Complete Exercise 1.2 Class 11 Solutions: Logic, Sets and Real Number System

Exercise 1.2 Solutions - Set Theory

1. If A ∩ B = ∅, prove that:

a) A ⊆ B'

To prove: A ⊆ B'

Let x ∈ A (arbitrary element)

Since A ∩ B = ∅, x ∉ B

Therefore, x ∈ B'

Since x was arbitrary, A ⊆ B'

b) B ∩ A' = B

To prove: B ∩ A' = B

We need to show: B ∩ A' ⊆ B and B ⊆ B ∩ A'

(i) B ∩ A' ⊆ B (by definition of intersection)

(ii) Let x ∈ B (arbitrary element)

Since A ∩ B = ∅, x ∉ A

Therefore, x ∈ A'

So x ∈ B and x ∈ A' ⇒ x ∈ B ∩ A'

Since x was arbitrary, B ⊆ B ∩ A'

Therefore, B ∩ A' = B

c) A ∪ B' = B'

To prove: A ∪ B' = B'

We need to show: A ∪ B' ⊆ B' and B' ⊆ A ∪ B'

(i) From part (a), A ⊆ B'

Also B' ⊆ B' (trivially)

Therefore, A ∪ B' ⊆ B'

(ii) B' ⊆ A ∪ B' (by definition of union)

Therefore, A ∪ B' = B'

2. If A ⊆ B, prove that:

a) B' ⊆ A'

To prove: B' ⊆ A'

Let x ∈ B' (arbitrary element)

Then x ∉ B

Since A ⊆ B, if x ∉ B then x ∉ A

Therefore, x ∈ A'

Since x was arbitrary, B' ⊆ A'

b) A ∪ B = B

To prove: A ∪ B = B

We need to show: A ∪ B ⊆ B and B ⊆ A ∪ B

(i) Let x ∈ A ∪ B

Then x ∈ A or x ∈ B

If x ∈ A, since A ⊆ B, then x ∈ B

If x ∈ B, then x ∈ B

Therefore, A ∪ B ⊆ B

(ii) B ⊆ A ∪ B (by definition of union)

Therefore, A ∪ B = B

c) A ∩ B = A

To prove: A ∩ B = A

We need to show: A ∩ B ⊆ A and A ⊆ A ∩ B

(i) A ∩ B ⊆ A (by definition of intersection)

(ii) Let x ∈ A (arbitrary element)

Since A ⊆ B, x ∈ B

Therefore, x ∈ A and x ∈ B ⇒ x ∈ A ∩ B

Since x was arbitrary, A ⊆ A ∩ B

Therefore, A ∩ B = A

3. Prove that:

a) B - A = B ∩ A'

To prove: B - A = B ∩ A'

Let x ∈ B - A (arbitrary element)

Then x ∈ B and x ∉ A

⇒ x ∈ B and x ∈ A'

⇒ x ∈ B ∩ A'

Therefore, B - A ⊆ B ∩ A'

Now let x ∈ B ∩ A'

Then x ∈ B and x ∈ A'

⇒ x ∈ B and x ∉ A

⇒ x ∈ B - A

Therefore, B ∩ A' ⊆ B - A

Hence, B - A = B ∩ A'

b) A - B' = A ∩ B

To prove: A - B' = A ∩ B

Let x ∈ A - B' (arbitrary element)

Then x ∈ A and x ∉ B'

⇒ x ∈ A and x ∈ B

⇒ x ∈ A ∩ B

Therefore, A - B' ⊆ A ∩ B

Now let x ∈ A ∩ B

Then x ∈ A and x ∈ B

⇒ x ∈ A and x ∉ B'

⇒ x ∈ A - B'

Therefore, A ∩ B ⊆ A - B'

Hence, A - B' = A ∩ B

c) A - B = B' - A'

To prove: A - B = B' - A'

Let x ∈ A - B (arbitrary element)

Then x ∈ A and x ∉ B

⇒ x ∉ B and x ∈ A

⇒ x ∈ B' and x ∉ A'

⇒ x ∈ B' - A'

Therefore, A - B ⊆ B' - A'

Now let x ∈ B' - A'

Then x ∈ B' and x ∉ A'

⇒ x ∉ B and x ∈ A

⇒ x ∈ A and x ∉ B

⇒ x ∈ A - B

Therefore, B' - A' ⊆ A - B

Hence, A - B = B' - A'

4. If A, B and C are subsets of a universal set U, prove that:

a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

To prove: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Let x ∈ A ∪ (B ∩ C)

Then x ∈ A or x ∈ (B ∩ C)

Case 1: If x ∈ A, then x ∈ A ∪ B and x ∈ A ∪ C

⇒ x ∈ (A ∪ B) ∩ (A ∪ C)

Case 2: If x ∈ (B ∩ C), then x ∈ B and x ∈ C

⇒ x ∈ A ∪ B and x ∈ A ∪ C

⇒ x ∈ (A ∪ B) ∩ (A ∪ C)

Therefore, A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C)

Now let x ∈ (A ∪ B) ∩ (A ∪ C)

Then x ∈ A ∪ B and x ∈ A ∪ C

If x ∈ A, then x ∈ A ∪ (B ∩ C)

If x ∉ A, then since x ∈ A ∪ B and x ∈ A ∪ C,

we must have x ∈ B and x ∈ C

⇒ x ∈ B ∩ C

⇒ x ∈ A ∪ (B ∩ C)

Therefore, (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C)

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

b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

To prove: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Let x ∈ A ∩ (B ∪ C)

Then x ∈ A and x ∈ (B ∪ C)

⇒ x ∈ A and (x ∈ B or x ∈ C)

⇒ (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)

⇒ x ∈ (A ∩ B) ∪ (A ∩ C)

Therefore, A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C)

Now let x ∈ (A ∩ B) ∪ (A ∩ C)

Then x ∈ A ∩ B or x ∈ A ∩ C

If x ∈ A ∩ B, then x ∈ A and x ∈ B

⇒ x ∈ A and x ∈ B ∪ C

⇒ x ∈ A ∩ (B ∪ C)

If x ∈ A ∩ C, then x ∈ A and x ∈ C

⇒ x ∈ A and x ∈ B ∪ C

⇒ x ∈ A ∩ (B ∪ C)

Therefore, (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C)

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

5. If A, B and C are subsets of a universal set U, prove that:

a) (A ∪ B)' = A' ∩ B'

To prove: (A ∪ B)' = A' ∩ B'

Let x ∈ (A ∪ B)'

Then x ∉ A ∪ B

⇒ x ∉ A and x ∉ B

⇒ x ∈ A' and x ∈ B'

⇒ x ∈ A' ∩ B'

Therefore, (A ∪ B)' ⊆ A' ∩ B'

Now let x ∈ A' ∩ B'

Then x ∈ A' and x ∈ B'

⇒ x ∉ A and x ∉ B

⇒ x ∉ A ∪ B

⇒ x ∈ (A ∪ B)'

Therefore, A' ∩ B' ⊆ (A ∪ B)'

Hence, (A ∪ B)' = A' ∩ B'

b) (A ∩ B)' = A' ∪ B'

To prove: (A ∩ B)' = A' ∪ B'

Let x ∈ (A ∩ B)'

Then x ∉ A ∩ B

⇒ x ∉ A or x ∉ B

⇒ x ∈ A' or x ∈ B'

⇒ x ∈ A' ∪ B'

Therefore, (A ∩ B)' ⊆ A' ∪ B'

Now let x ∈ A' ∪ B'

Then x ∈ A' or x ∈ B'

⇒ x ∉ A or x ∉ B

⇒ x ∉ A ∩ B

⇒ x ∈ (A ∩ B)'

Therefore, A' ∪ B' ⊆ (A ∩ B)'

Hence, (A ∩ B)' = A' ∪ B'

c) A - (B ∪ C) = (A - B) ∩ (A - C) = (A - B) - C

To prove: A - (B ∪ C) = (A - B) ∩ (A - C) = (A - B) - C

First, prove A - (B ∪ C) = (A - B) ∩ (A - C):

Let x ∈ A - (B ∪ C)

Then x ∈ A and x ∉ B ∪ C

⇒ x ∈ A and (x ∉ B and x ∉ C)

⇒ (x ∈ A and x ∉ B) and (x ∈ A and x ∉ C)

⇒ x ∈ (A - B) ∩ (A - C)

Therefore, A - (B ∪ C) ⊆ (A - B) ∩ (A - C)

Now let x ∈ (A - B) ∩ (A - C)

Then x ∈ A - B and x ∈ A - C

⇒ x ∈ A and x ∉ B and x ∈ A and x ∉ C

⇒ x ∈ A and (x ∉ B and x ∉ C)

⇒ x ∈ A and x ∉ B ∪ C

⇒ x ∈ A - (B ∪ C)

Therefore, (A - B) ∩ (A - C) ⊆ A - (B ∪ C)

Hence, A - (B ∪ C) = (A - B) ∩ (A - C)

Now prove A - (B ∪ C) = (A - B) - C:

Let x ∈ A - (B ∪ C)

Then x ∈ A and x ∉ B ∪ C

⇒ x ∈ A and x ∉ B and x ∉ C

⇒ x ∈ A - B and x ∉ C

⇒ x ∈ (A - B) - C

Therefore, A - (B ∪ C) ⊆ (A - B) - C

Now let x ∈ (A - B) - C

Then x ∈ A - B and x ∉ C

⇒ x ∈ A and x ∉ B and x ∉ C

⇒ x ∈ A and x ∉ B ∪ C

⇒ x ∈ A - (B ∪ C)

Therefore, (A - B) - C ⊆ A - (B ∪ C)

Hence, A - (B ∪ C) = (A - B) - C

Therefore, A - (B ∪ C) = (A - B) ∩ (A - C) = (A - B) - C

d) A - (B ∩ C) = (A - B) ∪ (A - C)

To prove: A - (B ∩ C) = (A - B) ∪ (A - C)

Let x ∈ A - (B ∩ C)

Then x ∈ A and x ∉ B ∩ C

⇒ x ∈ A and (x ∉ B or x ∉ C)

⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)

⇒ x ∈ (A - B) ∪ (A - C)

Therefore, A - (B ∩ C) ⊆ (A - B) ∪ (A - C)

Now let x ∈ (A - B) ∪ (A - C)

Then x ∈ A - B or x ∈ A - C

If x ∈ A - B, then x ∈ A and x ∉ B

⇒ x ∈ A and x ∉ B ∩ C

⇒ x ∈ A - (B ∩ C)

If x ∈ A - C, then x ∈ A and x ∉ C

⇒ x ∈ A and x ∉ B ∩ C

⇒ x ∈ A - (B ∩ C)

Therefore, (A - B) ∪ (A - C) ⊆ A - (B ∩ C)

Hence, A - (B ∩ C) = (A - B) ∪ (A - C)

Note: All proofs use the element method, which is the standard approach for proving set equalities. The key is to show that each set is a subset of the other.