Re: [math-fun] Fwd: [CASE:423927] GCD suggestions
GarethM> On 27/11/2013 21:30, 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. I'm unconvinced. In this context (multiplicative rather than additive) 0 is *larger* than everything else, not *smaller*, no? Well, we *are* looking for the "greatest".-) Do you propose a different answer, or deny there is one? And 0 emphatically doesn't "go into" anything "a whole number of times". -- g Touché. New wording: GCD(a,b):= the limit of the Euclidean process of iteratively subtracting the smaller from the larger. --rwg
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.
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. 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
At the very least, gcd(a,b) = m*a+n*b for some _integers_ m,n. The usual definition of gcd requires both m,n to be non-zero (actually gcd(m,n)=1). So gcd(a,0)=1*a+1*0. One could relax this to allow some kinds of algebraic integers -- e.g., Gaussian integers -- for Euclidean domains. But gcd(pi,e)=0=m*pi+n*e <=> m=n=0, so gcd(m,n)=0. At 06:21 PM 11/27/2013, Gareth McCaughan wrote:
Bill Gosper wrote:
If your developers are leery of GCD[1,Ï]â0, 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. 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.
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
On Thu, Nov 28, 2013 at 12:47 AM, Henry Baker <hbaker1@pipeline.com> wrote:
At the very least, gcd(a,b) = m*a+n*b for some _integers_ m,n.
The usual definition of gcd requires both m,n to be non-zero (actually gcd(m,n)=1).
It does? Then what is gcd(5,15)? I thought it was 5, but you can't write 5 as m*5 + n*15, with m and n nonzero. Maybe you meant "m and n not *both* 0"? Andy
Wait: 5 = 1*15 + (-2)*5 = 2*15 + (-5)*5 = 3*15 + (-8)*5 = . . . --Dan On 2013-11-28, at 6:39 AM, Andy Latto wrote:
On Thu, Nov 28, 2013 at 12:47 AM, Henry Baker <hbaker1@pipeline.com> wrote:
At the very least, gcd(a,b) = m*a+n*b for some _integers_ m,n.
The usual definition of gcd requires both m,n to be non-zero (actually gcd(m,n)=1).
It does? Then what is gcd(5,15)? I thought it was 5, but you can't write 5 as m*5 + n*15, with m and n nonzero.
Maybe you meant "m and n not *both* 0"?
Andy _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-2*5+1*15 On 2013-11-28 09:39, Andy Latto wrote:
On Thu, Nov 28, 2013 at 12:47 AM, Henry Baker <hbaker1@pipeline.com> wrote:
At the very least, gcd(a,b) = m*a+n*b for some _integers_ m,n.
The usual definition of gcd requires both m,n to be non-zero (actually gcd(m,n)=1).
It does? Then what is gcd(5,15)? I thought it was 5, but you can't write 5 as m*5 + n*15, with m and n nonzero.
Maybe you meant "m and n not *both* 0"?
Andy _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Here's a somewhat-principled generalized definition of GCD that gives rwg's GCD(1,pi)=0: GCD(a,b) := Inf( { m*a+n*b | m,n integers, m*a + n*b > 0 } ) --Michae;l On Wed, Nov 27, 2013 at 9:21 PM, Gareth McCaughan < gareth.mccaughan@pobox.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.
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.
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
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participants (7)
-
Andy Latto -
Bill Gosper -
Dan Asimov -
Gareth McCaughan -
Henry Baker -
Michael Kleber -
Mike Speciner