If P = {factors of 36} and Q = {factors of 48}, find :
(i) `P cup Q`, (ii) `P cap Q`, (iii) `Q - P` , (iv) `P' cap Q`

Answer

To solve the problem, we first need to identify the factors of 36 and 48, and then we will find the required set operations.

### Step 1: Identify the factors of 36 and 48
- **Factors of 36**: The factors of 36 are the numbers that divide 36 without leaving a remainder. They are:
- 1, 2, 3, 4, 6, 9, 12, 18, 36
- **Factors of 48**: The factors of 48 are the numbers that divide 48 without leaving a remainder. They are:
- 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Thus, we can define the sets:
- \( P = \{1, 2, 3, 4, 6, 9, 12, 18, 36\} \)
- \( Q = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48\} \)

### Step 2: Find \( P \cup Q \) (Union of P and Q)
The union of two sets includes all the elements that are in either set, without duplication.
- \( P \cup Q = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48\} \)

### Step 3: Find \( P \cap Q \) (Intersection of P and Q)
The intersection of two sets includes only the elements that are present in both sets.
- \( P \cap Q = \{1, 2, 3, 4, 6, 12\} \)

### Step 4: Find \( Q - P \) (Difference of Q and P)
The difference of two sets \( Q - P \) includes the elements that are in Q but not in P.
- \( Q - P = \{8, 16, 24, 48\} \)

### Step 5: Find \( P' \cap Q \) (Complement of P intersect Q)
The complement of P, denoted \( P' \), includes all elements not in P (considering the universal set as the set of all factors of 48).
- \( P' = \{8, 16, 24, 48\} \) (These are the factors of 48 that are not in P)
- Now, we find the intersection of \( P' \) and \( Q \):
- \( P' \cap Q = \{8, 16, 24, 48\} \)

### Summary of Results
1. \( P \cup Q = \{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48\} \)
2. \( P \cap Q = \{1, 2, 3, 4, 6, 12\} \)
3. \( Q - P = \{8, 16, 24, 48\} \)
4. \( P' \cap Q = \{8, 16, 24, 48\} \)