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How can $\text{lcm}\left(\dfrac{1}{x},\dfrac{1}{y}\right)$ be calculated? where x and y are integer numbers. Is it $\left(\text{lcm}(x,y)\right)^2$? Please provide proof if possible.

Let me explain by example what I am trying to find out:

let's suppose we have two periodic signals. The first signal peaks or triggers 3 times per second. The second signal triggers 7 times per second. We need to calculate when will the two signals trigger simultaneously in time, with the further restriction that the solution must be an exact (integer) number of seconds.

My intuition says: hey, let's use $\left(\text{lcm}(3,7)\right)^2$ = 441 ...Hey! It seems to work. On the 441th second, both signals do trigger at the same time. This follows from 441 exactly dividing $\dfrac{1}{3}$ and $\dfrac{1}{7}$. So now I want to know if this is a general solution to the problem and also if 441 is in fact the smallest interval of time where both signals trigger at the same time.

Hope it is clearer now.

carloscolombo
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  • @vadim: Given any $R$-module $M$ (such as $\mathbb{Q}$ viewed as a $\mathbb{Z}$-module) it makes sense to talk about the divisibility relation on elements of $M$ defined by $x \mid y$ iff $\exists r \in R : y = rx$. –  Oct 08 '15 at 19:31
  • Um, they trigger simultaneous after 1 second. – fleablood Oct 08 '15 at 19:44
  • The answers everyone gave before your explanation still solve your explanation. Hurkyl's answer is best and shows the will synch in 3*7/lcm[3,7] = 21/21 = 1 second. – fleablood Oct 08 '15 at 19:48
  • " the first signals triggers 3 times per second, so, it ticks at 0.3sec, 0.6sec, 0.9sec, 1.2sec, 1.5sec " Um, that is not 3 times a second. Three times a second is tick on 0.33333333, 0.6666666, and 1.0. If you really do mean it ticks on .3 and .6 etc. then you want lcm[3/10, 1/7] = 10 seconds. – fleablood Oct 08 '15 at 20:07
  • thanks @fleablood:

    Sorry , may be I am very poor at explaining the question. I'll improve on the example... the first signals triggers 3 times per second, so, it ticks at 0.3333sec, 0.6666sec, 0.9999sec, 1.3333sec, 1.6666sec ... I also forgot to explain that I am looking for an answer which is an exact number of seconds, not only simultaneity on time continum. I edited the example to explain better.

     – carloscolombo Oct 08 '15 at 20:09
  • Look. It ticks at 1 second, right? Now the other one ticks at 1/7, 2/7.... 6/7, 1. They both tick on 1 second. That is an exact number. not a "time continum". Why are you insistent that isn't an acceptable answer? The answer is 1. 441 in not the smallest. 1 is the smallest. – fleablood Oct 08 '15 at 20:13
  • ok, it was a stupid question. The answer is 1 second. I missed the fact that all my signals were exact divisors of 1 second! – carloscolombo Oct 08 '15 at 20:17
  • It was not a stupid question. – fleablood Oct 08 '15 at 20:32

4 Answers4

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1/x*(x/gcd(x,y)) = 1/gcd(x,y)

1/y*(y/gcd(x, y)) = 1/gcd(x,y) so 1/gcd(x,y) is a multiple.

If n = n'/x = (n'/x')(1/gcd(x,y)) where x' = x/gcd(x,y) and n = n~/y= (n~/y')(1/gcd(x,y)) where y' = y/gcd(x,y) is another, than n is a multiple of 1/gcd(x,y) so 1/gcd(x,y) is the least common multiple.

The LCM(a/b, c/d) = lcm(a,c)/gcd(b,d) = ac/gcd(a,c)gcd(b,d) (similar reason).

=== old incomplete and slightly wrong answer ====

1/x * x = 1. 1/y*y =1. So 1 is a common multiple. Is there positive integer smaller than 1?

fleablood
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  • Actually my answer is wrong. I assume lcm by definition was an integer. It need not be. so 1/gcd(x,y) is the right answer. – fleablood Oct 08 '15 at 20:36
  • yeah, I noticed, so I changed the correct answer to Theophile's. I upvoted yours, though, because it solved my particular problem. :) – carloscolombo Oct 08 '15 at 20:40
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According to this answer regarding the extension of lcm and gcd to rational numbers, we have:

$$\textrm{lcm}\left(\frac1x, \frac1y\right) = \frac{\textrm{lcm}(1,1)}{\textrm{gcd}(x,y)} = \frac{1}{\textrm{gcd}(x,y)}.$$

Edit: To address your specific example, we have $x=3, y=7$, so $$\textrm{lcm}\left(\frac13, \frac17\right) = \frac{\textrm{lcm}(1,1)}{\textrm{gcd}(3,7)} = \frac11 = 1.$$

Community
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Théophile
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It is simply $\dfrac1{\gcd(x,y)}$.

Bernard
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Recall that

$$ \mathrm{lcm}(ab, ac) = a \mathrm{lcm}(b, c) $$

so we can just multiply through by a common denominator to reduce it to an integer problem:

$$ \mathrm{lcm}(xy \cdot \frac{1}{x}, xy \cdot \frac{1}{y}) = xy \mathrm{lcm}(\frac{1}{x}, \frac{1}{y}) $$

or

$$ \mathrm{lcm}(\frac{1}{x}, \frac{1}{y}) = \frac{\mathrm{lcm}(y, x)}{xy} $$