Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

Then

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

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 ;-)

share|improve this question
add comment

1 Answer

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 $b$?

share|improve this answer
    
"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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.