If `A={x:x=3n,n epsilon Z}` and `B={x:x=4n,n epsilon Z}`, then find (`A nn B`).

Answer

To solve the problem, we need to find the intersection of the two sets A and B defined as follows:

1. **Define the Sets:**
- Set A is defined as \( A = \{ x : x = 3n, n \in \mathbb{Z} \} \)
- Set B is defined as \( B = \{ x : x = 4n, n \in \mathbb{Z} \} \)

2. **Identify the Elements of Set A:**
- The elements of set A can be generated by substituting integer values for \( n \):
- For \( n = 0 \): \( x = 3(0) = 0 \)
- For \( n = 1 \): \( x = 3(1) = 3 \)
- For \( n = -1 \): \( x = 3(-1) = -3 \)
- For \( n = 2 \): \( x = 3(2) = 6 \)
- For \( n = -2 \): \( x = 3(-2) = -6 \)
- Continuing this, we get: \( \ldots, -6, -3, 0, 3, 6, 9, 12, \ldots \)
- Thus, the elements of set A are all multiples of 3:
\[ A = \{ \ldots, -6, -3, 0, 3, 6, 9, 12, \ldots \} \]

3. **Identify the Elements of Set B:**
- The elements of set B can be generated similarly:
- For \( n = 0 \): \( x = 4(0) = 0 \)
- For \( n = 1 \): \( x = 4(1) = 4 \)
- For \( n = -1 \): \( x = 4(-1) = -4 \)
- For \( n = 2 \): \( x = 4(2) = 8 \)
- For \( n = -2 \): \( x = 4(-2) = -8 \)
- Continuing this, we get: \( \ldots, -8, -4, 0, 4, 8, 12, 16, \ldots \)
- Thus, the elements of set B are all multiples of 4:
\[ B = \{ \ldots, -8, -4, 0, 4, 8, 12, 16, \ldots \} \]

4. **Find the Intersection \( A \cap B \):**
- The intersection \( A \cap B \) consists of elements that are common to both sets A and B.
- We need to find common multiples of 3 and 4. The least common multiple (LCM) of 3 and 4 is 12.
- Therefore, the elements in the intersection will be all multiples of 12:
\[ A \cap B = \{ x : x = 12n, n \in \mathbb{Z} \} \]

5. **Conclusion:**
- The intersection of sets A and B can be expressed as:
\[ A \cap B = \{ x : x = 12n, n \in \mathbb{Z} \} \]