(a) If `N_(k)={kn:n inN}`, find `N_(3)capN_(5)andN_(4)capN_(6)`.
(b) If `N_(a)=(an: n inN},` describe `N_(4)capN_(6)andN_(3)capN_(5)`.

Answer

To solve the given problem, we need to find the intersections of specific sets defined by multiples of natural numbers.

### Part (a)

1. **Define the sets**:
- \( N_3 = \{ 3n : n \in \mathbb{N} \} \)
- \( N_5 = \{ 5n : n \in \mathbb{N} \} \)
- \( N_4 = \{ 4n : n \in \mathbb{N} \} \)
- \( N_6 = \{ 6n : n \in \mathbb{N} \} \)

2. **List the elements of each set**:
- \( N_3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, \ldots \} \)
- \( N_5 = \{ 5, 10, 15, 20, 25, 30, 35, \ldots \} \)
- \( N_4 = \{ 4, 8, 12, 16, 20, 24, 28, 32, \ldots \} \)
- \( N_6 = \{ 6, 12, 18, 24, 30, 36, \ldots \} \)

3. **Find \( N_3 \cap N_5 \)**:
- The common elements in \( N_3 \) and \( N_5 \) are the multiples of 15.
- \( N_3 \cap N_5 = \{ 15, 30, 45, 60, \ldots \} = N_{15} \)

4. **Find \( N_4 \cap N_6 \)**:
- The common elements in \( N_4 \) and \( N_6 \) are the multiples of 12.
- \( N_4 \cap N_6 = \{ 12, 24, 36, 48, \ldots \} = N_{12} \)

### Summary for Part (a):
- \( N_3 \cap N_5 = N_{15} \)
- \( N_4 \cap N_6 = N_{12} \)

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### Part (b)

1. **Define the general form**:
- \( N_a = \{ an : n \in \mathbb{N} \} \)

2. **Find \( N_4 \cap N_6 \)**:
- Since \( N_4 = \{ 4n : n \in \mathbb{N} \} \) and \( N_6 = \{ 6n : n \in \mathbb{N} \} \), the intersection will still yield multiples of 12.
- Thus, \( N_4 \cap N_6 = N_{12} \).

3. **Find \( N_3 \cap N_5 \)**:
- Similarly, \( N_3 = \{ 3n : n \in \mathbb{N} \} \) and \( N_5 = \{ 5n : n \in \mathbb{N} \} \) will yield multiples of 15.
- Thus, \( N_3 \cap N_5 = N_{15} \).

### Summary for Part (b):
- \( N_4 \cap N_6 = N_{12} \)
- \( N_3 \cap N_5 = N_{15} \)

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