relations 11th cbse 2.1

Question 1:

If, find the values of x and y.

ANSWER:

It is given that.

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore,  and.

 

∴ x = 2 and y = 1

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Question 2:

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?

ANSWER:

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B)

= (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Thus, the number of elements in (A × B) is 9.

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Question 3:

If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

ANSWER:

G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as

P × Q = {(p, q): p∈ P, q ∈ Q}

∴G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

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Question 4:

State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.

ANSWER:

(i) False

If P = {m, n} and Q = {n, m}, then

P × Q = {(m, m), (m, n), (n, m), (n, n)}

(ii) True

(iii) True

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Question 5:

If A = {–1, 1}, find A × A × A.

ANSWER:

It is known that for any non-empty set A, A × A × A is defined as

A × A × A = {(a, b, c): a, b, c ∈ A}

It is given that A = {–1, 1}

∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1),

(1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

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Question 6:

If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.

ANSWER:

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}

∴ A is the set of all first elements and B is the set of all second elements.

Thus, A = {a, b} and B = {x, y}

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Question 7:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

ANSWER:

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴L.H.S. = A × (B ∩ C) = A × Φ = Φ

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴L.H.S. = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To verify: A × C is a subset of B × D

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

We can observe that all the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D.

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Question 8:

Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

ANSWER:

A = {1, 2} and B = {3, 4}

∴A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

⇒ n(A × B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2m.

Therefore, the set A × B has 24 = 16 subsets. These are

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)},

{(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},

{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)},

{(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

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Question 9:

Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

ANSWER:

It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.

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Question 10:

The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

ANSWER:

We know that if n(A) = p and n(B) = q, then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒ n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),

(1, –1), (1, 0), and (1, 1)

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Question 1:

sets 11th cbse 1.4 1.5 1.7exercise full questions and answers explained

Question 1:

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 …}

As 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}

Question 2:

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

Question 3:

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.

Question 4:

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 ∪ C

(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}

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Question 5:

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

A ∩ B = Φ

Question 6:

If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find

(i) A ∩ B

(ii) B ∩ C

(iii) A ∩ C ∩ D

(iv) A ∩ C

(v) B ∩ D

(vi) A ∩ (B ∪ C)

(vii) A ∩ D

(viii) A ∩ (B ∪ D)

(ix) (A ∩ B) ∩ (B ∪ C)

(x) (A ∪ D) ∩ (B ∪ C)

Answer:

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11} = {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪ Φ = {7, 9, 11}

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

Question 7:

If A = {x: x is a natural number}, B ={x: x is an even natural number}

C = {x: x is an odd natural number} and D = {x: x is a prime number}, find

(i) A ∩ B

(ii) A ∩ C

(iii) A ∩ D

(iv) B ∩ C

(v) B ∩ D

(vi) C ∩ D

Answer:

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = {x: x is a even natural number} = B

(ii) A ∩ C = {x: x is an odd natural number} = C

(iii) A ∩ D = {x: x is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

Question 8:

Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}

(ii) {a, e, i, o, u}and {c, d, e, f}

(iii) {x: x is an even integer} and {x: x is an odd integer}

Answer:

(i) {1, 2, 3, 4}

{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ

Therefore, this pair of sets is disjoint.

Question 9:

If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},

C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find

(i) A – B

(ii) A – C

(iii) A – D

(iv) B – A

(v) C – A

(vi) D – A

(vii) B – C

(viii) B – D

(ix) C – B

(x) D – B

(xi) C – D

(xii) D – C

Answer:

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

Question 10:

If X = {a, b, c, d} and Y = {f, b, d, g}, find

(i) X – Y

(ii) Y – X

(iii) X ∩ Y

Answer:

(i) X – Y = {a, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

Question 11:

If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Answer:

R: set of real numbers

Q: set of rational numbers

Therefore, R – Q is a set of irrational numbers.

Question 12:

State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) {a, e, i, o, u } and {a, b, c, d} are disjoint sets.

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

Answer:

(i) False

As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) False

As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}

⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) True

As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) True

As {2, 6, 10} ∩ {3, 7, 11} = Φ

1.5 exercise

Ex 1.5 Class 11 Maths Question 1.
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.
Here 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’=U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
6, 7, 8, 9}

(ii) B’=U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= (1, 2, 3, 4, 5, 6}
(A∪C)’=U-(A∪C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
= {7, 8, 9}

(iv) A ∪ B = {1, 2, 3,4} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
(A∪B)’ = U – (A∪B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}

(v) We know that A’ = {5, 6, 7, 8, 9}
(A’)’ =U – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(B-C)’=U – (B-C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {1, 3, 4, 5, 6, 7, 9}.


Ex 1.5 Class 11 Maths Question 2.
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.
(i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e,f, g, h}

(ii) B’ = U – B = {a, b, c, d, e,f, g, h} – {d, e, f, g}
= {a, b, c, h}

(iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}.


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 ∈ W and 2x + 1 > 10}
Ex 1.5 Class 11 Maths Question 4.
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’
Ex 1.5 Class 11 Maths Question 5.
Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)’
(ii) A’∩B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Ex 1.5 Class 11 Maths Question 6.
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.
Here U = {x : x is a triangle}
A = {x: x is a triangle and has at least one angle different from 60°}
∴ A’ = U – A = {x : x is a triangle} – {x : x is a triangle and has atleast one angle different from 60°}
= {x : x is a triangle and has all angles equal to 60°)
= Set of all equilateral triangles.

1.5 Class 11 Maths Question 7.
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’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = φ
(iv) U’ ∩ A = φ ∩ A = φ

1.7 exercise

1. For any two sets A and B, prove that: A‘ – B‘ = B – A

Solution:

To prove, A’ – B’ = B – A

Firstly we need to show

A’ – B’ ⊆ B – A

Let, x ∈ A’ – B’

⇒ x ∈ A’ and x ∉ B’

⇒ x ∉ A and x ∈ B (since, A ∩ A’ = ϕ )

⇒ x ∈ B – A

It is true for all x ∈ A’ – B’

∴ A’ – B’ = B – A

Hence Proved.

2. For any two sets A and B, prove the following:
(i) A ∩ (A‘ ∪ B) = A ∩ B

(ii) A – (A – B) = A ∩ B

(iii) A ∩ (A ∪ B’) = ϕ

(iv) A – B = A Δ (A ∩ B)

Solution:

(i) A ∩ (A’ ∪ B) = A ∩ B

Let us consider LHS A ∩ (A’ ∪ B)

Expanding

(A ∩ A’) ∪ (A ∩ B)

We know, (A ∩ A’) =ϕ

⇒ ϕ ∪ (A∩ B)

⇒ (A ∩ B)

∴ LHS = RHS

Hence proved.

(ii) A – (A – B) = A ∩ B

For any sets A and B we have De-Morgan’s law

(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’

Consider LHS

= A – (A–B)

= A ∩ (A–B)’

= A ∩ (A∩B’)’

= A ∩ (A’ ∪ B’)’) (since, (B’)’ = B)

= A ∩ (A’ ∪ B)

= (A ∩ A’) ∪ (A ∩ B)

= ϕ ∪ (A ∩ B) (since, A ∩ A’ = ϕ)

= (A ∩ B) (since, ϕ ∪ x = x, for any set)

= RHS

∴ LHS=RHS

Hence proved.

(iii) A ∩ (A ∪ B’) = ϕ

Let us consider LHS A ∩ (A ∪ B’)

= A ∩ (A ∪ B’)

= A ∩ (A’∩ B’) (By De–Morgan’s law)

= (A ∩ A’) ∩ B’ (since, A ∩ A’ = ϕ)

= ϕ ∩ B’

= ϕ (since, ϕ ∩ B’ = ϕ)

= RHS

∴ LHS=RHS

Hence proved.

(iv) A – B = A Δ (A ∩ B)

Let us consider RHS A Δ (A ∩ B)

A Δ (A ∩ B) (since, E Δ F = (E–F) ∪ (F–E))

= (A – (A ∩ B)) ∪ (A ∩ B –A) (since, E – F = E ∩ F’)

= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)

= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)

= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)

= ϕ ∪ (A ∩ B’) ∪ ϕ

= A ∩ B’ (since, A ∩ B’ = A–B)

= A – B

= LHS

∴ LHS=RHS

Hence Proved

3. If A, B, C are three sets such that A ⊂ B, then prove that C – B ⊂ C – A.

Solution:

Given, ACB

To prove: C – B ⊂ C – A

Let us consider, x ∈ C–B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A

⇒ x ∈ C – A

Thus, x ∈ C–B ⇒ x ∈ C – A

This is true for all x ∈ C–B

∴ C – B ⊂ C – A

Hence proved.

4. For any two sets A and B, prove that
(i) (A ∪ B) – B = A – B

(ii) A – (A ∩ B) = A – B

(iii) A – (A – B) = A ∩ B

(iv) A ∪ (B – A) = A ∪ B

(v) (A – B) ∪ (A ∩ B) = A

Solution:

(i) (A ∪ B) – B = A – B

Let us consider LHS (A ∪ B) – B

= (A–B) ∪ (B–B)

= (A–B) ∪ ϕ (since, B–B = ϕ)

= A–B (since, x ∪ ϕ = x for any set)

= RHS

Hence proved.

(ii) A – (A ∩ B) = A – B

Let us consider LHS A – (A ∩ B)

= (A–A) ∩ (A–B)

= ϕ ∩ (A – B) (since, A-A = ϕ)

= A – B

= RHS

Hence proved.

(iii) A – (A – B) = A ∩ B

Let us consider LHS A – (A – B)

Let, x ∈ A – (A–B) = x ∈ A and x ∉ (A–B)

x ∈ A and x ∉ (A ∩ B)

= x ∈ A ∩ (A ∩ B)

= x ∈ (A ∩ B)

= (A ∩ B)

= RHS

Hence proved.

(iv) A ∪ (B – A) = A ∪ B

Let us consider LHS A ∪ (B – A)

Let, x ∈ A ∪ (B –A) ⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or x ∈ B and x ∉ A

⇒ x ∈ B

⇒ x ∈ (A ∪ B) (since, B ⊂ (A ∪ B))

This is true for all x ∈ A ∪ (B–A)

∴ A ∪ (B–A) ⊂ (A ∪ B)…… (1)

Conversely,

Let x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B

⇒ x ∈ A or x ∈ (B–A) (since, B ⊂ (A ∪ B))

⇒ x ∈ A ∪ (B–A)

∴ (A ∪ B) ⊂ A ∪ (B–A)…… (2)

From 1 and 2 we get,

A ∪ (B – A) = A ∪ B

Hence proved.

(v) (A – B) ∪ (A ∩ B) = A

Let us consider LHS (A – B) ∪ (A ∩ B)

Let, x ∈ A

Then either x ∈ (A–B) or x ∈ (A ∩ B)

⇒ x ∈ (A–B) ∪ (A ∩ B)

∴ A ⊂ (A – B) ∪ (A ∩ B)…. (1)

Conversely,

Let x ∈ (A–B) ∪ (A ∩ B)

⇒ x ∈ (A–B) or x ∈ (A ∩ B)

⇒ x ∈ A and x ∉ B or x ∈ B

⇒ x ∈ A

(A–B) ∪ (A ∩ B) ⊂ A………. (2)

∴ From (1) and (2), We get

(A–B) ∪ (A ∩ B) = A

Hence proved.