Question

If ${\text{aN = \\{ ax:x}} \in {\text{N\\} }}$ and ${\text{bN}} \cap {\text{cN = dN}}$ , where ${\text{b,c}} \in {\text{N,b}} \geqslant {\text{2,c}} \geqslant {\text{2}}$ are relatively prime, then write d in terms of b and c.

Answer 1

First, we’ll find the value of bN and cN that are the sets of all positive multiples of b and c respectively. Then will find dN which is the intersection of sets bN and cN but because of the given condition that b and c are relatively prime numbers. So we’ll find the value of d in terms of b and c following the conditions given in the question.

Complete step by step answer:

Given data: ${\text{aN = \\{ ax:x}} \in {\text{N\\} }}$
${\text{bN}} \cap {\text{cN = dN}}$ , ${\text{b,c}} \in {\text{N,b}} \geqslant {\text{2,c}} \geqslant {\text{2}}$ are relatively prime
From the set given we can say that,
${\text{bN = \\{ bx:x}} \in {\text{N\\} }} \\\ {\text{ = \\{ b,2b,3b,4b,}}.....{\text{\\} }} \\\$
Similarly, for $cN$

$${\text{cN = \\{ cx:x}} \in {\text{N\\} }} \\\ {\text{ = \\{ c,2c,3c,4c,}}.....{\text{\\} }} \\\$$

From the above equations, we can conclude that
bN $=$ set of positive multiples of ‘b’
cN $=$ set of positive multiples of ‘c’
${\text{bN}} \cap {\text{cN = dN}}$
Where dN $=$set of positive multiples of d
but b and c are relatively prime numbers
Therefore, for d to exist it should be the product of the b and c as any number can only be a multiple of two different prime numbers if and only if the when both the prime numbers are multiplied with each other.
i.e. $d = bc$

Note: We can also say that multiples of two relatively prime numbers are equal if and only if they are multiplied with each other. Here we can say that ${\text{bN}} \cap {\text{cN = dN}}$ will occur if $d = bc$ , else there isn’t any other condition possible.