One of the potential applications of this sort of thinking is what I might call 'perfect solutions to differential equations'. In particular, I'd like to figure out how to compute exact solutions to a simple DE -- e.g., simple harmonic motion -- such that the usual 'constants of motion' are truly constant. I'm willing to have delta t non-constant -- i.e., I may have to use relatively chaotic delta t's, but I'd like energy to be _perfectly preserved_ and _angular momentum_ to be _perfectly preserved_. It would also be nice if the denominators didn't grow too fast; I'd like to be ablt to pick delta t such that the denominators grow as slowly as possible (or even better, be bounded). At 08:58 AM 9/12/2013, Warren D Smith wrote:

In my sleep last night, I woke up from a dream having proved a theorem, thus proving that I am smarter when asleep than when awake.

THEOREM: Let f(z) be an analytic function. Then the set of z such that Re f(z), Im f(z) and |f(z)| all simultaneously are rational, is dense throughout the region of analyticity.

PROOF: As math-fun previously discussed ("pythagorean rationals") the set of points (x,y) on the perimeter of the unit circle x^2+y^2=1 with x,y simultaneously rational, is a dense multiplicative subgroup. Now the circle can be scaled up by any rational scaling factor leading to a dense set of such points in the whole x,y plane. That proves the theorem for f(z)=z. Since f() if analytic is a smooth continuous conformal mapping, locally invertible almost everywhere, the result follows. QED

EXTENSION: The theorem also is true if all rationals are demanded to be terminating decimals.

To see that note that (0.6, 0.8) lies on the unit circle, and all the integer powers of 0.6+0.8i form a dense subset of that circle-perimeter in the complex plane (a "Weyl sequence" in polar coordinates, using known fact arctan(3/4)/pi is irrational). And the circle can be scaled up by any terminating decimal scale factor to get density in the whole plane, etc.

The complex numbers z with Re z, Im z, and |z| all simultaneously rational form a dense multiplicative subgroup of the complex numbers (if terminating decimal, then dense subring).

What is we ask for terminating BINARY (not decimal)? Then my proof above does not work (thus proving the superiority of decimal versus binary) but the theorem still holds with a different proof, based on the fact you can scale the integer (x,y) with x^2+y^2=w^2, x,y,w integer, by any (positive or negative) power of 2.

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)