### Help with a lemma about rationals and coprime integers

Posted:

**Thu Jul 09, 2015 2:24 am UTC**For something I'm working on, I would like the following lemma to be true.

I'm reasonably certain that it is true, but it's one of those things where at first glance, you think "That's intuitive", but when you try to write down a proof, you realize that there are messy details to wade through.

Am I missing a clever approach by which you can prove it instantaneously? Or does it require a bit of wading through details?

The desired lemma:

(w,x,y,z) is a fixed 4-tuple of rational numbers. We consider all rational multiples of this 4-tuple, defined in the obvious way:

t(w,x,y,z) = (tw,tx,ty,tz) for any rational number t.

Some rational multiples of this rational 4-tuple will consist entirely of integers. Others will not.

My claim is that at most one rational multiple of the 4-tuple will consist of integers with no common divisor.

Example:

Maybe you can sort of see why I thought this might be "intuitive". However, writing a proof might be messy, and it feels like there should be an easier way.

Rough idea behind proof:

Write X = (a1/b1, a2/b2, a3/b3, a4/b4) where each of those fractions is in lowest terms

Let G = gcd(a1,a2,a3,a4) and let L = lcm(b1,b2,b3,b4)

Show that multiplying X by L/G would give you all integers

Show, very roughly speaking, that multiplying by "anything else" would either give you non-integers, or integers all having a common factor.

It seems like that would probably work, but it also seems like writing it all out would be a pain.

Am I missing a clever shortcut?

EDITED TO ADD:

Maybe the more "elegant" trick is to do something like this:

If X = (a1/b1, a2/b2, a3/b3, a4/b4), then certainly there exist some rational numbers t with the property that t*X consists only of integers (because for example we can choose t = b1b2b3b4).

Then maybe choose the infimum of all rational t such that t*X consists only of integers?

I'm reasonably certain that it is true, but it's one of those things where at first glance, you think "That's intuitive", but when you try to write down a proof, you realize that there are messy details to wade through.

Am I missing a clever approach by which you can prove it instantaneously? Or does it require a bit of wading through details?

The desired lemma:

(w,x,y,z) is a fixed 4-tuple of rational numbers. We consider all rational multiples of this 4-tuple, defined in the obvious way:

t(w,x,y,z) = (tw,tx,ty,tz) for any rational number t.

Some rational multiples of this rational 4-tuple will consist entirely of integers. Others will not.

My claim is that at most one rational multiple of the 4-tuple will consist of integers with no common divisor.

Example:

**Spoiler:**

Maybe you can sort of see why I thought this might be "intuitive". However, writing a proof might be messy, and it feels like there should be an easier way.

Rough idea behind proof:

Write X = (a1/b1, a2/b2, a3/b3, a4/b4) where each of those fractions is in lowest terms

Let G = gcd(a1,a2,a3,a4) and let L = lcm(b1,b2,b3,b4)

Show that multiplying X by L/G would give you all integers

Show, very roughly speaking, that multiplying by "anything else" would either give you non-integers, or integers all having a common factor.

It seems like that would probably work, but it also seems like writing it all out would be a pain.

Am I missing a clever shortcut?

EDITED TO ADD:

Maybe the more "elegant" trick is to do something like this:

If X = (a1/b1, a2/b2, a3/b3, a4/b4), then certainly there exist some rational numbers t with the property that t*X consists only of integers (because for example we can choose t = b1b2b3b4).

Then maybe choose the infimum of all rational t such that t*X consists only of integers?