Let `A={(n,2n):n epsilon N}` and B={(2n,3n):n `epsilon` N}. What is `A nnB` equals to ?

Options

A. `{(n,6n):n epsilon N}`
B. `{(2n,6n):n epsilon N}`
C. `{(n,3n):n epsilon N}`
D. `phi`

Answer

To find the intersection of the sets \( A \) and \( B \), we first need to define both sets clearly.

1. **Define Set A**:
- The set \( A \) is defined as \( A = \{(n, 2n) : n \in \mathbb{N}\} \).
- This means that for each natural number \( n \), the first element of the ordered pair is \( n \) and the second element is \( 2n \).
- For example, if we take \( n = 1, 2, 3 \), we get:
- For \( n = 1 \): \( (1, 2) \)
- For \( n = 2 \): \( (2, 4) \)
- For \( n = 3 \): \( (3, 6) \)
- Thus, a sample of set \( A \) is \( A = \{(1, 2), (2, 4), (3, 6), \ldots\} \).

2. **Define Set B**:
- The set \( B \) is defined as \( B = \{(2n, 3n) : n \in \mathbb{N}\} \).
- This means that for each natural number \( n \), the first element of the ordered pair is \( 2n \) and the second element is \( 3n \).
- For example, if we take \( n = 1, 2, 3 \), we get:
- For \( n = 1 \): \( (2, 3) \)
- For \( n = 2 \): \( (4, 6) \)
- For \( n = 3 \): \( (6, 9) \)
- Thus, a sample of set \( B \) is \( B = \{(2, 3), (4, 6), (6, 9), \ldots\} \).

3. **Finding the Intersection \( A \cap B \)**:
- The intersection \( A \cap B \) consists of all elements that are common to both sets \( A \) and \( B \).
- We need to find pairs \( (x, y) \) that can be expressed in both forms:
- From \( A \): \( (n, 2n) \)
- From \( B \): \( (2m, 3m) \)
- For an element \( (x, y) \) to be in both sets, we must have:
- \( x = n \) and \( y = 2n \) from set \( A \)
- \( x = 2m \) and \( y = 3m \) from set \( B \)

4. **Equating the Elements**:
- From \( A \): \( y = 2n \) implies \( n = \frac{y}{2} \).
- From \( B \): \( x = 2m \) implies \( m = \frac{x}{2} \).
- Now substituting \( n \) in terms of \( y \) into the equation from \( B \):
- We have \( x = n \) and \( y = 2n \).
- Therefore, \( n = \frac{y}{2} \) implies \( x = \frac{y}{2} \).

5. **Conclusion**:
- For \( (n, 2n) \) to equal \( (2m, 3m) \), we would need \( n = 2m \) and \( 2n = 3m \).
- Solving these equations leads to contradictions, meaning there are no common elements.
- Therefore, the intersection \( A \cap B \) is the empty set, denoted as \( \emptyset \).

**Final Answer**: \( A \cap B = \emptyset \)