Math Is Fun Forum

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Least common multiple of every natural number? :O

okay, so I proved in one of my homework problems that the intersection of the sets
{0, a, 2a, 3a, ... } and {0, b, 2b, 3b, ... } is { 0, u, 2u, 3u, ...} where u = LCM of a and b.

but in a later homework problem, it says:

let A[sub]n[/sub] = { kn ; k ∈ N }

(1) find the union of all the sets A[sub]n[/sub] from n = 1 to infinity
(2) find the intersection of all the sets A[sub]n[/sub] from n = 1 to infinity

1 is easy. Any element of A[sub]m[/sub] is also an element of A[sub]1[/sub]. So the union is just the natural numbers.

2, i don't know what to say. Given any finite upper limit M, we can find the intersection as
{0, u, 2u, 3u, ... } with u = lcm(1,2,3, 4, ... , M) but if we let M equal infinity, I don't know if u makes any sense.

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A logarithm is just a misspelled algorithm.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Least common multiple of every natural number? :O

Every positive integer k is missing from the set A[sub]k+1[/sub]. Hence, every positive integer k is missing from the intersection.

The intersection is therefore empty or {0}, depending on whether you consider 0 to be natural.

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Why did the vector cross the road?
It wanted to be normal.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Least common multiple of every natural number? :O

Thanks, mathsyperson!

That makes sense. On the otherhand, my alternate suggestion for how to build an infinite intersection also makes sense. Tricky thing about the infinite, so often two conflicting propositions seem correct.

I'll have to think about this for a while. Hmm.. maybe Anthony R. Brown can clear it up for me.

Last edited by mikau (2008-09-09 08:28:06)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Least common multiple of every natural number? :O

Don't try to think of things at infinity.  What it means to be in an infinite intersection is exactly what mathsyperson said, you must be in every single element of the collection.  If you can show that you aren't in at least one, then you aren't in the intersection.  You can think of this as "for all".  Also, for union, you can think of it as "there exists".

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."