RD Chapter 1- Sets Ex-1.1 |
RD Chapter 1- Sets Ex-1.2 |
RD Chapter 1- Sets Ex-1.3 |
RD Chapter 1- Sets Ex-1.4 |
RD Chapter 1- Sets Ex-1.5 |
RD Chapter 1- Sets Ex-1.7 |
RD Chapter 1- Sets Ex-1.8 |

**Answer
1** :

A ∪ {1, 2} = {1, 2, 3, 5, 9}

Elements of A and {1, 2} together give us the result

So smallest set of A can be

A = {1, 2, 3, 5, 9} – {1, 2}

A = {3, 5, 9}

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∩ (B – C) = (A ∩ B) – (A ∩ C)

(iv) A – (B ∪ C) = (A – B) ∩ (A – C)

(v) A – (B ∩ C) = (A – B) ∪ (A – C)

(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

**Answer
2** :

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Firstly let us consider the LHS

(B ∩ C) = {x: x ∈ B and x ∈ C}

= {5, 6}

A ∪ (B ∩ C) = {x: x ∈ A or x ∈ (B ∩ C)}

= {1, 2, 4, 5, 6}

Now, RHS

(A ∪ B) = {x: x ∈ A or x ∈ B}

= {1, 2, 4, 5, 6}.

(A ∪ C) = {x: x ∈ A or x ∈ C}

= {1, 2, 4, 5, 6, 7}

(A ∪ B) ∩ (A ∪ C) = {x: x ∈ (A ∪ B) and x ∈ (A ∪ C)}

= {1, 2, 4, 5, 6}

∴ LHS = RHS

Hence Verified.

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Firstly let us consider the LHS

(B ∪ C) = {x: x ∈ B or x ∈ C}

= {2, 3, 4, 5, 6, 7}

(A ∩ (B ∪ C)) = {x: x ∈ A and x ∈ (B ∪ C)}

= {2, 4, 5}

Now, RHS

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {2, 5}

(A ∩ C) = {x: x ∈ A and x ∈ C}

= {4, 5}

(A ∩ B) ∪ (A ∩ C) = {x: x ∈ (A∩B) and x ∈ (A∩C)}

= {2, 4, 5}

∴ LHS = RHS

Hence verified.

(iii) A ∩ (B – C) = (A ∩ B) – (A ∩ C)

B–C is defined as {x ∈ B: x ∉ C}

B = {2, 3, 5, 6}

C = {4, 5, 6, 7}

B–C = {2, 3}

Firstly let us consider the LHS

(A ∩ (B – C)) = {x: x ∈ A and x ∈ (B – C)}

= {2}

Now, RHS

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {2, 5}

(A ∩ C) = {x: x ∈ A and x ∈ C}

= {4, 5}

(A ∩ B) – (A ∩ C) is defined as {x ∈ (A ∩ B): x ∉ (A ∩ C)}

= {2}

∴ LHS = RHS

Hence Verified.

(iv) A – (B ∪ C) = (A – B) ∩ (A – C)

Firstly let us consider the LHS

(B ∪ C) = {x: x ∈ B or x ∈ C}

= {2, 3, 4, 5, 6, 7}.

A – (B ∪ C) is defined as {x ∈ A: x ∉ (B ∪ C)}

A = {1, 2, 4, 5}

(B ∪ C) = {2, 3, 4, 5, 6, 7}

A – (B ∪ C) = {1}

Now, RHS

(A – B)

A – B is defined as {x ∈ A: x ∉ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A – B = {1, 4}

(A – C)

A – C is defined as {x ∈ A: x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) ∩ (A – C) = {x: x ∈ (A – B) and x ∈ (A – C)}.

= {1}

∴ LHS = RHS

Hence verified.

(v) A – (B ∩ C) = (A – B) ∪ (A – C)

Firstly let us consider the LHS

(B ∩ C) = {x: x ∈ B and x ∈ C}

= {5, 6}

A – (B ∩ C) is defined as {x ∈ A: x ∉ (B ∩ C)}

A = {1, 2, 4, 5}

(B ∩ C) = {5, 6}

(A – (B ∩ C)) = {1, 2, 4}

Now, RHS

(A – B)

A – B is defined as {x ∈ A: x ∉ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A–B = {1, 4}

(A – C)

A – C is defined as {x ∈ A: x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) ∪ (A – C) = {x: x ∈ (A – B) OR x ∈ (A – C)}.

= {1, 2, 4}

∴ LHS = RHS

Hence verified.

(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}.

Firstly let us consider the LHS

A ∩ (B △ C)

B △ C = (B – C) ∪ (C – B) = {2, 3} ∪ {4, 7} = {2, 3, 4, 7}

A ∩ (B △ C) = {2, 4}

Now, RHS

A ∩ B = {2, 5}

A ∩ C = {4, 5}

(A ∩ B) △ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]

= {2} ∪ {4}

= {2, 4}

∴ LHS = RHS

Hence, Verified.

If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that:

(i) (A ∪ B)’ = A’ ∩ B’

(ii) (A ∩ B)’ = A’ ∪ B’

**Answer
3** :

(i) (A ∪ B)’ = A’ ∩ B’

Firstly let us consider the LHS

A ∪ B = {x: x ∈ A or x ∈ B}

= {2, 3, 5, 7, 9}

(A∪B)’ means Complement of (A∪B) with respect to universal set U.

So, (A∪B)’ = U – (A∪B)’

U – (A∪B)’ is defined as {x ∈ U: x ∉ (A∪B)’}

U = {2, 3, 5, 7, 9}

(A∪B)’ = {2, 3, 5, 7, 9}

U – (A∪B)’ = ϕ

Now, RHS

A’ means Complement of A with respect to universal set U.

So, A’ = U – A

(U – A) is defined as {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U – B

(U – B) is defined as {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∩ B’ = {x: x ∈ A’ and x ∈ C’}.

= ϕ

∴ LHS = RHS

Hence verified.

(ii) (A ∩ B)’ = A’ ∪ B’

Firstly let us consider the LHS

(A ∩ B)’

(A ∩ B) = {x: x ∈ A and x ∈ B}.

= {7}

(A∩B)’ means Complement of (A ∩ B) with respect to universal set U.

So, (A∩B)’ = U – (A ∩ B)

U – (A ∩ B) is defined as {x ∈ U: x ∉ (A ∩ B)’}

U = {2, 3, 5, 7, 9}

(A ∩ B) = {7}

U – (A ∩ B) = {2, 3, 5, 9}

(A ∩ B)’ = {2, 3, 5, 9}

Now, RHS

A’ means Complement of A with respect to universal set U.

So, A’ = U – A

(U – A) is defined as {x ∈ U: x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = U – A = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U – B

(U – B) is defined as {x ∈ U: x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∪ B’ = {x: x ∈ A or x ∈ B}

= {2, 3, 5, 9}

∴ LHS = RHS

Hence verified.

For any two sets A and B, prove that

(i) B ⊂ A ∪ B

(ii) A ∩ B ⊂ A

(iii) A ⊂ B ⇒ A ∩ B = A

**Answer
4** :

(i) B ⊂ A ∪ B

Let us consider an element ‘p’ such that it belongs to B

∴ p ∈ B

p ∈ B ∪ A

B ⊂ A ∪ B

(ii) A ∩ B ⊂ A

Let us consider an element ‘p’ such that it belongs to B

∴ p ∈ A ∩ B

p ∈ A and p ∈ B

A ∩ B ⊂ A

(iii) A ⊂ B ⇒ A ∩ B = A

Let us consider an element ‘p’ such that it belongs to A ⊂ B.

p ∈ A ⊂ B

Then, x ∈ B

Let and p ∈ A ∩ B

x ∈ A and x ∈ B

x ∈ A and x ∈ A (since, A ⊂ B)

∴ (A ∩ B) = A

For any two sets A and B, show that the following statements are equivalent:

(i) A ⊂ B

(ii) A – B = ϕ

(iii) A ∪ B = B

(iv) A ∩ B = A

**Answer
5** :

(i) A ⊂ B

To show that the following four statements are equivalent, we need to prove (i)=(ii), (ii)=(iii), (iii)=(iv), (iv)=(v)

Firstly let us prove (i)=(ii)

We know, A–B = {x ∈ A: x ∉ B} as A ⊂ B,

So, Each element of A is an element of B,

∴ A–B = ϕ

Hence, (i)=(ii)

(ii) A – B = ϕ

We need to show that (ii)=(iii)

By assuming A–B = ϕ

To show: A∪B = B

∴ Every element of A is an element of B

So, A ⊂ B and so A∪B = B

Hence, (ii)=(iii)

(iii) A ∪ B = B

We need to show that (iii)=(iv)

By assuming A ∪ B = B

To show: A ∩ B = A.

∴ A⊂ B and so A ∩ B = A

Hence, (iii)=(iv)

(iv) A ∩ B = A

Finally, now we need to show (iv)=(i)

By assuming A ∩ B = A

To show: A ⊂ B

Since, A ∩ B = A, so A⊂B

Hence, (iv)=(i)

For three sets A, B, and C, show that

(i) A ∩ B = A ∩ C need not imply B = C.

(ii) A ⊂ B ⇒ C – B ⊂ C – A

**Answer
6** :

(i) A ∩ B = A ∩ C need not imply B = C.

Let us consider, A = {1, 2}

B = {2, 3}

C = {2, 4}

Then,

A ∩ B = {2}

A ∩ C = {2}

Hence, A ∩ B = A ∩ C, where, B is not equal to C

(ii) A ⊂ B ⇒ C – B ⊂ C – A

Given: A ⊂ B

To show: C–B ⊂ C–A

Let us consider x ∈ C– B

⇒ x ∈ C and x ∉ B [by definition C–B]

⇒ x ∈ C and x ∉ A

⇒ x ∈ C–A

Thus x ∈ C–B ⇒ x ∈ C–A. This is true for all x ∈ C–B.

∴ A ⊂ B ⇒ C – B ⊂ C – A

7. For any two sets, prove that:

(i) A ∪ (A ∩ B) = A

(ii) A ∩ (A ∪ B) = A

Solution:

(i) A ∪ (A ∩ B) = A

We know union is distributive over intersection

So, A ∪ (A ∩ B)

(A ∪ A) ∩ (A ∪ B) [Since, A ∪ A = A]

A ∩ (A ∪ B)

A

(ii) A ∩ (A ∪ B) = A

We know union is distributive over intersection

So, (A ∩ A) ∪ (A ∩ B)

A ∪ (A ∩ B) [Since, A ∩ A = A]

A

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