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Revision as of 22:18, 6 May 2005 edit213.216.199.18 (talk) New section: Arguments against← Previous edit Revision as of 22:21, 6 May 2005 edit undoBradBeattie (talk | contribs)6,888 editsm Moved anon's commentNext edit →
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::: To be convinced that 0.999~ doesn't equal 1, you'd have to give me a proof. Any of the following methods are not acceptable: . --] 21:48, 6 May 2005 (UTC) ::: To be convinced that 0.999~ doesn't equal 1, you'd have to give me a proof. Any of the following methods are not acceptable: . --] 21:48, 6 May 2005 (UTC)

If for given "equation" <math>\frac{999\ldots}{1000\ldots} = 1</math>
then 999...=1000... (to get what?) that black is actually white? --Anon


== New section: Arguments against == == New section: Arguments against ==


I think the page needs a section on common misconceptions and flaws in the reasoning of these misconceptions. If the abracadabra section on the talk page says anything, it's that some people don't really get this. I think the page needs a section on common misconceptions and flaws in the reasoning of these misconceptions. If the abracadabra section on the talk page says anything, it's that some people don't really get this.

If for given "equation" <math>\frac{999\ldots}{1000\ldots} = 1</math>
then 999...=1000... (to get what?) that black is actually white?

Revision as of 22:21, 6 May 2005

Creation of this entry

I created this page in response to two threads I saw and the confusion that arose. Figured it was something worth noting. --BradBeattie 18:58, 6 May 2005 (UTC)

I think you are right. I submitted it first for deletion because the title looked a bit misleading. This is not a series of nines, the series is if you wish of
9 10 n {\displaystyle {\frac {9}{10^{n}}}}

Cheers, Oleg Alexandrov 19:01, 6 May 2005 (UTC)

True, the title was a little slap-dash. Thanks for the improvement. --BradBeattie 19:03, 6 May 2005 (UTC)

Abra-cadabra

"In mathematics, one could easily fall in the trap of thinking that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are."

0,999... is irrational and so is the article. Basis on "the proof" that 0.9999...=1 one could argue that irrational is rational which is simply jargon.

How about 999 1000 {\displaystyle {\frac {999\ldots }{1000\ldots }}} ? Might want to take a look at limits. --BradBeattie 20:16, 6 May 2005 (UTC)

If 0.999... is 1 then the whole basis of mathematics should be re-written. Mathematics is considered to be exact science. If 0.999... was EXACT 1 then it would not make any difference to say exempli gratia (for example) that domain is same than [0,1[ or 0.000...0001 is 0 which is the basis of differential calculus. One should not confuse the concept of irrationality with the concept rationality, or infinity with finity, or inexact with exact.

Could you please prove your statement? This page has a proof as to why 0.999~ = 1. Please provide your counter-proof. --BradBeattie 20:38, 6 May 2005 (UTC)

1/3 is often writen as 0.333... If you multiply 1/3 (or in your case referred as 0.333...) by 3 you get exact 1. It's not proofing. Is's abracadabra id est (that is) mumbo jumbo in magic industry. If you geometrically plot function Y=1/X where X= (instead of domain ]0,1]) what value do you get for Y when X=0 or how do you present it?

"~=" is different than "=" (equality)

What's your point? The number 0.9999... is not in

"What's your point? The number 0.9999... is not in [0, 1[. Oleg Alexandrov 21:05, 6 May 2005 (UTC)"

Well, 0.999... is not the zero, it's the other bound i.e. number of the domain/range [0,1[. It think you can guess which one of the bounds it is.
0,999... approaches its limit 1 BUT not equals 1. (anon forgot to sign)
Well, that's the very purpose of this article, to convince you that 0.999.. equals 1. You either show that this theorem is wrong, or believe it. :) Oleg Alexandrov 21:39, 6 May 2005 (UTC)
To be convinced that 0.999~ doesn't equal 1, you'd have to give me a proof. Any of the following methods are not acceptable: Alternative Proofs. --BradBeattie 21:48, 6 May 2005 (UTC)

If for given "equation" 999 1000 = 1 {\displaystyle {\frac {999\ldots }{1000\ldots }}=1} then 999...=1000... (to get what?) that black is actually white? --Anon

New section: Arguments against

I think the page needs a section on common misconceptions and flaws in the reasoning of these misconceptions. If the abracadabra section on the talk page says anything, it's that some people don't really get this.