I really need help with proving this problem:
For natural numbers k,n > 0 we define set M(k,n) = {k,2k,3k...nk}. Find out which elements are in following sets:
a) M(i,n) intersection M(j,n), where i,j,n are natural numbers and i, j are coprime numbers.
b) M(i,n) intersection M(j,n), where i,j,n are natural numbers and i, j are commensurable numbers.
Prove your answer.
My first thoughts are:
a) Thanks to that i and j are coprimes numbers that i would say that their intersection in this two sets would be their multiplication, and other elements are just multiples of this multiplication.
For example for M(15,n) and H(16,n): 15, 16 are coprimes. I think that M intersection H is: (1 * (15*16)=240;2 * 240;3 * 240...n * 240).
b) In this example i would use a procedure where i find prime factorization and least common multiple (LCM) would be our first element of the set. Next elements would be multiples of this least common multiple (LCM).
How can I prove that ?
We want to prove that $M(j, n) \cap M(i, n) = M(i \cdot j, m)$ where $m = \left[\frac{n\min(i, j)}{i\cdot j}\right]$. Suppose $i \cdot k_1 = j \cdot k_2$, since $(i, j) = 1$ then $j \mid k_1$. Then only numbers like $i \cdot j \cdot k$ are suitable. It is easy that all of them are suitable. I took $m = \left[\frac{n\min(i, j)}{i\cdot j}\right]$ since $i \cdot j \cdot m < \min(i,j) \cdot n$.
Use the same technique to prove that it will be $M\left(LCM(i, j), \left[\frac{n\min(i, j)}{LCM(i, j)}\right]\right)$.