# lcm, gcd proof correct?

• Feb 26th 2010, 02:43 AM
loopology
lcm, gcd proof correct?
I believe to have proven something that I need for an algorithm I'm developing.

The proposition is:

$\displaystyle lcm(b, c) / gcd(a, lm(b, c)) = lcm(a, b, c) / a;$

Proof:

$\displaystyle lcm(b,c) / gcd(a, lcm(b, c)) = lcm(a, lcm(b, c)) / a;$
Substitution: $\displaystyle k:= lcm(b, c);$
$\displaystyle k / gcd(a, k) = lcm(a, k) / a;$
$\displaystyle a \times k = lcm(a, k) \times gcd(a, k);$ // see Greatest common divisor - Wikipedia, the free encyclopedia

Is this correct?
Thank you
Bernhard
• Feb 26th 2010, 03:01 AM
Bacterius
Hello Bernhard,
Is the == sign a congruence ? $\displaystyle \equiv$ ? If yes, what is the modulus considered ? Because a congruence without modulus is meaningless.
(Funny like we all get the same idea when we lack an congruence sign on the computer :D)

Out of curiosity, what is the algorithm you are developing about, if not too indiscrete ?

EDIT : ah, no, I get it, this is the conditional equivalence sign in the C language. Well obviously, if $\displaystyle gcd(a, b, c) = 1$, then the equality holds, because $\displaystyle gcd(a, lcm(b, c)) = 1$, and $\displaystyle lcm(a, b, c) = a \times lcm(b, c)$. I don't really get your proof (what does it show ? a proof needs words), but I think you might get somewhere if you consider dividing the right hand side with the left hand side ...
• Feb 26th 2010, 06:32 AM
loopology
Hi Ray

Thanks for looking at my proof. Sorry for using a confusing notation (corrected it now).
I'm trying to prove that

$\displaystyle lcm(b, c) / gcd(a, lm(b, c)) = lcm(a, b, c) / a;$

holds, i.e. that $\displaystyle lcm(b, c) / gcd(a, lm(b, c))$ can be calculated by $\displaystyle lcm(a, b, c) / a;$

I believe to have proven the equivalence by applying what I hope are legal transformations to reach the equation

$\displaystyle a \times k = lcm(a, k) \times gcd(a, k);$

which is a valid connection between $\displaystyle lcm$ and $\displaystyle gcd$ (see Greatest common divisor - Wikipedia, the free encyclopedia) The cool thing about the substitution is that the equivalence applies for an arbitrary number of arguments to $\displaystyle lcm$, i.e. $\displaystyle lcm(b, c, d, e, f, g, ...)$ which happens to be what I actually need, not just two arguments (i.e. $\displaystyle lcm(b, c)$).

The algorithm is used for calculating BPMs to represent arbitrary rhythmic patterns. See Boss Dr.Beat DB-88