The discussion revolves around a problem in ring theory, specifically addressing the intersection of two sets of integers generated by positive integers m and n, and how this relates to their least common multiple (lcm). Participants are exploring the implications of the double inclusion method in proving that the intersection of these sets equals another set generated by the lcm.
There is ongoing exploration of the problem, with participants questioning the validity of their reasoning and seeking clarification on the use of mathematical properties such as the division algorithm and prime factorization. Some guidance has been offered regarding the need to demonstrate that any common multiple of m and n is also a multiple of their lcm.
Participants note the importance of ensuring that all integers involved are positive and that the definitions of the sets are clear. There is also mention of potential confusion regarding the notation and terminology used in the problem statement.
haruspex said:Obvious approach is to let x be a member of the set on the LHS and show it is therefore in kZ; then show the converse (which is trivial).
Zondrina said:Hmm okay. So :
We want to show for some positive integers, m and n, that [itex]m \mathbb{Z} \cap n \mathbb{Z} = k \mathbb{Z}[/itex] for some integer [itex]k \in \mathbb{Z} \space | \space lcm(m,n) = k[/itex]
Case : mZ[itex]\cap[/itex]nZ [itex]\subseteq[/itex] kZ
Choose any a[itex]\in[/itex]mZ[itex]\cap[/itex]nZ. We want to show a[itex]\in[/itex]kZ.
Observe that a is also a positive integer because mZ[itex]\cap[/itex]nZ is the set of all positive integers such that a[itex]\in[/itex]mZ and a[itex]\in[/itex]nZ.
Hence a = mx and a = ny for some integers x and y.
Is this on the right track? I'm getting a bit stuck.
Stuff I didn't quite know if it mattered :
---Now, since lcm(m,n) = k, we know that k is also a positive integer so that kZ is a set of positive integers---
Dick said:Ok, you've shown a is a common multiple of m and n. Can you show ANY common multiple of m and n is a multiple of the least common multiple?
Zondrina said:Hmmm, I think I see what you're getting at here.
Since lcm(m,n) = k, then lcm(m,n)a = ka for any positive integer a ( Any integer really ). I wouldn't understand what you mean by 'show' though.
Dick said:There is some content here. You need to show that if a is any common multiple of n and m then it is an integer multiple of lcm(n,m). Think about what the lcm means in terms of prime factorization. See http://en.wikipedia.org/wiki/Least_common_multiple
Zondrina said:We need to show that if a is any common multiple of m and n, then it is an integer multiple of lcm(m,n) = k.
That is, we must show a = kr for some integer r.
Since m|k and n|k, we have k = mq and k = np for integers q and p.
Is this on the right track?
Dick said:You haven't made any false statements. But you need to use another property of the integers to prove it. Like the division algorithm or prime factorization. Did you look at the reference I gave you and think about it?
Zondrina said:Yeah I read a bit of the page. I didn't really see how the prime factorization would be appropriate here when I could apply the D.A. I thought of a slightly better way to go about this I think.
We need to show that if a is any common multiple of m and n ( which means it is also an multiple of m and n which I showed ), then it is an integer multiple of lcm(m,n) = k.
That is, we must show a = qk for some integer k.
So we divide k by a so that by the DA we write a = qk + r, 0 ≤ r < k.
Since m|a, n|a, m|k and n|k, they also divide r = a − qk. So, r is a common multiple of m and n. But k is the least positive common multiple of m and n.
So, we would have a contradiction unless r = 0. So, we must have r = 0 and k|a so that a = qk as contended.
Dick said:Sure, that works.
Zondrina said:Some things I said were wrong in that post, so I edited it a bit. Like I got mixed up with a bunch of the letters, but hopefully that's right now. So i'll combine the work done in the posts so far into this one.
We want to show for some positive integers, m and n, that [itex]m \mathbb{Z} \cap n \mathbb{Z} = k \mathbb{Z}[/itex] for some integer [itex]k \in \mathbb{Z} \space | \space lcm(m,n) = k[/itex]
Case : mZ[itex]\cap[/itex]nZ [itex]\subseteq[/itex] kZ
Choose any a[itex]\in[/itex]mZ[itex]\cap[/itex]nZ. We want to show a[itex]\in[/itex]kZ.
Observe that a is also a positive integer because mZ[itex]\cap[/itex]nZ is the set of all positive integers such that a[itex]\in[/itex]mZ and a[itex]\in[/itex]nZ.
Hence a = mx and a = ny for some integers x and y. So that a is an integer multiple of m and n.
We need to show that if a is any common multiple of m and n ( which means it is also an multiple of m and n which I showed ), then it is an integer multiple of lcm(m,n) = k.
That is, we must show a = kq for some integer q.
So we divide a by k so that by the DA we write a = kq + r, 0 ≤ r < k.
Since m|a, n|a, m|k and n|k, they also divide r = a − kq. So, r is a common multiple of m and n. But k is the least positive common multiple of m and n.
So, we would have a contradiction unless r = 0. So, we must have r = 0 and k|a so that a = kq as contended.
Okay, so I've shown that a is both a integer multiple of m and n and also k. Since kZ is the set of all integer multiples of k, then it must be that a[itex]\in[/itex]kZ as desired which implies that mZ[itex]\cap[/itex]nZ [itex]\subseteq[/itex] kZ.
Lookin good so far?
Dick said:It looks good to me.
Zondrina said:Sweet, got to love the mornings. So now for the other way.
Case : mZ[itex]\cap[/itex]nZ [itex]\supseteq[/itex] kZ
I know there must be a trivial way to show this which doesn't require as much work as before. I believe this has something to do with the lcm(m,n) = k.
Hmm... Since lcm(m,n) = k, we know every member of kZ is an integer multiple of both m and n. Hence every element of kZ is also in mZ[itex]\cap[/itex]nZ.
Is it really that straightforward?
Dick said:Yes it is. That is the easy direction.