# Thread: LCM of 2 vulgar factions

1. ## LCM of 2 vulgar factions

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
$x(t)=4cos(2\varpi t)+cos(3\varpi t)$

2. Originally Posted by luhter
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
$x(t)=4cos(2\varpi t)+cos(3\varpi t)$
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.

3. 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
$
a=\frac{3}{2}
$

$
b=\frac{5}{8}
$

we make
$
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}
$

we multiply, simplify until we reach at the form
$
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}
$

$
a*n$
or $
b*m

$

4. 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
$
a=\frac{3}{2}
$

$
b=\frac{5}{8}
$

we make
--------> $
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}
$

we multiply, simplify until we reach at the form
$
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}
$

$
a*n$
or $
b*m

$

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

My method is based on the idea that for positive integers a,b,k, we have $\displaystyle\text{lcm}(a,b)=\frac{\text{lcm}(ka,k b)}{k}$. 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 $\displaystyle\text{lcm}(a,b)=\frac{\text{lcm}(ka,k b)}{k}$ is true; see Equation 19, the LCM is distributive. I suspect a proof would be pretty simple considering prime factorisations.

5. 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
$
a=\frac{3}{2}
$

$
b=\frac{5}{8}
$

we make
$
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}
$

we multiply, simplify until we reach at the form
$
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}
$

$
or $
It is proven that for positive integers a and b, $\displaystyle\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$. Writing $\displaystyle\frac{a}{b}$ and reducing to lowest terms results in $\displaystyle\frac{\frac{a}{\gcd(a,b)}}{\frac{b}{\ gcd(a,b)}}$; from this, it can be seen that taking either of the products you mentioned gives the lcm. Apparently this extends to the rational case.