If `aN = {ax:x in N} and bNnncN=dN,`where`b,cin N` are relatively prime, then show that `d=bc`.

Options

A. d = bc
B. c = bd
C. b = cd
D. None of these

Explanation

To solve the problem step by step, we need to analyze the sets involved and use the properties of relatively prime numbers. Let's break it down:

### Step 1: Understand the Definitions
We have:
- \( aN = \{ ax : x \in N \} \) which represents the set of all positive integer multiples of \( a \).
- \( bN = \{ bx : x \in N \} \) which represents the set of all positive integer multiples of \( b \).
- \( cN = \{ cx : x \in N \} \) which represents the set of all positive integer multiples of \( c \).
- \( dN = bN \cap cN \), which is the intersection of the sets of multiples of \( b \) and \( c \).

### Step 2: Determine the Intersection \( bN \cap cN \)
The intersection \( bN \cap cN \) consists of all numbers 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 given by:
\[ \text{lcm}(b, c) = bc \]
Thus, the set of all multiples of \( \text{lcm}(b, c) \) is:
\[ bN \cap cN = \{ k \cdot \text{lcm}(b, c) : k \in N \} = \{ k \cdot bc : k \in N \} \]

### Step 3: Relate the Intersection to \( dN \)
Since \( dN = bN \cap cN \), we can express \( dN \) as:
\[ dN = \{ k \cdot bc : k \in N \} \]
This means that \( dN \) consists of all positive integer multiples of \( bc \).

### Step 4: Conclude that \( d = bc \)
From the definition of \( dN \), we can conclude that:
\[ d = bc \]
This is because \( dN \) is the set of all positive integer multiples of \( d \), and since it matches the set of multiples of \( bc \), we have shown that \( d \) must equal \( bc \).

### Final Result
Thus, we have shown that:
\[ d = bc \]

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