How can LCM(1/x, 1/y) be calculated?

Question

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.

Answer 1

According to this answer regarding the extension of lcm and gcd to rational numbers, we have:

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

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

Answer 2

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

Answer 3

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?

Answer 4

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

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