Exercise 2.2

1. If A = {1, 2, 3, 4} and B = {1, 2, 3, 5, 6}, then find A ∩ B and B ∩ A. Are they equal?

Solution: Given: A = {1, 2, 3, 4} and B = {1, 2, 3, 5, 6}

Finding A ∩ B: A ∩ B represents elements that are .

Elements in A: , , ,

While Elements in B: , , , , .

Common elements: , ,

Therefore, A ∩ B = {1, 2, 3}

Finding B ∩ A: B ∩ A represents elements that are .

So, the common elements .

Therefore, B ∩ A = {1, 2, 3}

Are they equal?

Yes, they are equal. This demonstrates the commutative property of intersection: A ∩ B = B ∩ A

2. If A = {0, 2, 4}, find A ∩ φ and A ∩ A. What did you notice from the result?

Solution: Given: A = {0, 2, 4} Finding A ∩ φ: φ represents the set (or null set) which contains no elements.

A ∩ φ represents elements .

Since φ has no elements, there are no common elements. Therefore, A ∩ φ =

Finding A ∩ A: A ∩ A represents elements common to set A and .

Since every element of A is obviously in A, all elements of A are common. Therefore, A ∩ A = , ,

= A

Observations:

(1) A ∩ φ = φ: The intersection of any set with the empty set is always the set.

(2) A ∩ A = A: The intersection of any set with itself is always the set.

These are fundamental identity properties of set intersection.

3. If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A − B and B − A.

Solution:

Given: A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}

Finding A − B: A − B represents elements that .

Elements in A: , , , , Elements in B: , , , ,

Common element:

Elements in A but not in B: , , ,

Therefore, A − B = {2, 4, 8, 10}

Similarly, finding B − A: Elements in B but not in A: , , ,

Therefore, B − A = {3, 9, 12, 15}

Note: A − B ≠ B − A. Set difference is .

4. If A and B are two sets such that A ⊆ B then what is A ∪ B? Explain by giving an example.

Solution: If A ⊆ B, then A ∪ B =

Explanation: When A is a of B, element of A is in .

The A ∪ B contains elements from both sets, but since all elements of A are already in B, the union equals .

Example: Let A = {1, 3} and B = {1, 2, 3, 4, 5}

Verification that A ⊆ B: (1) 1 ∈ A and 1 ∈ and (2) 3 ∈ A and 3 ∈

Since all elements of A are in B, we have A ⊆ B.

Finding A ∪ B: A ∪ B = {1, 3} ∪ {1, 2, 3, 4, 5} A ∪ B = { , , , , } = B

This confirms that when A ⊆ B, then A ∪ B = B.

5. Let A = {x : x is a natural number}

B = {x : x is an even natural number}

C = {x : x is an odd natural number}

D = {x : x is a prime number}

Find A ∩ B, A ∩ C, A ∩ D, B ∩ C, B ∩ D and C ∩ D.

Solution: Set Definitions: A = {, , , , , , , , , , , , ...} (all natural numbers) while B = {, , , , , , ...} (all even natural numbers)

C = {, , , , , , ...} (all odd natural numbers) while D = {, , , , , , ...} (all prime numbers)

Finding A ∩ B: Since all even natural numbers also natural numbers, B A. Therefore, A ∩ B = = {x : x is an even natural number}

Finding A ∩ C: Since all odd natural numbers also natural numbers, C A. Therefore, A ∩ C = = {x : x is an odd natural number}

Finding A ∩ D: Since all prime numbers also natural numbers, D A. Therefore, A ∩ D = = {x : x is a prime number}

Finding B ∩ C: B contains only numbers, C contains only numbers. number can be both even and odd simultaneously. Therefore, B ∩ C = (empty set)

Finding B ∩ D: Even numbers in set D (prime numbers): only is the only even prime number. Therefore, B ∩ D = { }

Finding C ∩ D: Odd prime numbers: {, , , , , , , , ...}. All prime numbers except are odd. Therefore, C ∩ D = { , , , , , , , , ...} = {x : x is an odd prime number}

6. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and 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

Solution:

Given: A = { , , , , , , } ; B = { , , , , }

C = { , , , , , , , } ; D = { , , , }

(i) A − B: Elements in but not in : A ∩ B = { } (common elements). So, A − B = { , , , , , }

(ii) A − C: A ∩ C = { , }. So, A − C = { , , , , }

(iii) A − D: A ∩ D = { }. So, A − D = { , , , , , }

(iv) B − A: B ∩ A = { }. So, B − A = { , , , }

(v) C − A: C ∩ A = { , }. So, C − A = { , , , , , }

(vi) D − A: D ∩ A = { }. So, D − A = { , , }

(vii) B − C: B ∩ C = { , , , }. So, B − C = { }

(viii) B − D: B ∩ D = { }. So, B − D = { , , , }

(ix) C − B: C ∩ B = { , , , }. So, C − B = { , , , }

(x) D − B: D ∩ B = { }. So, D − B = { , , }

7. State whether the following statements is true or false. Justify your answers.

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

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

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

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

True

False

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

Solution: Two sets are disjoint if their intersection is (φ).

Finding intersection for the given sets: Common elements:

Intersection = {3} ≠ φ

Since the intersection is not empty, the sets are not disjoint.

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

Solution:

Finding intersection for the given sets: Common elements:

Intersection = {a} ≠ φ

Since the intersection is not empty, the sets are not disjoint.

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

Solution: Finding intersection for given sets: Common elements:

No number can be both even and odd simultaneously.

Intersection =

Since the intersection is empty, the sets are disjoint.

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

Solution:

Finding intersection: Common elements:

Intersection =

Since the intersection is empty, the sets are disjoint.

Chapter 2: Sets

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Exercise 2.2

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Chapter 2: Sets

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Glossary

Exercise 2.2