https://mathwithbaddrawings.com/2016/11/09/pick-a-truly-random-number/ On Wed, Dec 14, 2016 at 9:50 AM, Allan Wechsler <acwacw@gmail.com> wrote:
Of course pi fails Asimov's strict "normality" test. One of the P_k's is a rigorized equivalent to "x is the ratio of the circumference of a circle to its diameter". Almost all numbers fail this test; but pi and only pi passes it. So pi fails to be "strictly normal". In fact, every number that can be rigorously described or defined fails. While the vast preponderance of real numbers is "strictly normal", we will never be able to say anything interesting about any one of them. They are ineffable and unknowable.
On Wed, Dec 14, 2016 at 8:09 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's such a beautiful definition!
It depends on your language of propositions in an interesting way: if your propositions are first-order predicate logic over the language of rings, then 'random' = algebraic.
Probably you want second-order arithmetic? Reals are, a la descriptive set theory, sets of naturals.
-- APG.
Sent: Wednesday, December 14, 2016 at 8:23 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] record computation of Pi
Normality to a given base is surely a desirable thing to know about a number.
But my definition of a "random" number is as follows:
Let {P_k} be the set of all propositions in about individual real numbers (otherwise with no free variables)
such that
for each k, the set of real numbers not satisfying P_k has measure 0.
There are only countably many propositions P_k, so we may take the intersection S of all sets of numbers S_k where
S_k = {x in R | P_k(x) is true}
Since each one of these has full measure, the same is true of their countable intersection S.
Thus S is the set of all real numbers that satisfy all the propositions satisfied by almost all reals — and almost all reals lie in S.
Of course, all numbers in S are normal to every base, since almost all numbers must be that. But their lack of abnormality goes infinitely further.
—Dan
On Dec 13, 2016, at 5:56 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Plouffe's algorithm that gives the nth binary digit of pi without giving any others may lead to a proof of pi's normality in binary
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