Talk:0.999.../Arguments

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This page is for mathematical arguments concerning 0.999... Previous discussions have been archived from the main talk page, which is now reserved for editorial discussions. Before posting, you may want to read the FAQ on Talk:0.999... and the following FAQ.

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Frequently Asked Questions (FAQ)

Q: You guys talk a lot about real analysis, limits, and calculus; shouldn't this just be about arithmetic?

A: Unfortunately, in order to formally prove many qualities of numbers, one often has to resort to higher mathematics: real analysis in the case of real numbers, number theory in the case of integers, and so forth. The article and arguments page both aim to be understandable to all, but, since many skeptics ask for formal proofs, higher mathematics will inevitably come into play.

Q: Person X made a pretty convincing argument that $0.999... \ne 1$ ! Those who say otherwise are giving arguments I can't understand.

A: Before believing an argument, check sources, responses, and record. The main article is well-sourced, whereas arguments against generally cite (if anything) non-mathematical sources such as online message boards and dictionaries. Also, although most people are trying to write something everyone can understand, some arguments, in relying on higher mathematics, will not be easy to follow for all. Still, try to follow those you can. Finally, those who firmly believe that mainstream mathematics is mistaken will generally reveal their lack of rigor and/or contempt for experts and others who disagree with them. Before replying, read their other contributions to make sure you aren't siding with someone you yourself would not trust.

Q: I have a mathematical question.

A: Please check the FAQ on the talk page.

1 Trolls
2 Section Break
3 0.333...=1/3?
4 A few final questions
5 10c
6 This Might Help Skeptics
7 1/3 <> 0.333...
8 Keeping this simple
9 Explain this!!!

[edit] Trolls

The person who has been trolling this article for the past year is now posting with IP addresses that begin with 41.243. Please do not respond to him. He knows that what he posts is nonsense, so trying to talk sense into him is completely futile. -- Meni Rosenfeld (talk) 20:51, 1 December 2006 (UTC)

Good point. I suppose I thought it was worthwhile to help show others that this was the case, but hopefully that should be self-evident without any effort on my part or that of others. But what is the most succinct way of revealing trolls without wasting time feeding them and/or resorting to personal attacks? Calbaer 21:01, 1 December 2006 (UTC)

Deleting their posts, of course. But only for really serious cases, like this particular troll. Anyone who thinks this is inappropriate, probably just hasn't seen enough of his "work". -- Meni Rosenfeld (talk) 21:32, 1 December 2006 (UTC)

I am curious how you are capable of reading this editor's mind. --jpgordon^∇∆∇∆ 04:18, 2 December 2006 (UTC)

Occam's razor. -- Meni Rosenfeld (talk) 15:16, 2 December 2006 (UTC)

Well, there is that...but...I'm remembering when I was introduced to limits, epsilon proofs, and so on in my high school calculus class (back in the age when calculus really meant pebbles.) One of my significant "a-ha" moments was when limits "clicked" for me; it didn't make any sense, then suddenly it did...I'm also remembering a time a few years before that, when I was 11 or so, on a starry cold spring night in Denmark, discussing the stars with a friend, and suddenly "clicking" on the seeming infiniteness off the universe; my friend kinda freaked out, unable to grasp the concept of unendingness. I can easily see how someone could fail to have either of those epiphanies, but rather resist them, and thus be unable to understand simple things like "no, there's no infinity-th digit you can point to", and thus "no, there's no largest real number less than another real number." So for me, Occam suggests it's more likely that someone simply not comprehend a couple of key concepts than that they'd waste so much time and energy trying to drive a few very patient mathematicians to distraction. See Hanlon's razor. --jpgordon^∇∆∇∆ 15:46, 2 December 2006 (UTC)

When it didn't make any sense, did you log on to Wikipedia and declare to the world that basic mathematics was broken? Endomorphic 20:28, 5 December 2006 (UTC)

I very well might have had it been available; I, of course, was the smartest boy in the world. --jpgordon^∇∆∇∆ 15:34, 13 December 2006 (UTC)

And the Universe isn't unlimited; it merely loops back on itself. One cannot escape it within the bounds of three dimensions because it has follow-through walls... so to speak. If you're going fast enough it'd be like a chase-scene in a Hanna Barbera cartoon. ^_^ ~ SotiCoto 195.33.121.133 10:21, 30 January 2007 (UTC)

This may explain some of the apparent trolls. But not this one. I'm not sure I'll be able to convey this efficiently with words, but he's not some first-year student who's having trouble understanding limits. He has proven to be familiar with a lot of advanced mathematical topics (some of which I myself am not familiar with), and has always carefully chosen words which will be as irritating as possible while maintaining the impression that he knows what he is talking about (which he does, incidentally, only that what he knows and what he posts don't match). This cannot be "explained by stupidity", so Hanlon's razor doesn't apply. But I beg you to take a look at the archives for posts to this page by 41.243, by his previous incarnation 198.54.202.54, and by his other IPs which you will have no trouble recognizing by his malicious style. I have no doubt that you will reach the same conclusion I did. -- Meni Rosenfeld (talk) 16:10, 2 December 2006 (UTC)

Malice in mathematics. I love it... Yeah, you're probably right. I never stop being amazed at what wankers will do for entertainment. --jpgordon^∇∆∇∆ 16:21, 2 December 2006 (UTC)

BELOW IS UP TO DATE! I HAVE REMOVED THE 1/3 PROBLEM! AS THAT IS A FRACTION CALCULATION! AND REALLY NOTHING TO DO WITH A DECIMAL PROBLEM AS INFINITE 09 <> 1

INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,15/01/07.

The Variables below A & B are Both Single Start Values.

A = 1

B = 0.9

Below Proves A is Infinitely > B And Gives The Variable C This Value.

C = A / B = 1.1111111111......

Below is How Infinite 0.9 is Calculated.

0.9 x C = 0.9999999999 ......

Below Proves Infinite 0.9 is <>1 By Using The Original Variables A & B.

A x C = 1.1111111111 ......

B x C = 0.9999999999 ......

Your bottom section does not prove that 0.999... ≠ 1, the best it does is prove that 0.9 ≠ 1, which is pretty obvious. Maelin (Talk | Contribs) 13:21, 23 January 2007 (UTC)

Tried switching in 2 instead of 0.9? It doesn't involve 0.999... at all, and gets to 1 right at the B x C stage. In fact, try it with B as any number you like. It gives all kinds of bizarre results when you have to deal with obscure fraction to decimal conversions. As far as I'm concerned, where this fails is at the C = A / B stage, but since I don't know the higher mathematical language to express this in ways it would be understandable... I guess I can't. ~ SotiCoto 195.33.121.133 10:34, 30 January 2007 (UTC)

If a Number has to be Infinitely Multiplied! To Try and Reach a Target Value! Then the Target Value! Must be Greater than the Number being Infinitely Multiplied! Reason Being! If the Target Value was also Infinitely Multiplied! Then the Target Value would again be Greater! Number Infinitely Multiplied! = 0.9 Target Value! = 1 “

[Anthony.R.Brown 20/01/07].

Sorry, but I don't understand what you're talking about. What is "infinitely multiplied" supposed to mean? Do you speak of recurring decimals? Anyway, we all agree that 0.9 is less than 1. We also all agree that 0.999...<1.111... But why does that say anything about the relation between 0.999... and 1?

Even worse, consider the "proof" just removed from the article. There you said:

A = 1

B = 0.9

C = A / B = 1.1111111111......

B x C = 0.9999999999 ......

Now let's do a little algebra:

0.999... = B x C = B x (A/B) = (A x B)/B = A x (B/B) = A = 1.

So your proof does indeed show 0.999...=1, if anything. Yours, --Huon 13:22, 24 January 2007 (UTC)

Quote : "Now let's do a little algebra:

0.999... = B x C = B x (A/B) = (A x B)/B = A x (B/B) = A = 1. "

ARB

The problem with the Above Example algebra! you have put forward is,you have Made all the Variables into One single Calculation!!

The last Value for A (which is its start Value A x C) is A's Final Value! and not connected!

The last Value for B (which is its start Value B x C) is B's Final Value! and not connected!

A x C is One Calculation!

B x C is One Calculation!

If B was the same as A then it would = 1.111.. and Not 0.999.... —The preceding unsigned comment was added by 212.159.95.19 (talk • contribs) 10:51, January 25, 2007 (UTC)

I don't claim B=A. I also don't use A x C at all. Are you claiming that the values of A and/or B mysteriously change in mid-calculation? If so, when and why? If not, why shouldn't it be possible to use them all in the same calculation? Actually, didn't you use them all in the same calculation as well by introducing C = A / B?

I have some difficulties to understand what you mean. Please be more precise in your remarks. For example, I couldn't give any meaning whatsoever to the line "The last Value for A (which is its start Value A x C) is A's Final Value! and not connected!" --Huon 13:11, 25 January 2007 (UTC)

I----

It's Quite Simple!

A x C is One Calculation!(= 1.111...)

B x C is One Calculation!(= 0.999...)

If B is Infinitely Multiplied! it will still not = 1 (as in Infinite 0.9 <> 1)

If B started as = 1 Then it would have the Same Final Value as A

There is Only One way!! B Can = A and that is by more than One Calculation!! —The preceding unsigned comment was added by 212.159.95.19 (talk • contribs) 14:05, January 26, 2007 (UTC)

I can't follow your logic. To me, you defined B as 0.9, with just one nine after the decimal separator. A was defined to be 1. There is no way whatsoever how B can equal A, unless at some point at least one of these definitions is changed. Since C ≠ 0, A x C and B x C are unequal, too. We all agree on that, and the article doesn't claim otherwise, either.

Now to return to my algebra: You claimed that B x C = 0.999..., that is, that 0.9 x 1.111... = 0.999... I happen to agree. You also claimed that C = B / A, that is, that 1.111... = 1 / 0.9. Once again, I happen to agree. Combining this, we obtain

0.999... = 0.9 x 1.111... = 0.9 x (1 / 0.9) = 1 x (0.9 / 0.9) = 1.

Now tell me, what step don't you believe? I removed all As, Bs and Cs, so there is no disambiguity. --Huon 15:02, 26 January 2007 (UTC)

Umm, honestly, why are you bothering? This guy is either an idiot or a troll. Either way, there's no point in explaining anything to him. Dlong 16:25, 26 January 2007 (UTC)

WP:BITE. x42bn6 Talk 20:25, 26 January

2007 (UTC)

This Might help! try again!

The main problems are not saying at what stage A & B are at! i.e Single Start Values! or Calculated x C. I am going to Add some new Variables so everyone knows what is Happening!

...............................................................................................

INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,29/01/07.

The Variables below A1 & B1 are Both Single Start Values.

A1 = 1 "This has to be the Start Value for 1"

B1 = 0.9 "This has to be the Start Value for Infinite 0.9"

C = A1 / B1 = 1.1111111111...."There is only one Value & Variable for C"

A2 = A1 x C = 1.1111111111 ......"This is the same single Calculation as B2"

B2 = B1 x C = 0.9999999999 ......"This is now Infinite 0.9 and a single Calculation"

If B1 or B2 was at anytime to = A1 Then either would end up with the same Result as A2,but because it will take more than one single Calculation! for B1 or B2 to = A1 Then its Impossible from the Start for Infinite 0.9 to ever = 1 in a single Calculation. —The preceding unsigned comment was added by 212.159.95.19 (talk • contribs) 13:38, January 29, 2007 (UTC)

It's getting clearer, but the main problem remains. Indeed A1 is not equal to B1 (by definition). And indeed A2 is not equal to B2. But still we have:

0.999... = B2 
         = B1 x C 
         = B1 x (A1 / B1)
         = A1
         = 1

As I did before, I can get rid of variables and do it with numbers only. Where do you see any problems in this calculation? Which of the five lines do you doubt? --Huon 14:59, 29 January 2007 (UTC)

What you are doing above is the same trick everyone tries! you can make any Number = 1 e.g N1 = 0.12345612312 N2 = N1/N1 "N2 Now equals 1"

What you are not doing is making (B1 x C) = (A1 x C) in a Single Calculation! Because A1 is! and equals 1 and B1 is! and equals 0.9

A.R.B

I agree (as I said before) that 0.9 is not equal to 1. Neither is (B1 x C) equal to (A1 x C). But that's completely irrelevant to the article. The article doesn't claim 0.9 = 1, and it doesn't claim 0.999... = 1.111... either. What it does claim is 0.999... = 1, which, in your notation, can be written as (B1 x C) = A1. That happens to be a true statement, as shown above. Every single line in that proof is directly taken from your January 29 post. You still didn't tell me which of those lines you doubt. If you think they're all true, then obviously 0.999... = 1. --Huon 13:24, 1 February 2007 (UTC)

Below is as Clear as I can Make it!! A.R.B

INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,02/02/07.

( 1.1 ) x 0.9 = 0.99 " One Decimal Place = " 0.1 < 1

( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.1 < 1

( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.1 < 1

( 1 / 0.9 ) x 0.9 = 0.9..... " Infinite Decimal Place's = " 0.1 < 1 "

The 0.1 Difference Above is Permanent! Because it is an Infinite Difference!!

To Back up my Proof above! I would like to Make an Immortal Quote:

" IF TWO NUMBERS START WITH A DIFFERENCE IN THEIR VALUES! AND BOTH ARE MULTIPLIED BY THE SAME AMOUNT! ONE VALUE WILL ALWAYS BE GREATER THAN THE OTHER!! "

Hmm... not quite.

( 1.1 ) x 0.9 = 0.99 " One Decimal Place = " 0.1 < 1

0.99 is only 0.01 less than 1, not 0.1. Notice that you mulitply by 1.1 (1 1 after the decimal place) and the difference is off by 0.01 (1 zero before the 1)

( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.1 < 1

0.999 is only 0.001 less than 1. Notice that you mulitply by 1.11 (2 1's after the decimal place) and the difference is off by 0.001 (2 zeros before the 1)

( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.1 < 1

0.9999 is only 0.0001 less than 1. Notice that you mulitply by 1.111 (3 1's after the decimal place) and the difference is off by 0.0001 (3 zeros before the 1)

( 1 / 0.9 ) x 0.9 = 0.9..... " Infinite Decimal Place's = " 0.1 < 1 "

Following the pattern above, the number of 0's before the 1 (in the difference between your result and 1) should be equal to the number of 1's in the multiplier. Since 1/0.9 is equal to 1.1111... (with an infinite number of 1s after the decimal place) that means that our 0.000...1 should have an infinite number of 0s before the 1. And since it's impossible to have a number at the "end" of an infinitely long string of numbers (since by definition there is no end) it means that the difference between your result (0.999...) and 1 is 0.000... (with an infinitely long string of 0s, leaving no room to put a 1 on the "end" since there is no end). In other words 0.999... has no difference from 1, which must mean that it is 1.

Oh, and I'm not sure what to make of the quote, because it doesn't seem to apply to anything in this proof. You're not trying to multiple two different numbers by the same number at any time here... sorry. --Maelwys 13:41, 2 February 2007 (UTC)

And I'm still curious: Do you really claim that ( 1 / 0.9 ) x 0.9 is not equal to 1? You divide 1 by 0.9, then multiply by 0.9, and a little something gets lost in the process? --Huon 14:44, 2 February 2007 (UTC)

INFINITE 0.9 <> 1 PROOF : CONCLUSION By,Anthony.R.Brown,06/02/07. ................................................................................................

After all the Thoughts! and Arguments! the Solution is really Quite Simple!

V1 = 0.9999999999...." = The 0.9 Value from the Start Onwards! "

V2 = 0.0000000001...." = The < 1 Value from the Start Onwards! "

(a) The Above will always be True! as Calculated in the Equation Below

     ( V1 + V2 ) = 1

(b) There will always be Two Values from the Start Onwards

(d) There will always be an Infinite Difference ( V1 <> V2 )

More!!

V1 = 0.999999999999999999999999999..Infinite.0.9...." = The 0.9 Value from the Start Onwards! "

V2 = 0.000000000000000000000000000..Infinite.0.1...." = The < 1 Value from the Start Onwards! "

V1 = Infinite 0.9

V2 = The Infinite Difference

No Matter how Long V1 is! V2 will always be the same Length!

Because V1 starts as 0.9 and Because V2 starts as 0.1 The Difference will always be the same!

Sorry, there's still a couple problems with this. The main one is your definition of V2. You can't have a number that contains and infinite number of 0s, followed by a 1. Infinite means "never ending". That means that as long as you can put another number at the "end" of the string of numbers, it MUST be another zero (to fulfill the "never ending string" definition. So you can't possibly put a one at the "end" of the string, since there is no end. It seems that basically, your entire problem with accepting the proofs presented here is with understanding the definition of infinite. Infinite isn't a "really big" number. It's a limitlessly big number. So an infinite number of zeros doesn't mean that you just keep writing zeros for a long time and then eventually stop and put a 1 on the end. It means that you keep writing zeros forever. Forever and ever. You'll never be able to put a one anywhere. Therefor, there is no possible number that could be added to 0.999... to give 1. And THAT, is why 0.999... equals 1, because if there's no possible number that you could add to it to achieve 1, then there's no difference between the two, which means that they're the same number. --Maelwys 13:36, 6 February 2007 (UTC)

That's a completely different argument than what you said before, and at a first glance it sounds better. At least it can't immediately be turned into a proof of the equality. The problem here is that the "Infinite Difference" (shouldn't it rather be an infinitesimal difference?) V2 doesn't exist. There are no infinitesimals in the real numbers (unless you count zero). Worse, it does not have a decimal representation, since all digits would have to be zero. You seem to subscribe to the idea that there can be a 1 after infinitely many zeroes, but that's not true. Please read about decimal representations.

Finally, I find your utter lack of answers to my previous questions a little disappointing. I repeatedly turned your very own comments into a proof of the equality. You didn't point out where you believe me to be wrong. Couldn't you? --Huon 13:39, 6 February 2007 (UTC)

The reason he doesn't directly address any specific arguments is because he's trying to simultaneously argue this point in two different places, by posting the exact same posts in each place. So he's not responding personally to any counterarguments at all, just reformulating his theories and posting them on each board. Oh, and incidentally this present argument can indeed be turned to a proof of equality... he's correct that V2 is the only number that could possibly be added to V1 to equal 1. But since V2 is an impossible number (a unique digit at the end of an infinite string of digits isn't possible), that means that there is no number that could be added to V1 to equal 1, which must mean that V1 already equals 1. --Maelwys 13:47, 6 February 2007 (UTC)

A.R.B

To Maelwys

Quote " The main one is your definition of V2. You can't have a number that contains and infinite number of 0s, followed by a 1. Infinite means "never ending". That means that as long as you can put another number at the "end" of the string of numbers, it MUST be another zero "

A.R.B "There is No zero at the end of 0.999999......"

You are Missing the point of what I am saying!

We Cannot actually write down all the 0.9999's

All we know is that it starts 0.1 < 1 There is no reason possible! why at some stage Later! The missing 0.1 is Add to 0.9

Maelwys Quote

" Therefor, there is no possible number that could be added to 0.999... to give 1. And THAT, is why 0.999... equals 1, because if there's no possible number that you could add to it to achieve 1, then there's no difference between the two, which means that they're the same number."

A.R.B

We are not trying to find a Number that can be Add to 0.999... to give 1. we already know 0.999.. will always have it missing! and that's fine!

As I have already said " I am Quite Happy Knowing there is a Number Infinitely Smaller than 1

Anthony.R.Brown wrote:

To Maelwys

Quote 00.43.22

"The main one is your definition of V2. You can't have a number that contains and infinite number of 0s, followed by a 1. Infinite means "never ending". That means that as long as you can put another number at the "end" of the string of numbers, it MUST be another zero "

A.R.B "There is No zero at the end of 0.999999......"

I wasn't referring to 0.999... I was referring to your definition of V2 which you said was an infinite string of 0's, with a 1 at the end, which as I said is impossible.

Anthony.R.Brown wrote:

You are Missing the point of what I am saying!

We Cannot actually write down all the 0.9999's

We don't have to actually write them all down, we just have to accept that the 9's will continue forever.

Anthony.R.Brown wrote:

All we know is that it starts 0.1 < 1 There is no reason possible! why at some stage Later!

The missing 0.1 is Add to 0.9

Okay then, I agree, you're completely right, 0.9 is definitely 0.1 less than 1 and not equal to 1. Unfortunately you can't solve a number equality just by looking at the first couple digits of the number, especially not when it's infinitely long. You have to consider the entire number.

Anthony.R.Brown wrote:

A.R.B

We are not trying to find a Number that can be Add to 0.999... to give 1. we already know 0.999.. will always have it missing! and that's fine!

As I have already said " I am Quite Happy Knowing there is a Number Infinitely Smaller than 1

But you are trying to find a number that can be added to V1 to give 1, that's exactly what your argument was... that since V2 could be added to V1 to give 1, then V1 wasn't equal to 1. So my counter argument was that since V2 couldn't possibly exist as a number (or if it does, it's equal to 0), then that means V1 must be equal to 1. I know it's completely counterintuitive to say that a number starting with 0. is equal to a number starting with 1., but in this rare case, it's actually true. You just have to stop focusing on the first couple digits of the number and consider the entire thing (which involves wrapping your head around infinite, which is obviously also hard to do). --Maelwys 14:25, 6 February 2007 (UTC)

INFINITE 0.9 <> 1 PROOF : TRUTH TABLE : By,Anthony.R.Brown,07/02/07.

Below V1 = The Infinite 0.9 Sequence. : Below V2 = The < 1 Infinite Difference Sequence.

V1 = ( 0.9 ) + V2 = ( 0.1 ) = 1

V1 = ( 0.99 ) + V2 = ( 0.01 ) = 1

V1 = ( 0.999 ) + V2 = ( 0.001 ) = 1

V1 = ( 0.9999 ) + V2 = ( 0.0001 ) = 1

V1 = ( 0.99999 ) + V2 = ( 0.00001 ) = 1

V1 = ( 0.999999 ) + V2 = ( 0.000001 ) = 1

V1 = ( 0.9999999 ) + V2 = ( 0.0000001 ) = 1

V1 = ( 0.99999999 ) + V2 = ( 0.00000001 ) = 1

V1 = ( 0.999999999 ) + V2 = ( 0.000000001 ) = 1

V1 = ( 0.9999999999 ) + V2 = ( 0.0000000001 ) = 1

The Table above clearly shows no matter how long V1 is! it will always need the algebra + symbol for V1 to equal 1 ( V1 + V2 ) = 1

This proves Infinite 0.9 <> 1 because if ( V1 + V2 ) = 1 is allowed! it will be a contradiction to the term Infinite 0.9.

—The preceding unsigned comment was added by 212.159.95.19 (talk • contribs).

Regarding the 0.0...1 number, there is no last "1" in essence. There are an infinite number of zeroes - there is no, if you like, room, for the 1. x42bn6 Talk 15:05, 7 February 2007 (UTC)

If you assume the difference 1 - 0.999... to be non-zero (and I believe that's what you do), is there a natural number n such that 1/n is less than 1 - 0.999...? If not, have a look at the Archimedean property, which would be violated. If so, I'll be able to easily construct a contradiction, such as 0.999...9 with some finite number of nines being greater than 0.999... with infinitely many nines. Thus, the difference 1 - 0.999... has to be zero. --Huon 15:46, 7 February 2007 (UTC)

Quote: from above!

" :Regarding the 0.0...1 number, there is no last "1" in essence. "

Sorry to Disappoint you! there will always be a last "1"

1 - 0.9 = 0.1 1 - 0.99 = 0.01 1 - 0.999 = 0.001

We don't need Anymore examples! it is an Infinite Calculation!!

Anthony.R.Brown 08/02/07

Please read the definition of Infinite, since you seem to be misunderstanding that concept. If you have an infinitely long string of 9s, then you also have an infinitely long string of 0s, and by definition you cannot define the "end" of an infinitely long string, nor can there be a "last 1". That means that your 1-0.999... = 0.000... (which is simply 0, meaning there is no difference between 1 and 0.999..., meaning that they're the same number). --Maelwys 13:51, 8 February 2007 (UTC)

1 - 0.9 = 0.1 " Notice this is a number! with a 1 on the end " " If it's not? then funny if you add " 0.1 to 0.9 you get 1

1 - 0.99 = 0.01 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.01 to 0.99 you get 1

1 - 0.999 = 0.001 " Notice this is a number! with a 1 on the end " " If it's not? then funny! if you add " 0.001 to 0.999 you get 1

The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!

Because 1 - ( Another Number < 1 ) Will always have an Infinite Difference!!!

P.s the 1 "at the End!" and the 9 "at the End!" is a term to decribe! at the End of an Infinite Sequence! for as far as we have Calculated! and can see on paper! We all know it's not an actual End!! but in the Above three examples,we can see them at the End Three Times!!

Anthony.R.Brown 09/02/07

I agree with your first three statements. All of those numbers do add together to get 1. Your fourth statement however ("The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!") is a fallacy. Your pattern holds for every number of a finite series only. Any defined finite number approaching infinite, that is <1 will have another number defining the difference between the first number and 1. However, infinite numbers are a special case where this pattern doesn't apply. I also have a problem with your first statement ("Because 1 - ( Another Number < 1 ) Will always have an Infinite Difference!!!"). Any 1 - (another number < 1) will always have a finite difference, that much is true. However again a/Infinitely repeating decimals are special, and so must be defined specially. And b/You're "proving" that 0.999... is less than 1 by saying "1 - another number < 1 has a difference, so since 0.999... is < 1, there must be a difference, so since there's a difference, 0.999... must be < 1". This is circular logic, because you're starting out by assuming your proof is correct, and then using that assumption to prove that it is correct. Finally, in response to your PS, again you can't "describe the end of an infinite sequence" because by the very definition of "infinite" ("Boundless, endless, without end or limits, uncountable, innumerable"; from wiktionary:infinite) there IS no end. Again, I strongly suggest you read the article at infinite because the main problem here seems to be that you don't properly understand that the concept of infinite isn't "a really big number" but it is "an endlessly big number". Hopefully reading the article will help you understand that infinitely long decimal numbers must be handled differently than finitely long decimal numbers, and not all the same rules apply when you're adding or subtracting them. --Maelwys 13:54, 9 February 2007 (UTC)

To Maelwys

Quote: "I agree with your first three statements. All of those numbers do add together to get 1. Your fourth statement however ("The Above is Only a Small sample from the Start Onwards! But I can Assure you! is Infinitely possible!") is a fallacy. Your pattern holds for every number of a finite series only."

If my pattern only holds for every number of a finite series only ?

Can you Give me the ( Maximum Possible finite series for 0.9 ) onwards!

and Can you Give me the ( Maximum Possible finite series for 0.1 ) onwards!

Anthony.R.Brown 12/02/07

No, I can't give the maximum possible number, because it's a finite series that approaches infinite and so it's not possible to write out the number. However, there is a difference between a number approaching infinite in it's number of digits (extremely large, but still definable) and a number that has an infinite number of digits (indefinably large, with no possible end). Your solution above works for any number with a finite number of digits, no matter how large a number of digits it is (you could have 5 trillion and one 9s, and then add to it a number with 5 trillion 0s and a 1 in the 5 trillionth+1 space). But it doesn't work for the infinite case, where there is no "+1 space" available to put that 1. --Maelwys 14:55, 12 February 2007 (UTC)

INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07

A = 1 " Single Start Value For 1 "

B = 0.9 " Single Start Value For 0.9 "

C ( A/B ) x B " Infinite 0.9 Value "

D ( A - C ) " Infinite < 1 Value "

C <> ( C + D ) " INFINITE 0.9 <> 1 "

If: D = A - C; Then: C + D = C + A - C; C + D = A
If: C = A/B * B; Then: C = A
If: C + D = A and C = A; Then: C + D = C
Therefor: Since A = 1; Then: C = 1, D = 0
--Maelwys 14:43, 12 February 2007 (UTC)

To Maelwys

Quote:

(1) If: D = A - C (2) Then: C + D = C + A - C (3) C + D = A (4) If: C = A/B * B (5) Then:C = A (6) If: C + D = A and C = A; (7) Then: C + D = C (8) Therefor: Since A = 1; (9) Then: C = 1 and D = 0

A.R.B

(1) " Yes D = 0.001...etc (2) " Yes C + D = 1 " = " Yes C + A - C = 1 " (3) " Yes C + D = A " equals 1 (4) " Yes C = A/B x B " equals the same as 0.999...etc (5) " Wrong! Then:C = A " C = 0.999...etc A = 1 (6) " Yes C + D = A " equals 1

   " Wrong! and C = A " C = 0.999...etc  A = 1

(7) " Wrong! Then: C + D = C " C + D = 1 and C = 0.999...etc (8) " Yes Therefor: Since A = 1 " (9) " Wrong! Then: C = 1 and D = 0 " C = 0.999...etc and D will always = ( A - C )

      0.001...etc < 1

Anthony.R.Brown 15/02/07

INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07

A = 1 " Single Start Value For 1 "

B = 0.9 " Single Start Value For 0.9 "

C ( A/B ) x B " Infinite 0.9 Value "

D ( A - C ) " Infinite < 1 Value "

C <> ( C + D ) " INFINITE 0.9 <> 1 "

Can you explain why 5, 6, 7, and 9 are wrong? Without using numbers, I mean. I'm just looking at the letter equations, and solving them without substituting in the values yet. For example, line 5 is simply solving for line 4 (C = A/B x B), which you agree with. Using basic rules of multiplication, A/B x B = A, there's no other thing that it COULD equal. If I said A = 4 and B = 2, would you argue that 4/2 x 2 DOESN'T equal 4? Or if A = 5 and B = 7, would you argue that 5/7 * 7 doesn't equal 5? A/B x B will always equal A (anything divided by a number and then multiplied by the same number equals the original number, it's basic logic). So if A/B x B = A, then C = A, hence line 5. The rest is just subsitution of C = A into the other formulas, that follows out to show A = C = 1, and D = 0. --Maelwys 14:28, 15 February 2007 (UTC)

You seem to get your symbolic calculations wrong when you claim that (A/B) x B is not equal to A. By definition, A/B is the number that, when multiplied by B, gives A. Then (A/B) x B equals A as a direct consequence.

That alone is enough to show that C=A and thus D=0. All your formulas don't show otherwise, since you seem to assume (probably "by inspection") that there is a non-zero difference between 0.999... and 1 and then use that non-zero difference to "show" they're unequal - circular reasoning. --Huon 14:36, 15 February 2007 (UTC)

To Maelwys

First it's important to look at the Calculated Facts in my,

INFINITE 0.9 <> 1 PROOF : FORMULA : 12/02/07

( f1 ) A " Will always be > than B " " Because A - B = 0.1... "

( f2 ) A " Will always be > than C " " Because A - C = 0.001...etc "

( f3 ) A " Will always be > than D " " Because A - D = 0.999...etc "

( f4 ) A " Will always only be = to A and ( C + D ) " " Because A = A and A = ( C + D ) "

I only answered the way you wrote it down! thinking you understood what was happening in the Calculations! that's why it's important to stick with the original Formula including the brackets ()

Quote: " 4 Using basic rules of multiplication, A/B x B = A, "

A.R.B

C ( A/B ) x B " Infinite 0.9 Value "

C = ( 1 / 0.9 ) = ( 1.111...etc ) x ( 0 .9 ) = 0.999...etc

Quote : ( 5 ) " Wrong Because of ( f4 ) above "

Quote : ( 6 ) " Wrong Because of ( f4 ) above "

Quote : ( 7 ) " Wrong Because of ( f4 ) above "

Quote : ( 9 ) " Wrong Because of ( f4 ) and ( f3 ) above "

Anthony.R.Brown 19/02/07

Maelwys

I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A. Ignoring the values that you've assigned to A and B, even with the brackets there, it's basic math. If I ask you what (A/F) x F is, without knowing the value of F, the answer is still A. By definition, division and multiplication are the inverses of each other, and order of operations is irrelevant (so (A/B) x B = (AxB)/B = A either way).

The other problem with the above is that you're starting with an assumption, and then trying to prove it using your initial assumption. You're starting with the assumption that C < A, and then using that proof (f2, f3) to prove (f4) that A cannot equal C. But A can equal C for cases where D = 0, and since you're starting with A and B and solving for C and D, you can't automatically assume that C < A and D > 0, you have to assume nothing about them, and THEN prove that it's true. So ignoring your initial assumptions, we can see that A/B x B = A, therefor C = A, therefor since A = C + D, D = 0, and all the rest of the formulae fit into place.

To Maelwys

Quote:

"I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A "

A.R.B

It is impossible for (A/B) x B = A because there is an Infinite Difference!

from my Formula C = (A/B) x B = 0.999....etc

A - C will always have an Infinite Difference! 0.001...etc

The rest of what you have put forward is Answered in my Formula!

The Infinite Difference's can never be made up without the use of the Algebra + sign.

Anthony.R.Brown 19/02/07

You're still suffering the same problem. You're starting with the assumption that C < A, and then using that assumption to justify why C < A. If you want to make a proof, you have to start without any assumptions, and then get to the answer that you need. When the algebraic proof shows that A = C, you can't discount that because you want C to be less than A, and then say that it's proof that C is less than A. (A/B) x B = A, every time. I don't care if A is 43 and B is 762, (A/B) x B still equals A. Your argument is basically the same as if I said "X=2, Y=1, X-1=Y, but because I want to prove that 1=2, X=Y, so -1=0 must also be true, and since -1=0, if I add 2 to both sides I get 1=2, which proves that it!". The logic doesn't work if you start with your assumption, and then try to prove it using your conclusion as proof of itself. --Maelwys 15:37, 19 February 2007 (UTC)

May I suggest that your friend read and study Logical implication before attempting further proofs? Tparameter 16:28, 19 February 2007 (UTC)

Let's have a look at why there will always be an Infinite Difference between A and C and why the Difference Remains the same from the start Onwards in my Formula! And why the Algebra Sign + (Difference Value) is the Only way to make up the Difference's!

A = 1 " Single Start Value For 1 " B = 0.9 " Single Start Value For 0.9 " C ( A/B ) x B " Infinite 0.9 Value " D ( A - C ) " Infinite < 1 Value " C <> ( C + D ) " INFINITE 0.9 <> 1 "

I am going to give three examples below from the start onwards! and then the Infinite example!

Example (1) ex1A ( A - C) " Stage one/Decimal place " = ( 1 ) -( 0.9 ) = ex1D ( 0.1 )

Example (2) ex2A ( A - C) " Stage Two/Decimal place's " = ( 1 ) -( 0.99 ) = ex2D ( 0.01 )

Example (3) ex3A ( A - C) " Stage Three/Decimal place's " = ( 1 ) -( 0.999 ) = ex3D (0.001)

In the above Examples! for A to be able to Equal 1 again using the Algebra Sign + (Difference Value) the Anwers are shown Below!

Example (1) ex1A + ex1D = 1

Example (2) ex2A + ex2D = 1

Example (3) ex3A + ex3D = 1

Now below I will give the Infinite Example!

Example (I) exIA ( A - C) " Infinite Decimal place's " = ( 1 ) -( Infinite 0.9 ) = exID ( Infinite 0.1 )

Example (I) exIA + exID = 1

In the above Infinite Example! the amount of .9's and .1 (Difference) will always be the Same Length as each other from the Start onwards! And more Important the Acutual Value Difference will always be the same!! What I mean by this is shown Below!

0.999999999999999999999999999999999999999999999999999999999999999.......... 0.000000000000000000000000000000000000000000000000000000000000001..........

For the above Example we can remove the .9's and the .0's After Stage one/Decimal place!

Because they are the same lenghth! and show the Difference is Permanent!

What we end up with is the Same Two Values! as from the Start Onwards!!

0.9 0.1

The two Final Calculations Below are the Same!!

Example (1) ex1A + ex1D = 1 Example (I) exIA + exID = 1

Anthony.R.Brown 21/02/07

INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07

A = 1 " Single Start Value For 1 "

B = 0.9 " Single Start Value For 0.9 "

C ( A/B ) x B " Infinite 0.9 Value "

D ( A - C ) " Infinite < 1 Value "

C <> ( C + D ) " INFINITE 0.9 <> 1 "

Since this conversation is just between you an I anyway, I'm limiting it to just one place, so I don't have to keep making the same arguments (that you're ignoring) in two places at once. So now I'm only posting these arguments at the Mathisfun board. --Maelwys 14:37, 21 February 2007 (UTC)

                       RECURRING DEFINITION PROOF 03/03/07  by Anthony.R.Brown.

The Definition for Recurring is given below..............................................................................

“ A Number that repeats itself Endlessly! Continuously repeating the same Number! “

Below are examples of Recurring Numbers! They are of course Endless! Length is example only!

Recurring ( 0.1 ) = .111111111111111111111111111111111111111111111111111111111111111

Recurring ( 0.2 ) = .222222222222222222222222222222222222222222222222222222222222222

Recurring ( 0.3 ) = .333333333333333333333333333333333333333333333333333333333333333

Recurring ( 0.4 ) = .444444444444444444444444444444444444444444444444444444444444444

Recurring ( 0.5 ) = .555555555555555555555555555555555555555555555555555555555555555

Recurring ( 0.6 ) = .666666666666666666666666666666666666666666666666666666666666666

Recurring ( 0.7 ) = .777777777777777777777777777777777777777777777777777777777777777

Recurring ( 0.8 ) = .888888888888888888888888888888888888888888888888888888888888888

Recurring ( 0.9 ) = .999999999999999999999999999999999999999999999999999999999999999

The above examples are shown to be True! in the Calculations below! According to the Definition for Recurring!

( 0.1 ) Recurring < 1 and <> 1

( 0.2 ) Recurring < 1 and <> 1

( 0.3 ) Recurring < 1 and <> 1

( 0.4 ) Recurring < 1 and <> 1

( 0.5 ) Recurring < 1 and <> 1

( 0.6 ) Recurring < 1 and <> 1

( 0.7 ) Recurring < 1 and <> 1

( 0.8 ) Recurring < 1 and <> 1

( 0.9 ) Recurring < 1 and <> 1

INFINITE 0.9 <> 1 PROOF : THE INFINITE 10 % PERCENT DIFFERENCE 03/03/07 by Anthony.R.Brown.

One of the most Accurate and sound ways to prove Infinite 0.9 <> 1 is to give the Infinite Percent Difference Values from the Start Onwards!

The example below is for the first Ten Decimal Place's,which clearly shows the count increasing as more and more Infinite .9's are looked at! But remain a constant Difference of 10% in relation to how far the example has traveled.

Count = ....... ( 1 )..( 2 )..( 3 )..( 4 )..( 5 )..( 6 )..( 7 )..( 8 )..( 9 )..( 10 )

Infinite 10 % Difference =....(10%)..(10%)..(10%)..(10%)..(10%)..(10%)..(10%)..(10%)..(10%)..(10%)

Infinite 0.9 = ( 0.9 )( 1.8 )( 2.7 )( 3.6 )( 4.5 )( 5.4 )( 6.3 )( 7.2 )( 8.1 )( 9 )

.............. ( x 1 )( x 2 )( x 3 )( x 4 )( x 5 )( x 6 )( x 7 )( x 8 )( x 9 )( x 10 )

Infinite 0.9 Onwards! there will always be a 10 % Difference making it Impossible for Infinite 0.9 to ever = 1

INFINITE 0.9 <> 1 PROOF : FORMULA : By,Anthony.R.Brown,12/02/07

A = 1 " Single Start Value For 1 "

B = 0.9 " Single Start Value For 0.9 "

C ( A/B ) x B " Infinite 0.9 Value "

D ( A - C ) " Infinite < 1 Value "

C <> ( C + D ) " INFINITE 0.9 <> 1 "

As stated above, I've already responded to these at the Mathisfun board and will continue this conversation there only. --Maelwys 13:51, 6 March 2007 (UTC)

To Maelwys

First of all Congratulations! With your persistence's in trying to understand the problems! Associated with Infinite/Recurring 0.9 Everyone will all Benefit from it! I have not been able to explain fully! What I'm trying to say, because we have only been concentrating on Infinite/Recurring 0.9 as if it is unique in the world of Infinite/Recurring Number/Values. I have always known this not to be true! All Infinite/Recurring Number/Values behave in the same way according to the type!! They belong to,there are Normal Number/Values and Infinite/Recurring Number/Values and Infinite/Recurring Difference Number/Values.

Example “ Normal 0.1 “ = 0.1

Example “ Infinite/Recurring 0.1 “ = 0.111...

Example “ Infinite/Recurring Difference 0.1 “ = 0.001...

With the three examples above! They can all equal the Value 1

“ Normal 0.1 “ + “ Normal 0.9 “ = ( 0.1 + 0.9 ) = 1

“ Infinite/Recurring 0.1” + “ Infinite/Recurring 0.9” = ( 0.111... + 0.999... ) = 1

With the second example above! Because the 0.1 and 0.9 are Infinite/Recurring the same Start Values are being repeated over and over again!and actually equal ( 0.1 + 0.9 ) and not ( 0.111... + 0.999... ) as if the Values are getting larger which of course would be 1.111...etc.

“ Infinite/Recurring Difference 0.1 “ + “ Infinite/Recurring 0.9 “ = ( 0.001... + 0.999... ) = 1

With the third example above! Because the 0.1 is the Infinite/Recurring Difference we Know that the Value has come from ( 1 - Infinite/Recurring 0.9 ) there will always be a .1 Difference the .00 is the Decimal Point Shift!

The above Shows how there can be Three Different Types of 0.1 and why its Important to know which one is being used in any Calculations!

[edit] Section Break

To Maelwys Examples for 0.8

Again you are Comparing Normal Number Calculations with Infinite/Recurring Number Values!

8 / 10 = 0.8

8 / 9 = 0.888...

For 0.9 The Normal Calculations above give Different Results! And also for some other 0.1 to 0.9 Values!

9 / 12 = 0.75

9 / 11 = 0.8181818...

9 / 10 = 0.9

9 / 9 = 1

9 / 8 = 1.125

So the above shows using the Same Normal Calculations! There is no Infinite/Recurring 0.9 What we need is a more Precise way for Calculating Any Infinite/Recurring Number Values!

The best way I know is Multiplying the Original Start Value! By Infinite 1.111... This will then give a True Infinite/Recurring Number Value! For Any 0.1 to 0.9 Value.

Now going back to your original question how can 8 / 10 be equal to 8 / 9 the Answer is they are not! Concerning Normal Number Calculations! Even though they both have a 0.8 Firstly the 0.8 for 8 / 10 is 0.8 and not the Start! For 8 / 9 it does have a 0.8 at the Start! But as I have pointed out,it is not possible with some other 0.1 to 0.9 Infinite/Recurring Number Values! Calculated in the way you have put forward!

THE DEFINITIVE REASON INFINITE/RECURRING 0.9 IS < 1 AND <> 1 10/03/07 by Anthony.R.Brown

( 1 ) If any Number Starts or has a Zero followed by a Decimal point at the beginning! it will always be < 1 and <> 1 if no additional Math is applied to the Number!

( 2 ) Numbers that conform to ( 1 ) as examples are 0.1 to 0.9

( 3 ) Infinite/Recurring Numbers have no end! the Numbers repeat themselves endlessly!

( 4 ) Because Infinite/Recurring Numbers are always the same Number being repeated,we don't need to try and write down many as examples,because the Number is always the same.

( 5 ) An example for ( 4 ) is Infinite/Recurring ( 0.9 ) = 0.9 the reason this is True is because the Number never changes!

( 6 ) The same can be applied to any Infinite/Recurring ( N ) = N

( 7 ) From ( 5 ) and ( 6 ) we can prove Infinite/Recurring ( 0. 1 ) to ( 0.9 ) < 1 and <> 1

1 - ( 0.1 ) = 0.9 and <> 1 1 - ( 0.2 ) = 0.8 and <> 1 1 - ( 0.3 ) = 0.7 and <> 1 1 - ( 0.4 ) = 0.6 and <> 1 1 - ( 0.5 ) = 0.5 and <> 1 1 - ( 0.6 ) = 0.4 and <> 1 1 - ( 0.7 ) = 0.3 and <> 1 1 - ( 0.8 ) = 0.2 and <> 1 1 - ( 0.9 ) = 0.1 and <> 1

THE INFINITE/RECURRING ( N ) = N PROOF 11/03/07 by Anthony.R.Brown

Infinite Recurring ( n ) = n because! ( n ) will always = n

( 1 ) is wrong. If it were true, you could argue that 0.999... is a number beginning by a 0 followed by a decimal separator and thus has to be less than 1. Even if it were correct, you would have to prove it.

If I understand ( 4 ) and ( 5 ) correctly, you claim that 0.9999...=0.9? That, especially, 0.9999...< 0.91? Then what about 0.9999...-0.9? I would claim that the difference amounts to 0.0999..., but according to you, it's zero? I don't believe you can give any sort of reference for this rather unconventional definition recurring decimals, can you? Besides, the equality 0.9999...=0.9 would also contradict most of your previous "proofs".

Oh, another problem I came up with: I assume that, say, 0.141414... is equal to 0.14, because the 14 recurs? Then how about ( 0.99 ), where the 99 recurs? I would claim that for an infinite string of nines, it doesn't matter whether I group them as single or double digigts, but for you, that suddenly and mysteriously makes a difference of 0.09?

In conclusion, I'm sorry to say that you have much more basic problems than whether 0.999...=1 or not. You should have a look at basic maths, logic, and arithmetics. --Huon 20:10, 12 March 2007 (UTC)

A.R.B

To Huon!

what you have to remember! and understand! is that all Numbers/Values have a Starting point!

somehow any Infinite/Recurring Number Value,has to be Calculated! it does not just appear!

Infinite/Recurring 0.9 ( one decimal place ) Starts as 0.9 and therefore 1 - 0.9 will always have an  Infinite Recurring Difference of 0.1 and so  Infinite Recurring 0.9 will always be < 1 and <> 1 (Infinite decimal places! give the same difference! )

Quote: by Huon " Then what about 0.9999...-0.9? "

A.R.B the above calculation conforms to my answer at the Start! and equals 0 the 0.999... is the same single value being Infinitely repeated!

Quote: by Huon "I assume that, say, 0.141414... is equal to 0.14, because the 14 recurs? "

A.R.B 0.141414... EQUALS 0.14 because it's where the sequence changes! 0.90909... equals 0.90 for the same reason!

Quote: by Huon " Then how about ( 0.99 ), where the 99 recurs? I would claim that for an infinite string of nines, it doesn't matter whether I group them as single or double digits, but for you, that suddenly and mysteriously makes a difference of 0.09? "

A.R.B Infinite/Recurring ( 99 ) equals 99 Infinite/Recurring ( 999 ) equals 999 and so on.... because the Number Values are always being repeated over! and over again!

I have never heard of numbers' "starting points" before. Can you give a precise definition (a general one, please, not just a few examples)? Preferably quoting a book where the concept is explained?

In contrast, to me a recurring decimal would be the limit of a corresponding sequence of finite decimals. That's (my favourite version of) the standard definition, and with slight variations (due to authors' tastes) it should be found in every book on basic analysis. As an example, I just found it in Griffiths/Hilton, A Comprehensive Textbook of Classical Mathematics, chapter 24, especially paragraph 24.5. Griffiths and Hilton don't use the limit notation but one equivalent to it, and they explicitly discuss the case of recurring nines.

Unless we can agree on some basic definitions, arguing about 0.999... and 1 is useless. And currently I not only don't understand your definitions; I'm not even sure there is something to understand. So please enlighten me, preferably with sources. Yours, Huon 17:05, 21 March 2007 (UTC)

I've got a couple problems with your whole recurring=starting number theory. First:

"0.90909... equals 0.90"

But if that's true... and we know that 0.90 = 0.9 because trailing 0s at the end of the decimal number are irrelevent, and you earlier claimed that 0.999... equals 0.9, so now 0.999... = 0.909090... ?

"Infinite/Recurring ( 99 ) equals 99 Infinite/Recurring ( 999 ) equals 999 and so on.... because the Number Values are always being repeated over! and over again!"

So 0.999... = 0.(9) = 0.9; and 0.999... = 0.(99) = 0.99; and 0.999... = 0.(999) = 0.999; That means that 0.9 = 0.99 = 0.999 and any other following 9s are irrelevent? --Maelwys 17:22, 21 March 2007 (UTC)

To Maelwys again?

a Starting point for any Number is the beginning of the Calculation that made the Number! so 3 x 3 the starting point would be 1 x 3 etc.... 0.90.... is different to 0.99.... you can always take one away from the other to show the difference! from decimal point stage one onwards! the same can be done to all your other calculations! to show the difference!

Anthony.R.Brown 02/04/07

"Starting point for a number"? "Calculation that made the number"? What do you mean here? None of this is standard terminology, so if you use your own terminology you should explain it. Thanks. --Kprateek88(Talk | Contribs) 14:16, 2 April 2007 (UTC)

Don't bother trying to understand his terminology. Or his logic for that matter. I've read all this guys posts and he's beyond being able to be reasoned with. He can be ignored unless you're looking for an easy laugh (the only reason I even read his posts). Dlong 02:32, 3 April 2007 (UTC)

The fools are back again above! they really should have a section for them only! rather than coming on this post with comments only!

Anthony.R.Brown 03/04/07

Unless you can explain your terminology or, even better, give sources, Dlong has a point. Can you? --Huon 17:11, 3 April 2007 (UTC)

I agree with Huon and Dlong. Except for it being a laugh; "from decimal point stage one onwards!" and the like are just painfull. Didn't Maelwys move the diologue to www.mathisfun.com forum or suchlike?

PS before anyone throws WP:BITE at me, Anthony's not new, and WP:FAITH no longer applies. Endomorphic 21:47, 3 April 2007 (UTC)

Indeed. I am reminded of an item on the talk archive for WP:FAITH: "Remember that at least trolls know they're trolls; the dedicated crank doesn't understand they're a crank." One of the surest sign of being a crank is to call those who disagree with you — even (and especially) if everyone disagrees with you — fools. The exclaimation marks and zealously self-attributed "proofs" are a bit of a tip-off, too. Calbaer 21:54, 3 April 2007 (UTC)

Yeah, I spent a couple months trying to discuss this issue and see Anthony's point of view. During that time he was mirroring the entire conversation both here and on a thread at the mathisfun.com forums. Since there were several participants (including he and I) at mathisfun, and only he and I here, I moved the conversation to where the action was, so I didn't have to keep responding to the same posts in two places. Unfortunately, after almost 3 months of trying to wrap my mind around his definitions of "single start value" and "infinite 0.9 equals 0.9" and other such things, I was forced to drop out of the conversation for fear my head might explode. And since I think I was the last holdout still taking part in the conversation and making an attempt to understand Anthony's point of view (long since given up simply making him understand mine, since he had no interest in even trying to), the conversation basically died when I stopped responding. Then I guess he decided to come back here and try to start it up all over again... good luck to whomever attempts to decode these proofs next, and beware exploding heads. ;-) --Maelwys 15:36, 4 April 2007 (UTC)

[edit] 0.333...=1/3?

Does 0.333...=1/3 it's my belief that 0.333... does not exactly equal 1/3 because there is no point at which 0.333... becomes 1/3. It gets closer and closer as it extends but it never exactly reaches 1/3. What do you think?

Firstly, 0.333... is a number, not a process. It doesn't "get closer and closer" to 1/3; its value doesn't change over time. Secondly, 1/3 is a real number and as such has a decimal representation. 0.333... is the only candidate. --Huon 18:36, 6 January 2007 (UTC)

0.333... is proved to exactly equal 1/3 by long division. It's a rational number because it be represented in decimal form. It's all due to base-10 (See: logarithm). And if you start to question the credibility of long division, I suggest passing 4th grade before you continue to respond. Thanks and good day! --Dabigkid 22:57, 8 January 2007 (UTC)

You say it's the only candidate but it's still only an aproximation. It's not exact, is it and if so prove it.

It's exact, not an approximation. I say two things:

1/3 has a decimal representation.
No decimal representation but 0.333... represents 1/3.

That allows me to conclude that indeed 0.333...=1/3. Admittedly, while 2. is obvious, 1. is a lot trickier (although that's one of the strong points of decimal representations: While they're not necessarily unique, every real number has one). Thus, for a full proof, I would employ other methods, probably geometric power series and their convergence, analogous to the article's section. --Huon 19:58, 6 January 2007 (UTC)

Representation I never assumed for an instant to be the same as identicality. 0.333... is as close to 1/3 as one can get with a decimal representation, but why should that automatically mean they are the same thing? ~ SotiCoto 195.33.121.133 10:55, 30 January 2007 (UTC)

Umm, there's a proof right below your post (now mine), that you apparently decided to skip. Dlong 13:41, 30 January 2007 (UTC)

I'm not a mathematician. While there once was a time that I might have been able to follow that (like 5 years ago), I'd struggle now... and I don't doubt that there is some axiom somewhere in there that I would contest. Thats usually the issue I have with such things, as opposed to the process itself (which is consistant).

{EDIT} Ah... I get it now. The capital sigma and all that is basically implying from n=1 to infinity the whole 3/10 + 3/100 + 30/1000 ... etc... right? And the issue I personally have with it is regarding the whole Infinity - 1 = Infinity thing. It would basically need the mathematical definition of "Infinity" to convince me of why this would be so, because I believe I am thinking of it as being something rather different. ~ SotiCoto 195.33.121.133 12:33, 31 January 2007 (UTC)

Again, the ∞ symbol in summations is just a shorthand notation which I have explained in one of the other threads. As for your other question, I have given the analogy somewhere - I say "4" when I actually mean "the real number represented by the numeral 4", and I also say "0.333..." when I mean "the real number represented by the decimal expansion 0.333...", which is 1/3. "representation" here is in exactly the sense you have given, that you can get arbitrarily close to 1/3 by taking some finite part of 0.333... .

To prove that 0.333... = 1/3, consider that the definition of an infinite decimal expression is the infinite sum of the corresponding finite decimal numbers. That is, $0.333\dots = \sum_{n=1}^{\infty} \frac{3}{10^n}$ . From that, it follows that:

$\begin{align} 0.333\dots &= \frac{3}{10} + \sum_{n=2}^{\infty} \frac{3}{10^n} \\ 10\times 0.333\dots &= 3 + \sum_{n=2}^{\infty} \frac{10\times 3}{10^n} \\ &= 3 + \sum_{n=2}^{\infty} \frac{3}{10^{n-1}} \\ &= 3 + \sum_{m=1}^{\infty} \frac{3}{10^m} \\ &= 3 + 0.333\dots \\ Therefore \\ 10\times 0.333\dots &= 3 + 0.333\dots \\ 9\times 0.333\dots &= 3 \\ 0.333\dots &= \frac{3}{9} = \frac{1}{3} \end{align}$

Thus 0.333... = 1/3 by definition of how to calculate an infinite decimal expression.

I was going to start typing something on why it is the only decimal expression for 1/3, but unfortuantely I'm out of time. :( If someone else would like to include a proof that 0.333... is the unique decimal expression for 1/3, please feel free. Dugwiki 00:18, 9 January 2007 (UTC)

Let $a_1,a_2,\ldots$ decimal digits be given such that $1/3 = 0.a_1a_2a_3\ldots$ . Let $\mathcal{P}(n)$ be the assertion that $a_1, a_2, \ldots, a_n$ are all 3. $\mathcal{P}(0)$ holds vacuosly. Let $n \in \mathbb{N}$ and assume $\mathcal{P}(n)$ . If

a n + 1 > 3

, then $1/3 = 0.a_1a_2\ldots a_na_{n+1}\ldots \geq 0.a_1a_2\ldots a_na_{n+1} \geq 0.333333\ldots 34 > 1/3$ . Similarly if

a n + 1 < 3

, then $1/3 = 0.a_1a_2\ldots a_na_{n+1}\ldots < 0.a_1a_2\ldots a_n(a_{n+1}+1) \leq 0.33\ldots 33 < 1/3$ . Hence, we must have

a n + 1 = 3

, so $\mathcal{P}(n+1)$ holds. By induction, every digit is 3. 24.46.154.178 09:52, 13 January 2007 (UTC)

The above proof is a step in the right direction, but it kind of glosses over the difference between 1/3 having a unique decimal representation (0.333...) and, for example, 1/5 having two representation: 0.2000... and 0.1999... . I think to tighten it up you'd want to give mathematical reasoning for why the above type of proof doesn't apply in the same way to 1/5. Dugwiki 18:49, 15 January 2007 (UTC)

Along the same lines, there appears to be a technical error in part of your proof. You say that "Similarly if

a n + 1 < 3

, then $1/3 = 0.a_1a_2\ldots a_na_{n+1}\ldots < 0.a_1a_2\ldots a_n(a_{n+1}+1)$ ." However, as in the case of 1/5, 0.2000 = 0.1999... . So it might be that, for example,

a n + 1 = 1

and

a n + k = 9

for all k>1. In that case, it would follow that $0.a_1a_2\ldots a_na_{n+1}\ldots = 0.a_1a_2\ldots a_n(a_{n+1}+1)000...$ . The strictly less than inequality does not always apply in cases where you have decimals which end in "000..." and "999...".

Dugwiki 19:02, 15 January 2007 (UTC)

The technical error is well-spotted, but at that point one could without the slightest problems substitute "<=" for "<", for the important "<" comes a little later, when 0.333...3<1/3. I consider it a good proof, showing that there is no decimal representation for 1/3 but 0.333... The difference to the 1/5 case is well-hidden indeed; it's P(0), which no longer holds vacuously for 1/5 (or, say, 1.000...) and relies on the fact that 0<1/3<1 (and thus makes sure of the one non-3 digit). Yours, Huon 20:44, 15 January 2007 (UTC)

Showing 1 / 3 = 0.3... is wrong. You have to carry the one when working it out.

It should be:

1 / 3 = 0.3... r 0.0...1

You try working it out on paper without having to carry the one forever. So there will always be a remainder.

60.226.58.94 17:06, 6 March 2007 (UTC)

No. Dlong 17:49, 6 March 2007 (UTC)

The fact that there is "always a remainder" is the whole reason why the decimal representation of 1/3 is infinite. There is no "end" to place the remainder. Leebo⁸⁶ 18:09, 6 March 2007 (UTC)

A + B + C = 1

A = (Decimal Value) B = (Decimal Value) C = (Decimal Value)

Using the Three Decimal Values A + B + C above must equal 1

The Three Decimal Values must have the Same Values!

It's impossible to give 3 single Decimal Values for A,B and C

Only fractions can give True 1/3 Values as one 3rd

by Anthony.R.Brown 21/03/07

[edit] A few final questions

Does anyone have any proof that 0.999... even exists? Sorry for being so annoying, but I'd like to understand it, and frankly the article and FAQ are not too helpful for me. Calculus just seems to reflect on my eyes without actually being detected properly, or maybe it enters my brain and gets lost and eaten by [Grue (monster)|grue]....

Either way, could someone supply proof that 0.999... is even possible? Preferably with an explanation backing it up that's light on the calculus and links to pages that are calculus level? (if you don't use too many calculus terms, and tried to explain it out, I'd have a much better time understanding it!) Personally, I don't see how it could be considered to exist but not an [infinitesimal], after all, both operate on the fact that there can be an infinite number of numbers after the decimal point. While the latter outright needs to have an infinite amount of distance from the decimal point, the former instead outright assumes that there cannot be any other numbers after it, which seems a bit impossible to me, and would hypothetically have tot "end" in an infinitesimal (but then again so would any number, so that doesn't say much.)

Another thing... assume 0.999... is a₁+a₂+..., where a_x is equal to 9/10^x. But assume you multiply 0.999... by ten. Obviously, the 9s don't stop, but at the same time I don't see what's preventing there from being a 0 after it. Sure, it's an [infinitesimal] to some extent (at the very least, it has a similar idea behind it), but am I incorrect in thinking that the number of nines after the decimal point in 0.999... are infinte? Am I incorrect in thinking that an infinitesimal is just 1/infinite? Why would one be allowed but the other shunned? At first glance, it very much seems like a childish attempt to prove that 0.999...=1, to cast one part of infinity aside yet require the other.

And finally, what makes 0.999... to be 1 anyway? If you were to take a 10 and a 9, and you were allowed to add a 9 to the end of the 9, but at the cost of adding a 0 to the end of the 10, would you ever be able to allow them to become the same number? I don't think so, although that example is not mathematical in nature, and you'd never reach infinity this way, since infinity > x where x is any number, and yet there's always an x+1. But what causes this addition to be different?

First person who responds to any of these questions in a way that I understand and agree to gets 20 points! Extra incentive to respond, I hope! (not that these points are worth anything, but if I ever actually USE my account here, I might put you in my user page as a legend. Not that it would be that impossible to get me to believe 0.999...=1, it just takes explaining it to me, rather then just throwing out proofs and definitions. It might be easier to do that in real time, I suppose, so if anyone wants to contact me through such a way to attempt to finally shut me up, then I'd be happy to try to obtain such contact.) Thank you to all who respond! 67.83.72.38 22:35, 28 February 2007 (UTC)

0.999... exists in the same sense as sqrt(2) exists.It is a set of symbols with a definition attached. If we don't define what we mean by a repeating decimal, like 0.999..., then it might as well not exist. However, there is a definition in use for 0.999..., and that definition is the limit as n tends to infinity of 9/10 + 9/100 + 9/1000 + ... + 9/(10^n). Take that as a string of words for the moment, I'll come back to what I mean by that. The reason we have this definition is because it means we get a lot of nice properties from decimals. It means we can write any real number as a decimal - if we didn't use this definition we couldn't represent 1/3 as 0.333... . Sure, this definition breaks some things, namely it means that some numbers have two ways to write them. But on the whole, not defining recurring decimals has worse drawbacks - you get lots of numbers that don't have a decimal representation. And defining them differently tends to do funny things to operations like subtraction.

So I've thrown a definition at you and told you that there's a lot of good reasons for this definition. But the definition itself probably isn't too illuminating - we need to go back a step and ask what a limit of a sequence as n tends to infinity is. Firstly, a sequence is just a list of numbers, like 1,1,1,1,1,1,... or 1/2,1/4,1/8,1/16... and so on; we can label each term of a series and so point at an nth term, and we consider that we know a series if I have a rule which means that if you give me a number n, I can tell you what that term in the sequence is. For example, the rule for the first sequence there is: the nth term is 1. For the second, it's the nth term is (1/2^n).

Now for the heart of the matter. The definition of 'a sequence tends to a limit l as n tends to infinity' means that, if you give me any positive number e, I can give you a number N, and any term in the sequence after the Nth will be within e of the limit l. Feel free to read over that a few times. It's important. Again, we didn't have to define things this way, but when we do lots of things work out well when we do and I think you'd be hard pressed to come up with one which works better. Notice in the definition infinity is never mentioned. The 'as n tends to infinity' is just a nice phrase we like to use. Infinity has nothing to do with what we actually mean when we say it.

Some sequences never satisfy this definition for any limit l, for example 1, -1, 1, -1... However, we can prove that the ones were dealing with here do work for some l. As a warm up, let's apply the definition to the first example sequence. We think 1,1,1,1,1,... should probably tend to 1. And it does. Whatever e you give me, I can reply with an N of 0; any term from 1 onwards is clearly within e of one - it *is* one.

Now let's go for the big one. I claim that the limit as n tends to infinity of the sequence whose nth term is 9/10 + 9/100 + ... + 9/(10^n) exists and is 1. Firstly, look at the difference between one and the nth term. This is 1 - 9/10 - 9/100 - ... -9/(10^n) = 1/(10^n). Now if you give me an e, I will be able to find an N which works. Any N greater than log (base 10) of 1/e will work. Another way to find one is to count the digits after the decimal point before the non-zero numbers start - N is the number of digits. We notice that the difference of each term from 1 is less than that of the one before it, so once we've found one term which works,all the ones after it work too, satisfying that part of the definition.

And with that, we're done. To summarize, we've given a reasonable definition of what 0.999... is (the limit of a sequence) and a reasonable definition of what a limit is which doesn't appeal to any wooly concepts like infinity. We've also shown that within these definitions 0.999... = 1. You could use different definitions, but these are the ones currently in use in the mathematical community, and they're what the article is based on. As I said, different definitions often break other things.I won't respond in detail to the rest of your questions (it's late) but would say in general that it's best not to think of 0.999... intuitively as an 'infinite' string of nines, because people's intuitive ideas of inifinity vary. Almost everyone would agree 1+1=2, so we can use intuition there, but some would say infinity + 1 = infinity, others wouldn't. To sort things out when not everyone's intution says the same you need to go back to definitions. The digit manipulation proofs you appear to disagree with are valid but rely on several properties of real numbers which are intuitively right for some people but aren't that easy to prove (though can be proven). If your intution doesn't agree to that of those proofs then you need to look a little deeper, namely at analysis and limits, to see why 0.999..., as we define it, equals 1.

If there's any part of this post you think seems like it doesn't follow then I can fill in gaps with more detail as needed and prove why things I've assumed are in fact OK. I may start mentioning things like least upper bounds if you poke at the right places, but they're really not that bad and are quite handy. Hope this helps. 131.111.8.98 00:14, 1 March 2007 (UTC)

I think I can cover at least one of the things you left off, on digit manipulation. I'm not certain it's accurate, so correct me if I'm wrong, but it's part of what makes this make sense to me. First, imagine your infinite decimal - whatever it is you imagine when you think of that. Now, consider a specific one, the "infinitesimal". You're probably imagining something like 0.00...0001, am I right? Good. Now, what comes next? (It sounds silly, but it's the best thing I know of, because it applies at the point of action.) What comes next? Let me back up.

The whole point of an infinite decimal expansion was that, no matter how many digits you've written, there's still more. In fact, as best I can tell, that's the entire intuitive (and possibly formal) definition of infinity - there's always more. More minutes, more feet, more dots, more steps in a process especially. For any point along it, there's another beyond, and another. Even if you've managed to go infinitely far along it (that is, so far that no matter how far back you tried to go, there'd still be more), there's still more. That's what infinity does. Or at least, we can agree that whatever else it does, it has at least this property. Let me know if it seems like infinity doesn't have this property, I'll try to phrase it better.

So, lets say you've managed to write infinitely many zeros. (I think that's not allowed in a decimal, but let's say.) You've found the last one. Great. What comes next? Apparently, infinitely many more digits, since the existence of each requires the next.

The reason it works for me, and the reason it might work for you, is that it fixes most conceptual problems I have with infinity. I've been given many phrases trying to expain why I can't put 0.000...0001, chief among them "there is no last digit", but this makes it make sense. It's not that the number ends an infinite distance away, but that it never ends. For each, there's a next.

This clarifies why Infinity-1=Infinity, for instance. Infinity is either a Category or a Property (dending on the context). As a category, it contains all things that answer "Yes!" to the question, "Is there always more of you?" As a property, it applies to all things that answer yes to that question. That's why people keep saying infinity "isn't a number", by the way, because it's not. It's not even a number, since many demonstrably different amounts are called infinite. Take a block of gold. If you carve off a corner, is the remaining block still gold? Of course. Something's goldness or not-goldness doesn't change when you cut a piece off. Similarly, consider the natural numbers. Instead of counting from 0, count from 1, or 20, or 100. Since you've removed 1, 20, or 100 numbers, this would correspond reasonably to Infinity-(1,20,100) in the way people usually mean. Now, consider the question, "Is there always more?" Sure is. So, it's still infinite. Still gold. You might even say "Infinity-1=Infinity", if you wanted to pretend Infinity was a single unique value.

To return to my point, the definition of 0.999... and the undefined state of 0.000...0001 are entirely consistent, in fact they seem unavoidable to me. Exactly because there're infinitely many 0s, it's meaningless to tack on a 1 after the last one - it's meaningless, that is, to even talk about a 0 with no 0 after it. That's what you meant by "infinitely many 0s" in the first place. You could talk about a list of 0s where each has another after it, except one of them, where it stops. Or you could talk about a list of 0s where each has another after it, period. In the first case, you're quite right that this stuff wouldn't work. But you're talking about the second. So, you can only talk about 0s with more 0s after them. That means you could still pick any spot to stick the 1 in, but it wouldn't do what you're trying for, since it would be just another digit, with more digits after it, without the distinction of being lonely at the end.

I'm not sure how this connects into your question about 1/infinity. As best I understand that one, the problem is that there's no such thing. If you reduce something in absolute magnitude without bound, you have no choice but to wind up with 0. I'd rather stick to the digit manipulation, though, and leave division for someone else.

Oh, one problem I noticed: "For each, there's a next" isn't the best description of infinity in all cases. For instance, unlike the natural numbers, the rational numbers, or the digits of a decimal expansion, the real numbers themselves can't easily be listed in a clear each-next order. I think it still captures the spirit, though. Black Carrot 10:15, 7 March 2007 (UTC)

[edit] 10c

10*0.999 does not equal 9.999 as shown in the article, rather 9.99

Yes, that's right. But the article doesn't say that 10*0.999=9.999, it says that 10*0.999...=9.999... (note infinitely repeating digits).—M_C_Y_1008 (^talk/_contribs) 04:08, 10 March 2007 (UTC)

Ie, Once one decides to escape to infinity there is no coming back... Almost like its in its own numerical space :-D Ansell 05:44, 19 March 2007 (UTC)

[edit] This Might Help Skeptics

I'm a graduate student in mathematics, but I too once struggled with the notion that 0.999...=1. I understood the various proofs, but something about it bothered me nonetheless. Finally it all clicked during a course in real analysis when I understood the notion that a real number does not have a UNIQUE decimal representation. Once I understood and believed this, I no longer had any trouble believing that 0.999... is EXACTLY EQUAL TO 1. Consider that a word in the English language is written using a finite combination (technically a permuation) of letters. However, we ascribe an abstract meaning to the written symbols. The same is true for numbers. We use decimals as a way of representing numbers, and there is nothing wrong with doing so. However, remember that a written decimal representation of a given real number is a way of expressing the abstract notion of what that number really is. I hope that helped. Mickeyg13 21:14, 11 March 2007 (UTC)

That's one step, and a critical one. The thing is, that's not the only problem people have with it. People also have a problem, for instance, with limits, which are central to the argument. Recognizing that it could have both names doesn't convince people it does, you see. For that, they need to understand how infinite sequences converge. Black Carrot 08:03, 12 March 2007 (UTC)

Also, the point is emphasized in both the article and the FAQ (question 4). I'm not quite sure how/whether to emphasize it even more. Calbaer 21:35, 12 March 2007 (UTC)

This article has convinced me of the truth (deserves the FA status it's got). As Black Carrot said, limits were my problem as well, as in, I didn't understand infinite. I thought that there must be a last 9, but of course, when there's an infinite amount of 9s, there isn't a last one. Also, if 0.999...=1 then shouldn't 0.999... redirect to 1. Sorry, I had to put that joke in... Cream147 _{Shout at me for doing wrong} 00:53, 17 March 2007 (UTC) __________________________________

I understood right away that the same number could be represented in different ways - that was never my objection. I'm also aware of other counterintuitive mathematical results such as the Monty Hall Problem - that article was well written and easily changed my mind about how to see it's problem. But this .999... article does nothing to demonstrate it's claim. Instead:

1) Proofs fall back on other dubious claims and ultimately circular logic. and appeals to authority.

2) The article evades it's own subject by insisting that the dialog must only be discussed in the context of the Real Set, when that set is defined as being unable to contain this subject. It's like using whole number theory to prove that 1/2 can't exist.

3) The fundamental ingredient of .999..., the Infinitesimal is featured right here in a respected article.

4) Even stranger concepts like the square root of negative one are accepted mathematics. Does one have to pass a beauty contest to get such concepts in?

5) Any further questioning lead to rudeness

6) Arguments state what happens at the "End of infinity" and then deny that anything else can go on there.

7)There are different kinds of infinity - why not different kinds of Zero as their reciprocal?

The fact that the writers here don't even seem to understand my objections makes me doubtful that they know more, rather then less then I do. I'd be interested in doing a poll to see how many new readers actually accept what this article has to say.- Algr

You indicate that no one here can understand you and that makes you conclude that they know less than you do? Occam's razor would seem to indicate that it's you having trouble communicating, rather than everyone else being too stupid to understand you. Nevertheless, I'll address your objections:

1) The logic isn't circular; if you think it is, please explain how. "Appeals to authority" are the basis of Wikipedia; see WP:RS and WP:NOR for why.

2) Real numbers are being discussed because real numbers (and their sub- and supersets) are what infinite field mathematics is concerned with in over 99.99% of its applications. You can always makes something up and call it "math" (if it's internally consistent), but, if it's useless, it's not usually instructive to study.

3) This is wrong. As stated in the article, the FAQ, the discussion, and even the link you give, there is no nonzero constant infinitesimal in the real numbers.

4) The square root of negative one is very, very useful. Number systems constructed to force 0.999… to be something other than 1 are very, very useless. This is largely because adding i to real numbers does not change their properties, while insisting that 0.999… be something other than 1 does.

5) I'm not sure how you feel that people claiming 0.999…=1 are rude. Certainly some will call a troll a troll and a crank a crank, but generally such trolls and cracks are ruder than those responding to them.

6) There is no "end of infinity." Period.

7) Infinity is not a real number. It is a concept useful in discussing the real numbers, but not a real number itself, so discussing its reciprocal is meaningless, formally speaking.

Clearly the above won't convince you that 0.999…=1. But if you have the humility to perceive your own fallibility and the limits of your knowledge, hopefully understanding the above will put you in a better position to understand 0.999… and mathematics in general. By the way, no one ever claimed that your objection was that no two representations could correspond to the same number, just that that was an obstacle for some people. With apologies to Tolstoy, everyone who understands 0.999…=1 understands the same thing, while everyone who doesn't understand it doesn't understand it in his or her own way. Calbaer 17:26, 17 March 2007 (UTC)

Infinity exists in mathematics but is not a real number in the same way that sets exist in mathematics but are not real numbers. Eyu100^{(t|fr|Version 1.0 Editorial Team)} 16:59, 18 March 2007 (UTC)

Re: 7. The infinite cardinals aren't in the Reals. Adding any of them breaks the group and field operations. A "different kinds" of zero would itself instantly break the algebra. To get an idea of how bad "breaking" an algebra is: imagine your bank account contains $4293 and you depositing another $214. You get a friendly message from your bank: "Your account has been terminated because your account balance was not a valid number." You don't get any money back because the bank can't fingure out how much to give you. We have to be able to do stuff with our numbers and know that we still have numbers; throwing infinities into the mix makes that impossible. Endomorphic 22:53, 18 March 2007 (UTC)

One final thing: the first line of the second paragraph of infintesimal is "Nonzero infinitesimals are not members of the set of real numbers." Also: when people say infantesimals, the vast majority of the time they really mean differentials. Endomorphic 23:14, 18 March 2007 (UTC)

But money is defined as having two decimal places so you do not run into the problem of infinitesimals. Also, practicality does not neccesitate that .999... be defined as 1; it just requires that it be treated as such in practice. Math should be "pure", meaning that it should not tied down by practicality but represent reality regardless of if it is only relevant in theory. Science has to be concerned with physical constraints but math does not.--JEF 00:05, 19 March 2007 (UTC)

The bank error is an illustration of broken algebra, regardless of whether or not it was broken by infintesimals. You suggest considering only those numbers for which the algebra is not broken - no infintesimals. That's exactly what the reals are. Secondly, practicality *does* neccesitate that .999... = 1. This is not a definition but a consequence. The argument does not invoke infintesimals at any stage. Endomorphic 02:47, 19 March 2007 (UTC)

[edit] 1/3 <> 0.333...

I've asked Uni professors, students and maths teachers. 0.9 recurring does not equal 1. 0.3 recurring is an estimate of 1/3, the closest you can come with decimals, but is not exactly 1/3, that is why you use 1/3 in equations. therefore, every single proof is wrong as they all rely on 1/3 being 0.3 recurring.—The preceding unsigned comment was added by 60.224.230.251 (talk • contribs).

If you don't believe 1/3 = 0.333..., then please read 0.999...#Algebraic proof, which doesn't deal with 1/3 at all. --Maelwys 13:26, 21 March 2007 (UTC)

Even so, 1/3 does = 0.333... it's not an approximation. Leebo ^T/C 13:28, 21 March 2007 (UTC)

I sincerely doubt you've asked any university professors, at least, not any math ones. Otherwise you would have gotten the exact opposite answer. (This assumes competence, which may be a bad assumption).Dlong 13:31, 21 March 2007 (UTC)

Or, perhaps, he did ask competent mathematics university proffesors, but either did not ask the right questions, or did not interpret the answers correctly. That happens when trying to delve into a subject one does not understand. -- Meni Rosenfeld (talk) 15:29, 21 March 2007 (UTC)

Not to be cynical, but the unsigned says that he asked professors and teachers, then concluded that 0.999... isn't 1, not that the professors and teachers actually told him as much.... Calbaer 16:12, 21 March 2007 (UTC)

Then again, this is the same fellow who blanked the article, replacing it with a diatribe beginning, "You guys are complete idiots," so we should probably neither assume good faith nor waste our time with the user's vague and potentially deceitful concerns. Calbaer 18:27, 21 March 2007 (UTC)

[edit] Keeping this simple

I know it's been said before, but the arguments for 0.999... really could be kept simpler and perhaps be more effective.

Almost every argument against it (beyond the obviously trivial ones) comes down to misunderstanding that the "..." means the limit as n approaches infinity in the sum. That fact alone pretty much ends the discussion on both sides.

However, the arrogance and superiority on one side of this debate seems to want to avoid the easier explanation, apparently because it seems as if it is a compromise. The simple explanation perhaps appears to concede that the arguments against 0.999... = 1 have sound philosophical grounds, but just misunderstand the meaning of some of the symbols, and hence is perhaps less attractive than the proofs that instead allow you to just be right and them stupid.

Maybe just show a graph plotting the curve of 0.9, 0.99, ... , and plotting a line at 1. The curve obviously approaches the line, and explain that the line it approaches is exactly what the "..." portion of 0.999... is referring to. The '9' symbols may only reference a number closer and closer to 1, but the "..." symbol makes it 1. Draw a little arrow pointing to the line at 1 and explain that this is exactly what "..." is referring to. Here on this curve are 0.9, and 0.99, and 0.999, and so on, and it looks like it's getting closer to this line up here, and incidentally that is what "..." means. That's how infinity works when you stick it on the end of a number. That is the definition of "..." I don't know if I know how to create such an image and make it available here myself, but I'll give it a whirl if no one else already has.

By using the notion of upper bounds and approach, you're at least speaking the language that non-believers seem to want to think in. You're conceding that this number will often, unfortunately, be thought of conceptually as a process and not a number. I think believers, however, consider this relatively simple explanation to be somehow conceding the notion that it doesn't really equal it, just approaches it, or something similarly patronizing, and hence avoid it and thereby attract more of the argument they're trying to defeat.

Like the famous algebraic proof. Pretty much every line in that proof has the "..." in it, which is the very thing the non-believer misunderstands. So it doesn't really help much. --The Yar 15:24, 23 March 2007 (UTC)

You could try to create such an image, but, honestly, I think you're being naïve. The main problem with people reading the article and still refusing to believe

0.999... = 1

is not that they don't believe that $0.999... = \lim_{n = \{1,2,...\}} \sum_{i=1}^{n} 9 (10^{-i})$ but that they don't believe that $1 = \lim_{n = \{1,2,...\}} \sum_{i=1}^{n} 9 (10^{-i})$ . They understand the general concept of a limit, but don't understand that there are no nonzero infinitesimals in the real numbers. Your image, in fact, would only fuel their objections: "See? There's always a gap between the terms and 1. That's the gap I was talking about that assures that 0.999... isn't 1!" That may sounds silly and wrongheaded to you and me, but, if you're trying to convince the hold-outs, I doubt your idea would help. (I suppose you could always post to the talk page first and see if any of the aforementioned hold-outs are convinced. Or if any are convinced by the above.) Calbaer 16:01, 23 March 2007 (UTC)

I agree with your prediction as to what the response to such an image would be, but I would disagree somewhat with where the dissenters generally go wrong. I think that they agree that

$1 = \lim_{n = \{1,2,...\}} \sum_{i=1}^{n} 9 (10^{-i})$ , and that

$0.999... = \sum_{i=1}^{\infty} 9 (10^{-i})$ , but deny that

$\sum_{i=1}^{\infty} 9 (10^{-i}) = \lim_{n = \{1,2,...\}} \sum_{i=1}^{n} 9 (10^{-i})$ , claiming that the use of a limit in the last equality is somehow an "approximation", despite its being (to my understanding) the very definition of the sum of an infinite series.--Trystan 21:18, 23 March 2007 (UTC)

I don't think anyone's being arrogant. An image would be interesting, and I agree that people most often have trouble with accepting the initial definitions - but time and time again, people explain that 0.999... by definition means the limit, and they will still either refuse that definition, or instead say "But it never actually reaches 1!" By all means, if you can think you can make the article simpler to understand, then please go ahead and try. Mdwh 00:55, 24 March 2007 (UTC)

There is additionaly subtelty here. Using a standard construction, Real numbers truly are processes on Rational numbers; a real number may be defined as an equivalence class of cauchy sequecnes of rational numbers. The Dedikind Cuts option has real numbers defined as sets of rational numbers rather than sequences. Neither of these help people who expect Pi to be fundamentally the same sort of creature as 1/4. Endomorphic 22:09, 25 March 2007 (UTC)

[edit] Explain this!!!

1 - 0.999... = 0.000...1
(1-0.999...)(1-0.999...) = (0.000...1)(1-0.999...)
1^2 - 2(1)(0.999...) + (0.999...)^2 = (0.000...1)(1) + (0.000...1)(0.999...)
1^2 - (1)(0.999...) - (0.000...1)(0.999...) = (1)(0.999...) - (0.999...)^2 - (0.000...1)(0.999...)
1(1 - 0.999... - 0.000...1) = 0.999...(1 - 0.999... - 0.000...1)
1 = 0.999...
0.999... - 0.999... = 0.000...1
0.999... = 0.999...1
...and just for show...
0.999... + 0.000...1 = 0.999...1 + 0.000...1
0.999...1 = 0.999...2
???

1 can not equal 0.999... — Ian Lee (Talk - Contribs - Sign - Gimme!) 10:41, 27 March 2007 (UTC)

Well, your first problem is with your definition of 0.000...1. What is 0.000...1 as a number? It's an infinitely long string of 0s, with a 1 on "the end". But if it's infinitely long, there is no end. So where's the 1? It can't exist, meaning that your formula 1 - 0.999... = 0. And if there's a difference of 0 between the numbers, then they're obviously the same, so 1 = 0.999...

Your second problem is with the contradiction in lines 1, 6 and 7. In line 1 you start off by saying that 1 and 0.999... are not equal (because there's a difference between them). Then in line 6 you say they are equal. Then in line 7 you say there's a difference between a number and itself (so now 0.999... and 0.999... aren't even equal). You have to get your equalities straight, what's equal to what here?

And your third problem is one that many people come across in writing a disproof for this concept. You set out to prove that 1 <> 0.999... and the very first thing you do is state that 1 <> 0.999... You can't start a proof by assuming that you're right, you have to start by assuming nothing, and then prove that you're right. --Maelwys 11:14, 27 March 2007 (UTC)

Maelwys is correct, and there are further problems.

In line 3, there's a wrong sign; it should be - (0.000...1)(0.999...) in the end. That doesn't matter since you correct it in the next line; probably just a typo.
The step from line 5 to line 6 consists of a division by 1-0.999...-0.000...1, where by your first line 1-0.999...=0.000...1, so in effect you divide by zero. With that, I could "prove" anything, including 1=2.
If I understand the step from line 7 to line 8 correctly, you add 0.999... on both sides. Then line 8 should again read 0.999...=1; what is 0.999...1 supposed to be?
Consider it the other way around. You set out with a difference 0.000...1 between 0.999... and 1, and you were able to "conclude" that 0.999...=1 (except for the division by zero error). That would only show that they must be equal since assuming otherwise leads to a contradiction.

So in effect, if your calculations were correct you would have shown 0.999...=1. Unfortunately they're wrong, but that doesn't mean that the conclusion is wrong too. Yours, Huon 13:21, 27 March 2007 (UTC)

The ellipsis stands for infinitely repeating decimal places. You can't "end" it with a number like .000...1. That number has no meaning. Leebo ^T/C 13:29, 27 March 2007 (UTC)

Everyone, please, quit repeating this line about not being able to put a one after infinitely many zeroes "because infinity has no end". It might possibly convince some people who otherwise wouldn't be convinced, but the cost is too high; in fact doing infinitely many things, and then doing something else, is commonplace in mathematics. See transfinite induction and ordinal number.

It's a correct statement, of course, that this putative "0.000...1" is not the decimal representation of any real number. But the reason is (slightly) subtler than just "infinity has no end". --Trovatore 15:26, 27 March 2007 (UTC)

Also, a simple "read the FAQ" would be a better response to a "proof" in which the first line is explained as wrong in the FAQ itself. But, yes, we should be careful about the use of the word, "infinity." More accurate language is in the FAQ:

Q: Can't "1 - 0.999..." be expressed as "0.000...1"?

A: No. "0.000...1" is not a meaningful string of symbols because, although a decimal representation of a number has a potentially infinite number of decimal places, each of the decimal places is a finite distance from the decimal point; the meaning of digit d being k places past the decimal point is that the digit contributes d · 10^-k toward the value of the number represented. It may help to ask yourself how many places past the decimal point the "1" is. It cannot be an infinite number of places, because all places must be finite. Also ask yourself what would be the value of $\frac{0.000\dots1}{10}$ . If a real number divided by 10 is itself, then that number must be 0.

Calbaer 15:47, 27 March 2007 (UTC)

Hm, I see a problem with the language there. The decimal representation of a real number has an actually infinite number of decimal places, not potentially infinite. --Trovatore 16:01, 27 March 2007 (UTC)

Perhaps the term "significant" needs to be inserted; I think the distinction is between 1.25 and 1.24999... --jpgordon^∇∆∇∆ 16:27, 27 March 2007 (UTC)

Well, 1.25 is really 1.25000..., so even there you have an actually infinite number of decimals. (And they're all significant; zero is just as significant as any other digit.) --Trovatore 17:36, 27 March 2007 (UTC)

1.25 may be 1.25000..., but I don't think it's "wrong" to consider the decimal places of 1.25 as 1., 0.1, and 0.01. Certainly there are implicit zeros at 0.001 and below, but there are also implicit zeros at 10. and above. So there are multiple ways of defining a "decimal," and I don't think any is canonical. However, if you think it's confusing, you can modify the FAQ to say that "although a nonterminating decimal representation of a number has an infinite number of decimal places" which will be technically correct for any of the various definitions. Calbaer 17:57, 27 March 2007 (UTC)

If we go along with this and assume the algebra (such as it is) is valid, then as far as I can see, what we have here is a proof-by-contradiction. But it proves that 1-0.999...≠0.000...1, not that 1≠0.999...

0.999...=1 is arrived at from the assumption that 0.000...1≠0, as is 1≠0.999... -- so the assumption is false. So if 0.000...1 = 0 then 1-0.999...=0 so 0.999...=1. I fail to see why I should be concerned about this. Andrew 15:46, 29 March 2007 (UTC)

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