
#1
May2804, 02:11 PM

P: 88

Very common debate: is 0.999999 repeating 1?
Opinions? 



#2
May2804, 02:16 PM

P: 88

didn't realize that there was anther exactly the same question in logics thread.




#3
May2804, 02:21 PM

P: 225

What is the difference, meaning subtract .999... from 1.000.... The difference would be 0.000...1. But it's not valid to put something after the "..." That's asking what comes after infinity, which isn't a valid question. The expression .000...1 is a typographic error, and not something that is even defined in the set of real numbers. 



#4
May2804, 02:54 PM

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P: 9,398

Is 0.999repeating 1?
It is not an opinion that they are equal, it is a very easy provable fact and only cranks who don't understand the way mathematics work insist they are different after it has been patiently explained to them.
We mean base ten, work out what the infinite sum 0.999... is, if that doesn't convince you then you need to look up the definitions you don't understand in the phrase: they represent the same equivalence class in the cauchy sequences of rationals modulo convergence that define the real number system. 



#5
May2804, 03:01 PM

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P: 2,226

Well I'm 99.99999.....% certain it's equal to 1 :D




#6
May2804, 03:03 PM

P: 617

This has to be the most asked question on this forum.




#7
May2804, 03:08 PM

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P: 2,226

In fact I now propose jcsd's theorum: 



#8
May2804, 03:10 PM

P: 617

.999… not equal to 1? That’s kiddy stuff, just watch me argue that .3333… is not equal to 1/3




#9
May2804, 03:13 PM

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Corollary to JCSD's theorem:
every bulletin board etc attracts an idiot, a troll, or possibly both. 



#10
May2804, 05:01 PM

P: 43

Why should .9999... equal 1 and not .9999...? Trying to get from .9999... to 1 is just like trying to accelerate your spaceship to the speed of light. You keep getting closer, but you can't get that last little bit.




#11
May2804, 05:14 PM

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P: 7,291

I knew it.^~ A last little bit poster would have to show up. Do we try to explain it to him?
That "last little bit" is [tex] \frac 1 \infty [/tex]. By the definition of infinity, that last little bit is zero. So essentially this is true by definition, but beyond that it is completely consistent and provable in many different manners. There is no law that says each point on the real number line must have a unique representation. In fact just the opposite is true, every point on the real number line has many (perhaps an infinite) number of different ways to represent it. 



#12
May2804, 05:28 PM

P: 43





#13
May2804, 05:45 PM

P: 132

Eyes can trick your mind.




#14
May2804, 05:48 PM

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P: 7,291

My apologies, having been involved in this same discussion on several different forums over the last 2 or 3 years I do not recall anyone ever saying "oh I see" so perhaps am a bit cyncial about the whole issue.




#15
May2804, 05:50 PM

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P: 2,226

let x = 0.9999... => 10x = 9.99999... 10x  x = 9x = 9 => x = 1 All we are really saying is: [tex]\sum_{n=1}^{\infty} \frac{9}{10^n} = 1[/tex] 



#16
May2804, 05:52 PM

Emeritus
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P: 734

I'm sorry for the harsh response to your question, Grizzlycomet. We generally try to not be hard on people because of the questions they ask; the problem is that this particular topic is visited way too often by people trying to push their "new math", "theory of infinity" and whatnot, instead of trying first to understand how standard math deals with the issue.




#17
May2804, 05:55 PM

P: 43





#18
May2804, 06:06 PM

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P: 7,291

Read this page. There are 2 proofs, the first simply uses the sum of an infinite series formula. The second is my version of how a Mathematician approaches the problem. I feel that it also gives a very intuitive feel for why equality holds.



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