**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 ≤ (x^{2} + 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 = 2^{4 }= 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 = 2^{4} = 16

**Solution 10(9):**

d. D = {x|x ∈ Z, x is neither negative nor positive}

A = {x| x ∈ R, x^{2} – x = 0}

∴ A = {0, 1}

B = {x| x ∈ N, 2x = 3}

∴ B = Ø

C = {x| x ∈ R, x^{2} = – 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 x
^{2}– 1 = 0 and x ∈ N.

Thus, the set B has only one element 1.

B = {x | x ∈ N, x^{2}– 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 = 2^{n} = 2^{3} = 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, x^{2} – x = 0}

x^{2} – 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, x^{3} = x}

Given, x^{3} = x

∴ x^{3}– x = 0

∴ x(x^{2} – 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, x^{2} = x}

Given, x^{2} = x

∴ x^{2} – 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, x^{2} = x}

Given, x^{2} = x

∴ x^{2} – 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)