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In various comments to answers of this question, the following claim is made without proof. It seems to be true based on a few examples, but I can't quite prove it.

Suppose $\ a,b\in\mathbb{Q}^+.\ $ Then $\ \exists\ c,d,k\in\mathbb{Z}^+,\ $ such that $\ a = \frac{c}{k},\ $ and $\ b = \frac{d}{k},\ $ and $\ \gcd(c,d,k) = 1.$

Here is my attempt to prove it:

There exists unique $\ w,x,y,z \in \mathbb{Z}^+\ $ such that $\ a = \frac{w}{x},\ b = \frac{y}{z},\ $ and $\ \gcd(w,x) = \gcd(y,z) = 1.$

Then, I think that

$$\ a = \frac{ w\cdot \frac{z}{\gcd(x,z)} }{ x\cdot \frac{z}{\gcd(x,z)} },\quad b = \frac{ y\cdot \frac{x}{\gcd(x,z)} }{ z\cdot \frac{x}{\gcd(x,z)} } $$

solves the problem, because for some reason we should have:

$$ \gcd\left( \frac{wz}{\gcd(x,z)}, \frac{xz}{\gcd(x,z)}, \frac{yz}{\gcd(x,z)}\right) = 1.\quad (1) $$

However, I don't see an obvious way to prove $\ (1).\ $ I have observed that each of the terms of $\ (1)\ $ can be manipulated using the identity: $\ xz = \gcd(x,z) \cdot \text{lcm}(x,z),\ $ but I don't quite see how to make progress with this.

I also feel I am overlooking an alternative, more straightforward approach to the original claim...

Adam Rubinson
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  • If $a = \frac{m}{n}$ and $b = \frac{p}{q}$, then $a = \frac{mq}{nq}$ and $b = \frac{pn}{nq}$. Is this what you're looking for? – kieransquared Jan 03 '23 at 18:21
  • @kieransquared but then $\ \gcd(mq,nq,pn)\ $ is not necessarily $\ 1.\ $ For example, consider $\ a = \frac{5}{6},\ b = \frac{5}{9}.\ $ Then your approach gives us $\ \frac{45}{54}\ $ and $\ \frac{30}{54},\ $ all of which have $\ 3\ $ as a common factor. – Adam Rubinson Jan 03 '23 at 18:24
  • @AdamRubinson: Nobody is requiring that the expressions be in least terms. In fact, usually they won't be. There is no way to write both $\frac{1}{2}$ and $\frac{1}{3}$ with a common denominator and in least terms. – Arturo Magidin Jan 03 '23 at 18:27
  • Sorry, I forgot to add that into the original question. I've added that in now. Although after a hasty closing of my question, I'm not sure it matters... Can you please re-open my question. Sorry, I forgot to add in that condition when I originally wrote my question. – Adam Rubinson Jan 03 '23 at 18:28
  • $\text{gcd}(c,d,k) = 1$ is an important addition! – kieransquared Jan 03 '23 at 18:31
  • Yes I thought I wrote that when I first wrote the question, but I didn't. Sorry – Adam Rubinson Jan 03 '23 at 18:31
  • For your edit: the least common denominator $,k,$ satisfies the gcd constraint (else $,1<p\mid c,d,k,$ so $p$ can be cancelled from both fractions $,a = (c/p)/(k/p),\ b = (d/p)/(k/p),,$ contra minimality of $k)\ \ $ – Bill Dubuque Jan 03 '23 at 18:52
  • Update: now the dupe applies to the edited question also. I added the prior comment to the linked dupe, so now it also applies to your edit (this is also mentioned elsewhere but likely difficult to find by search). But it fits naturally there too. – Bill Dubuque Jan 03 '23 at 19:44

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