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Disclaimer: I'm an engineer, not a mathematician

I have a set of three fractions (a/b, c/d, e/f). I can multiply them all by another fraction, so that their mutual ratios remain the same. I want to end with natural numbers (i, j, k) where

$$\gcd(i, j, k) = 1$$

I tried the following:

$$ \dfrac{g}{h} = \gcd\left(\dfrac{a}{b}, \dfrac{c}{d}, \dfrac{e}{f} \right) $$


$$ \begin{cases} i = \dfrac{a \cdot h}{b \cdot g} \\ \\ j = \dfrac{c \cdot h}{d \cdot g} \\ \\ k = \dfrac{e \cdot h}{f \cdot g} \end{cases} $$

seems to work, but I can't prove it's always true. Is this a valid conjecture?

Another problem I ran into: I needed the denominator of a reduced fraction, and I couldn't find it! There sure must be a function $f$ where

$$f\left(\dfrac{a}{b}\right) = b$$

for the reduced fraction $\dfrac{a}{b}$?

I'm not a mathematician, so please type slowly ;-)

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Answer to your first question:

This is true by the definition of the rational GCD. Firstly it assures that for your pairs $\left(\dfrac{a}{b}, \dfrac{c}{d}, \dfrac{e}{f} \right)=\left(x, y, z \right)$ we have with $r=\gcd\left(x, y, z \right)$ that $\left(\dfrac{x}{r}, \dfrac{y}{r}, \dfrac{z}{r} \right)$ are all integers and that $r$ is maximal rational number with that property.

Assume $\gcd\left(\dfrac{x}{r}, \dfrac{y}{r}, \dfrac{z}{r} \right)$ is not $1$, then you can multiply $r$ by that number and maintain the property that those numbers are integer. Therefore you get a contradiction to the maximality of $r$.

Answer to your second question:

If the fraction $\dfrac{a}{b}$ is reduced you have that $\gcd(1,\dfrac{a}{b})=\dfrac{1}{b}$ by the above properties. Can you use this to get a formula for $b$?

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"and that r is maximal rational number". Is $r$ rational, or natural? – stevenvh Jun 11 '12 at 9:10
I use the extension of the GCD from mathworld to rational numbers. $r$ is actually rational and represents your $\dfrac{g}{h}$ – Listing Jun 11 '12 at 9:11
Ah, the $x$, $y$ and $z$ are not my natural number result then? Because if they were, their $\gcd$ would also be natural. – stevenvh Jun 11 '12 at 9:13
As I wrote I use $x=\dfrac{a}{b}, y=\dfrac{c}{d}, z=\dfrac{e}{f}$ – Listing Jun 11 '12 at 9:14
I saw that, but misinterpreted. Sorry about that. – stevenvh Jun 11 '12 at 9:28

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