Talk:0.999.../Archive 8

message: This is a trivial point to the article, but I will address it here anyway. The word doppelgänger should generally not be capitalized, with obvious exceptions such as when it is used as the first word in a sentence, or as the title of an article or book, etc. It should not be capitalized in the context in which it is currently used in this article. Please see the rules of capitalization at dictionary.com and grammarbook.com. It is true that doppelgänger is not native to English; however, this does not imply that it should always be capitalized. Please see the list of foreign words and phrases at infoplease.com. Most of these generally should not be capitalized. The only notable exception is when a word is derived from a proper noun. By examining the etymology of doppelgänger, it can be seen that this is not the case. Rishodi 20:01, 3 May 2006 (UTC)
replies: text: The article on doppelgänger is not a helpful guide; it uses both capitalizations seemingly at random! I will defer to Wikipedia convention on this, whatever that happens to be. But if lower-case is used, write the link as I have in this note. --KSmrqT 20:57, 3 May 2006 (UTC)
replies: text: Yes, I looked at the Wikipedia article on doppelgänger and it improperly capitalizes the word in many places. I'm going to change the reference in this article back to lowercase, and at some point in the future I may have to go through the doppelgänger article and correct the instances of the word that are incorrectly capitalizated. Rishodi 21:58, 3 May 2006 (UTC)

text: Regardless of what the article does, in general (aside from the cases of beginning a sentence, etc.) doppelgänger is left uncapitalized in English, while in German it is always capitalized. -- He Who Is[ Talk ] 02:06, 23 June 2006 (UTC)
thread_title:  Capitalization of "Doppelgänger" 

message: This is good material for kids like myself to prove to other kids that they're smarter in the endless debate of who's smarter. Other than that, I don't see much use to this. Not very sure if I got dumber or smarter after reading this but it was a good article with a correct theory. Case Closed! —The preceding unsigned comment was added by 70.127.34.98 (talk • contribs) 07:39, 11 May 2006  (UTC)
:Actually, I think it's more important than just showing off your smarts - it's showing that some things that are mathematically provable that at first seem counterintuitive: 0.999... initially looks like it's strictly less than 1, but the article proves that they're equal. It just goes to show that you can't just look at a result and say "that's right", you have to be able to prove or disprove it before you can make any use of it. Confusing Manifestation 11:43, 11 May 2006 (UTC)
replies: text: And conversely, many facts that mathematicians take for granted have been disproved over the centuries, although at a glance they seem to be easily verifiable. For instance, the product of the squared roots of to numbers equals the squared root of the product of those numbers. This has been proven to be true only over the principal branch of the real axis, and otherwise led to the assumption that any number is its own opposite. The axiom of choice is another example. -- He Who Is[ Talk ] 17:44, 23 June 2006 (UTC)
replies: text: The axiom of choice was disproved?? Any example of a fact "mathematicians take for granted" over the last century that has since been disproven? (Cj67 05:07, 25 June 2006 (UTC)
replies: text: From the article on AC: "However, there are schools of mathematical thought, primarily within set theory, that either reject the axiom of choice or investigate consequences of axioms inconsistent with AC." While it has never been conclusively proven or disproven, in many mathematical disciplines \neg AC is taken to be true. -- He Who Is[ Talk ] 22:01, 27 June 2006 (UTC)
replies: text: And there's a proof that the axiom of choice cannot be proven (or disproven), which most of these schools accept. Septentrionalis 23:19, 27 June 2006 (UTC)
replies: text:  A far cry from "ha[s] been disproved."  In fact, it has been proven, as mentioned above, that the axiom cannot be disproven (within ZF).  Still waiting for examples of disproved things.  You are simply wrong, and you should admit it.  I think you should also be more reluctant to consider yourself to be an "advanced mathematician" (although maybe you are joking about that). (Cj67 18:10, 28 June 2006 (UTC)
replies: text: Yes, it was in fact a joke. But as for the examples you wanted: In my oiginal post, I did provide one when I mentioned the square root of a product. It was also once thought that any polynomial equation with algebraic coefficients would have algebraic roots and thus could be solved via some algorithm. This is now known to be untrue, thanks to Galois groups. It was also once beleived by many mathematicians that all real numbers could be represented as a fraction of integers and were thus rational. This is now known to be untrue as well. Moreover, similar beleifs were once held pretaining to algebraic numbers. In fact, it might even be a good idea to include some of this in the article in order to clarify the significance of this. -- He Who Is[ Talk ] 18:53, 28 June 2006 (UTC)
replies: text: Huh? The roots of a polynomial over the algebraic numbers are algebraic. Septentrionalis 01:24, 29 June 2006 (UTC)

text: I had restricted to within the last century, because of course there is a difference between now and the pre-rigor days.  Over-and-out. (Cj67 00:13, 29 June 2006 (UTC)

text: Ah. I hadn't noticed. But in my earlier reply, I said over 'the centuries,' so I daresay this doesn't actually make my point any less true, just not as important in the present context. But I do wish do know, do you think the results presented in the article really are of any consequence, and just disagreed with my assertions, or did you disagree with the point that I used them to reinforce as well? Septentrionalis, see quintic. The roots of algebraic polynomialss degrees 0-4 are all all agebraic, but fo instance, the roots of the quintic polynomial x4 − x + 1 = 0 are not. For this reason, it has been proven that no {Quintic Formula" analogous to the quadratic, cubic, and quartic formulas can exist, nor can exist for degree higher than five. That is not to say no quintic polynomials have algebraic roots, though. It depends on a given polynomials Galois group. -- He Who Is[ Talk ] 02:51, 30 June 2006 (UTC)
replies: text: You seem to be using a different definition of algebraic. An algebraic number is defined to be any root of a polynomial with integer coefficients, so the roots of x4 − x + 1 = 0 are per definition algebraic. I think you mean "solvable by radicals". -- Jitse Niesen (talk) 03:13, 30 June 2006 (UTC)

text: Is that true?  I am not aware of a result which says that if the roots are rational, then it is solvable by radical, though the roots can be found by exhaustion using Viète's formulas. -lethe talk [ +] 03:45, 30 June 2006 (UTC)

text: If you meant to say "not all rational polynomials are solvable by radical", then you are of course correct.  But that's not what you said, even on your second try. -lethe talk [ +] 14:19, 30 June 2006 (UTC)
thread_title: Does this matter?

message: This article needs a section Entitled
== Significance ==
Or
== Real life Applications ==
Has there been any applications of this proof?  Has anything been built of of this proof or is it a dead end?  
Also for those of us that use "0.999 repeating" as a physical representation of an abstract number "approaching but not equal to" one we need a new variable.  Has a new variable been given (No pun in tended).
For instant although .999¯ equals one ... when we graph the number line (-∞,1),(1,∞) or (x≠1) How do we illustrate the "conceptual number" before and after 1 or any other whole number.  Before we proved this we just used .999¯ and .000¯1 to say this concept.  Now that we Proved this number actaully does not exist we need a new way of saying what we have to say.  This Proof is hard to Grasp becose we don't have an alternitive to issustrate the idea.  
Remember that "1" is a symbol for a number. And if ".999¯" equals one then it is just a #REDIRECT 1.  We now need a new Title for XYZ.--E-Bod 22:38, 11 June 2006 (UTC)
thread_title:  Significance 

message: n/9
/   	.
0/9 	0.000
1/9 	0.111
2/9 	0.222
3/9 	0.333
4/9 	0.444
5/9 	0.555
6/9 	0.666
7/9 	0.777
8/9 	0.888
9/9 	0.999
10/9	1.111
n/9
/   	.
0/9 	0.000
1/9 	0.111
2/9 	0.222
3/9 	0.333
4/9 	0.444
5/9 	0.555
6/9 	0.666
7/9 	0.777
8/9 	0.888
9/9 	1.000
10/9	1.111
If you make a Table it makes sense.  If you try with a proof people will try to think it is a 1=2 proof.  Somebody still needs to address what to call the number that approaches 1 but isn't.  Just like X^0 is one not zero if you make a table and look at it.
:That's the point of the proof: The number that "approaches 1 but isn't" does not exist (at least, not in the standard reals we usually employ). As to the significance, I can't think of any "real world applications" right now. This proof should just show the non-existence of such an "approaching number" and thus clarify our mental image of the reals. If you want to discuss the mathematical details, we should probably do so on the arguments page. Yours, Huon 09:04, 12 June 2006 (UTC)
replies: text: Sorry Ill move it to arguments]] now.  I was Actually Suggesting We move the Charts into The Article.  Or a Variant of the chart.  I will Further the Discussion under arguments.  However I'm not arguing the proof is wrong I was suggesting Article Content.--E-Bod 23:27, 12 June 2006 (UTC)
replies: text: Note the word "if" in Huon's comment. As long as the discussion is about improvements to the article, it can be held here. -- Meni Rosenfeld (talk) 15:37, 13 June 2006 (UTC)
thread_title:  Table 

message: User:Hamish2k added a  tag to the top of the article without rationale.  I disagree and have of now removed the tag.  Perhaps someone could say what exactly needs to be wikified in this article? —Mets501 (talk) 12:35, 19 June 2006 (UTC)
thread_title:  Wikify tag 

message: The Mathematical Gazette joke was published in 1954. The link, however, states that it is "[to be read in the voice of Fractured Fairy Tales]". Fractured Fairy Tales did not air until 1959.
Since there is no way we can edit the quote on the linked site, there's nothing we can do about this mistake ... but I wanted to mention it somewhere, just so I can feel smug. :-) Rpresser 16:05, 26 June 2006 (UTC)
thread_title:  Trivial trivia about the linked joke 

message: If .999 = 1, it follows then that 1/(1-.999...) = 1/0, so it is possible to divide by zero?
//// Pacific PanDeist * 05:40, 3 July 2006 (UTC)
replies: text: You might be more interested in the Division by zero article. As for .999..., we can also make the observation that since sqrt 4 = 2, it follows that 1/(2-sqrt 4) = 1/0; it doesn't seem too helpful to me. Am I misunderstanding you? Melchoir 06:22, 3 July 2006 (UTC)
replies: text: [Edit conflict] See division by zero. Your argument is absurd in any case; it is the same as saying, "If 1 = 1, it follows then that 1/(1-1) = 1/0, so it is possible to divide by zero?". Specifically, why did you decide it was possible to calculate 1 / (1-0.999...)? This is just as possible as calculating 1 / 0 - which depends on our definition of division. -- Meni Rosenfeld (talk) 06:25, 3 July 2006 (UTC)
replies: text: To be precise, it is possible to divide by zero in some cases. For instance, the Projected reals allow this because they add an unsigned infinity element, so it can represent both positive and negative infinity, both limits to zero of 1/n. -- He Who Is[ Talk ] 03:37, 4 July 2006 (UTC)

text: Well, since 0.999... = 1, and thereby 1 - 0.999... = 0, it is of course impossible for 1 / (1 - 0.999...) to be different from 1 / 0. Both (if, as in the real projective line, they are considered to equal ∞) are the type of infinity that "is just there" - the idea of chasing something down an endless path is a common misconception regarding these issues, and is generally something we wish to avoid in order to keep things mathematically accurate. -- Meni Rosenfeld (talk) 13:28, 4 July 2006 (UTC)

text: Since this discussion does not belong on the article talk page, I replied on the user talk page. In the interests of keeping traffic on this page focused on improving the article, please redirect any further responses. Thanks. --KSmrqT 19:54, 4 July 2006 (UTC)
replies: text: Thank you, I'll keep my responses there. //// Pacific PanDeist * 07:04, 5 July 2006 (UTC)
thread_title:  division by zero?