y = {1, 2, 3}
(ii) A = {a, e, i, o, u}; B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3} B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = {x: x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}; B = Φ
Answer:
(i) X = {1, 3, 5} Y = {1, 2, 3}
X ∪ Y= {1, 2, 3, 5}
(ii) A = {a, e, i, o, u} B = {a, b, c}
A ∪ B = {a, b, c, e, i, o, u}
(iii) A = {x: x is a natural number and multiple of 3} = {3, 6, 9 …}
B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}
A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}
∴ A ∪ B = {x: x = 1, 2, 4, 5 or a multiple of 3}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
∴ A∪ B = {x: x ∈ N and 1 < x < 10}
(v) A = {1, 2, 3}, B = Φ
A ∪ B = {1, 2, 3}
23.. Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Answer:
Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
A ∪ B = {a, b, c} = B
24. If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Answer:
If A and B are two sets such that A ⊂ B, then A ∪ B = B.
25. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find
(i) A ∪ B
(ii) A UC
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Answer:
A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
26. Find the intersection of each pair of sets:
(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = {a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
B = {x: x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ
Answer:
(i) X = {1, 3, 5}, Y = {1, 2, 3}
X ∩ Y = {1, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
A ∩ B = {a}
(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}
B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}
∴ A ∩ B = {3}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
= {7, 8, 9}
A ∩ B = Φ
(v) A = {1, 2, 3}, B = Φ.
So, A ∩ B = Φ