Timeline for Base change for $\sqrt{2}.$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 7, 2014 at 16:44 | comment | added | Igor Rivin | @NikitaSidorov I, for one, am completely agnostic on the subject, and have absolutely no clue why anyone would believe what you say everyone believes. I am, of course, just a caveman. | |
| Jul 7, 2014 at 16:42 | comment | added | Igor Rivin | @NikitaSidorov And of course, we all speak for ourselves. | |
| Jul 7, 2014 at 16:27 | comment | added | Nikita Sidorov | Of course, we all believe it is transcendental. And of course, we all know that this is way beyond our reach at present. Why ask, then? | |
| Jul 7, 2014 at 12:58 | comment | added | Igor Rivin | @AnthonyQuas I cannot actually understand your comment... | |
| Jul 7, 2014 at 12:42 | comment | added | Pietro Majer | Since the digits are the same but the base is larger, in a certain sense, $\theta_{11}$ has a quicker rational approximations than $\sqrt 2$. Can this fact affect the irrationality measure of $\theta_{11}$? | |
| Jul 7, 2014 at 8:57 | comment | added | Stefan Kohl♦ | @AnthonyQuas: Are there reasons to believe in this independency, besides mere numerical observations? | |
| Jul 7, 2014 at 4:27 | comment | added | Anthony Quas | Algebraic irrationality "is independent of" base expansions; there are countably many algebraic irrationals; Therefore "with probability 1", all algebraic irrationals are normal to all bases. | |
| Jul 7, 2014 at 1:49 | comment | added | Igor Rivin | @GeraldEdgar What is the justification for the well-known conjecture? | |
| Jul 6, 2014 at 22:21 | comment | added | Gerald Edgar | $\theta_{11}$ is certainly irrational. There is a well-known (but still far from proved) conjecture that all algebraic irrationals are normal in all bases. So (of course) your number is transcendental. But your question (whether there is an easy proof of it) is not answered by this. | |
| Jul 6, 2014 at 15:36 | history | asked | Igor Rivin | CC BY-SA 3.0 |