Timeline for Why does an argument similiar to 0.999...=1 show 999...=-1?
Current License: CC BY-SA 3.0
60 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2020 at 10:38 | history | edited | CommunityBot |
Commonmark migration
|
|
| May 13, 2016 at 6:12 | answer | added | AJY | timeline score: 1 | |
| Apr 3, 2016 at 12:17 | comment | added | Soham | math.stackexchange.com/questions/1231061/is-0-9999-equal-to-1 | |
| Feb 14, 2016 at 12:53 | comment | added | Aerinmund Fagelson | @CommonToad Just to make a quick remark on your comment "--- I accept that two numbers can have the same supremum ---" Numbers are fixed, it makes no sense to talk about the supremum of a number. The supremum is a concept defined only for sets of numbers :) | |
| Feb 14, 2016 at 7:55 | audit | First posts | |||
| Feb 14, 2016 at 7:56 | |||||
| Jan 30, 2016 at 23:45 | comment | added | Tavian Barnes | Related: matheducators.stackexchange.com/questions/4351/… | |
| Jan 28, 2016 at 14:45 | comment | added | Eric Wilson | Thanks for sharing this, it provides an excellent illustration of the importance of understanding these repeated decimals in terms of limits. | |
| Jan 28, 2016 at 4:38 | history | protected | CommunityBot | ||
| Jan 28, 2016 at 4:22 | answer | added | Ben | timeline score: 3 | |
| Jan 27, 2016 at 17:15 | answer | added | zrice03 | timeline score: 6 | |
| Jan 26, 2016 at 19:05 | comment | added | Kyle Strand | Note that if you use the summation notation to define $x$ as in your edit and then perform the subtraction $0.1x - x$ term-wise, you end up with $\sum^{\infty}_{k=0}{8.1 \cdot 10^k}$, which no longer appears to converge to $0.9$. | |
| Jan 26, 2016 at 0:06 | answer | added | Joshua | timeline score: 1 | |
| Jan 25, 2016 at 23:41 | answer | added | Steven Alexis Gregory | timeline score: 4 | |
| Jan 25, 2016 at 22:34 | comment | added | Steven Alexis Gregory | You can't say $x = \dots999$ because the left side is a variable that represents a real number while the right side is not a real number. Hence you can't assign a value of true or false to any conclusions based on that statement. | |
| Jan 25, 2016 at 20:07 | comment | added | Patrick Roberts | This logic is as flawed as saying 0.1 ∞ - ∞ = 0.9. The result of the subtraction is indeterminant, not finite. | |
| Jan 25, 2016 at 17:39 | history | tweeted | twitter.com/StackMath/status/691676828436664321 | ||
| Jan 25, 2016 at 17:19 | comment | added | Gus | related videos: youtube.com/watch?v=PS5p9caXS4U youtube.com/watch?v=XFDM1ip5HdU | |
| Jan 25, 2016 at 17:09 | comment | added | drewdles | Related: math.stackexchange.com/questions/1567307/… | |
| Jan 25, 2016 at 3:49 | comment | added | Slipp D. Thompson | @Cornstalks Aha. Okay, your comment makes more sense now. | |
| Jan 25, 2016 at 3:47 | comment | added | Cornstalks |
@SlippD.Thompson: Not on the original post. On the original post it went straight from 10x = 9.999... to x = 1.
|
|
| Jan 25, 2016 at 3:02 | comment | added | Slipp D. Thompson | @Cornstalks The way CommonToad did the calculations, only one step was missing: $9x = 9$. It took me a second also, but I've never considered myself great at math (only good enough for programming). | |
| S Jan 25, 2016 at 0:59 | history | suggested | ruakh | CC BY-SA 3.0 |
added one step in the proof, to make it easier to connect the dots; and, improved formatting a bit
|
| Jan 25, 2016 at 0:47 | comment | added | ruakh | Re: "two numbers can have the same supremum depending on how you generate a decimal representation": It would be more correct to say "some real numbers have two decimal representations". | |
| Jan 25, 2016 at 0:40 | review | Suggested edits | |||
| S Jan 25, 2016 at 0:59 | |||||
| Jan 24, 2016 at 21:44 | comment | added | reuns | don't be upset, and note my answer in the bottom which is highly related to your comment :) I don't really care of weird number fields/rings, I'm only trying to understand, as everyone does on that forum | |
| Jan 24, 2016 at 21:41 | comment | added | Thomas Andrews | But in that ring, $0.999\dots$ and $\dots999$ are not $0$, and nobody represents formal powers series $\dots999.999\dots$, so I don't see what your point is, other than, "If I misinterpret your comment, I can come up with a counter-example." @user1952009 | |
| Jan 24, 2016 at 21:39 | comment | added | Thomas Andrews | @user1952009 So you have a (partial) function from the ring of formal Laurent series to the complex number, $e_{10}:R\to\mathbb C$, which evaluates the analytic continuation of a series at $10$, if it is defined there. But a homomorphism like this is not the same thing as saying in the ring that $\sum 9\cdot 10^k=0$. | |
| Jan 24, 2016 at 21:36 | comment | added | reuns | I was speaking of $\sum_{k=0}^\infty 9 z^k = \frac{9}{1-z}$ and $\sum_{k=-1}^{-\infty} 9 z^k = -\frac{9}{1-z}$ in their domain of convergence, I don't see the problem to think to formal Laurent series as a ring where $\sum_{k=-\infty}^\infty z^k = 0$, I personnally often use the fact that $\int_0^\infty x^s dx = 0$ in the ring of Mellin transform/Dirichlet series, which is nearly exactly the same (and yes it means something and it is very useful) | |
| Jan 24, 2016 at 21:33 | comment | added | Thomas Andrews | Show me how $\sum_{k=-\infty}^{\infty} 9\cdot 10^k$ is defined in that ring. You are talking about an evaluation function from that ring to $x=10$, but in that ring itself, there is no topological definition of convergence that gives you such a result, no. @user1952009 | |
| Jan 24, 2016 at 21:03 | comment | added | reuns | @thomas Andrews : analytic continuation of Laurent series don't form a ring ? | |
| Jan 24, 2016 at 20:49 | vote | accept | CommonToad | ||
| Jan 24, 2016 at 20:47 | comment | added | Thomas Andrews | Of course, if $....999$ is $-1$ and $0.999\dots=1$ then $\dots999.999\dots=0$. :) That's not true in any ring. There are some (topological) rings where $\dots 999=-1$ and some where $0.999\dots=1$ but no ring where both make sense. | |
| Jan 24, 2016 at 14:57 | review | Suggested edits | |||
| Jan 24, 2016 at 15:16 | |||||
| Jan 24, 2016 at 13:19 | answer | added | image357 | timeline score: 10 | |
| Jan 24, 2016 at 12:44 | history | edited | CommonToad | CC BY-SA 3.0 |
added 20 characters in body
|
| Jan 24, 2016 at 12:41 | comment | added | Hagen von Eitzen | You start with $999\ldots$, but seem to continue with $999\ldots9$? I suppose you rather mean $\ldots999$ | |
| Jan 24, 2016 at 12:38 | history | edited | CommonToad | CC BY-SA 3.0 |
spacing
|
| Jan 24, 2016 at 12:32 | comment | added | Masked Man | Isn't 999...9.0 = ∞? | |
| Jan 24, 2016 at 10:25 | history | edited | wythagoras | CC BY-SA 3.0 |
added 5 characters in body; edited title
|
| Jan 24, 2016 at 2:14 | answer | added | Simply Beautiful Art | timeline score: 6 | |
| Jan 24, 2016 at 0:43 | answer | added | Milo Brandt | timeline score: 23 | |
| Jan 24, 2016 at 0:42 | answer | added | Christopher King | timeline score: 6 | |
| Jan 23, 2016 at 23:48 | comment | added | Cornstalks | To those (like me) who read the OP's proof of 0.999... = 1 and thought "Well that's not very convincing..." just realize some steps were elided (full digit manipulation proof can be found on Wikipedia). | |
| Jan 23, 2016 at 22:31 | answer | added | chepner | timeline score: 4 | |
| Jan 23, 2016 at 20:28 | comment | added | MJD | Closely related: Divergent series and $p$-adics | |
| Jan 23, 2016 at 20:15 | answer | added | reuns | timeline score: 6 | |
| Jan 23, 2016 at 20:07 | history | edited | zyx |
edited tags
|
|
| Jan 23, 2016 at 20:03 | answer | added | Henry Swanson | timeline score: 46 | |
| Jan 23, 2016 at 20:01 | comment | added | Daniel Fischer | I see you have a Stack Overflow account. Then you may be somewhat familiar with two's complement representation. If you can imagine a two's complement representation with infinitely many bits, you won't be surprised that in that representation the number with all bits $1$ represents $-1$. Analogously, in an infinite ten's complement representation, the number with all digits $9$ represents $-1$. It's not the usual representation of numbers (where an infinite string of nines before the decimal yields an invalid representation), but it makes sense. It's the $10$-adic numbers, as quid mentions. | |
| Jan 23, 2016 at 19:51 | answer | added | Aerinmund Fagelson | timeline score: 219 | |
| Jan 23, 2016 at 19:47 | history | edited | CommonToad | CC BY-SA 3.0 |
deleted 3 characters in body
|
| Jan 23, 2016 at 19:46 | answer | added | quid | timeline score: 117 | |
| Jan 23, 2016 at 19:44 | history | edited | CommonToad | CC BY-SA 3.0 |
deleted 3 characters in body
|
| Jan 23, 2016 at 19:38 | history | edited | CommonToad | CC BY-SA 3.0 |
deleted 47 characters in body
|
| Jan 23, 2016 at 19:36 | comment | added | the_candyman | It seems you are defining $$\sum_{n=0}^{+\infty}9 \cdot 10^n = -1.$$ It remembers me that $$\zeta(-1) = \sum_{n=1}^{+\infty}n = -\frac{1}{12}$$. [See this][1] to understand what is wrong with your reasoning. [1]: en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF | |
| Jan 23, 2016 at 19:36 | history | edited | CommonToad | CC BY-SA 3.0 |
added 1 character in body
|
| Jan 23, 2016 at 19:31 | comment | added | Pierpaolo Vivo | Your questions 'What mathematics is this touching on?' and 'what the below argument ... is about?' are unclear. Could you please edit your post and specify exactly what you would like us to do? | |
| Jan 23, 2016 at 19:28 | comment | added | Wojowu | $999\dots$ is not a real number. And as soon as there is a rightmost $9$, it is a finite, positive real number. | |
| Jan 23, 2016 at 19:25 | review | First posts | |||
| Jan 23, 2016 at 19:31 | |||||
| Jan 23, 2016 at 19:24 | history | asked | CommonToad | CC BY-SA 3.0 |