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Geek Culture / Proofs that 0.999... = 1

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Shadowtroid
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Posted: 7th May 2010 01:46
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Quote: "Its not a concept its a hard cold fact. Numbers CANNOT end."


Yes, but infinity IN OF ITSELF is a concept. Numbers never end, but infinity is not an actual number.

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General Jackson
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Posted: 7th May 2010 01:47
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Well that's correct. lol.

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Slow Programmer
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Posted: 7th May 2010 02:50
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I agree that 1 = 0.9999... for all reasonable purposes. However, consider this. If you plot number lines at 0.9999... and at 1 on the same graph they will never converge into a single line.
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bitJericho
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Posted: 7th May 2010 03:14 Edited at: 7th May 2010 06:55
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Yes, but plotting 1 and 3/3 will

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Diggsey
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Posted: 7th May 2010 04:04 Edited at: 7th May 2010 04:06
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@Slow Programmer
They will be the same line. It's not like there's a small difference they are actually the same number

Quote: "The equality 0.999…=1 has long been accepted by mathematicians and taught in textbooks. In the last few decades, researchers of mathematics education have studied the reception of this equality among students. Some reject it due to their intuitions that each number has a unique decimal expansion, that nonzero infinitesimal numbers should exist, or that the expansion of 0.999… eventually terminates. These intuitions fail in the real numbers, but alternate number systems can be constructed bearing some of them out. Indeed, some settings contain numbers that are "just shy" of 1; these are generally unrelated to 0.999…, but they are of considerable interest in mathematical analysis."


Just because you write it differently doesn't make it a different number.

It's exactly the same as 1.0 being the same number as 1. The '.0' part is not necessary but it is still the same number.


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jrowe
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Posted: 7th May 2010 05:03 Edited at: 7th May 2010 05:04
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If you're arguing that 0.999... and 1 are not equal, then you're arguing that:
1 - 0.999... = 0.00...01

The number on the right would require an infinite number of zeroes before the "1". So in fact, the "1" wouldn't be there...

EDIT: d'oh, didn't see p2.

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Neuro Fuzzy
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Posted: 7th May 2010 05:28
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I was trying to figure out how this was wrong for like 10 seconds until I realized that this works because .999... * 10 = 9.999..., instead of 9.99, and that is the key to why this proof works. Just me being stupid, but i'll post for crystal clarity's sake.


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BMacZero
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Posted: 7th May 2010 06:35 Edited at: 7th May 2010 06:35
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Calculus says this is true. Here's a more-understandable version of Zoto's proof #4 (I find it more convincing than the others because they have the ring of manipulation that leads to things like that 1=2 "proof")




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Oneka
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Posted: 7th May 2010 09:42 Edited at: 7th May 2010 09:50
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0.9 repeating is not 1

case and point,

1=1
0.9 repeating = 0.9 repeating

0.9 repeating < 1
Reason I say this is because 0.9 repeating is 0.x1 less than 1
where x = infinite amount of 0s between, but its still less than 1, if 0.9 = 1 then it wouldn't be 0.9, it would just be 1

I dont think that just because it came out algebraically its right, I think it falls more with just a problem with how we deal with numbers, because atleast the way I think of it, 0.9r and 1 are two different numbers

personally, I just see them as two different numbers, but the raw end of it, it doesnt matter because its so close to 1 it doesnt make a difference, but I still believe it is less than one, even with the who limit concept in mind, I still think that way, just like every number is always raised by 1, multiplied by 1, divided by 1, etc


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Neuro Fuzzy
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Posted: 7th May 2010 09:51
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Quote: "0.9 repeating is not 1

case and point,"


NOOOOO!!!! we had made so much progress


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Oneka
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Posted: 7th May 2010 09:55
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lol,

When I look at it, I think solving it atleast by the most simple method

1/3 = 0.333r ruins it, because converting it from 1/3 to 0.3r we already have loss accuracy since its now being represented in a finite form

so 1/3 * 3 = 1, but 0.3r * 3 != 1, this is just not even a mathematic look to it, its just something just as easy to do, 3/3 = 1 but 0.3r * 3 = 0.9r

Accuracy is lost once we change form so I still dont think that 1=0.9r


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sealclubber
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Posted: 7th May 2010 10:48
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The issue here is that when people think of 0.999..., they intuitively think of the sequence of numbers {0.9, 0.99, 0.999,...} which clearly approaches 1. From here, it's (naively) logical to assume that since every number in the sequence is less than 1, 0.999... must be too.

But understand that you have given the number 0.999... no clear definition--it is certainly not equal to ANY of those numbers in the sequence. Indeed we define 0.999... as the limit of that sequence, which again we've already agreed upon as 1. That is, by definition alone 0.999... = 1. This is not a special case; it is how we define any infinite sum.

And again one must understand that this definition is not illogical. Indeed it is the only logical choice, for no matter what number you choose less than 1, we can always go far enough out in the sequence of 0.9's to find a number bigger than your chosen one.
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ionstream
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Posted: 7th May 2010 11:25
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Ooh there was a quote I liked from another site, that said that if two numbers are different, then there is a number between them, like their average: but what number could possibly go between 0.9~ and 1?

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Zotoaster
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Posted: 7th May 2010 11:29
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Quote: "1/3 = 0.333r ruins it, because converting it from 1/3 to 0.3r we already have loss accuracy since its now being represented in a finite form"

We have lost no accuracy and it's not being represented in a finite form. There are an infinite number of 3s, so it's 100% accurate.

"everyone forgets a semi-colon sometimes." - Phaelax
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Dia
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Posted: 7th May 2010 14:27 Edited at: 7th May 2010 14:39
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It is a concept in Maths (ok, and philosophy) - there is no ACTUAL number with the value of infinity, so talking about 0.999... is a purely theoretical number anyway.

but it does then beg the question:

if 1=0.999...

then

1-1=1-0.999...

0.000...000 = 0.000...001

hence we have successfully proven that zero = one!

Take that MRS smith! (my third grade maths teacher)

this is the kind of problem you run into when you start treating infinity as a number

This is not the Sig you are looking for....
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David R
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Posted: 7th May 2010 14:39 Edited at: 7th May 2010 14:43
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Quote: "1-1=1-0.999...
0.000...000 = 0.000...001
"


No - that'd be zero.

0.999' = 1

.'. 1-1 = 0

You can't say

1-0.999... = 0.000...000 = 0.000...001

because 0.999' is 1 (a synonym if you will)

Your statement implies that 1-1 != 1-1 which is a pretty bizarre contradiction

EDIT:
Quote: "hence we have successfully proven that zero = one!"


Even if you could say that it's 0.000000' [...] 1 then that's in the same situation as 0.999' - it's infinitely long and ends up as just 0. That does not in anyway imply 1=0

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Green Gandalf
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Posted: 7th May 2010 16:35
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If 0.9999999... = 1 what does 0.8888888... equal?

[Retreats to a safe distance. ]
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Neuro Fuzzy
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Posted: 7th May 2010 17:03
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Yeah, what about the limit of a sequence that is also a non-terminating decimal?


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Oneka
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Posted: 7th May 2010 17:32 Edited at: 7th May 2010 17:33
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Quote: "We have lost no accuracy and it's not being represented in a finite form. There are an infinite number of 3s, so it's 100% accurate."


What I meant by that is when you divide 1 by 3, you cant mutliply it out and get a perfect 1, you get 0.3r, and when you multiple by 3, you get 0.9r, what I was saying is that if you leave it in the form 1/3 and multiply by 3 you get 3/3 which is equal to 1, not 0.9r, I know that as it approaches infinity it equals 1, thats easy limit, but analyzing it logically to me atleast, but I think its just an issue of writing it out is all, I think that best form to do that equation isnt to right a decimal, but to leave it as a fraction, because its not being put into a close to finite form atleast (Its still infinite, but the concept is easier to see for me atleast as 1/3)

and if 0.9r = 1, then, pi = 3 nuff said


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Diggsey
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Posted: 7th May 2010 19:50 Edited at: 7th May 2010 19:53
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@Oneka
Quote: "What I meant by that is when you divide 1 by 3, you cant mutliply it out and get a perfect 1"


Yes you can...

a = 1/3
3a = 3*1/3
3a = 3/3
3a = 1

Therefore, 3*0.333... = 1

There isn't really any question about it. The rules which define how we write down numbers state that 0.999... = 1. Just accept it already

@GG
0.888... equals 8/9.

The fact is that our method of writing tries to reduce the amount of writing necessary. Of course, we could write every number as a recurring decimal. (ie. one as 0.999... two as 1.999...) but for some numbers it defines a shortened form to represent it.


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Oneka
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Posted: 7th May 2010 23:32 Edited at: 7th May 2010 23:34
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I believe you are missing the point I am saying, I understand where you coming from, I am simply explaining how I view how someone can think and me how 0.9r does not = 1, because with your statement

Quote: "a = 1/3
3a = 3*1/3
3a = 3/3
3a = 1"


youve said that a = 1/3 which when multiplied by 3, equals 3/3 which is one,

but when you right it out as 0.3r and then multiply by 3 you get 0.9r,

All I am saying is that the perfect form to right it with the best accuracy is 1/3 in my mind and not 0.3r because when you right it as 0.3r you cant perfectly bring it back to 1, you get 0.9r instead
where as 1/3 + 2/3 = 3/3 you get 1, do you understand what I am saying now?

EDIT:
I am not saying that its not right, I am just debating on the way we represent it, thats all


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Zotoaster
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Posted: 7th May 2010 23:44
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The problem with your statement is that you're assumine 1/3 and 0.333... are different, in that 0.333... is somehow less accurate than 1/3, but in fact it's not. It's just, again, a different representation.

"everyone forgets a semi-colon sometimes." - Phaelax
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Oneka
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Posted: 7th May 2010 23:53 Edited at: 7th May 2010 23:58
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I dont think that 1/3 and 0.3r are necessarily different, I just think that when I say the accuracy is lost is that when you go to do math using the decimal form 0.3r and not the fraction that you have loss accuracy in the sense of working it out, not actual number loss

For example in the situation I would not do math by 0.3r + 0.3r + 0.3r = 1, I would always write it as 1/3, because that way I would by hand get a perfect 1 out and not 0.9r, And so for by the other logic then, 0.89r would be equal to 0.9?

EDIT:
to put it simple in just the way I view it, I dont think that 1/3 = 0.3r in that sense then, I think that 0.3r is the closest approximation to it but I think it is best written as 1/3, thats just my concept of it


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Green Gandalf
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Posted: 8th May 2010 00:08
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@Diggsey

Quote: "Of course, we could write every number as a recurring decimal."


How about pi or sqrt(2)? IIRC a recurring decimal is necessarily a rational number.

[Gets ready to retreat to an even safer distance.]
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BMacZero
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Posted: 8th May 2010 00:11
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Quote: "If 0.9999999... = 1 what does 0.8888888... equal?"

Applying the method I used in my last post, that is (0.8)/(1-0.1) = 8/9 (as Diggsey said ).


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Peter H
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Posted: 8th May 2010 00:37
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This thread...


MEGAFAIL

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Neuro Fuzzy
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Posted: 8th May 2010 00:54 Edited at: 8th May 2010 08:29
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Quote: "How about pi or sqrt(2)? IIRC a recurring decimal is necessarily a rational number."

Ah! My favorite numbers are the transcendental numbers!!! I love 'em!!! Especially how they can relate to each other to form integers.

[IE e^(i*pi), i is not transcendental. e^(n) where n is a complex number (and NOT transcendental) can never be a complex number.]

[edit]
^above statement is incorrect. True statement:
[ e^(n) where n is a non-transcendental complex number can never be a non-transcendental number. e can be substituted for any transcendental number and this statement remains true.]


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AndrewT
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Posted: 8th May 2010 01:52
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Whoa. I thought we had gotten past this, like, thirty years ago.

Guess not.

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sealclubber
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Posted: 8th May 2010 02:13
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Quote: "e^(n) where n is a complex number (and NOT transcendental) can never be a complex number."


This is certainly not correct.
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Neuro Fuzzy
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Posted: 8th May 2010 08:27
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Quote: "This is certainly not correct. "


Ah, I see the error of my ways, I meant non-transcendental. Forgetting what is a subset of what X_X


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sealclubber
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Posted: 8th May 2010 08:51
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In any case, the issue here is that when people think of 0.999..., they intuitively think of the sequence of numbers {0.9, 0.99, 0.999,...} which clearly approaches 1. From here, it's (naively) logical to assume that since every number in the sequence is less than 1, 0.999... must be too.

But understand that you have given the number 0.999... no clear definition--it is certainly not equal to ANY of those numbers in the sequence. Indeed we define 0.999... as the limit of that sequence, which again we've already agreed upon as 1. That is, by definition alone 0.999... = 1. This is not a special case; it is how we define any infinite sum.

And again one must understand that this definition is not illogical. Indeed it is the only logical choice, for no matter what number you choose less than 1, we can always go far enough out in the sequence of 0.9's to find a number bigger than your chosen one.as
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