If ξ = {x : x ∈ N, x < 25} and A = {x : x is a composite number}, then find A′ in set builder form and also in roster form.
Answer
ξ = {x : x ∈ N, x < 25}
ξ = {1, 2, 3, ..., 24}
A = {x : x is a composite number}
A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {1, 2, 3, ..., 24} - {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}
A′ = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23}
As A′ contains 1 and all the prime numbers less than 25,
Hence, A′ = {x : x ∈ N, x = 1 or x is a prime number less than 25} and A′ = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23}.
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find A ∪ B and A ∩ B.
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
B = {-4, -2, 0, 2, 4}
A ∪ B - contains all the elements in set A and B.
A ∪ B = {-5, -3, 0, 3, 5} ∪ {-4, -2, 0, 2, 4}
A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {-5, -3, 0, 3, 5} ∩ {-4, -2, 0, 2, 4}
A ∩ B = {0}
Hence, A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5} and A ∩ B = {0}.
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then verify that n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
B = {-4, -2, 0, 2, 4}
From part (i), A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5} and A ∩ B = {0}.
Here n(A) = 5, n(B) = 5, n(A ∪ B) = 9 and n(A ∩ B) = 1.
n(A) + n(B) - n(A ∩ B) = 5 + 5 - 1 = 9 = n(A ∪ B)
Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find A′ and verify that n(A) + n(A′) = n(ξ).
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-5, -3, 0, 3, 5}
A′ = {-4, -2, -1, 1, 2, 4}
Here n(A) = 5, n(A′) = 6 and n(ξ) = 11.
n(A) + n(A′) = 5 + 6 = 11 = n(ξ)
Hence, A′ = {-4, -2, -1, 1, 2, 4} and n(A) + n(A′) = n(ξ).
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find B′ and verify that n(B′) = n(ξ) - n(B).
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
B = {-4, -2, 0, 2, 4}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-4, -2, 0, 2, 4}
B′ = {-5, -3, -1, 1, 3, 5}
Here n(B) = 5, n(B′) = 6 and n(ξ) = 11.
n(ξ) - n(B) = 11 - 5 = 6 = n(B′)
Hence, B′ = {-5, -3, -1, 1, 3, 5} and n(B′) = n(ξ) - n(B).
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find A ∪ A′, A ∩ A′, B ∪ B′ and B ∩ B′.
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
B = {-4, -2, 0, 2, 4}
From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.
A ∪ A′ - contains all the elements in set A and A′.
A ∪ A′ = {-5, -3, 0, 3, 5} ∪ {-4, -2, -1, 1, 2, 4}
A ∪ A′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} = ξ
A ∩ A′ - contains all the common elements in set A and A′.
A ∩ A′ = {-5, -3, 0, 3, 5} ∩ {-4, -2, -1, 1, 2, 4}
A ∩ A′ = φ
B ∪ B′ - contains all the elements in set B and B′.
B ∪ B′ = {-4, -2, 0, 2, 4} ∪ {-5, -3, -1, 1, 3, 5}
B ∪ B′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} = ξ
B ∩ B′ - contains all the common elements in set B and B′.
B ∩ B′ = {-4, -2, 0, 2, 4} ∩ {-5, -3, -1, 1, 3, 5}
B ∩ B′ = φ
Hence, A ∪ A′ = ξ, A ∩ A′ = φ, B ∪ B′ = ξ and B ∩ B′ = φ.
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find (A ∪ B)′ and A′ ∩ B′. Are they equal?
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
B = {-4, -2, 0, 2, 4}
From part (i), A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}.
From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.
(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.
(A ∪ B)′ = ξ - (A ∪ B)
(A ∪ B)′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {-5, -4, -3, -2, 0, 2, 3, 4, 5}
(A ∪ B)′ = {-1, 1} ...............(1)
A′ ∩ B′ - contains all the common elements in set A′ and B′.
A′ ∩ B′ = {-4, -2, -1, 1, 2, 4} ∩ {-5, -3, -1, 1, 3, 5}
A′ ∩ B′ = {-1, 1} ...............(2)
From (1) and (2), we can see that
(A ∪ B)′ = {-1, 1} and A′ ∩ B′ = {-1, 1}. Hence, they are equal.
If ξ = {x | x ∈ I, -5 ≤ x ≤ 5}, A = {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then find (A ∩ B)′ and A′ ∪ B′. Are they equal?
Answer
ξ = {x | x ∈ I, -5 ≤ x ≤ 5}
ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
A = {-5, -3, 0, 3, 5}
B = {-4, -2, 0, 2, 4}
From part (i), A ∩ B = {0}.
From part (iii), A′ = {-4, -2, -1, 1, 2, 4} and from part (iv), B′ = {-5, -3, -1, 1, 3, 5}.
(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.
(A ∩ B)′ = ξ - (A ∩ B)
(A ∩ B)′ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} - {0}
(A ∩ B)′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} ...............(1)
A′ ∪ B′ - contains all the elements in set A′ and B′.
A′ ∪ B′ = {-4, -2, -1, 1, 2, 4} ∪ {-5, -3, -1, 1, 3, 5}
A′ ∪ B′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} ...............(2)
From (1) and (2), we can see that
(A ∩ B)′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5} and A′ ∪ B′ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}. Hence, they are equal.
If ξ = {x | x ∈ N, x ≤ 12}, A = {prime numbers} and B = {odd numbers}, then
(i) find A ∪ B and A ∩ B
(ii) verify that n(A ∪ B) + n(A ∩ B) = n(A) + n(B)
(iii) find A′ and B′
(iv) find (A ∪ B)′ and verify that n(A ∪ B) + n((A ∪ B)′) = n(ξ)
(v) find (A ∩ B)′ and A′ ∪ B′. Are they equal?
Answer
ξ = {x | x ∈ N, x ≤ 12}
ξ = {1, 2, 3, ..., 12}
A = {prime numbers}
A = {2, 3, 5, 7, 11}
B = {odd numbers}
B = {1, 3, 5, 7, 9, 11}
(i) A ∪ B - contains all the elements in set A and B.
A ∪ B = {2, 3, 5, 7, 11} ∪ {1, 3, 5, 7, 9, 11}
A ∪ B = {1, 2, 3, 5, 7, 9, 11}
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {2, 3, 5, 7, 11} ∩ {1, 3, 5, 7, 9, 11}
A ∩ B = {3, 5, 7, 11}
Hence, A ∪ B = {1, 2, 3, 5, 7, 9, 11} and A ∩ B = {3, 5, 7, 11}.
(ii) Here n(A) = 5, n(B) = 6, n(A ∪ B) = 7 and n(A ∩ B) = 4.
n(A ∪ B) + n(A ∩ B) = 7 + 4 = 11
n(A) + n(B) = 5 + 6 = 11
Hence, n(A ∪ B) + n(A ∩ B) = n(A) + n(B).
(iii) A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {1, 2, 3, ..., 12} - {2, 3, 5, 7, 11}
A′ = {1, 4, 6, 8, 9, 10, 12}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {1, 2, 3, ..., 12} - {1, 3, 5, 7, 9, 11}
B′ = {2, 4, 6, 8, 10, 12}
Hence, A′ = {1, 4, 6, 8, 9, 10, 12} and B′ = {2, 4, 6, 8, 10, 12}.
(iv) (A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.
(A ∪ B)′ = ξ - (A ∪ B)
(A ∪ B)′ = {1, 2, 3, ..., 12} - {1, 2, 3, 5, 7, 9, 11}
(A ∪ B)′ = {4, 6, 8, 10, 12}
Here n(A ∪ B) = 7, n((A ∪ B)′) = 5 and n(ξ) = 12.
n(A ∪ B) + n((A ∪ B)′) = 7 + 5 = 12 = n(ξ)
Hence, (A ∪ B)′ = {4, 6, 8, 10, 12} and n(A ∪ B) + n((A ∪ B)′) = n(ξ).
(v) (A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.
(A ∩ B)′ = ξ - (A ∩ B)
(A ∩ B)′ = {1, 2, 3, ..., 12} - {3, 5, 7, 11}
(A ∩ B)′ = {1, 2, 4, 6, 8, 9, 10, 12} ...............(1)
A′ ∪ B′ - contains all the elements in set A′ and B′.
A′ ∪ B′ = {1, 4, 6, 8, 9, 10, 12} ∪ {2, 4, 6, 8, 10, 12}
A′ ∪ B′ = {1, 2, 4, 6, 8, 9, 10, 12} ...............(2)
From (1) and (2), we can see that
(A ∩ B)′ = {1, 2, 4, 6, 8, 9, 10, 12} and A′ ∪ B′ = {1, 2, 4, 6, 8, 9, 10, 12}. Hence, they are equal.
If ξ = {x : x ∈ N, x ≤ 12}, A = {x : x ≥ 7} and B = {x : 4 < x < 10}, then find :
(i) A′
(ii) B′
(iii) A ∪ B
(iv) A ∩ B
(v) A - B
(vi) B - A
(vii) (A ∪ B)′
(viii) A′ ∩ B′
Also verify that :
(i) (A ∪ B)′ = A′ ∩ B′
(ii) A - B = A ∩ B′
(iii) n(A ∪ B) + n((A ∪ B)′) = n(ξ)
(iv) n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)
Answer
ξ = {x : x ∈ N, x ≤ 12}
ξ = {1, 2, 3, ..., 12}
A = {x : x ≥ 7}
A = {7, 8, 9, 10, 11, 12}
B = {x : 4 < x < 10}
B = {5, 6, 7, 8, 9}
(i) A′
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {1, 2, 3, ..., 12} - {7, 8, 9, 10, 11, 12}
Hence, A′ = {1, 2, 3, 4, 5, 6}.
(ii) B′
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {1, 2, 3, ..., 12} - {5, 6, 7, 8, 9}
Hence, B′ = {1, 2, 3, 4, 10, 11, 12}.
(iii) A ∪ B
A ∪ B - contains all the elements in set A and B.
A ∪ B = {7, 8, 9, 10, 11, 12} ∪ {5, 6, 7, 8, 9}
Hence, A ∪ B = {5, 6, 7, ..., 12}.
(iv) A ∩ B
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {7, 8, 9, 10, 11, 12} ∩ {5, 6, 7, 8, 9}
Hence, A ∩ B = {7, 8, 9}.
(v) A - B
A - B - contains all the elements in set A but not in B.
A - B = {7, 8, 9, 10, 11, 12} - {5, 6, 7, 8, 9}
Hence, A - B = {10, 11, 12}.
(vi) B - A
B - A - contains all the elements in set B but not in A.
B - A = {5, 6, 7, 8, 9} - {7, 8, 9, 10, 11, 12}
Hence, B - A = {5, 6}.
(vii) (A ∪ B)′
(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.
(A ∪ B)′ = ξ - (A ∪ B)
(A ∪ B)′ = {1, 2, 3, ..., 12} - {5, 6, 7, ..., 12}
Hence, (A ∪ B)′ = {1, 2, 3, 4}.
(viii) A′ ∩ B′
A′ ∩ B′ - contains all the common elements in set A′ and B′.
A′ ∩ B′ = {1, 2, 3, 4, 5, 6} ∩ {1, 2, 3, 4, 10, 11, 12}
Hence, A′ ∩ B′ = {1, 2, 3, 4}.
Verification :
(i) From part (vii), (A ∪ B)′ = {1, 2, 3, 4} ...............(1)
From part (viii), A′ ∩ B′ = {1, 2, 3, 4} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, (A ∪ B)′ = A′ ∩ B′.
(ii) From part (v), A - B = {10, 11, 12} ...............(3)
A ∩ B′ - contains all the common elements in set A and B′.
A ∩ B′ = {7, 8, 9, 10, 11, 12} ∩ {1, 2, 3, 4, 10, 11, 12}
A ∩ B′ = {10, 11, 12} ...............(4)
From (3) and (4), we can see
LHS = RHS
Hence, A - B = A ∩ B′.
(iii) Here n(A ∪ B) = 8, n((A ∪ B)′) = 4 and n(ξ) = 12.
n(A ∪ B) + n((A ∪ B)′) = 8 + 4 = 12 = n(ξ)
Hence, n(A ∪ B) + n((A ∪ B)′) = n(ξ).
(iv) Here n(A - B) = 3, n(B - A) = 2, n(A ∩ B) = 3 and n(A ∪ B) = 8.
n(A - B) + n(B - A) + n(A ∩ B) = 3 + 2 + 3 = 8 = n(A ∪ B)
Hence, n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B).
If n(ξ) = 50, n(A) = 15, n(B) = 13 and n(A ∩ B) = 10, find n(A′), n(B′) and n(A ∪ B)
Answer
Given:
n(ξ) = 50
n(A) = 15
n(B) = 13
n(A ∩ B) = 10
∵ n(A) + n(A′) = n(ξ)
Putting the values, we get
⇒ 15 + n(A′) = 50
⇒ n(A′) = 50 - 15
⇒ n(A′) = 35
∵ n(B) + n(B′) = n(ξ)
Putting the values, we get
⇒ 13 + n(B′) = 50
⇒ n(B′) = 50 - 13
⇒ n(B′) = 37
∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ n(A ∪ B) = 15 + 13 - 10
⇒ n(A ∪ B) = 28 - 10
⇒ n(A ∪ B) = 18
Hence, n(A′) = 35, n(B′) = 37 and n(A ∪ B) = 18.
If n(ξ) = 60, n(A) = 35, n(B′) = 36 and n((A ∩ B)′) = 51, then find :
(i) n(B)
(ii) n(A ∩ B)
(iii) n(A ∪ B)
(iv) n(A - B)
Answer
Given:
n(ξ) = 60
n(A) = 35
n(B′) = 36
n((A ∩ B)′) = 51
(i) ∵ n(B) + n(B′) = n(ξ)
Putting the values, we get
⇒ n(B) + 36 = 60
⇒ n(B) = 60 - 36
⇒ n(B) = 24
Hence, n(B) = 24.
(ii) ∵ n(A ∩ B) + n((A ∩ B)′) = n(ξ)
Putting the values, we get
⇒ n(A ∩ B) + 51 = 60
⇒ n(A ∩ B) = 60 - 51
⇒ n(A ∩ B) = 9
Hence, n(A ∩ B) = 9.
(iii) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ n(A ∪ B) = 35 + 24 - 9
⇒ n(A ∪ B) = 59 - 9
⇒ n(A ∪ B) = 50
Hence, n(A ∪ B) = 50.
(iv) ∵ n(A - B) = n(A) - n(A ∩ B)
Putting the values, we get
⇒ n(A - B) = 35 - 9
⇒ n(A - B) = 26
Hence, n(A - B) = 26.