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Math Help - LCM of 2 vulgar factions

  1. #1
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    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)
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  2. #2
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    Quote Originally Posted by luhter View Post
    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.
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  3. #3
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    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
    <br />
a=\frac{3}{2} <br />
    <br />
b=\frac{5}{8}<br />
    we make
    <br />
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}<br />
    we multiply, simplify until we reach at the form
    <br />
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}<br />
    and the answer is either
    <br />
a*n or <br />
b*m<br /> <br />
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by luhter View Post
    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
    <br />
a=\frac{3}{2} <br />
    <br />
b=\frac{5}{8}<br />
    we make
    --------> <br />
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}<br />
    we multiply, simplify until we reach at the form
    <br />
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}<br />
    and the answer is either
    <br />
a*n or <br />
b*m<br /> <br />

    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.
    Last edited by undefined; June 27th 2010 at 11:24 AM.
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by luhter View Post
    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
    <br />
a=\frac{3}{2} <br />
    <br />
b=\frac{5}{8}<br />
    we make
    <br />
o=\frac{a}{b} \text{ ; n,m are } \mathbb{Q}<br />
    we multiply, simplify until we reach at the form
    <br />
o=\frac{m}{n} \text{ ; n,m are } \mathbb{N}<br />
    and the answer is either
    <br />
a*n or <br />
b*m<br /> <br />
    I think this can be proven to work because:

    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.
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