If `aN = {ax|x in N}` and `bN nn cN = dN`, where `b,c epsilon N`, then

Options

A. `d = bc`
B. `c = bd`
C. `b = cd`
D. `d = LCM (b,c)`

Explanation

To solve the problem, we need to analyze the sets defined in the question and derive the relationship between them.

### Step-by-Step Solution:

1. **Understanding the Sets**:
- We are given that \( aN = \{ ax | x \in N \} \). This means \( aN \) is the set of all positive integer multiples of \( a \).
- Similarly, we can define \( bN = \{ bx | x \in N \} \) and \( cN = \{ cx | x \in N \} \), where \( b \) and \( c \) are also natural numbers.

2. **Intersection of Sets**:
- The intersection \( bN \cap cN \) consists of all elements that are common to both \( bN \) and \( cN \). This means we are looking for numbers that are multiples of both \( b \) and \( c \).

3. **Finding the Intersection**:
- The common multiples of \( b \) and \( c \) can be expressed in terms of their least common multiple (LCM). Therefore, we have:
\[ bN \cap cN = \{ k \cdot \text{lcm}(b, c) | k \in N \} \]
- This means that the intersection set consists of all positive integer multiples of \( \text{lcm}(b, c) \).

4. **Setting the Intersection Equal to \( dN \)**:
- We are given that \( bN \cap cN = dN \). This implies:
\[ dN = \{ dk | k \in N \} \]
- Since \( dN \) is also a set of multiples, we can equate the two sets:
\[ \{ k \cdot \text{lcm}(b, c) | k \in N \} = \{ dk | k \in N \} \]

5. **Equating the Multiples**:
- For the two sets to be equal, \( d \) must be equal to \( \text{lcm}(b, c) \). Thus, we conclude:
\[ d = \text{lcm}(b, c) \]

### Final Answer:
The relationship is given by:
\[ d = \text{lcm}(b, c) \]