GSEB Solutions for Class 9 Mathematics – Set Operations (English Medium)
GSEB SolutionsMathsScience
Exercise 1:
Solution 1:
Given,
A = {-1, 5}
B = {-2, -1, 0, 1, 2}
C = {2, 4, 6, 8, 10, 12}
D = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7}
∴ A ⊂ D [Each and every element of A is present in D]
And, B ⊂ D [Each and every element of B is present in D]
Solution 2:
A = {x| x ∈ Z, 4 ≤ (x2 + 1) <25}
∴ A = {-5, -4, -3, 1, 2, 3}
Given, B = {-2, -1, 0, 1, 2}
A ∪ B = {-5, -4, -3, 1, 2, 3} ∪ {-2, -1, 0, 1, 2}
∴ A ∪ B = {-5, -4, -3, -2, -1, 0, 1, 2, 3}
A ∩ B = {-5, -4, -3, 1, 2, 3} ∩ {-2, -1, 0, 1, 2}
∴ A ∩ B = {1, 2}
Solution 3:
For any two sets A and B, if A ⊂ B then A ∪ B = B and A ∩ B = A.
Given, A ⊂ B and A ≠ Ø.
∴ A ∩ B = A ≠ Ø.
Therefore, A and B can never be disjoint sets.
Solution 4:
U = {x| x ∈ N, x ≤ 20}
∴ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
A = {x| x ∈ N, x is factor of 18}
∴ A = {1, 2, 3, 6, 9, 18}
A’= {x| x ∈ U, x ∉ A}
∴ A’ = {4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20}
B = {x| x ∈ N, x is a multiple of 3, x < 20}
∴ B = {3, 6, 9, 12, 15, 18}
B’ = {x| x ∈ U, x ∉ B}
∴ B’ = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}
Verification of De Morgan’s laws
- To prove: (A ∪ B)’ = A’ ∩ B’A ∪ B = {1, 2, 3, 6, 9, 12, 15, 18}∴ (A ∪ B)’= {4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20} …(i)A’ ∩ B’ = {4, 5, 7, 8, 10, 11,13, 14, 15, 16, 17, 19, 20} …(ii)
∴ (A ∪ B)’ = A’ ∩ B’ from (i) and(ii)
A’ ∩ B’ = {4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20} ∩ {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}
(A ∪ B)’= {x| x ∈ U, x ∉ (A ∪ B)}
A ∪ B = {1, 2, 3, 6, 9, 18} ∪ {3, 6, 9, 12, 15, 18} - To prove: (A ∩ B)’ = A’ ∪ B’∴ (A ∩ B) = {3, 6, 9, 18}(A ∩ B)’ = {1, 2, 4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20} …(i)A’ ∪ B’ = {1, 2, 4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20} …(ii)
(A ∩ B)’ = A’ ∪ B’ … from (i) and (ii)
A’ ∪ B’ = {4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20} ∪ {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}
(A ∩ B)’ = {x| x ∈ U, x ∉ (A ∩ B)}
A ∩ B = {1, 2, 3, 6, 9, 18} ∩ {3, 6, 9, 12, 15, 18}
Solution 5:
Given, U = {1, 2, 3, ……., 20}
A = {1, 2, 3, 4, 5, 6, 7, 8}
B = {x| x ∈ N, x <10}
∴ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
C = {x| x ∈ N, 3x < 20}
∴ C = {1, 2, 3, 4, 5, 6}
To verify the distributive laws,
- A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
- A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Now, we have to show that
- A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(B ∩ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∩ {1, 2, 3, 4, 5, 6}
∴ (B ∩ C) = {1, 2, 3, 4, 5, 6}
A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∪ {1, 2, 3, 4, 5, 6}
∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6, 7, 8} …(i)
(A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8} ∪ {1, 2, 3, 4, 5, 6, 7, 8, 9}
∴ (A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∪ {1, 2, 3, 4, 5, 6}
∴ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8}
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∩ {1, 2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} …(ii)
∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) …from (i) and (ii) - A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∪ {1, 2, 3, 4, 5, 6}
∴ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ∩ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9}
∴ A ∩ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} …(i)
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9}
∴ (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8}
A ∩ C = {1, 2, 3, 4, 5, 6, 7, 8} ∩ {1, 2, 3, 4, 5, 6}
∴ (A ∩ C) = {1, 2, 3, 4, 5, 6}
(A ∩ B) ∪ (A ∩ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∪ {1, 2, 3, 4, 5, 6}
∴ (A ∩ B) ∪ (A ∩ C) = {1, 2, 3, 4, 5, 6, 7, 8} …(ii)
∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) …from (i) and (ii)
* The question given in the textbook has been rectified and the solution has been provided accordingly.
Solution 6:
Given U = {1, 3, 4}
A = {1}
A’ = {x| x ∈ U, x ∉ A}
∴ A’ = {3, 4}
Solution 7:
Given, U = Z
U = {… -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 …}
- A = {2x| x ∈ Z}
∴ A = {……. -4, -2, 0, 2, 4, …..}
A’ = {x| x ∈ U, x ∉ A}
∴ A’ = {…. -5, -3, -1, 1, 3, 5, ……}
∴ A’= {2x – 1 | x ∈ Z }
But, B = {2x – 1 | x ∈ Z }
∴ A’ = B - B = {2x – 1 | x ∈ Z }
∴ B = {……. -5, -3, -1, 1, 3, 5, …..}
B’ = {x| x ∈ U, x ∉ B}
∴ B’ = {…. -4, -2, 0, 2, 4, ……}
∴ B’= {2x| x ∈ Z }
But, A = {2x – 1 | x ∈ Z }
∴ B’ = A
Solution 8:
Set A contains 4 elements.
∴ Total number of subsets of A = 24 = 16.
Thus, there are 16 subsets of A which can be written as follows.
Ø, {-1}, {0}, {1}, {2}, {-1, 0}, {-1, 1}, {-1, 2}, {0, 1}, {0, 2}, {1, 2}, {-1, 0, 1}, {-1, 0, 2}, {-1, 1, 2}, {0, 1, 2}, {-1, 0, 1, 2}
Solution 9:
Given, A = {6, 8, 10, 12, 14}
B = {8, 9, 10, 11, 12, 13}
C = {7, 8, 9, 10, 12, 14}
To prove that (A ∩ C) ∪ B = (A ∪ B) ∩ (B ∪ C)
A ∩ C = {6, 8, 10, 12, 14} ∩ {7, 8, 9, 10, 12, 14}
∴ A ∩ C = {8, 10, 12, 14}
(A ∩ C) ∪ B = {8, 10, 12, 14} ∪ {8, 9, 10, 11, 12, 13}
∴ (A ∩ C) ∪ B = {8, 9, 10, 11, 12, 13, 14} …(i)
(A ∪ B) = {6, 8, 10, 12, 14} ∪ {8, 9, 10, 11, 12, 13}
∴ (A ∪ B) = {6, 8, 9, 10, 11, 12, 13, 14}
(B ∪ C) = {8, 9, 10, 11, 12, 13} ∪ {7, 8, 9, 10, 12, 14}
∴ (B ∪ C) = {7, 8, 9, 10, 11, 12, 13, 14}
(A ∪ B) ∩ (B ∪ C) = {6, 8, 9, 10, 11, 12, 13, 14} ∩ {7, 8, 9, 10, 11, 12, 13, 14}
∴ (A ∪ B) ∩ (B ∪ C) = {8, 9, 10, 11, 12, 13, 14} …(ii)
∴ (A ∩ C) ∪ B = (A ∪ B) ∩ (B ∪ C) …from (i) and (ii)
Solution 10(1):
C. {3, 4}
U = {x|x ∈ N, x < 5}
∴U = {1, 2, 3, 4}
A = {x|x ∈ N, x ≤ 2}
∴ A = {1, 2}
A’ = {x|x ∈ U, x ∉ A}
∴ A’ = {3, 4}
Solution 10(2):
a. ⊂
The null set Ø is subset of every set.
Therefore, Ø ⊂ {Ø}.
Solution 10(3):
a. {1, 2, 3, 4, 5}
A ∪ B = {1, 2, 3} ∪ {3, 4, 5}
∴ A ∪ B = {1, 2, 3, 4, 5}
Solution 10(4):
b. ⊂
A = {x |x ∈ N, x ≤ 7}
∴ A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 4, 6}
Each and every element of set B is present in set A.
Therefore, B ⊂ A.
Solution 10(5):
b. {2}
A = {1, 2, 3}
B = {2, 3, 4}
C = {3, 4, 5}
C’ = {x| x ∈ U, x ∉ C}
∴ C’ = {1, 2}
U = {1, 2, 3, 4, 5}
(A ∩ B) = {1, 2, 3} ∩ {2, 3, 4}
∴ (A ∩ B) = {2, 3}
(A ∩ B) ∩ C’ = {2, 3} ∩ {1, 2}
∴ (A ∩ B) ∩ C’ = {2}
Solution 10(6):
a. equal
A = {x| x ∈ N, x ≤ 3}
∴ A = {1, 2, 3}
B = {1, 2, 3}
∴ A = B
If A = {x| x ∈ N, x ≤ 3}, B = {1, 2, 3}, U = N, then A and B are equal sets.
Solution 10(7):
d. {3, 4} ⊂ A
3 ∉ A is incorrect because 3 ∈ A.
{1} ∈ A is incorrect because 1 ⊂ A.
{2} ∈ A is incorrect because 2 ⊂ A
Solution 10(8):
d. 16
A = {1, 2, 3, 4}
Set A has 4 elements.
∴ Number of subsets of A = 24 = 16
Solution 10(9):
d. D = {x|x ∈ Z, x is neither negative nor positive}
A = {x| x ∈ R, x2 – x = 0}
∴ A = {0, 1}
B = {x| x ∈ N, 2x = 3}
∴ B = Ø
C = {x| x ∈ R, x2 = – 4}
∴ C = Ø
D = {x| x ∈ Z, x is neither negative nor positive}
∴ D = {0}
Therefore, D is a singleton set.
Solution 10(10):
c. C
A = {0, 1, 2, 4}
B = {1, 3, 5, 7, 9}
C = {0, 1, 4, 3, 9}
A ∩ B = {0, 1, 2, 4} ∩ {1, 3, 5, 7, 9}
∴ A ∩ B = {1}
(A ∩ B) ∪ C = {1} ∪ {0, 1, 4, 3, 9}
∴ (A ∩ B) ∪ C = {0, 1, 4, 3, 9}
But, C = {0, 1, 4, 3, 9}
Therefore, (A ∩ B) ∪ C = C
Solution 10(11):
d. A = Ø and B = Ø
If A ∪ B ≠ Ø, then A = Ø and B = Ø.
Solution 10(12):
c. {0, 1, 2, 3, 4}
A = {x| x ∈ N, x ≤ 4}
∴ A = {1, 2, 3, 4}
B = {-1, 0, 1, 2, 3}
C = {0, 1, 2}
(A ∪ B) = {1, 2, 3, 4} ∪ {-1, 0, 1, 2, 3}
∴ (A ∪ B) = {-1, 0, 1, 2, 3, 4}
(A ∪ C) = {1, 2, 3, 4} ∪ {0, 1, 2}
∴ (A ∪ C) = {0, 1, 2, 3, 4}
(A ∪ B) ∩ (A ∪ C) = {-1, 0, 1, 2, 3, 4} ∩ {0, 1, 2, 3, 4}
∴ (A ∪ B) ∩ (A ∪ C) = {0, 1, 2, 3, 4}
Solution 10(13):
c. {7}
U = {1, 2, 3, 4, 5, 6, 7}
A = {1, 2, 3, 4}
A’ = {x| x ∈ U, x ∉ A}
∴ A’ = {5, 6, 7}
B = {2, 4, 5, 6}
B’ = {x| x ∈ U, x ∉ B}
∴ B = {1, 3, 7}
A’ ∩ B’ = {5, 6, 7} ∪ {1, 3, 7}
∴ A’ ∩ B’ = {7}
Solution 10(14):
a. Ø
Ø ∩ U’ = Ø
∵ U’ = Ø
We get,
Ø ∩ Ø’ = Ø
Solution 10(15):
b. A’ ∪ B
By De Morgan’s law,
(A ∩ B’)’ = A’ ∩ (B’)’
A’ ∩ (B’)’ = A’ ∪ B [∵(B’)’ = B]
Exercise 1.1:
Solution 1:
(a)
- x = -1 is the only value which satisfies x + 1 = 0 and x ∈ Z.∴ A = {x | x ∈ Z, x + 1 = 0} is a singleton set.
Thus, the set A has only one element -1.
∴ A = {x | x ∈ Z, x + 1 = 0} is a singleton set. - x = 1 is the only value which satisfies x2 – 1 = 0 and x ∈ N.
Thus, the set B has only one element 1.
B = {x | x ∈ N, x2 – 1 = 0} is a singleton set. - There are no prime numbers between 13 and 17.
Thus, C = {x | x ∈ N, x is a prime number between 13 and 17} is an empty set.
(b)
- A = {x | x ∈ N, x ≤ 7}
∴ A = {1, 2, 3, 4, 5, 6, 7}
B = {x | x ∈ Z, -3 ≤ x ≤ 3}
∴ B = {-3, -2, -1, 0, 1, 2, 3}
Both the sets have equal number of elements butnot the same elements.
Therefore, set A and B are equivalent sets. - A = {x | x ∈ N, x is a multiple of 2, x < 10}
∴ A = {2, 4, 6, 8}
B = {x | x ∈ N, x is an even natural number with a single digit}
∴ B = {2, 4, 6, 8}
Both the sets have same elements.
Therefore, set A and B are equal sets.
Solution 2:
Set A = {1, 2, 3} has three elements.
The number subsets of A = 2n = 23 = 8
The list of all subsets of A can be written as follows:
Ø , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
Solution 3:
A = {x | x ∈ Z, x2 – x = 0}
x2 – x = 0
∴ x (x – 1) = 0
∴ x = 0 or x – 1 = 0
∴ x = 0 ∈ Z or x = 1 ∈ Z
∴ A = {0, 1}
B = {x | x ∈ N, 1 ≤ x ≤ 4}
∴ B = {1, 2, 3, 4}
Element 0 is present in set A but is not present in set B.
Therefore, A is not subset of B.
So, we cannot say that A ⊂ B.
Solution 4:
Given,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 4, 6, 8}
A’ = {3, 5, 7, 9, 10}
A ∪ A’ = {1, 2, 4, 6, 8} ∪ {3, 5, 7, 9, 10}
A ∪ A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = U
Hence, A ∪ A’ = U
Solution 5:
The non-empty subsets of A are given below.
{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}
The non-empty subsets of B are given below.
{3}, {4}, {6}, {3, 4}, {3, 6}, {4, 6} and {3, 4, 6}
- Non-empty subsets of X satisfying X ⊂ A, X ⊄ B are given below.
{1}, {2}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}
- Non-empty subsets of X satisfying X ⊂ B, X ⊄ A are given below.
{4}, {6}, {1, 2}, {3, 4}, {4, 6} and {3, 4, 6}
- Non-empty subset of X satisfying X ⊂ A, X ⊂ B is {3}.
Solution 6:
- {1, 2, 3} ⊂ {1, 2, 3} ⇒ True
For every set A, A ⊂ A
- {a, b} ⊄ {b, c, a} ⇒ False
Each and every element of {a, b} is present in {b, c, a}
Therefore, {a, b} ⊂ {b, c, a}
- Ø ∉ {Ø} Þ False
Ø is null set
Therefore, Ø is the subset of set {Ø}.
Therefore, Ø ⊂ {Ø}
- {3} ⊂ {1, 2, {3}, 4} Þ False
Set {3} is an element of set {1, 2, {3}, 4} Therefore, {3} ∈ {1, 2, {3}, 4}
Exercise 1.2:
Solution 1:
set A = {1, 2, 3, 4}
set B = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6} …(1)
B ∪ A = {3, 4, 5, 6} ∪ {1, 2, 3, 4}
B ∪ A = {1, 2, 3, 4, 5, 6} …(2)
∴ A ∪ B = B ∪ A … from (1) and (2)
Solution 2:
A = {x| x ∈ N is a factor of 12}
∴ A = {1, 2, 3, 4, 6, 12}
B = {x| x ∈ N, 2 < x < 7}
∴ B = {3, 4, 5, 6}
A ∩ B = {1, 2, 3, 4, 6, 12} ∩ {3, 4, 5, 6}
∴ A ∩ B = {3, 4, 6}
Solution 3:
The distributive of union over intersection:
For any three sets A, B and C
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
We have,
A = {2, 3, 4, 5}, B = {4, 5, 6} and C = {1, 3, 5, 7}
B ∩ C = {4, 5, 6} ∩ {1, 3, 5, 7} = {5}
A ∪ (B ∩ C) = {2, 3, 4, 5} ∪ {5} = {2, 3, 4, 5}
A ∪ (B ∩ C)= {2, 3, 4, 5} …(1)
A ∪ B = {2, 3, 4, 5} ∪ {4, 5, 6} = {2, 3, 4, 5, 6}
A ∪ C = {2, 3, 4, 5} ∪ {1, 3, 5, 7} = {1, 2, 3, 4, 5, 7}
(A ∪ B) ∩ (A ∪ C) = {2, 3, 4, 5, 6} ∩ {1, 2, 3, 4, 5, 7}
(A ∪ B) ∩ (A ∪ C) = {2, 3, 4, 5} …(2)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) From (1) and (2)
Hence, the distributive of union over intersection is verified.
Solution 4:
A = {x| x ∈ N, x is a prime factor of 12}
∴ A = {2, 3}B = {x| x ∈ n, x is a prime factor of 20}
∴ B = {2, 5}
A ∩ B = {2, 3} ∩ {2, 5}
Solution 5:
A = {x| x ∈ N, x < 10}
∴A = {1, 2, 3, 4, 4, 5, 6, 7, 8, 9}
B = {x| x ∈ N, x is multiple of 3}
∴B = {3, 6, 9, 12}
C = {x| x ∈ Z, – 4 < x < 4}
∴ C = {- 3, – 2, – 1, 0 , 1, 2, 3}
We have,
A = {1, 2, 3, 4, 4, 5, 6, 7, 8, 9}, B = {3, 6, 9, 12} and C = {- 3 , – 2, – 1, 0 , 1, 2, 3}
B ∪ C = {3, 6, 9, 12} ∪ {- 3, – 2, – 1, 0 , 1, 2, 3}
B ∪ C = {- 3, – 2, – 1, 0, 1, 2, 3, 6, 9, 12}
A ∩ (B ∪ C) = {1, 2, 3, 4, 4, 5, 6, 7, 8, 9} ∩ {- 3, – 2, – 1, 0, 1, 2, 3, 6, 9, 12}
A ∪ (B ∩ C) = {1, 2, 3, 6, 9} …(1)
A ∩ B = {1, 2, 3, 4, 4, 5, 6, 7, 8, 9} ∩ {3, 6, 9, 12}
A ∩ B = {3, 6, 9}
A ∩ C = {1, 2, 3, 4, 4, 5, 6, 7, 8, 9} ∩ {- 3, – 2, – 1, 0, 1, 2, 3}
A ∩ C = {1, 2, 3}
(A ∩ B) ∩ (A ∩ C) = {3, 6, 9} ∩ {1, 2, 3}
∴(A ∩ B) ∩ (A ∩ C) = {1, 2, 3, 6, 9} …(2)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) From (1) and (2)
Solution 6:
A = {1, 2, 3, 4}
B = {x| x ∈ N, 4 ≤ x ≤ 6}
∴ B = {4, 5, 6}
A ∩ B = {1, 2, 3, 4} ∩ {4, 5, 6} = {4} …(1)
[For any two non-empty sets A and B, if A ∩ B = Ø, then the sets A and be are disjoint sets]
A ∩ B = Ø … From (1)
Hence, A and B are disjoint sets.
Exercise 1.3:
Solution 1:
U = {x| x ∈ N, x < 10}
∴ U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2x | x ∈ N, x < 5}
∴ A = {2, 4, 6, 8}
B = {1, 3, 4, 5}
Now, A ∪ B = {2, 4, 6, 8} ∪ {1, 3, 4, 5} = {1, 2, 3, 4, 5, 6, 8}
∴ (A ∪ B)’ = {x | x ∈ U, x ∉ (A ∪ B)} = {7, 9}
A ∩ B = {2, 4, 6, 8} ∩ {1, 3, 4, 5} = {4}
∴ (A ∩ B)’ = {x| x ∈ U, x ∉ (A ∩ B)} = {1, 2, 3, 5, 6, 7, 8, 9}
Solution 2:
U = {a, b, c, d, e, f, g, h}
P = {b, c, d, e, f}
∴ P’ = {x| x ∈ U, x ∉ P} = {a, g, h}
Q = {a, c, d, e, g}
∴ Q’ = {x| x ∈ U, x ∉ Q} = {b, f, h}
Verification of De Morgan’s laws:
- To prove that (P ∪ Q)’ = P’ ∩ Q’
P ∪ Q = {b, c, d, e, f} ∪ {a, c, d, e, g}
∴ P ∪ Q = {a, b, c, d, e, f, g}
(P ∪ Q)’ = {x| x ∈ U, x ∉ (P ∪ Q)}
∴ (P ∪ Q)’ = {h} …(i)
P’ ∩ Q’ = {a, g, h} ∩ {b, f, h}
∴ P’ ∩ Q’ = {h} …(ii)
∴ (P ∪ Q)’ = P’ ∩ Q’ …from (i) and (ii) - To prove that (P ∩ Q)’ = P’ ∪ Q’
P ∩ Q = {b, c, d, e, f} ∩ {a, c, d, e, g}
∴ P ∩ Q = {c, d, e}
(P ∩ Q)’ = {x| x ∈ U, x ∉ (P ∩ Q)}
∴ (P ∩ Q)’ = {a, b, f, g, h} …(i)
P’ ∪ Q’ = {a, g, h} ∪ {b, f, h}
∴ P’ ∪ Q’ = {a, b, f, g, h} …(ii)
∴ (P ∩ Q)’ = P’ ∪ Q’ …from (i) and (ii)
Solution 3:
Given, U = N
Let A = {1, 3, 4, 5} and B = {3, 5}
∴ B ⊂ A
A’ = {x| x ∈ U, x ∉ A}
∴ A’ = {2, 6, 7, 8, 9, 10……} …(i)
B’ = {x| x ∈ U, x ∉ B}
∴ B’ = {1, 2, 4, 6, 7, 8, 9, 10……}
A’ ∩ B’ = {2, 6, 7, 8, 9, 10……} ∩ {1, 2, 4, 6, 7, 8, 9, 10……}
∴ A’ ∩ B’ = {2, 6, 7, 8, 9, 10…}
∴ A’ ∩ B’ = A’
Therefore, if B ⊂ A, then A’ ∩ B’ = A’
Solution 4:
A = {x| x ∈ Z, x3 = x}
Given, x3 = x
∴ x3– x = 0
∴ x(x2 – 1) = 0
∴ x(x – 1)(x + 1) = 0
∴ x = 0 or (x – 1) = 0 or (x + 1) = 0
∴ x = 0 ∈ Z or x = 1 ∈ Z or x = -1 ∈ Z
∴ A = {-1, 0, 1}
B = {x| x ∈ Z, x2 = x}
Given, x2 = x
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x – 1 = 0
∴ x = 0 or x = 1
∴ x = 0 ∈ Z or x = 1 ∈ Z
∴ B = {0, 1}
C = {x| x ∈ N, x2 = x}
Given, x2 = x
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x – 1= 0
∴ x = 0 or x = 1
∴ x = 0 ∉ N or x = 1 ∈ Z
∴ C = {1}
U = {-1, 0, 1, 2} …(Given)
- (B ∪ C)’ = B’ ∩ C’
B ∪ C = {0, 1} ∪ {1} = {0, 1}
(B ∪ C)’ = {x | x ∈ U, x ∉ (B ∪ C)}
∴ (B ∪ C)’ = {-1, 2} …(i)
B = {0, 1} - ∴ B’ = {x| x ∈ U, x ∉ B} = {-1, 2}
C = {x| x ∈ U, x ∉ C}
∴ C’ = {-1, 0, 2}
B’ ∩ C’ = {-1, 2} ∩ {-1, 0, 2} = {-1, 2} …(ii)
∴ (B ∪ C)’ = B’ ∩ C’ …from (i) and (ii) - C = {1}
∴ C’ = {x| x ∈ U, x ∉ C}
∴ C’= {-1, 0, 2}
∴ (C’)’ = {x| x ∈ U, x ∉ C’}
∴ (C’)’ = {1}
But, C = {1}
∴ (C’)’ = C - B ∩ C = {0, 1} ∩ {1} = {1}
∴ (B ∩ C)’ = {x| x ∈ U, x ∉ (B ∩ C)}
∴ (B ∩ C)’ = {-1, 0, 2} …(i)
B’ ∪ C’ = {-1, 2} ∪ {-1, 0, 2} = {-1, 0, 2} ….(ii)
∴ (B ∩ C)’ = B’ ∪ C’ from (i) and (ii)
Solution 5:
U = {x| x ∈ N, (x + 1)2 < 40}
∴ U = {1, 2, 3, 4, 5}
A = {x| x ∈ N, x < 4} = {1, 2, 3}
B = {2x| x ∈ N, x < 4} = {2, 4}
A’ = {x| x ∈ U, x ∉ A} = {4, 5}
B’ = {x| x ∈ U, x ∉ B}’ = {1, 3, 5}
Verification of De Morgan’s laws:
- To prove that (A ∪ B)’ = A’ ∩ B’
A ∪ B = {1, 2, 3} ∪ {2, 4}
∴ A ∪ B = {1, 2, 3, 4}
(A ∪ B)’ = {x| x ∈ U, x ∉ (A ∪ B)}
∴ (A ∪ B)’ = {5} …(i)
A’ ∩ B’ = {4, 5} ∩ {1, 3, 5}
∴ A’ ∩ B’ = {5} …(ii)
(A ∪ B)’ = A’ ∩ B’ from (i) and (ii) - To prove that (A ∩ B)’ = A’ ∪ B’
A ∩ B = {1, 2, 3} ∩ {2, 4}
∴ A ∩ B = {2}
(A ∩ B)’ = {x| x ∈ U, x ∉ (A ∩ B)}
(A ∩ B)’ ={1, 3, 4, 5} …(i)
A’ ∪ B’ = {4, 5} ∪ {1, 3, 5}
A’ ∪ B’ = {1, 3, 4, 5} …(ii)
(A ∩ B)’ = A’ ∪ B’ from (i) and (ii)