# LCM of 2 vulgar factions

• Jun 27th 2010, 08:34 AM
luhter
LCM of 2 vulgar factions
(Headbang) Pff i've lost a 2h on internet trying to find how do you compute the LCM of 2 fractions

For example LCM of 2/3 and 1/4 is 2 , while LCM of 3/2 and 1/4 is 3/2

Why i need this?
Because i need to find the period of this
$\displaystyle x(t)=4cos(2\varpi t)+cos(3\varpi t)$
• Jun 27th 2010, 08:56 AM
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Quote:

Originally Posted by luhter
(Headbang) Pff i've lost a 2h on internet trying to find how do you compute the LCM of 2 fractions

For example LCM of 2/3 and 1/4 is 2 , while LCM of 3/2 and 1/4 is 3/2

Why i need this?
Because i need to find the period of this
$\displaystyle x(t)=4cos(2\varpi t)+cos(3\varpi t)$

Hmm, I've never seen the LCM defined for rationals like that. Anyway, you seem to have provided all the info necessary for a definition and its implications, and what I come up with is:

Suppose we have two positive rationals a'/b' and c'/d'. First reduce to lowest terms to get a/b and c/d. Then let e=lcm(b,d). Multiply both rationals by e to get integers f=ae/b and g=ce/d. The let h=lcm(f,g). Then the final answer is h/e.

I haven't tested this thoroughly or proven it, so if someone sees a problem with it, please say so. It "seems right" for the time being.
• Jun 27th 2010, 09:43 AM
luhter
Yep, it seems to work fine. Thanks alot! :)
Curious how you rationalized it .

I found it how to make it another way

let's say we have
$\displaystyle a=\frac{3}{2}$
$\displaystyle b=\frac{5}{8}$
we make
$\displaystyle o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}$
we multiply, simplify until we reach at the form
$\displaystyle o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}$
$\displaystyle a*n$ or $\displaystyle b*m$
• Jun 27th 2010, 10:07 AM
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Quote:

Originally Posted by luhter
Yep, it seems to work fine. Thanks alot! :)
Curious how you rationalized it .

I found it how to make it another way

let's say we have
$\displaystyle a=\frac{3}{2}$
$\displaystyle b=\frac{5}{8}$
we make
--------> $\displaystyle o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}$
we multiply, simplify until we reach at the form
$\displaystyle o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}$
$\displaystyle a*n$ or $\displaystyle b*m$

I think when you wrote $\displaystyle \text{n,m are } \mathbb{Q}$ you meant $\displaystyle \text{a,b are (in) } \mathbb{Q}$

My method is based on the idea that for positive integers a,b,k, we have $\displaystyle \displaystyle\text{lcm}(a,b)=\frac{\text{lcm}(ka,k b)}{k}$. I haven't proven this but it seems true. Then extend it allowing a,b to be positive rationals, while keeping k a positive integer.

Edit: MathWorld says that $\displaystyle \displaystyle\text{lcm}(a,b)=\frac{\text{lcm}(ka,k b)}{k}$ is true; see Equation 19, the LCM is distributive. I suspect a proof would be pretty simple considering prime factorisations.
• Jun 27th 2010, 09:05 PM
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Quote:

Originally Posted by luhter
Yep, it seems to work fine. Thanks alot! :)
Curious how you rationalized it .

I found it how to make it another way

let's say we have
$\displaystyle a=\frac{3}{2}$
$\displaystyle b=\frac{5}{8}$
we make
$\displaystyle o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}$
we multiply, simplify until we reach at the form
$\displaystyle o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}$
$\displaystyle a*n$ or $\displaystyle b*m$
It is proven that for positive integers a and b, $\displaystyle \displaystyle\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$. Writing $\displaystyle \displaystyle\frac{a}{b}$ and reducing to lowest terms results in $\displaystyle \displaystyle\frac{\frac{a}{\gcd(a,b)}}{\frac{b}{\ gcd(a,b)}}$; from this, it can be seen that taking either of the products you mentioned gives the lcm. Apparently this extends to the rational case.