User talk:72.87.210.78

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[edit] July 2007

Welcome to Wikipedia. It might not have been your intention, but your recent contribution removed content from 0.999.... Please be more careful when editing pages and do not remove content from Wikipedia without a good reason, which should be specified in the edit summary. Take a look at the welcome page to learn more about contributing to this encyclopedia. If you would like to experiment again, please use the sandbox. Thank you. SMC89 14:11, 9 July 2007 (UTC)

If this is a shared IP address, and you didn't make the edit, consider creating an account for yourself so you can avoid further irrelevant notices.

Please refrain from making unconstructive edits to Wikipedia, as you did to 0.999.... Your edits appear to be vandalism and have been reverted. If you would like to experiment, please use the sandbox. Thank you. Andrij Kursetsky ⊗ 14:22, 9 July 2007 (UTC)

If this is a shared IP address, and you didn't make any unconstructive edits, consider creating an account for yourself so you can avoid further irrelevant warnings.
If you dispute the statement that the real number 0.999... is equal to 1, take it to Talk:0.999.../Arguments. Note, though, that this equality is a proven theorem, and some of the people who will respond there actually know mathematics, so keep your cool and present your arguments in an orderly and civil way. -- Meni Rosenfeld (talk) 14:37, 9 July 2007 (UTC)

That's 6 unconstructive edits in half an hour, which not only can be seen as vandalism but also violates WP:3RR. If you continue, you will be blocked from editing Wikipedia. -- Meni Rosenfeld (talk) 14:42, 9 July 2007 (UTC)

This may help you understand why .9999... = 1: [1]. It's a pretty understandable bunch of proofs of why this is so. Gscshoyru 14:43, 9 July 2007 (UTC)

That Web page is obviously bullshit.
It isn't, though. If you can find a flaw in any of the arguments, then you have a right to say that, but unless you start pointing out flaws, you can't dispute their mathematical validity. Gscshoyru 14:53, 9 July 2007 (UTC)
The flaw is obvious. You are basically saying that infinitesimals do not exist, when they are in fact the reciprical of infinity.
If 0.999... < 1, then what is the average of the two? (the number midway between them) --Maelwys 14:58, 9 July 2007 (UTC)
The average is 0.999…
But if the average of 0.999... and 1 is 0.999..., then the difference (1 - 0.999...) = 2 * (0.999... - 0.999...) (twice the distance from one number to the average of the two numbers), so 1 - 0.999... = 2 * (0), 1 - 0.999... = 0, 1 = 0.999... --Maelwys 15:03, 9 July 2007 (UTC)
[Edit conflict] There are no nonzero real infinitesimals. Other structures like the hyperreal numbers have infinite quantities and infinitesimals, but they are not the subject of our discussion.
Anyway, do you agree that the average of a and b is \frac{a+b}{2}? If so, your assertion that the average of 0.999... and 1 is 0.999... can be written as 0.999...=\frac{1+0.999...}{2}. Multiply by 2 and subtract 0.999..., and you get 0.999... = 1. -- Meni Rosenfeld (talk) 15:06, 9 July 2007 (UTC)


This is an infinitesimal being dealt with, a reciprical of infinity.

(unindent) Infinity is not an actual number, therefore there is no reciprocal of it.

But since you like infinity, so much... do you agree that .999... = .9 + .09 + .009 ...? Gscshoyru 15:15, 9 July 2007 (UTC)

Yes, of course. And infinity is an actual number, even if it's impossible to count to (because there is no end to it). Therefore the reciprocal of something that goes on forever and encompasses everything would have to be something that is infinitely small.
You have been blocked from editing for a period of 31 hours in accordance with Wikipedia's blocking policy for persistent disruption and vandalism. Once the block has expired, you are welcome to make constructive contributions. If you believe this block is unjustified, you may contest the block by adding the text {{unblock|your reason here}} below.

Клоун 14:52, 9 July 2007 (UTC)

This blocked user (block log | autoblocks | rangeblocks | unblock | contribs | deleted contribs) has asked to be unblocked, but an administrator has reviewed and declined this request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy). Do not remove this unblock request while you are blocked.

Request reason: "I was adding a {{POV}} template to 0.999... because that article obviously violates Wikipedia:Neutral point of view, so I was blocked by a biased admin improperly accusing me of vandalism when he himself is actually the vandal."

Decline reason: "3RR is 3RR. When you return, feel free to discuss this at the appropriate place. — jpgordon∇∆∇∆ 15:02, 9 July 2007 (UTC)"

Please make any further unblock requests by using the {{unblock}} template. However, abuse of the template may result in your talk page being protected.

So, bias is apparently acceptable at Wikipedia, then.

It's not bias if it's a proven fact... under normal circumstances what you did might make sense... but in mathematics, there is no POV. And true or not, you violated the 3RR. Gscshoyru 15:09, 9 July 2007 (UTC)

[edit] 0.999... and infinitesimals

The initial part of the article concerns definitions made with the real number system (ie. standard analysis), for which the proofs are entirely valid. Later on, it also deals with alternate number systems, including the infinitesimals, for which the symbol 0.999... has alternate interpretations. Going into an edit war because it fails to consider your favourite number system from the very off is hardly the way to go about addressing any bias which you perceive. In particular, trying to get the page deleted[2] is, at best, an infringement of WP:POINT;at worst, it's outright vandalism.

If you see yourself having a productive future editing articles on Wikipedia, you need to develop a more level-headed approach to things. And you might want to take a look at WP:3RR; talk pages are the way to go in content disputes, not edit wars. Mark H Wilkinson 15:58, 9 July 2007 (UTC)