Topposting gets me confused :-S so I reordered this a bit
On Jul 21, 2011, at 7:44 AM, Júlio Hoffimann wrote:
Hi Thomas,
Thank you for the intention, but as far as i know, GCD and LCM makes no sense in the rational field. Any rational number is divisible by all rationals. For instance, given two rationals b, r in Q, r != 0, you can write:
b = (b * r^-1) * r
and then b is divisible by r.
2011/7/21 Christopher Henrich
GCD and LCM can be defined for rationals, if one takes "divisible" to mean "divisible with an integer quotient" and, similarly, "multiple" to mean "multiple by an integer." Then, for example, GCD(1/2, 1/3) = 1/6 and LCM(1/2, 1/3) = 1.
Hi Christopher,
This is the first time i heard about this interpretation. For me, still makes no sense, but if people thinks is useful, ok.
Basically I used the approach desribed by Christopher. (see also http://en.wikipedia.org/wiki/Least_common_multiple#Rational unfortunately it does not cite anything for this paragraph (there even is a similar approach for modulo calculation http://de.wikipedia.org/wiki/Division_mit_Rest#Verallgemeinerung:_Reelle_Zah... - sorry German only)) Especially a lcm makes sense in my opinion, as any two numbers have a lcm; the question only is what base set [Zahlenraum] one uses. (the posted implementation gives rational lcms (base set = R), but it could be modified to only give integer lcms (base set = Z)). The gcd is certainly trickier as it appears to me to be less intuitive. However the best way to describe what it does is probably what I am using it for: Consider a 2d vector v=(a,b). Now we don't care about the length |v| but only for the direction, which is defined by the factor between a and b. Lets further assume that both a and b are rational. Multiplying both a and b with gcd(a,b) [c=a*gcd(a,b) and d=b*gcd(a,b)] will give v'=(c,d), where both c and d are integers and furthermore the smallest integers that represent the direction (or factor between a and b). HTH, Thomas
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