if a N=`{a x:x in N} and b N cap c N = dN ,` where b,c in N are relatively prime , then

Options

A. `d=bc`
B. `c=bd`
C. `b=cd`
D. `a=bcd`

Answer

To solve the given problem step by step, we will analyze the information provided and derive the conclusion logically.

### Step-by-Step Solution:

1. **Understanding the Sets**:
- We are given that \( A_N = \{ ax : x \in \mathbb{N} \} \), which means \( A_N \) is the set of all positive integer multiples of \( a \).
- Similarly, we can define:
- \( B_N = \{ bx : x \in \mathbb{N} \} \) (the set of all positive integer multiples of \( b \))
- \( C_N = \{ cx : x \in \mathbb{N} \} \) (the set of all positive integer multiples of \( c \))

2. **Intersection of Sets**:
- The intersection \( B_N \cap C_N \) consists of all positive integers that are multiples of both \( b \) and \( c \).
- Since \( b \) and \( c \) are relatively prime, the least common multiple (LCM) of \( b \) and \( c \) is \( bc \).
- Therefore, we can express the intersection as:
\[ B_N \cap C_N = \{ k : k \text{ is a multiple of } bc \} = \{ d x : x \in \mathbb{N} \} \text{ where } d = bc \]

3. **Equating the Intersection to \( D_N \)**:
- According to the problem, we have:
\[ B_N \cap C_N = D_N \]
- From our previous step, we concluded that:
\[ D_N = \{ dx : x \in \mathbb{N} \} \]
- Thus, we can equate:
\[ D_N = \{ k : k \text{ is a multiple of } bc \} \]

4. **Conclusion**:
- Since \( D_N \) is defined as the set of multiples of \( d \), and we have established that \( d = bc \), we conclude:
\[ d = bc \]

### Final Answer:
Thus, the correct conclusion is that \( d = bc \).