Timeline for Conjecture on irrational algebraic numbers
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2015 at 20:01 | comment | added | Charphacy | @BillMance You say “there is a much stronger conjecture that every irrational algebraic real number is normal in every base”, but can you give a citation for this conjecture? Does it have a name? The best I have is “Hypothesis A” from Bailey & Crandal's On the Random Character of Fundamental Constant Expansions (2001). | |
| Nov 9, 2014 at 6:35 | vote | accept | barak manos | ||
| Jul 7, 2014 at 19:45 | comment | added | Bill Mance | @barakmanos A real number $x$ is normal in base $b$ if for all $k$ and all blocks of digits $B$ of length $k$, the asymptotic frequency of $B$ in the $b$-ary expansion of $x$ is $b^{-k}$. Equivalently, the sequence $(b^n x)$ is u.d. mod $1$. en.wikipedia.org/wiki/Normal_number I didn't put it as an answer because I felt that what Anthony Quas wrote was far better. But this conjecture is far stronger than the one you are asking about and many people believe it to be likely to be true. Although as far as I know it is nowhere near being settled. | |
| Jul 6, 2014 at 20:05 | comment | added | barak manos | @BillMance: What does normal mean? If it does indeed "cover" the conjecture I wrote, then I suppose that in a certain way it may be a good-enough answer for the first part of my question (which is pretty much the only part yet to be answered here). | |
| Jul 6, 2014 at 19:46 | comment | added | Bill Mance | I just want to remark since no one has mentioned it yet: there is a much stronger conjecture that every irrational algebraic real number is normal in every base. | |
| Jul 6, 2014 at 16:32 | comment | added | barak manos | @Joël: I just checked on Wikipedia to freshen up my memory, and this Chaitin's Number is tightly related to the Halting Problem, hence to my Turing Machine comment above. | |
| Jul 6, 2014 at 16:27 | comment | added | Joël | @barak: no, nobody is talking about Turing machines here. | |
| Jul 6, 2014 at 14:38 | comment | added | barak manos | @WillSawin: In the comment about Turing machine above? | |
| Jul 6, 2014 at 14:31 | comment | added | Will Sawin | @barakmanos You're thinking of Chaitin's number, I think? | |
| Jul 6, 2014 at 13:07 | answer | added | Anthony Quas | timeline score: 14 | |
| Jul 6, 2014 at 11:51 | review | Close votes | |||
| Jul 6, 2014 at 14:19 | |||||
| Jul 6, 2014 at 9:55 | comment | added | Geoff Robinson | The Liouville's constant I refer to is $\sum_{n=1}^{\infty}10^{-n!}.$ | |
| Jul 6, 2014 at 9:12 | answer | added | P Vanchinathan | timeline score: 2 | |
| Jul 6, 2014 at 9:10 | comment | added | barak manos | @StefanKohl: You mean, I can simply take any irrational algebraic number $q$ and any natural number $b$, and divide $q$ by $b$ over and over? Hmmmm... good one, thanks. | |
| Jul 6, 2014 at 9:07 | comment | added | Stefan Kohl♦ | Assuming you are talking about initial sequences of digits: for the second point, as for any $b$ and any $n$ there are irrational algebraic numbers whose representation in base $b$ starts with $n$ zeros, there is no such bound. | |
| Jul 6, 2014 at 9:04 | comment | added | barak manos | @GeoffRobinson: Liouville's constant? Is that from the proof of the fact that some problems cannot be solved on a Turing machine? Isn't that number given in base $2$ (in which case, it does contain all the digits)? | |
| Jul 6, 2014 at 9:00 | comment | added | Geoff Robinson | For the last point: yes, as you seem to know, it fails for transcendental numbers: for example, Liouville's constant is transcendental, but only has $1$'s and $0$'s in its decimal expansion. For more information, read about general Liouville numbers. | |
| Jul 6, 2014 at 8:59 | history | edited | barak manos | CC BY-SA 3.0 |
added 120 characters in body
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| Jul 6, 2014 at 8:52 | history | asked | barak manos | CC BY-SA 3.0 |