Just in case it wasn't clear, my post was entirely in jest. I respect the use of smileys in many situations, but in other cases it ruins a joke, like someone poking you in the ribs after making a pun. So I opted to omit the smiley in this case, trusting that those who know my math-fun persona would recognize the (attempt at) humor. Jim Propp On Thursday, November 28, 2013, James Propp <jamespropp@gmail.com> wrote:
While we're discussing such matters, perhaps we should consider whether the "lowest common denominator" of two or more fractions has been properly defined.
Most math teachers will tell you that the lowest common denominator of the fractions 5/12 and 11/18 is 36. But what about -36, which is lower? Or -72, which is lower still?
Clearly the lowest common denominator of any set of fractions is minus infinity.
And don't tell me that -72 is "higher" than -36, citing Joe Silverman or some other deluded arithmetician; any Alaskan meteorologist can tell you that -72 is much lower than -36.
Jim Propp
On Thursday, November 28, 2013, Bill Gosper <billgosper@gmail.com> wrote:
Bill Gosper wrote:
If your developers are leery of GCD[1,π]→0, I think I can muster authoritative corroborators, if not corroborative literature. Motivation: The gcd of two real quantities is the largest quantity that goes into each a whole number of times.
[me:]
I'm unconvinced. In this context (multiplicative rather than additive) 0 is *larger* than everything else, not *smaller*, no?
[Bill:]
Well, we *are* looking for the "greatest".-) Do you propose a different answer, or deny there is one?
I deny there is one, or at least that there is any number that has a good claim.
Please name one.
And 0 emphatically doesn't "go into" anything "a whole number of times".
Touché. New wording: GCD(a,b):= the limit of the Euclidean process of iteratively subtracting the smaller from the larger.
The *limit* of doing this for commensurable quantities is zero. The gcd is the last thing you get immediately before 0.
n[755]:= NestList[ Simplify[{Max[#] - Min[#], Min[#]}] &, {GoldenRatio - 1/2, 1/√5}, 6]
Out[755]= {{-(1/2) + GoldenRatio, 1/√5}, {3/(2 √5), 1/√5}, {1/(2 √5), 1/√5}, {1/(2 √5), 1/(2 √5)}, {0, 1/(2 √5)}, {1/(2 √5), 0}, {1/(2 √5), 0}}
So GCD[GoldenRatio - 1/2, 1/√5] is 1/(2√5)
In[758]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {69, 105}, 16]
Out[758]= {{69, 105}, {36, 69}, {33, 36}, {3, 33}, {30, 3}, {27, 3}, {24, 3}, {21, 3}, {18, 3}, {15, 3}, {12, 3}, {9, 3}, {6, 3}, {3, 3}, {0, 3}, {3, 0}, {3, 0}}
So GCD[69,105] is 3.
In[761]:= NestList[Simplify[{Max[#] - Min[#], Min[#]}] &, {1, π}, 3 + 7 + 15 + 1]
Out[761]= {{1, π}, {-1 + π, 1}, {-2 + π, 1}, {-3 + π, 1}, {4 - π, -3 + π}, {7 - 2 π, -3 + π}, {10 - 3 π, -3 + π}, {13 - 4 π, -3 + π}, {16 - 5 π, -3 + π}, {19 - 6 π, -3 + π}, {22 - 7 π, -3 + π}, {-25 + 8 π, 22 - 7 π}, {-47 + 15 π, 22 - 7 π}, {-69 + 22 π, 22 - 7 π}, {-91 + 29 π, 22 - 7 π}, {-113 + 36 π, 22 - 7 π}, {-135 + 43 π, 22 - 7 π}, {-157 + 50 π, 22 - 7 π}, {-179 + 57 π, 22 - 7 π}, {-201 + 64 π, 22 - 7 π}, {-223 + 71 π, 22 - 7 π}, {-245 + 78 π, 22 - 7 π}, {-267 + 85 π, 22 - 7 π}, {-289 + 92 π, 22 - 7 π}, {-311 + 99 π, 22 - 7 π}, {-333 + 106 π, 22 - 7 π}, {355 - 113 π, -333 + 106 π}}
In[762]:= N[%]
Out[762]= {{1., 3.14159}, {2.14159, 1.}, {1.14159, 1.}, {0.141593, 1.}, {0.858407, 0.141593}, {0.716815, 0.141593}, {0.575222, 0.141593}, {0.433629, 0.141593}, {0.292037, 0.141593}, {0.150444, 0.141593}, {0.00885142, 0.141593}, {0.132741, 0.00885142}, {0.12389, 0.00885142}, {0.115038, 0.00885142}, {0.106187, 0.00885142}, {0.0973355, 0.00885142}, {0.0884841, 0.00885142}, {0.0796327, 0.00885142}, {0.0707813, 0.00885142}, {0.0619298, 0.00885142}, {0.0530784, 0.00885142}, {0.044227, 0.00885142}, {0.0353756, 0.00885142}, {0.0265241, 0.00885142}, {0.0176727, 0.00885142}, {0.00882128, 0.00885142}, {0.0000301444, 0.00882128}}
So GCD[1, π] is going to 0.
Gareth>I agree that gcd(1,pi) can be thought of as a certain sort of
limit of real numbers that tend to 0 -- but I don't think the usual sort of limit is the right one, because the metric it implicitly invokes isn't the right one. --
g
What's the right one?
The smallest square you can make by dividing a rectangle into
squares by the greedy algorithm is GCD(length,width). What could
be more natural?
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun