Timeline for Why does an argument similiar to 0.999...=1 show 999...=-1?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2016 at 20:41 | comment | added | Thomas Andrews | @PyRulez No, the series doesn't converge in the $p$-adic numbers, except when $p=2,5$. And $0.99999\dots$ doesn't converge in $10$-adic numbers or any $p$-adic numbers. | |
| Jan 24, 2016 at 20:37 | comment | added | Thomas Andrews | Or even in the $2$-adic or $5$-adic numbers... | |
| Jan 24, 2016 at 16:05 | comment | added | Christopher King | @quid (I suppose that works with $.999 \dots$ as well. (You gettings $0$ if you subtract $1$.)) | |
| Jan 24, 2016 at 16:04 | comment | added | quid | @PyRulez yes the proof is valid. One can also add $1$ to the series and note that one gets $0$. | |
| Jan 24, 2016 at 16:02 | comment | added | Christopher King | @CommonToad One cool thing is, I'm pretty sure you proof is valid in p-adic numbers. Like, you can use your proof or something like it to proof this identity. | |
| Jan 24, 2016 at 10:54 | comment | added | CommonToad | Thanks for this gives me a nice thing to look up. @SpamIAm - nice book - bought :) | |
| Jan 24, 2016 at 2:29 | comment | added | Christopher King | @Ant Yeah, it isn't necesarrily a full answer, but a good one. The OP wasn't necessarily fimilar with real numbers. | |
| Jan 23, 2016 at 20:55 | comment | added | quid | @Ant this is true in a way, but this answer does not live in a vacuum. When I gave it there already was an answer, now deleted, and comments that explained the problem with the original calculation. Thus I did not reiterate it. Moreover it also used to ask "What mathematics is this touching on?" | |
| Jan 23, 2016 at 20:42 | comment | added | Ant | this is a nice answer but honestly, it does not address the OP problem. First one should explain why what the OP is doing "naively" does not work, and why p-adic numbers seems to work instead | |
| Jan 23, 2016 at 20:14 | comment | added | BlueRaja - Danny Pflughoeft | Ramanujan famously got a college rejection letter from Prof. M.J.M. Hill due to his proof that $1+2+3+...=-1/12$ which used similar reasoning to OP's proof. Euler also tinkered with proofs like this, and argued they should be considered valid. See the article on Divergent sums for more information. | |
| Jan 23, 2016 at 20:03 | comment | added | quid | @MarkS. The completion of the rational numbers with respect to the $10$-adic distance. It is true it is not a field but I do not see why this would be a problem in this context. | |
| Jan 23, 2016 at 20:00 | comment | added | reuns | if $k$ is not prime, $\mathbb{Q}_k$ is a commutative ring instead of a commutative field | |
| Jan 23, 2016 at 19:59 | comment | added | Viktor Vaughn | Just as an FYI to the OP: Fernando Gouvêa plays with similar seemingly divergent sums in the introduction of his book, p-adic Numbers. I think you would enjoy reading it. | |
| Jan 23, 2016 at 19:54 | comment | added | Mark S. | What do you mean by $\mathbb{Q}_{10}$? I would think it wouldn't be defined because $\mathbb{Z}_{10}$ has zero divisors. | |
| Jan 23, 2016 at 19:46 | history | answered | quid | CC BY-SA 3.0 |