Timeline for If I exchange infinitely many digits of $\pi$ and $e$, are the two resulting numbers transcendental?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| May 31, 2017 at 2:07 | comment | added | user10290 | If one of them turns out to be algebraic, I would love to see that polynomial. | |
| Mar 24, 2017 at 14:46 | vote | accept | CommunityBot | moved from User.Id=10290 by developer User.Id=35285 | |
| Mar 24, 2017 at 14:46 | comment | added | user10290 | Thanks Joel. You did show that we do get continuum many transcendentals in this way. Seems this is the best that can be done for now. | |
| Mar 23, 2017 at 17:14 | comment | added | user10290 | There is a forcing extension where continuum many transcendentals are coded by $e$ and $\pi$. | |
| Mar 23, 2017 at 12:59 | comment | added | Joel David Hamkins | If there are infinitely many digits of disagreement, then that forcing is equivalent to adding a Cohen real, and indeed you will get two transcendental numbers generically that way, since it is dense to avoid any particular algebraic number. | |
| Mar 23, 2017 at 5:52 | comment | added | user10290 | I have an idea where the forcing conditions are finite exchanges, and we already know these are still transcendental. Then, we get them both transcendental in the forcing extension. | |
| Mar 23, 2017 at 5:38 | comment | added | user10290 | Thank you Joel! I was going to say that next, that there would be uncountably many transcendental numbers this way! We have been looking for them! | |
| Mar 23, 2017 at 3:30 | comment | added | Christian Remling | I thought this ($\pi - e$ rational?) is a famous open problem. Apparently it's not famous, though it may well be open, and at least wikipedia thinks it is: en.wikipedia.org/wiki/… | |
| Mar 23, 2017 at 3:11 | comment | added | Joel David Hamkins | Oh, I had assumed that this was known. I have edited my answer. | |
| Mar 23, 2017 at 3:10 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Mar 23, 2017 at 3:02 | comment | added | Andy Putman | I was pretty sure this was known, but I am not an expert and could be mistaken. | |
| Mar 23, 2017 at 2:53 | comment | added | Timothy Chow | In any case, it's clear that sometimes you get transcendental numbers. Just swap the digits in all but finitely many places. | |
| Mar 23, 2017 at 2:50 | comment | added | Timothy Chow | @AndyPutman : Really? I'm pretty sure it is not known that $\pi - e$ is irrational. | |
| Mar 23, 2017 at 2:28 | comment | added | Gerhard Paseman | I don't know either, but if not, then pi - e is rational, and I think both pi and e would have the same irrationality measure. I believe one for e is known and the one for pi is not. Gerhard "Indulging In Irrational Guessing Now" Paseman, 2017.03.22. | |
| Mar 23, 2017 at 2:21 | comment | added | Andy Putman | @Dirk: If they differed in only finitely many digits, then their difference would be rational. This is known not to be the case. | |
| Mar 23, 2017 at 2:11 | comment | added | Dirk | Why is it clear that there are really a continuum many different numbers? Not that I think differently, but I don't know an argument why $\pi$ and $e$ have infinitely many different digits. | |
| Mar 23, 2017 at 1:52 | comment | added | Myself | Could $\pi - e$ be a rational number with a power of 10 in the denominator? Or weird stuff like that? (In that case I think the trick doesnt work but any digit exchange will still do since adds a rational number.) | |
| Mar 23, 2017 at 1:48 | comment | added | Konstantinos Kanakoglou | Wow! that was nice! | |
| Mar 23, 2017 at 1:13 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |