Question

If \(aN=\{ax:x\in N\}\) and \(bN\cap cN=dN\) where \(b,c\in N\) are relatively prime then

• A. \(d=bc\)
• B. \(c=bd\)
• C. \(b=cd\)
• D. \(a=bd\)

Hint:

The intersection of multiples of two numbers is the set of multiples of their LCM.

Answer 1

The Correct Option is A

Concept:
The set: \[ aN=\{ax:x\in N\} \] means the set of all multiples of \(a\). Similarly: \[ bN=\text{set of multiples of }b \] and \[ cN=\text{set of multiples of }c \] The intersection: \[ bN\cap cN \] means numbers that are multiples of both \(b\) and \(c\).

Step 1: Understand the intersection of multiples.
Common multiples of \(b\) and \(c\) are multiples of: \[ \operatorname{lcm}(b,c) \] So: \[ bN\cap cN=\operatorname{lcm}(b,c)N \]

Step 2: Use relatively prime condition.
Given that \(b\) and \(c\) are relatively prime. This means: \[ \gcd(b,c)=1 \] For relatively prime numbers: \[ \operatorname{lcm}(b,c)=bc \]

Step 3: Compare with given form.
Given: \[ bN\cap cN=dN \] But: \[ bN\cap cN=bcN \] Therefore: \[ dN=bcN \] So: \[ d=bc \] Hence, the correct answer is: \[ \boxed{(A)\ d=bc} \]