Exercise 1.5
Q1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find
(i) A′
(ii) B′
(iii) (A ∪ C)′
(iv) (A ∪ B)′
(v) (A′)′
(vi) (B – C)′
Solution: Given that
U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }.
(i) A' = {5, 6, 7, 8, 9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4, 5, 6}
Therefore, (A ∪ C)′ = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4, 6, 8}
Therefore, (A ∪ B)′ = {5, 7, 9}
(v) A' = {5, 6, 7, 8, 9}
(A')' = A = {1, 2, 3, 4}
(vi) B - C = {2, 8}
(B - C)' = 1, 3, 4, 5, 6, 7, 9}
Q2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = { f, g, h, a}
Solution: Given that
U = { a, b, c, d, e, f, g, h}
(i) A = {a, b, c}
A' = {d, e, f, g, h}
(ii) B = {d, e, f, g}
B' = {a, b, c, h}
(iii) C = {a, c, e, g}
C' = {b, d, f, h}
(iv) D = { f, g, h, a}
D' = {b, c, d e}
Q3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}
(ii) { x : x is an odd natural number }
(iii) {x : x is a positive multiple of 3}
(iv) { x : x is a prime number }
(v) {x : x is a natural number divisible by 3 and 5}
(vi) { x : x is a perfect square }
(vii) { x : x is a perfect cube}
(viii) { x : x + 5 = 8 }
(ix) { x : 2x + 5 = 9}
(x) { x : x ≥ 7 }
(xi) { x : x ∈ N and 2x + 1 > 10 }
Solution: Given that U = { 1, 2, 3, 4, 5, 6, 7 ....}
(i) Let A = {x : x is an even natural number}
Or A = {2, 4, 6, 8 .....}
A' = { 1, 3, 5, 7 .....}
= {x : x is an odd natural number}
(ii) Let B = { x : x is an odd natural number }
Or B = { 1, 3, 5, 7 .....}
B' = {2, 4, 6, 8 .....}
= {x : x is an even natural number}
(iii) Let C = {x : x is a positive multiple of 3}
Or C = {3, 6, 9 ....}
C' = {1, 2, 4, 5, 7, 8, 10 .....}
= {x: x N and x is not a multiple of 3}
(iv) Let D = { x : x is a prime number }
Or D = {2, 3, 5, 7, 11 ... }
D' = {1, 4, 6, 8, 9, 10 ...... }
= {x: x is a positive composite number and x = 1}
(v) Let E = {x : x is a natural number divisible by 3 and 5}
Or E = {15, 30, 45 .....}
E' = {x: x is a natural number that is not divisible by 3 or 5}
(vi) Let F = { x : x is a perfect square }
F' = {x: x N and x is not a perfect square}
(vii) Let G = {x: x is a perfect cube}
G' = {x: x N and x is not a perfect cube}
(viii) Let H = {x: x + 5 = 8}
H' = {x: x N and x ≠ 3}
(ix) Let I = {x: 2x + 5 = 9}
I' = {x: x N and x ≠ 2}
(x) Let J = {x: x ≥ 7}
J' = {x: x N and x < 7}
(xi) Let K = {x: x N and 2x + 1 > 10}
K = {x: x N and x ≤ 9/2}
Q4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that
(i) (A ∪ B)′ = A′ ∩ B′
(ii) (A ∩ B)′ = A′ ∪ B′
Solution:
(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.
(A ∪ B)′ = A′ ∩ B′
A ∪ B = {2, 3, 4, 5, 6, 7, 8}
LHS = (A ∪ B)′ = {1, 9} ...(i)
RHS = A′ ∩ B′
= {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}
= {1, 9} .... (ii)
LHS = RHS
Hence Verified.
Solution:
(ii) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.
(A ∩ B)′ = A′ ∪ B′
A ∩ B = {2}
LHS = (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9 }
RHS = A′ ∪ B′
= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9 }
LHS = RHS
Hence Verified
Q5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′,
(ii) A′ ∩ B′,
(iii) (A ∩ B)′,
(iv) A′ ∪ B′
Solution:
(i) (A ∪ B)′
Venn diagram of (A ∪ B)′
(ii) A′ ∩ B′,
Venn diagram of A′ ∩ B′
Note: Venn diagram of A′ ∩ B′ will be same as (A ∪ B)′
Because (A ∪ B)′ = A′ ∩ B′
(iii) (A ∩ B)′
Venn diagram of (A ∩ B)′
(iv) A′ ∪ B′
Venn diagram of A′ ∪ B′
Q6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?
Solution:
A = {the set of all triangles with at least one angle different from 60°}
A' = {the set of all equilateral triangles}
Q7. Fill in the blanks to make each of the following a true statement :
(i) A ∪ A′ = . . .
(ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . .
Solution:
(i) A ∪ A′ = U
(ii) φ′ = U
Therefore φ′ ∩ A = U ∩ A = A
so, φ′ ∩ A = A
(iii) A ∩ A′ = φ
(iv) U′ ∩ A = φ