Talk:0.999.../Arguments/Archive 7

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Archive 1

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Archive 5

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

(c) Neither of the Individual Values will ever Equal 1

(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:

" 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

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!

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)

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 "

INFINITE/RECURRING : THE FINAL CONCLUSION : By Anthony.R.Brown 01/06/07.

(1) All Infinite/Recurring Number/Values have a Start Value the reason being it has to be known what are being Infinitely/Recurred before a Number/Value can be Infinitely/Recurred.

(2) All Infinite/Recurring Number/Values that Start with Zero then a Decimal point will always have a Value less than One Example: 0.(n) < 1

(3) All Infinite/Recurring Number/Values that Start with One or Greater than One! then a Decimal point will always have a Value Greater than One Example: 1.(n) > 1

(4) Exception to (2) & (3) above All Infinite/Recurring Number/Values that Start with One then a Decimal point will always have a Value Equal to 1 where (n) has Infinite/Recurring Zero's Example: 1.(000...) = 1

(5) The Infinite/Recurring Number/Value for (n) regarding size! Does not decide if the Infinite/Recurring Number/Value is > 1 or < 1 except (4) above where (n) are Zero’s

(6) Infinite/Recurring Example ( 1 / 0.9 ) x ( 0.9 ) = 0.( 999...) or as 0.(n) < 1

GENIUS ANSWER TO THE INFINITE/RECURRING 0.9 PROBLEM By Anthony.R.Brown 04/06/07.

TV = " The total value for the amount of Infinite/Recurring 0.9's from the Start Onwards "

TD = " The amount of Decimal places Infinite/Recurring 0.9 has Travelled in the Sequence "

TV / TD Will always = 9

Example1 9 / 1 = 9 Example2 18 / 2 = 9 Example3 8991 / 999 = 9

Infinite TV / Infinite TD = 9

The above Proves TV can never Equal 1

This is wrong in several places. Most importantly, (2) and (5) would require a proof - that are the statements we're interested in. They are wrong; just as there's one exception to (3), there's one exception to (2), namely 0.999...

(6) is wrong even without any recurring decimals: Obviously (1/0.9)*0.9 = (1/(9/10)) * (9/10) = (10/9) * (9/10) = 90/90 = 1, all safely in the realm of the rationals without any recurring and/or infinite decimals.

The "genius answer" is probably largely correct, but irrelevant: The "total value" seems to be the sum of the digits which, when divided by the number of digits, indeed equals nine for any finite substring of 0.999... But where is a connection between the "total value" and the number? You implicitly use uniqueness of representation (and representations aren't unique) if you claim that because 0.(9) and 1 have different total values, they must be different.

I would be most interested in your reasoning concerning (6). The equations I gave above all use basic definitions and basic algebra of rational numbers - do you really disagree? If so, where? --Huon 17:12, 5 June 2007 (UTC)

(2)Has to be correct! Unless you can show how you get rid of the Decimal point 0.

(5)Has to be correct! Unless you can show a number with a Decimal point Start 0. > 1

(6)This is all WRONG!! (1/0.9)*0.9 = (1/(9/10)) * (9/10) = (10/9) * (9/10) = 90/90 = 1 Because it's one unbroken Calculation! I could now add again on the end 1/0.9)*0.9 = 0.999...

What you have done above is the same old Rounding up trick! where it's possible to make any Number = 1 Example N = 9.123456789 and now N / N = 1

The only Calculation allowed! is the Number we are talking about ie. Infinite/Recurring 0.9 it cannot be broken down into other numbers and changed! and just as important! is that the Infinite/Recurring 0.9 Problem is a Decimal problem ony! no FRACTIONS!

So remember the Only Number to be Calculated as Infinite/Recurring is 0.9

(1 / 0.9 )this is the Value that Calculates the Infinite/Recurring! x ( 0. 9 ) now being done!!

Anthony.R.Brown 07/06/07

Example 0.(TV) < 1 Because (TV) Never Ends (TD) Must Always Be The Same Length As (TV)

We Know That TV / TD = 9 Is True So 0.(TV) = 0.(9) Where 0.(9) < 1

I am not Calculating Infinity / Infinity or Infinity x Infinity

TV and TD are Infinitely Known Values! nothing Changes them! thats Why TV / TD = 9

Anthony.R.Brown 07/06/07

Sorry, but I don't follow your reasoning concerning (6). You say that for every number N, we have N/N=1. I agree with that. You also claim that 0.999... = (1/0.9) * 0.9; again I agree.

But do you doubt that (1/0.9) * 0.9 = 0.9 / 0.9 (which would, as a special case of N/N, be equal to 1)? Just to make this step more explicit, I did the calculation you claimed was "wrong because it's unbroken". I could break it up into lots of separate steps, so what? You claim I "round up". Where? Is (1/0.9) * 0.9 = (1/(9/10)) * 9/10, or was some rounding done here? Is (1/(9/10)) * 9/10 = 10/9 * 9/10, or not? How about 10/9 * 9/10 = 90/90? If I rounded somewhere, it must have been at one of these steps. Which one? The first just substitutes 9/10 for 0.9. Do you doubt 0.9=9/10? In that case, you should better give a definition of 0.9. The second consists of forming the reciprocal value of a fraction and can probably be found in school textbooks. The third is multiplication of fractions, again middle school-level math.

If we can't even agree on this basic level, arguing higher maths will obviously be pointless, so I won't comment on your other claims right now. --Huon 14:59, 7 June 2007 (UTC)

It's Quite Simple! for Infinite/Recurring 0. 9 you can only Calculate the 0.9 as Infinite/Recurring

As soon as you do any other Calculations! you are changing the Nature of the Problem! 90/90 = 1 is the same as 123 / 123 = 1 in Rounded Value!

(1 / 0.9 )this is the Value that Calculates the Infinite/Recurring! x ( 0. 9 ) now being done!!

My example above is 0.9 having one Calculation done to it! which is the Answer!

This is the same ( 1.111... ) x ( 0.9 ) = 0.999...

Anthony.R.Brown 08/06/07

We seem to be talking at cross purposes. We seem to agree that 90/90=1 (I suppose you don't claim that this equality is correct only up to rounding, with 1 being a little larger than 90/90, or do you?), and I take it we also agree that (1/0.9)*0.9=0.999... Now I claimed that (1/0.9)*0.9=90/90, and I gave what I consider an extremely detailed proof. This claim does not involve any infinite and/or recurring decimals. Those only appear in the one equation we already seem to agree about. With my claim I am not "changing the nature of the problem"; I set out to prove 0.999...=1, and I do exactly that by several intermediate steps.

I am not arguing along the lines of "0.999.../0.999...=1, and so 0.999... must equal 1." What I show is that 0.999...=90/90, and so 0.999... must equal 1.

Let's do it step by step. Do you claim that (1/0.9)*0.9 ≠ (1/(9/10))*(9/10)? If so, do you claim that 0.9 ≠ 9/10, or that they're equal but that for some reaon I can't fathom I still cannot substitute one term for the other in a given formula without significant change?

As a postscriptum, I would really be grateful if you could devote a little more care to your use of English. I'm not a native speaker, so I'll gladly concede that my command of English is far from perfect, but sometimes your sentences seem so garbled that it's getting difficult to determine what you mean. And precision is crucial here. --Huon 13:45, 8 June 2007 (UTC)

His argument is basically that although 90/90 = 1 and (1/0.9)*0.9 = 0.999..., there's apparently some unwritten rule that he's trying to enforce that says that you can't equate an infinitely recurring decimal to a fraction. So if you're going to use a set of numbers to calculate an infinite decimal (such as he's done above), that's all you're allowed to do with it and you can't manipulate it in any way that doesn't create an infinite decimal. Nobody is sure where he got this rule from, but he's quite insistent about following it, and so refuses to be swayed by any argument that requires any other math to be performed, as you've tried to do above. Apparently by this same rule we're not allowed to show that (1/0.9) * 0.1 = 1/9 either, because it must equal 0.111... and no other number, so I guess 1/9 can't equal 0.111... according to this logic. Fractions as we know them are useless in this twisted world. Go figure. --Maelwys 14:04, 8 June 2007 (UTC)

Mr. Brown continues

THE REASON INFINITE/RECURRING 0.9 IS CALCULATED WRONG By Anthony.R.Brown 09/06/07.

The reason Infinite/Recurring 0.9 is Calculated wrong! By everyone that thinks Infinite/Recurring 0.9 = 1 is because of one important thing they are all doing wrong!

And that is by applying multiple Calculations to the 0.9 in various different ways!

Whether it may be the Classic rounding up method ( 90 / 90 ) = 1 which is the same as getting rid of any Number with a decimal point and making the numbers equal 1 ( 123.99 / 123.99 ) = 1 or ( 0.123 / 0.123 ) = 1

And any other multiple calculations! (1 / 0.9 ) x 0.9 = (1 / (9 / 10 )) x ( 9 / 10 ) = ( 10 / 9 ) x ( 9 / 10 ) = 90 / 90 = 1 "Here again (1 / 0.9 ) x 0.9 = 0.999... can be done again! but the whole Example is wrong!

For Infinite/Recurring 0.9 you can only Calculate the 0.9 as one single Calculation!

As soon as you do any other Calculations! you are changing the Nature of the problem!

( 1 / 0.9 ) this is the Value that Calculates the Infinite/Recurring! x ( 0. 9 ) now being done!

My example above is 0.9 having one Calculation done to it! which is the Answer!! This is the same ( 1.111... ) x ( 0.9 ) = 0.999...

So there we have it! the Problem has been found! ..............................................

Anthony.R.Brown 09/06/07

For Infinite/Recurring 0.9 you can only Calculate the 0.9 as one single Calculation --says who? --jpgordon^∇∆∇∆ 14:20, 9 June 2007 (UTC)

Not says who? The original Problem is " CALCULATE INFINITE/RECURRING 0.9 "

You will NOTICE! above a Single Request/Task Emphasised on the 0.9 only!

Anthony.R.Brown 11/06/07

Still, if the problem is to "calculate infinite/recurring 0.9" (though I cannot remember that being the "original problem", stated by whom? Where?), why should I not be allowed to use more than a single equation?

To me, the original problem is to "prove (or disprove) that 0.999...=1", and even if that's a single task, why shouldn't multiple steps be allowed as long as their combination solves the task at hand? And even if that's disallowed, what would keep me from subdividing the task into many sub-tasks, each of which is solvable by a single calculation, and whose combination, again by single-calculation tasks, finally yields 0.999...=1? --Huon 14:09, 11 June 2007 (UTC)

Because by allowing Multiple Calculations! the problem can be continuously Changed!

If I prove something is the Colour GREEN! there is no need to try and mix other Colours to make GREEN again! or change it to any other Colour!

e.g As soon as Multiple Calculations Show 0.999... = 1 I then Continue ( 1 / 0.9 ) x ( 0.9 ) = 0.999...which takes us back to the original Start of the Problem!............

Anthony.R.Brown 12/06/07

If I can show (by multiple calculations or otherwise) that 0.999...=1, what's wrong with noting that (1/0.9)*0.9 =0.999...? I will also have shown that (1/0.9)*0.9 equals 1, but that's no surprise, is it? (Actually, I expect to have shown that beforehand in order to show that 0.999...=1.) Sure, I must be careful not to employ circular reasoning (ie I show that A is a consequence of B, and B is true because it follows from A). But I'm not aware of you ever claiming something like that.

By the way, I saw you claim that (1/0.9) * 0.9 = 0.999... and that 1.111... * 0.9 = 0.999... Isn't that an example of multiple calculations as well? --Huon 11:53, 12 June 2007 (UTC)

To Huon! and Everyone else! Please take Note! ( ONE CALCULATION ONLY ON THE 0.9 )

( 1 / 0.9 x 0.9 ) = " The Infinite Multiplication " " Now Times the 0.9 Only " = 0.999...

So now You must do the Same ( ONE CALCULATION ONLY ON THE 0.9 ) to try and prove what you are saying that somehow it equals 1 ?

Below is the False Example doing Multiple Calculations on The 0.9

(1 / 0.9 ) x 0.9 =

(1 / (9 / 10 ) " The Second Calculation on the 0.9 ? "

 x ( 9 / 10 ) " The third Calculation on the 0.9 ? "
 = ( 10 / 9 ) " The fourth Calculation on the 0.9 " 
 x  ( 9 / 10 ) " The fifth Calculation on the 0.9 ? "
 = 90 / 90 = 1 " And now the Final Sixth Calculation!

which Rounds up the Previous Decimal Point Values! into an Integer Whole Number/Value!! "

So The True Infinite/Recurring 0.9 Problem Solution is!...

(INFINITE/RECURRING MULTIPLICATION) x (0.9) = (RESULT 0.999...)

Anthony.R.Brown 22/06/07

Well, I'm convinced. In fact, I'd go so far as to say that your last post clears up any confusions regarding this issue, and that any further discussion is unnecessary.--Trystan 16:00, 22 June 2007 (UTC)

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:

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

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)

First you prove that 10 * 0.(3) = 3.(3), then you get 9*0.(3) = 3 from nowhere just as 3*0.(3) is 1 in your mind? Doesnt look like you proved anything, its just another fancy way of saying 1/3 = 0.(3) and not 1/3 = 0.(3)3 + 0.(0)1/3 ? 80.202.226.35 22:55, 6 May 2007 (UTC)

Let X denote the number 0.(3). From 10 × X = 3 + X the author of the proof infers that 9 × X = 3. This is pretty standard high-school algebra. It looks like mathematics may not not quite be your forte, so perhaps you should be careful before you start attacking the proofs of others. --Lambiam ^Talk 00:04, 7 May 2007 (UTC)

Let

a_{1},a_{2},\ldots

decimal digits be given such that

1/3=0.a_{1}a_{2}a_{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_{1}a_{2}\ldots a_{n}a_{n+1}\ldots \geq 0.a_{1}a_{2}\ldots a_{n}a_{n+1}\geq 0.333333\ldots 34>1/3

. Similarly if

a_{n+1}<3

, then

1/3=0.a_{1}a_{2}\ldots a_{n}a_{n+1}\ldots <0.a_{1}a_{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_{1}a_{2}\ldots a_{n}a_{n+1}\ldots <0.a_{1}a_{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_{1}a_{2}\ldots a_{n}a_{n+1}\ldots =0.a_{1}a_{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

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....

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)

Quote: Above " Does anyone have any proof that 0.999... even exists? "

It Exists! ( 1/ 0.9 ) x ( 0.9 ) = 0.999...

Anthony.R.Brown 07/06/07

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)

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.--Jorfer 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)

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)

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)

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\dots 1}{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)

Semantically, 0.999... = 1 is nonsense

I said this a while back, and I'll say it again, this time in simpler terms.

One of the most basic fundamentals (I would call it an axiom) of mathematics is the concept of digits--different places representing different values.

In the number "394" (assuming base 10), we have three "100s" units, nine "10s" units, and four "1s" units. This semantic definition is IMPORTANT. Take away these definitions, and the number "394" becomes entirely meaningless.

Thus:

1.000... <= the "1" means that we have one "1s" unit.

0.999... <= the "0" means that we have NO "1s" units. Period.

1.blahblahblah will ALWAYS be greater than 0.blahblahblah, period, as a matter of simple semantics. If the mathematic properties of infinitesimals violate this semantic rule, then "0.999..." (and all other infinitesimals) need to be given different terminology or different symbology (like "i" represents the square root of -1).

Sheesh. It's like watching people try to prove that zebra should come before apple in dictionary because zebras are just so goddamn awesome. Newsflash: the word "zebra" is arbitrary; the spelling order of the dictionary (and the meaning of a decimalized number) is not. Violating the agreed-upon semantics of the system doesn't make any sense when you can simply change the label. Admittedly, this is probably more symbolic of the fact that our semantic system is flawed (plenty of rational numbers--such as 1/3--have no non-infinitesimal decimal representation) but that doesn't mean it's ok to go ahead and violate it without qualifications.

And before you mention it, yes I've considered the possibility that the "..." itself is the qualification, but if that's the case I feel it needs to be stressed much more strongly just how big of a qualification it is. Instead of saying "the '...' means that the 9s go on forever!" (this isn't said so much as implied, and the article's picture definitely doesn't help), we should instead say "ok, this number really can't be represented in decimal form without breaking some rules. This is the best we can do and yes, it behaves a little differently and should be interpreted differently." That's the real reason why so many people have a problem with this proof--without qualifications, it's semantically absurd. It's like saying "1 = 0" and neglecting to mention that the "0" is, in this case, a special number that doesn't obey the normal laws of decimals.

I'm tempted to post this to the talk page, but I'll try it out here first. Just try to realize that my argument isn't really mathematical, it's semantic. The infinitesimal symbology is extremely confusing, more confusing than if we used the "+" sign to denote addition OR subtraction, and you guys are NOT making matters better by throwing a bunch of equations. The mathematical proofs do not trump the semantic flaw--emphasis that infinitesimals aren't true-blue law-abiding decimals (which doesn't mean that they aren't true-blue law-abiding real rational numbers) can. --Lode Runner 01:13, 15 April 2007 (UTC)

Your zebra analogy is in fact the wrong way around. There are formal and entirely unambiguous rules for sorting things alphabetically, just as there are formal and entirely unambiguous rules for defining real numbers by decimal expansions (a consequence of which is that 0.999... = 1 ). It's more like people arguing that "kitten" should come before "cat" in the dictionary because they have an intuitive notion that kittens precede cats, but the formal rules of alphabetical sorting dictate that "cat" must come before "kitten". Maelin (Talk | Contribs) 03:14, 15 April 2007 (UTC)

Saying that 1 shouldn't equal 0 isn't an "intuitive notion." Let me stress again that I'm not arguing with the mathematics behind infinitesimals. Read my reply to Calbaer below. --Lode Runner 07:14, 17 April 2007 (UTC)

Infinitely long decimals are not defined in the same way as that "394" example given above. Instead, they are defined to be the limits of certain sequences - and the sequence (0.9, 0.99, 0.999, ...) whose limit 0.999... is defined to be happens to converge to 1, making 0.999...=1 by uniqueness of limits. --Huon 09:16, 15 April 2007 (UTC)

Precisely my point. Infinitely long decimals violate the basic semantics of decimals. --Lode Runner 07:11, 17 April 2007 (UTC)

Your claim that different places represent different values is true. Your conclusion, however, that this must mean that different digits result in different values, is false. The fundamental misunderstanding of your logic is that all numbers must have unique representations, but nowhere in the definitions, semantic or otherwise, is that asserted. It happens to be true for terminating decimals, but that does not mean it is true for all decimals. Consider the Roman numerals for an example where this isn't even true for integers. As in decimals, the placing of a numeral changes the value. If an "I" is before a "V" or a "X," it represents the value, "-1," whereas, otherwise, it is "+1." And, sure enough, "IIII" is the same value as "IV." This is a slightly different phenomenon, but it is the simplest example of a number system in which two representations correspond to the same value. Another are the fractions, in which 1/2 is identical to 2/4, 3/6, 4/8, etc. Or time, in which 0:00 often written 12:00 am or 24:00. Or in the polar coordinate system, where (x, 0°) is the same as (x, 360°) for any x and (0, y) is the same as (0, z) for any y and z. Or IEEE 754 under the normal rules of math, in which +0 and -0 are equal. In all of these, different places represent different values, yet it is obvious that the same number can have different representations. Saying otherwise would be like saying that you couldn't call a zebra "Hippotigris," because it already has the name "zebra" and that should be the only thing you refer to it as. Calbaer 18:35, 15 April 2007 (UTC)

"Your conclusion, however, that this must mean that different digits result in different values, is false." I drew no such conclusion. I don't have a problem with a number having more than one representation--that is not the issue here. The issue is a VIOLATION of the agreed-upon demical semantics. To borrow your Roman numeral analogy, it would be like saying IIIV (which, by the rules of Roman numerology, should be 2 or possibly 8) is equal to IV (which is 4.) Regardless of the mathematic reasoning behind it, it's an extremely bad representation. It's not simply unintuitive--it's fundementally WRONG. It's not because mathematics cares--it doesn't. 7 and 8 (the symbols) could switch places every third Tuesday of the month and the pure mathematics wouldn't be affected at all, but does that make it right, consistent, or sane?

There's no sense in arguing that the mathematicians should change their symbology (though I can think of a number of acceptable alternatives--a fraction with an extra symbol of some sort to denote it as an infinitesimal form, or somehow marking the leading digit (in this case "0") to indicate that it's not meant to be interpreted as a normal decimal would), but I think that we can at least word the article better. I don't have a problem with the concept of "being infinitely close to something is the same as being equal to that something"--I have a problem with people saying that 1 = 0, and I think this is the same problem most others are having. It's not the mathematics--it's the semantics. Whoever invented infinitesimal notation was a dumbass and while it isn't wikipedia's place to fit his mistake, we can at least try to minimize the confusion it causes. --Lode Runner 07:11, 17 April 2007 (UTC)

Nobody is claiming 0=1, neither semantically nor otherwise. And even worse, that you keep mentioning infinitesimals suggests that you got something wrong mathematically - no infinitesimals should appear here. Of course the 0 in 0.999... means the same thing as that in, say, 0.5 - that you don't add any multiples of 10°=1. But the rest of the stuff you add to make up 0.999... happens to equal 1 when added together - so in effect, while 1.000...=1+0, you have 0.999...=0+1.

Besides, I fail to see what you propose. Should, in your opinion, 0.999... denote some other number? If so, which one? If not, what's your point? --Huon 09:18, 17 April 2007 (UTC)

And even worse, that you keep mentioning infinitesimals suggests that you got something wrong mathematically - no infinitesimals should appear here. Fine. I was under the impression that the number "0.999..." was called an infinitesimal, or at least was infinitesimally close to the number "1.000..." I don't care what you call it, then, because this is not a mathematic argument.

But the rest of the stuff you add to make up 0.999... happens to equal 1 when added together - so in effect, while 1.000...=1+0, you have 0.999...=0+1. I accept that explanation on a mathematic level. I think that it's extremely murky on a semantic level, and I think that this murkiness is contributing to the controversy and confusion regarding this concept. My motivation here is not to change the notation (since that's definitely not wikipedia's place) nor make any sort of mathematical argument. I am merely suggesting that the article clarify that the number "0.999..." subverts the basic, axioms of decimal notation, because the leading zero clearly (classically) indicates "we have no ones units, therefore our number must be smaller than one." --Lode Runner 06:18, 18 April 2007 (UTC)

So addition is "a VIOLATION of the agreed-upon demical semantics" because 6+6 clearly has a value of 0 for it's tens units, and so must be smaller than 11? Endomorphic 21:31, 18 April 2007 (UTC)

I don't think having a number with a 0 in the ones place that equals 1 is any more a violation of semantics than saying "I have zero dollars and one hundred cents." It's easier just to say that one has one dollar, just as it's easier to write 1 rather than 0.999..., but neither is incorrect, inconsistent, or unclear.--Trystan 17:13, 17 April 2007 (UTC)

An interesting, but misleading example. Instead of thinking about real-life dollars and cents (which are physical objects), consider numbers in a spreadsheet where dollars and cents are completely fungible. Does it make sense to say "I don't have a dollar--I have 100 cents!" in that context? No, it does not. In the world of our spreadsheet, 100 cents IS one dollar, and it's therefore bad form to state that you do not have a dollar (by analogy, the "0" in "0.999...") when you clearly do. --Lode Runner 06:18, 18 April 2007 (UTC)

Actually, a better way to put it is to say, "I have more than 90 cents, more than 99 cents, more than 99.9 cents, etc., but I have no more than a dollar." Clearly, then, you have a dollar. Calbaer 19:51, 17 April 2007 (UTC)

My point exactly. --Lode Runner 06:18, 18 April 2007 (UTC)

Although Lode Runner seems a bit confused (what with all the talk of "infinitesimal notation"), I think he or she is trying to say that all decimals except terminating decimals are too confusing to use, that 0.333... shouldn't have a meaning but we should rather say that 1/3 doesn't have a decimal representation, for example. He or she — and I'm glad to be corrected if wrong about this — is trying to say that having an infinite number of digits is a violation of his or her notion of decimals. However, this article isn't about the most limited notion of decimal notation, but about the most general notation. So the limited notion of decimals Lode Runner subscribes to is irrelevant. Perhaps the article could be made more accessible to those ignorant or skeptical of repeated decimal notation, though presumably they'd go to the article on repeated decimals anyway. Any advice on how to improve the article — which, although correct, isn't perfect — is appreciated.

So the limited notion of decimals Lode Runner subscribes to is irrelevant. This isn't limited notion, it's *correct* notion. If leading "0" (in a positive number) is changed into a "1", your number gets bigger. This is a simple, straightforward consequence of basic decimal notation, and to my knowledge repeating decimals like "0.999..." are the only numbers that break it.

Perhaps the article could be made more accessible to those ignorant or skeptical of repeated decimal notation That was my intent, yes, but that's a rather condescending way to put it. Conventional decimal notation predates repeated decimal notation by several thousand years, it is unchallenged in the vast majority of mathematic fields and (probably) all practical applications, and yet here we have repeated decimal notation blatantly contradicting conventional decimal notation by implying that a "0" digit is, in fact, nonzero.

Again, I stress I'm not arguing against the mathematics (and I don't care whether infinitesimal is the right term or not.) I am merely showing why this article has been so controversial and why math students have such a hard time accepting the concept. It's a sound theory, but it's_bad_form. Better form would be, I dunno, drawing a square around the leading 0 to emphasize it's not meant to be taken literally/conventionally, but we can at least clarify the situation by illustrating the contradiction with conventional notation and explaining why it's a necessary violation. --Lode Runner 06:18, 18 April 2007 (UTC)

By the way, with regard to "infinitesimals" and "being infinitely close to something is the same as being equal to that something," that's semantically wrong. Rather, the correct phrasing is, "If two numbers have a difference with an absolute value that is smaller than any positive number, then that difference is zero and thus the two numbers are identical." That concept is free of infinities and infinitesimals, and has the side benefit of being correct. Calbaer 19:51, 17 April 2007 (UTC)

"1.blahblahblah will ALWAYS be greater than 0.blahblahblah, period, as a matter of simple semantics"
"Whoever invented infinitesimal notation was a dumbass"

Load Runner, please try to keep your comments constructive. We get our share of trolls and cranks around here, and it would be a shame to misjudge someone with genuine issues.

Do you have semantic problems with the statement "999 + 4 = 1003" ? The left hand side has no thousands units, whereas the right hand side does; the same objection you had to 0.999...=1. A common objection is that "999 + 4" involves a manipulation (addition) whereas "0.999..." is just a number. Thing is, all numbers imply manipulations in their representations; as you note in "the number "394" (assuming base 10), we have three "100s" units, nine "10s" units, and four "1s" units." The decimal 0.999... does indeed have a ones unit when the implicit manupulations are performed. Endomorphic 21:45, 17 April 2007 (UTC)

We get our share of trolls and cranks around here, and it would be a shame to misjudge someone with genuine issues. Then don't? I didn't insult anyone except the dumbass who invented the notation. If the dumbass himself shows up here (likely he won't, because he's most likely long dead) I'll apologize. No one else has any right to take offense, and I have been nothing but constructive.

Do you have semantic problems with the statement "999 + 4 = 1003" ? The left hand side has no thousands units, whereas the right hand side does; the same objection you had to 0.999...=1. Um, no. The left side has a value for its thousands units, and its value is "0". 0 < 1 -- no problem there. Leading zeros are usually truncated for the simple reason that it's a tad cumbersome to preface every single number with an infinite number of zeros. --Lode Runner 05:44, 18 April 2007 (UTC)

Are you saying that 999+4 is not equal to 1003? On the grounds that 999+4 has a value of "0" for it's thousands units, whereas 1003 has a thousands unit value of "1" ? Endomorphic 21:01, 18 April 2007 (UTC)

As indicated many times in the talk page (and implicitly in the FAQ and article), decimal numbers don't have axioms. Numbers have axioms. Decimal numbers, as their representation, have certain properties as a result of what they're representing. But they don't have axioms. If they did, you'd be able to list them, and we'd be able to derive all properties of numbers from them without contradiction. But you can't, because there's no such list. This misunderstanding — or in some cases the refusal to accept this — is one of the main roadblocks to understanding 0.999....

For example, you say that 0.999... violates the concept that changing a '0' to '1' in a digit place increases the number. It doesn't violate this. 0.999... is less than 1.999.... Similarly, 0.999... is greater than 0.98999.... If you want to change multiple digits, however, the story is a bit more complicated, as those introduced to the concept of 0.999... find out. And there's no axiom or property that says that 1.xxx... must be greater than 0.yyy... for all x and y. None. You may have surmised it from the properties of terminating decimals, but it was never a theorem in a math book that applied to general decimals. Because it's untrue for them, just as the number-theoretic result "There is no integer between 0 and 1" ceases to apply if we replace "integer" with "real number."

In the case of semantics and definitions, 0.999... is not an infinitesimal and neither is 1-0.999...; the former value is equal to 1 and thus the latter 0. Also, your argument as to the age of number systems, whether or not it is factual, is fatuous. It would be like saying that because radioactivity was discovered long after Newton, we needn't worry about who gets a hold of nuclear weapons; Newton's properties should be the only ones that matter! Calbaer 06:47, 18 April 2007 (UTC)

Inactivity

Is this page finally dead? It was fun to watch all the trolls and cranks, but I guess this is still a cause for celebration! Quendus 09:30, 3 May 2007 (UTC)

Give it a while and it'll flare up again, regular as clockwork. Maelin (Talk | Contribs) 10:06, 3 May 2007 (UTC)

Modest proposal

Want to argue? http://www.google.fi/search?q=1-0.99999999999999999999999999999999999999999999999999999999999999999999999999999

By the way, don't blame _me_ if I broke the template, blame Wikipedia for blacklisting TinyURL, those little ignorant bastards. Just use the preview subdomain with the TinyURL URL if you don't want your nuts busted. --88.193.241.224 16:16, 6 May 2007 (UTC)

Is that supposed to be some sort of "proof by calculator"? While I find Google's defects in that regard a little surprising, I see no relevance to this article. --Huon 16:22, 6 May 2007 (UTC)

Is it really surprising? Google seems to use ordinary double-precision floating point arithmetic. Asking it for 1-0.999999999999999 is enough. -- Meni Rosenfeld (talk) 08:18, 8 May 2007 (UTC)

JUST GET ON WITH IT

0.999..... doesn't equal 1!

Proof: there is at least 0.000..... (at the end there is a 1) difference between the two!!

GET ON WITH IT! —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs).

You have a very unique argument that has never been raised before in the history of mathematics. Thank you for your useful contribution. Dlong 04:55, 8 May 2007 (UTC)

They would probably argue that 0.000......0001 (with an infinite number of zeros between 0.000... and ...0001) equals 0. Don’t you think? --Van helsing 08:07, 8 May 2007 (UTC)

You have not given any mathematical argument, so I cannot respond with one. Think of this, however: "Infinite" is something which is not finite, that is, has no end. How, then, can there be an infinite number of zeroes and then 1 at "the end"? -- Meni Rosenfeld (talk) 08:15, 8 May 2007 (UTC)

I get the feeling that the contributing value of my statement above converges to zero when I read the rest of this page (for the first time :-)) --Van helsing 08:27, 8 May 2007 (UTC)

Meni, I wish you would stop repeating this "no end to infinity" nonsense. I'm pretty sure you know better and are just looking for a quick way to the end of the argument, but since it's not a valid argument, I'm not sure what you gain. One more time for everyone: There is nothing whatsoever wrong with considering infinitely many zeroes, followed by a one at the end. It would (in the simplest case) have length ω+1, the second smallest infinite ordinal number. That string just does not happen to be the decimal representation of any real number. --Trovatore 18:42, 8 May 2007 (UTC)

Sure, if you're talking about denumerable strings of digits. But we're talking about a real number. Since that "last" 1 has no arithmetic value (although I'm certain someone will say it's 1^-∞), it can't contribute to the final sum of the digits. So, yeah, it's correct that arithmetically there is no end to an infinite sequence/sum of digits. — Loadmaster 17:30, 9 May 2007 (UTC)

I think that's pretty much what I said. There is no problem, formally, with doing infinitely many things, and then doing something else. It just doesn't have any interpretation, in any obvious way, as a real number. --Trovatore 17:34, 9 May 2007 (UTC)

I have no problem to cease using this argument. I would appreciate it, though, if you could explain what, in your opinion, is the best response to comments such as the one at the beginning of the thread. -- Meni Rosenfeld (talk) 19:22, 8 May 2007 (UTC)

How about "The second entry in the FAQ explains why your reasoning is incorrect"? I mean, there's nothing the anon asks for that isn't answered by that entry. Calbaer 20:16, 8 May 2007 (UTC)

It doesn't, because i don't see why if 0.999... was k, there would be a difference between 10^-k. —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs).

Sure, 0.999...9 with k nines is smaller than 1 by a difference of 10^-k. But 0.999... doesn't have just some large but finite number of nines, but infinitely many. 10^-(infinity) is only defined as a limit, and that limit is zero. It does not correspond to some decimal with infinitely many zeroes followed by a "1" - that's not a well-defined (representation of a) real number.--Huon 00:17, 9 May 2007 (UTC)

Responding to Meni -- Calbaer's right that the FAQ entry is pretty good. But I think if we want to be honest with our readers, we need to level with them that it's not really a question of whether their intuitions can be formalized. A great many things can be formalized, but not all are of equal value. So we could emphasize things like the following: It's easy to see that the decimal representation of 1/3, if it's to have one of length ω cannot be anything other than 0.333.... So if the equality 0.999... = 1 is to be avoided, then one of the following must happen:

1. 1/3 does not have any decimal representation (of length ω) at all, or

2. 1/3 has the decimal representation 0.333..., but it is not possible to multiply 0.333... by 3 in the obvious way.

In case (1), we might ask, what is the point of using decimal representations, if they are not capable of representing such a prosaic and commonplace real as 1/3? In case (2), what is the countervailing advantage for interpreting decimal representations in such a way that the obvious multiplication algorithm does not always work?

Also we might point out the radix-specificity of their pleading for special treatment for "terminating" decimals (that is, representations that end in an infinite string of zeroes). Following their reasoning, there is a number infinitesimally different from 1/10, but not from 1/3; in base three it would be totally different. Contrasted with the radix-indifference of the standard approach, this is good evidence that the standard approach better reflects the underlying Platonic ideal object.

I know that this sort of argumentation is unsatisfying to many mathematicians whose conceit it is that they don't deal in metaphysics, but only in formal proof. Personally, I think it's a good object lesson that formal proof is not always enough. Be that as it may, I simply can't accept lying to our readers by saying you can't put anything after an infinite string. --Trovatore 00:34, 9 May 2007 (UTC)

1/3 doesn't equal .333... it doesn't have a decimal representation. If that is to be true, then 0.999... also doesn't have a decimal representation. So 0.999... isn't a real number.

And, if .999... equals 1, then .888... must equal .9, and .0999... must equal .10. Correct? Then that won't be true if the other equations don't work. —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs).

I don't understand what you mean by "0.999... doesn't have a decimal representation". It is a decimal representation by definition. But .8999... equals .9, and once more I don't understand what you mean by "Then that won't be true if the other equations don't work." What other equations? What won't be true, exactly, and why? Yours, Huon 08:32, 9 May 2007 (UTC)

nevermind. i meant that 0.999... isn't 1 if 0.899... isn't 0.9. —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs).

Well, yes, you're right about that, much in the way that water isn't wet if fire isn't hot. --Trovatore 22:04, 9 May 2007 (UTC)

If you know the candle is fire, then the meal was cooked a long time ago --Maelwys 11:14, 10 May 2007 (UTC)

That "unsigned" note is a little strange. It was added by the original aouthor (so why not sign properly, using four tildes, ~~~~), but the editor isnt't even 70.187.134.109, but rather 75.17.175.154. I would ask you to sign your own IP and not some other in the future, or better yet, create an account of your own. Thanks. Huon 22:11, 9 May 2007 (UTC)

Sorry Huon, I just had to access this page on different computers, and each had a different IP adress.

70.187.134.109 04:20, 10 May 2007 (UTC) Another thing. If your school year was forever, and you got 10 zeroes on your first 10 tests, then no matter how many 100%'s you get on all your other tests, your average still isn't 100% Same goes with 0.999... Get it? 70.187.134.109 04:23, 10 May 2007 (UTC)

That school year example is useless because there infinity is divided by infinity, which isn't well-defined, and the natural approach of limits indeed gives an average of 100%. Besides, I see no reason why that average should be 0.999... - if we were to define some notion of average by which the result isn't 100% (automatically making it a non-real number), it should be greater than that for, say, 100 failed tests. Which of those almost-100% averages is supposed to be 99.999...%, and how can I write the others? -- Huon 08:21, 10 May 2007 (UTC)

Keep this simple. Think about this. 1 - 0.9 = 0.1. Correct? Then 1 - 0.99 = 0.01. 1 - 0.999 = 0.001. 1 - 0.9999 = 0.0001. So obviously, 1 - 0.999... is 0.000.....0001! —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs).

First, patterns of symbols can't make for a proof; mathematical axioms, theorems, and other properties can. Second, even if you were to allow for "proof by symbology," you're inserting the ... at the end of the left-hand side and the middle of the right-hand side. Clearly you wouldn't say that 22 = 22 implies that 224 = 242, would you? Order counts. Calbaer 02:47, 11 May 2007 (UTC)

Then if order counts, how is 0.999... equal 0.999...? Certainly 0.999... does equal 0.999... but in your terms, 0.999... also equals 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...999999999999999998 which means that 0 equals 1. if 0.999... equals 1, then 0.999... (8 at the end) equals 0.999... which equals 1. So if you keep going, you would say that 0.999... (7 at the end) equals 0.999... (8 at the end) which equals 0.999...which equals 1. That also means that 0.9 equals 1, and that 0.7 equals 1, and that 0 equals 1! Also, 1 would equal 1000, 0 would equal 1000000000, and -infinity equals infinity! Does 0 equal 1000? NO! So 0.999... does not equal 1.

70.187.134.109 21:19, 13 May 2007 (UTC)

I never said that identical order and/or symbols are necessary for equality; in fact, I've said the exact opposite at several points in the archive and in the second question's explanation in the FAQ. I said that "order counts," that is, the order cannot be ignored in considering equality. You ignore it in your "proof" that 1 – 0.999... is 0.000...0001, juggling the digits and the "...", as Melchoir illustrates below in showing that, according to "symbol math," 0.999...+0.000...0001 should be 0.999...0001. By the way, if you believe that 0.000...0001 represents an actual number, what is one half of 0.000...0001? Calbaer 21:58, 15 May 2007 (UTC)

oh, and on that problem, i mean that 1 - 0.999... equals 0.000..001.

70.187.134.109 21:25, 13 May 2007 (UTC)

Assuming that such things work, wouldn't 0.999... + 0.000...001 = 0.999...001 and not 1? Melchoir 21:46, 13 May 2007 (UTC)

No, because if you add properly, 9 + 1 equals 10, which you carry on infinitly until you get 0.9 + 0.1, which equals 1.

See. If we have 0.999999 + 0.000001, then the answer is 1. 0.000009 + 0.999991 is 1.000000... and so you are incorrect. —The preceding unsigned comment was added by 70.187.134.109 (talk • contribs) 23:53, 13 May 2007 (UTC)

But you never have a 9 and a 1 in the same place, so 9 + 1 never comes up:

  0.999...
+ 0.000...001
-------------
= 0.999...001

Melchoir 02:25, 14 May 2007 (UTC)

Actually, at the end of infinity, you have a nine and a one, not a zero and a one.

The only possible way that your equation can happen is if there's a zero at the end of 0.999... but there isn't. There's a 9.

 0.999...9
+0.000...1
----------
 1.000...0

—The preceding unsigned comment was added by Goldkingtut4 (talk • contribs).

There is no "last 9" (or, in the context of sequences such as the digits of real numbers, "end of infinity"). Think about it: Try dividing 0.999...9 by 10, then adding 0.9. Do you get 0.999...99? What makes that any different from 0.999...9? If it's the same, then 0.999...9=1.0 using the above arguments. If it's different, how would you tell the difference?

In any event, each digit in a valid decimal must correspond to a finite distance from the decimal point, and thus to that number times a positive power of ten. If you insist that the "last 9" is infinitely far from the decimal point, then 0.999...9 isn't a valid real number (unlike 0.999...). Calbaer 02:36, 21 May 2007 (UTC)

If 0.999... equals 1, than 0.999...8 equals 0.999... and 0.999...6 equals 0.999... and so 0.5 equals 0.999... and 0 equals 0.999... and so 0 equals 1 and so -OO equals OO (as OO being infinity) and does 0 equal 1? No, so 0.999... doesn't equal 1.—The preceding unsigned comment was added by Goldkingtut4 (talk • contribs).

I really wish people would read the page before bringing up the same tired objections that have been debunked countless times. Dlong 04:34, 23 May 2007 (UTC)

Well, people can read the page and still not "get it." However, refusals to answer questions about their logic processes seem to be good indications that they don't want to "get it."... Calbaer 05:35, 23 May 2007 (UTC)

Gold, you have indeed derived something absurd, which indicates one or more of your assumptions is incorrect. The faulty premise, however, isn't that 0.999... = 1, it's that 0.999...8 is a meaningful representation of a number. Besides, what's the next number down from 0.999...0?--Trystan 15:17, 23 May 2007 (UTC)

0.8999......

But surely you'd agree that 0.8999... < 0.91 and 0.91 < 0.999...0, right? If 0.8999... is the next number down from 0.999...0, how can there be a number 0.91 between them? This is another contradiction that demonstrates that your assumptions are invalid. Maelin (Talk | Contribs) 04:11, 1 July 2007 (UTC)

so then what's the next number down from 1? also, in response to Huon's {infinity divided by infinity} thing, ok... so what's 1 divided by 0.999...? of course its not 1 because only 1 divided by 1 is 1

There is no "next number down from 1", which especially means that 0.999... is not this "next number". As for the division, since 0.999...=1, we have 1/0.999... = 1/1 = 1. --Huon 08:16, 3 July 2007 (UTC)

fine, what's 0.999... x 10? — [Unsigned comment added by 68.5.252.152 (talk • contribs).]

9.999... (which incidentally, also equals 10, by the same logic that 0.999... = 1) --Maelwys 16:07, 4 July 2007 (UTC)

What is 0.999... plus 0.999... plus 0.999... plus 0.999... (and keep going)???????? endlessly you would end with a whole number —The preceding unsigned comment was added by 75.17.175.154 (talk • contribs) 23:08, July 30, 2007 (UTC).

No, you would end with infinity, just as when adding any other positive number infinitely often. --Huon 03:02, 31 July 2007 (UTC)

Proof by sequences

How valid is proof by the sum to infinity equation?

If we define 0.(9) as the sum to infinity of a series a=0.9 and r=0.1 we get the following

0.(9) = a/(1-r)
      = 0.9/(1-0.1)
      = 0.9/0.9
      = 1

I assume you mean $0.(9)=\sum _{i=0}^{\infty }{\frac {r}{a^{i}}}={\frac {r}{1-a}}$ . Your proof is perfectly valid except for the convergence of the geometric series itself, which should be first-year calculus. Those who doubt the equality will probably either disagree with your definition of 0.999... or will doubt the convergence. We've seen both. Yours, Huon 12:05, 9 May 2007 (UTC)

Thanks for that 222.154.170.110

Just to Show how wrong the above is!

You can make any Number equal 1

0.123/0.123 = 1

I wish people would stop using the above method in their calculations to try and prove any Infinite/Recurring Numbers equal 1

( 1 / 0.9 ) x 0.9 = Infinite/Recurring 0.999... "ONE CALCULATION ONLY PLEASE! ON THE 0.9"

Anthony.R.Brown 24/08/07 —Preceding unsigned comment added by 213.151.217.130 (talk) 11:33, August 24, 2007 (UTC)

Can you rephrase that to make some sense? What is it you object to; the sum for an infinite geometric series? Oli Filth 12:02, 24 August 2007 (UTC)

Unfortunately, Anthony R. Brown is a crank who defines 0.999... as (1/0.9) x 0.9 (as he does above) but disagrees with the law of associativity and doubts (1/0.9) x 0.9 = 0.9/0.9. Arguing with him is rather useless; see Talk:0.999.../Arguments#Trolls and the following sections. --Huon 12:30, 24 August 2007 (UTC)

Two questions

(where ~ means repeating) Does -.9~ = -1 and does 3.9~ = 4 64.236.245.243 18:19, 10 May 2007 (UTC)

Yep! 0.999... = 1, 10.999... = 11, -0.999... = -1, 3.999... = 4, 5.234999... = 5.235, etc. --Maelwys 19:04, 10 May 2007 (UTC)

Random Thought

It's not really an argument so much as it is a question, but here goes.... Now, I could very well be false, but am I to understand that 0.999... is for all mathematical purposes equal to an infinite set or whatever of 0.9*10^(0-n) (aka 0.9+0.09+0.009+0.0009...)

$0.999...=\sum _{n=1}^{\infty }0.9\times 10^{(0-n)}$

Given that, say, in this headline, the number/variable/whatever e is equal to an infinite set of n, where n is the item in the set. Really, it can be anything so long as the sum of items in the set from 1 to n increases as n increases, for any value of n. But I'll go by that definition, since that works quite effectively.

$e=\sum _{n=1}^{\infty }n$

With all that, we now have e, which is every positive integer added together. It's effectively infinite. Say we take 1/e or even 10^(0-e). That would be an infinitesimal.

Now say we subtract 1 by 1/e. What would I get, exactly? Also, what would I get from 1-10^(0-e)? Would 1/e and 1^(0-e) effectively be 0s, like how 0.999... is effectively 1, even though personal cognition says otherwise? Does that mean that 1.000...1 (where the number of 0s in between the decimal point and the second 1 is equal to e; basically formed from 1+10^(0-e)) is equal to 1? I could buy that as being as true as 0.999... is 1. Basically, it's true because infinitesimals don't exist in the real number system, and since the difference between the two numbers is an infinitesimal, there is no real difference, making them the same. But like I said, this is more of a question then any sort of argument, so, I must ask, am I right?

This makes be wonder what would happen if I added 1/e to 0 e times. I suppose that'd be impossible to do though, in the real number system, since if 1/e is 0, then e, being 1/(1/e), would be effectively 1/0, and therefore impossible (but maybe under other number systems, as well as the stream of thought that if division is simply counting how many of one thing is in another thing, it would still effectively be infinity; it just can't, in the real number system, be used as such.) Also, all this means that 0.999...931384917 is equal to 1, so long as the ... represents an infinite number of nines. Haha, it's almost kind of like everything after the ... in 0.999... is tossed into the garbage as far as the real number system is concerned. It kind of reminds me of a purely integer, computer-based number system where 8.5 would happily become 8. Is that the sort of train of thought I have to have to think 0.999...=1? 67.83.72.38 20:24, 6 June 2007 (UTC)

Oh man, but I just realised that I effectively killed the possibility of e existing in the real number system, even though it's a perfectly possible set. So troublesome. I guess infinite sets are typically impossible to use as numbers properly, with a few exceptions. Maybe. 67.83.72.38 20:28, 6 June 2007 (UTC)

You're having difficulting grasping .999... = 1 because, in your mind, there could be stuff after the ... ? Well... that's not the case. If there were other numbers, you'd write the whole thing out, because the "infinite string of 9's" would no longer be infinite when your other numbers begin. .999... means nothing but 9's after the ...

I realize others have said that this is not true all the time in some other number system, and I got scolded last time, but .999...931384917 (or any other numbers after the ...) is not a meaningful string of numbers in the context of the article. Leebo ^T/C 20:37, 6 June 2007 (UTC)

See how that's sort of like with purely integer number systems, where there is no decimal point at all, rendering everything supposedly after it completely nonexistant, as well as meaningless? And out of curiosity, is the article about 0.999... or 0.999... (real number)? If it's SOLELY about the real number, perhaps it should be given a more accurate title, and we could give each encyclopedia-worthy number system its own page for 0.999..., eventually. I don't mean to be mean towards you, it's just that that's a bit buggy. I ought to get off my lazy butt and see if I can find a few sources that use 0.999... to mean a number directly below 1, that way I can easily, without argument to the contrary, add into the article somewhere that 0.999... is sometimes used in less mathematic oriented conversations as a number that is mind numbingly close to 1, but still less then it. 67.83.72.38 20:46, 6 June 2007 (UTC)

I don't think it's rendering anything non-existant or meaningless. There are 9's. Infinitely. We can't write infinite 9's, so we use the repeating notation. It wouldn't be "repeating" if there were other numbers after it, so there's nothing else to render meaningless.

I didn't take what you said as mean. If you find sources, we can discuss them. Leebo ^T/C 20:51, 6 June 2007 (UTC)

There is no end to the nines, correct, but that doesn't mean there can't be something after a mass of unending nines. Well, it does in the real number system, but I DID say that the bunch of numbers after the 0.999... part would be nonexistant. Think about it this way though. If you COULD write an infinite amount of nines, and then you still had writing space left, there's certainly no reason you can't put a 3 at the end, now is there? That, simplified, would end up being 0.999...993, where there are an infinite number of nines.

Indeed, also, in a non-real number system, which allows for infinitesimals (which is just 1/infinity, as I described above), you would need to do nothing more then subtract 1 by an infinitesimal twice to get 0.999...998. The problem with doing that in the real number system is that infinitesimals don't really exist, and therefore, much like how decimals don't exist in a purely integer system, they simply aren't there.

To fully understand my analogy, you have to advance yourself beyond the real number system (although to which, I'm unsure of) where it IS possible to have infinitesimals, and look at the system that lacks infinitesimals and compare it to the system that lacks decimals. Hopefully it'll make sense once that's done. ^_^; 67.83.72.38 21:21, 6 June 2007 (UTC)

The problem is that what you've described, although imaginable, corresponds to no mathematical construct. Assign a value to 0.999...998 and you might as well assign a value to 0.$@)=-^Z~. I know of no number system — with or without infinitesimals — where 0.999...998 corresponds to something meaningful. One of the problems of many people who don't understand 0.999... is to presume that every string they can imagine has a mathematical meaning. But math arises from mathematical operations, not notational imaginations, so this type of representation-centric thinking is a barrier to understanding of mathematics. It may be fun to play with just notions, but it goes against the spirit of mathematics, where notation follows meaning, not vice versa. Calbaer 21:46, 6 June 2007 (UTC)

Look above to where I defined e using a set. Take 10^(0-e). That will be referred to as 0.000...001 (which in the real number system would merely refer to as 0.) This number has an infinite zeroes after the decimal point but before the 1. Subtract 1 by this number twice. Since 1 could be written as 1.000...000, it would therefore become 0.999...998. I thought I already explained that above. In fact, I clearly read an explanation almost exactly the same as the one above. Maybe I shouldn't've taken a question to a page that has Arguments in the name; instead of answers, I've been getting arguments and "no that doesn't work, idiot!"s. Back to the point, if you still don't understand it, then let me point it out to you. ... is there to generally assume it repeats, sometimes indefinitely. Therefore, 0.999...998 repeats the 999 bit until it hits 998. Because the ... could be only a finite number of repeating in that syntax, I suppose it'd be wise to say that the 999 repeats indefinitely. If my mathematical terms are incorrect, then I'd very much like to know, rather then be called an idiot for NOT knowing. Also, I've never really done that. While I do believe, following basic algebra, any string of letters could easily be a variable and therefore can be assigned a number... I don't really buy that things like 0.$@)=-^Z is a number, and I doubt you'll find many people who think otherwise. 67.83.72.38 01:08, 7 June 2007 (UTC)

If you COULD write an infinite amount of nines, and then you still had writing space left, there's certainly no reason you can't put a 3 at the end, now is there? Wait... what? How is there writing space? The 9's don't end. There's never any space left. If you write a 9, you've got to write another. As soon as you write a 3, the 9's weren't infinite. Leebo ^T/C 01:31, 7 June 2007 (UTC)

Uffa, this again. There is nothing whatsoever wrong with writing infinitely many 9s, and then a 3. If you're worried about space, well, just make each successive 9 half the size of the previous one; the sum of their sizes is now a convergent series, so you'll have space after it in which to write the 3. The digits will then have order type ω+1, where ω is the smallest infinite ordinal number.

What you don't get that way, of course, is the decimal representation of any real number. --Trovatore 01:42, 7 June 2007 (UTC)

I think you were the one who told me not to respond that way last time. You guys can take it, I always leave something out and get corrected. Leebo ^T/C 01:57, 7 June 2007 (UTC)

But it's not that you left something out; it's that your argument is completely wrong from the start. There is nothing right about it. So why did you repeat it? --Trovatore 02:11, 7 June 2007 (UTC)

Okay. Wow. You don't have to worry, I'm just going to lurk, no need to pile it on. Leebo ^T/C 03:04, 7 June 2007 (UTC)

I realize I'm being defensive, I guess I had just hoped that pointing out that I'd been corrected in the past (in my statement higher in the thread) would prevent the blunt "You are wrong, don't do it again" response. I'd like to discuss the topic, but it's clear that I don't have the capability. Leebo ^T/C 03:23, 7 June 2007 (UTC)

Sorry, I certainly don't want to come across as authoritarian, that's not the point at all. I get irritated with this particular argument because it essentially denies the validity of my entire academic field (set theory). I'd rather have people believe the wrong thing about 0.999... (which is, after all, not really very important), than have them believe the right thing for an invalid reason, when that invalid reason would block them from understanding transfinite ordinals.

The "nothing after infinity" argument is tempting because it looks like a fast way to shut off debate, but in fact it's just pulling a fast one on the people trying to learn. There are good reasons the real numbers are defined the way they are, but those reasons take much more time than that to explain. The best response, I think, is to explain that we are speaking from the standardly accepted viewpoint, but that if they can make their notions precise, and come up with a body of results as rich and useful as the current theory (or even a thousandth as rich and useful), then that's a perfectly legitimate line of inquiry, and they should by all means pursue it and publish it. It goes without saying that none of the objectors here appears to be anywhere near that path. --Trovatore 04:12, 7 June 2007 (UTC)

I understand what you're saying. If it wasn't clear by now, I'm not really all that familiar with more than simple math, so I really shouldn't be trying to explain things. I should be asking questions. Thanks for taking the time to explain your thought process. Leebo ^T/C 05:19, 7 June 2007 (UTC)

This universe

(Header added by Meni Rosenfeld)

Could it be that 0.999...=1 is linked to the existence of this universe? —The preceding unsigned comment was added by Nikusor665 (talk • contribs).

I doubt it. -- Meni Rosenfeld (talk) 19:39, 8 July 2007 (UTC)

On the other hand, if 0.999... != 1, it could be linked to the destruction of the universe. See The Nine Billion Names of God. --jpgordon^∇∆∇∆ 19:54, 8 July 2007 (UTC)

But could a universe in wich 1=2 , exist? Oops...that`s 2 universes cause 1=2...oh no.. thats 1 universe cause 2=1 .... Well, could it exist? Nikusor665 14:15, 9 July 2007 (UTC)

That's a tough one. That can depend on your mathematical philosophy viewpoint. If you mean "can a universe exist where real numbers aren't that important, but other structures, such as the reals modulo 1 (where 1=2), are very important", I think the answer is a definite yes. If you mean "can a universe exist where defining real numbers like we do here will lead to the conclusion that the real numbers 1 and 2 are equal", I guess the answer is no. -- Meni Rosenfeld (talk) 14:30, 9 July 2007 (UTC)

Ok, than I ask this. Can THIS universe exist if 1=2 ?

Well, 1=2 is a contradiction (assuming classical logic, that we're talking about real numbers, etc.), and since anything follows from a contradiction, if 1=2 then this universe can exist, but also, if 1=2 then this universe cannot exist (hence the name "contradiction"). Some might say I am mixing apples and oranges here, but as you no doubt have gathered, it is hard to give meaningful answers to such questions. -- Meni Rosenfeld (talk) 14:59, 9 July 2007 (UTC)

Is there a simple mathematical way, starting from 1 is not equal to 0.999... to reach a contradiction ? (something like 10 doesnt equal 10) Nikusor665 05:23, 11 July 2007 (UTC)

That depends on the amount of mathematical rigor you want to have. The easiest methods would probably involve something along the lines of the digit manipulation proof:

We have 10*0.999... - 0.999... = 9, so 9*0.999... = 9. If 0.999...is not equal to 1, you can divide by 0.999... (since it's clearly greater than zero) to obtain that 9 doesn't equal 9. Similarly you can turn every proof given in the article into a proof by contradiction. --Huon 05:49, 11 July 2007 (UTC)

But in your argument, dont you already start with 1=0.999... ? Or maybe I dont understand...Nikusor665 06:07, 11 July 2007 (UTC)

Or help me with this .... like in the digit manipulation proof, is there a simple mathematical way, starting from c=1, to reach c=0,999... Nikusor665 06:17, 11 July 2007 (UTC)

No, I don't start with 1=0.999...; we have 10*0.999...=9.999... (multiplication by 10 shifts the decimal separator), and 9.999...-0.999...=9.

Starting with c=1 is less obvious than starting with c=0.999..., because we'd need some way of establishing that a number is equal to 0.999... (compare "9c=9, so c equals 1"). You could use the definition of 0.999... as the limit of the sequence (0.9, 0.99, 0.999, ...) and go on to show that 1 is equal to that limit, but it would seem more natural to write it down the other way round. In general, I believe mathematicians tend to start with something unknown (or not well understood) and show that it's equal to something known (or better understood). --Huon 08:39, 11 July 2007 (UTC)

But dont start with 10*0,999...-0,999...=9 Just start with 1 does not equal 0,9... and so on. But how is it that 0,999... is not well understood ? It seems that you understand it as well as 1.

So if now you understand it as well as 1, just proove it viceversa.

Oh, and dont get me wrong, I`m not on a crusade against 1=0,9...Nikusor665 09:28, 11 July 2007 (UTC)

We can prove that 0.999... = 1 with or without the assumption that

0.999...\neq 1

. The article presents a number of way to prove that. Once this is done, supposing that

0.999...\neq 1

will immediately result in a contradiction, since 0.999... = 1 and

0.999...\neq 1

.

0.999... is well understood by the mathematical community. If you ask why is it not well understood by the general population, this is because decimal expansions, real numbers and mathematics in general are not well understood by the general population. -- Meni Rosenfeld (talk) 11:40, 11 July 2007 (UTC)

Assume that 1 does not equal 0.999...,
then 9 does not equal 9*0.999...,
then 9 does not equal 10*0.999...-0.999...,
then 9 does not equal 9.999...-0.999...,
then 9 does not equal 9.

And admittedly from a purely mathematical perspective 1 and 0.999... are equally well understood, since we know they're equal. But if we want to prove this equality, we'll have to use definitions of 0.999... and of 1. Giving a formal definition for 0.999... that does not a priori involve the equality to 1 will be a little difficult, and probably lead to something along the limit definition I mentioned above. --Huon 13:05, 11 July 2007 (UTC)

Actually the whole problem of 0.999...=1 arises from a restriction in mathematical syntax. With the one in use 0.999... really equals 1, though this just grounds in the mentioned problem and isn't the case. Richh.r 14:17, 17 July 2007 (UTC)

Care to explain more clearly what you meant? -- Meni Rosenfeld (talk) 14:44, 17 July 2007 (UTC)

Does 1/oo = zero?

Does 1/oo = zero ? 79.113.32.66 17:54, 7 August 2007 (UTC)

Well... infinity isn't actually a number. So an expression like that can't exist in mathematics. But, if you have a limit of 1 over some function that a approaches infinity, then yes, the limit approaches 0. i.e. the limit of 1/x as x increases without bound is 0, because 1/x gets closer and closer to 0 as x increases. I hope that answers your question... Gscshoyru 17:59, 7 August 2007 (UTC)

I have to disagree with the claim that "an expression like

{\tfrac {1}{\infty }}=0

cannot exist in mathematics". True, it cannot appear in the context of real numbers, as ∞ is not a real number, but there is nothing stopping us from discussing other structures where infinity or infinite quantities exist. Such is the real projective line, where an object denoted by ∞ exists and, indeed,

{\tfrac {1}{\infty }}=0

. -- Meni Rosenfeld (talk) 18:09, 7 August 2007 (UTC)

Really? I did not know that. Infinity is not a number in the context he was discussing, though, as you said. But thanks for pointing that out, I'm used to teachers explaining that it's a concept, not a number. Gscshoyru 18:13, 7 August 2007 (UTC)

It is not at all clear what was the context of this question (or its relevance to the article, for that matter). It is true that infinity is first and foremost a concept which can have different interpretations in different situations, but again, there is no problem with manifesting it as a concrete object in suitable structures. It also depends on what you mean by "number" - most people think about real numbers or complex numbers, but you can pretty much take any generalization of them and call it a number. -- Meni Rosenfeld (talk) 18:22, 7 August 2007 (UTC)

You may remember me expressing at the refdesk my skepticism of teachers teaching that

{\tfrac {x}{0}}=\infty

. My problem was not so much with the equality itself (again in the real projective line, this is true for any

x\neq 0

) but rather with the obvious fact that the teachers have never heard about the real projective line, so they have no idea what they are teaching, thus giving the impression that ∞ might be a real number.

Note that I am equally dissatisfied with teachers saying that "1/0 is undefined, period". Undefined only means that it has not yet been defined, a situation for which there is often a remedy (extending the definition). -- Meni Rosenfeld (talk) 18:35, 7 August 2007 (UTC)

The problem with this, of course, is the reason why encyclopedia articles are so bad for teaching concepts... when you learn a concept you learn the overall idea, and then the exceptions. But you can't explain things like that in an encyclopedia article. The problem is, the fact that 1/0 is undefined, period, is the general idea of the concept that the students need to know, and they'll worry about the exceptions later. There's an essay on wikipedia somewhere that says what I'm saying in greater detail, though I know not where... in any case, it'd be better if one could say, hey, there's exceptions, but don't worry about them yet.

I wonder if what I just said makes any sense... yay for going off on a tangent! (I edit conflicted, I'm gonna answer the querent, now, discussion later) Gscshoyru 18:56, 7 August 2007 (UTC)

Perfect sense. But there isn't really a need to mention "exceptions", only to specify context. The statement "in the real numbers, division by zero is undefined" is fine; the statements "division by zero is undefined" or "if you divide by zero, the universe implodes" are not. -- Meni Rosenfeld (talk) 19:29, 7 August 2007 (UTC)

Point taken. I see what you're saying. The essay I read used the filling of electron orbitals as an example, though, and it's more understandable in that case than in this... but teachers should always tell the truth, so they should mention the cases in which what they are saying is true is true. Gscshoyru 19:32, 7 August 2007 (UTC)

Well, 1/oo has to be zero, otherwise 0.999... doesnt equal 1. No ?

Explain your reasoning. Gscshoyru 18:56, 7 August 2007 (UTC)

We.., 1/oo must be equal to zero, cause if it would be >0, 1/oo would be a number between 0.999... and 1 . No ? 79.113.32.66 19:05, 7 August 2007 (UTC)

It could also not be a number at all. In the mathematics of real numbers, 1/infinity makes no sense because infinity is not a real number...

Also, .5 is greater than 0, but not between 1 and .999... so I don't understand why 1/infinity would have to be... perhaps you meant 1- 1/infinity? If you did, then a way you could do what you want to do is a limit, and in that case, limit of 1 - 1/x as x goes to infinity is 1, because 1/x goes to 0 as x goes to infinity. Gscshoyru 19:10, 7 August 2007 (UTC)

So 1/oo is .5 ? :)) If 1/oo aint a number, what is it? Isnt it a notation for zero, just as 0.999... is another notation for 1 ?

No, I was pointing out a misunderstanding in what you said... you said that if 1/infinity was greater than 0, then 1/infinity is between 1 and .999... which is untrue. .5, for instance, is greater than 0, but not between .999... and 1. I was pointing out a counterexample to what you said. You probably meant that 1 - 1/infinity would be between the two. Am I right? Gscshoyru 19:19, 7 August 2007 (UTC)

Oh, yeah, sorry, its late here in Romania. Anyway, why isnt 1/oo another notation for zero,just as 0.999... is another notation for 1 ?

Because in normal, regular, average real numbers, oo is not a number. Therefore, in normal, average, regular real numbers, 1/oo is nonsensical, because oo is not a number. Keep in mind in other systems it can be a number, as discussed above, and it is also true that in other systems 1 > .999... But, in the normal, regular, average real numbers, oo is not a number and .999... = 1. Gscshoyru 19:27, 7 August 2007 (UTC)

I would not say in any case that

{\tfrac {1}{\infty }}

is a notation for 0. Rather, it is a notation for 1 divided by infinity. It then depends on your definitions of "1", "divided by", and "infinity". In the real numbers, which is what we should be discussing, there aren't any definitions to deal with this, so the expression is meaningless. -- Meni Rosenfeld (talk) 19:29, 7 August 2007 (UTC)

If infinite is not a number, and 0.999... has an infinite number of 9`s , then what the bip makes 0,999... a number?

The operator "+" is not a number, and yet "3+4" is a number. You can use objects which are not numbers when defining numbers. -- Meni Rosenfeld (talk) 19:32, 7 August 2007 (UTC)

This discussion is incomplete without a mention of structures like the hyperreal numbers, where different magnitudes of infinite quantities exist. -- Meni Rosenfeld (talk) 19:33, 7 August 2007 (UTC)

Well, I can divide 1 to infinity just as you construct 0.999... from an infinite number of 9`s .

In case you are trying to defend you observation that

{\tfrac {1}{\infty }}

is a notation for 0, I will say that I also do not call 0.999... a notation for 1. Rather, 0.999... is a notation for a number which has a decimal expansion with zeroes at every place to the left of the decimal point and 9 at every place to the right. In the context of real numbers (pretty much the only context in which decimal expansions are meaningful), this number happens to be 1. For a simpler example: "3+4" is not a notation for 7, but rather a notation for the sum of the numbers 3 and 4. In the context of integers, this happens to be 7.

This is all just my interpretation though, other people might have different opinions on what can be considered a notation for what. -- Meni Rosenfeld (talk) 20:09, 7 August 2007 (UTC)

Actually there's a significant difference between "infinitely many" and "infinity". There are countably infinitely many nines in 0.999...: There's one for every natural number, ie a bijection between the set of natural numbers and the set of decimal places where 0.999... has a "9". But that definition of "infinitely many" does not make use of oo in any way, and it's well-defined even in contexts where oo is not. --Huon 22:09, 7 August 2007 (UTC)

Well, if there are countably infinitely many 9`s in 0.999... I`ll define 1/oo as 1/(111...)and there are countably infinitely many numbers in the denominator. No?

Or 1/999... if you dont like how 1/111... looks like :) .

Well ?

Interesting way of putting it... there's a difference between a decimal expansion and a really big number...

Look, the limit of 1/x as x increases towards infinity is 0, ok? Your idea of what 1/oo is results in 0. It doesn't exactly work that way, but if we make it work the way you want it to work, yes, 1/oo is 0. Ok? It's just that you need limits to do it. Gscshoyru 10:23, 8 August 2007 (UTC)

Another question. Some guy on a forum defined 999... (infinitely many 9`s) as [sum k=0 to infinite, 9x10^k] . As in 9+90+900+9000 ........ and so on. No ? But every element is a natural number. 9 is natural, 90 is natural etc. No ? But if we add a bunch of natural numbers, shouldnt we get also a natural number ?

Define bunch... a finite sum of finite numbers will give you a real number. And infinite sum of finite numbers may not. Gscshoyru 12:35, 8 August 2007 (UTC)

Formula usable to show that any number equals 1

I don't know about the others, but you can use the first formula to prove that ANY NUMBER is equal to one. Although I'm not mathematically proficient enough to know why this formula is invalid, the counterexample is proof that it is. Pieisangry 00:22, 18 August 2007 (UTC)

Which formula? Oli Filth 00:24, 18 August 2007 (UTC)

Assuming that you mean the first proof in the article, would you care to show how? 137.222.42.77 07:58, 20 August 2007 (UTC)