Talk:0.999.../Arguments/Archive 4

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Why was my section deleted?

(Personal attack deleted by SCZenz 17:13, 9 November 2006 (UTC)) 41.243.17.117 10:41, 9 November 2006 (UTC)

If you'd have a look in the history instead of posting insults and being generally obnoxious, you'd see that it was reverted because you deleted part of the Deep thought section. I dare you to stop being rude.

As a side question to other editors, are we allowed to delete things with this kind of obnoxiousness? Maelin (Talk | Contribs) 13:08, 9 November 2006 (UTC)

Maelin is correct above on all points. However, the "too true" conspiracy theory above his post has formidable comedic merit. Tparameter 15:57, 9 November 2006 (UTC)

(Personal attack deleted by SCZenz 17:13, 9 November 2006 (UTC)) 41.243.52.149 17:07, 9 November 2006 (UTC)

Semantic Objection

I haven't studied anything beyond precalc/trig, so I don't know if any of this can be formalized but I do believe that this is a very compelling semantic argument against 1 = 0.999...

While I'm not sure if it's an axiom per se, a fundemental assumption with our number system is that figures to the left always always ALWAYS represent larger values than figures to the right. For instance, in the number 349,108 the 3 represents 300,000 and is greater than the rest of the number, 4 represents 40,000, which is greater than the following figures, etc.

As a result, you can take any two numbers and truncate them at any arbitrary place (ones, tens, millions, millionths, hundreths, etc.) and if an inequality holds true with the truncated numbers, it will also hold true with the full numbers. I can look at the number 4,***,***,***,***,***,*** and know for a fact that it's smaller (<) than 5,***,***,***,***,***,***. It simply doesn't matter what the missing digits are--this inequality will always be true because an inequality using the leftmost digits will always translate out to the same inequality for the full numbers as well.

Thus, saying 1 = 0.999... is akin to saying 1 = 0. This shouldn't be a difficult concept to grasp, people. I think I learned it first in kindergarten--the "0" in "0.999..." means "there are no ones units." However, the "1" in the lefthand side of the equasion means that there is one "ones" unit. Therefore, just as 1 > 0, it must also be true that 1 > 0.999...

I accept that due to the quirks of the decimal system as applied to surreal numbers, 1 might very well have to equal 0.999... but I feel that this statement violates some pretty basic (and important) rules of arithmetic. 0.999... simply is not a number like 1 is a number. It can't be fully and accurately represented on paper (an infinite number of 9s) and possibly can't be truly comprehended at all by the human mind. It simply doesn't follow the same set of rules that "1" does, so I'd tentatively suggest that it's not entirely *fair* to compare an integer with its surreal equivalent. Much like the imaginary number i, which (IIRC) must be kept separate from the "real" part of the figure, I think that infinitesimals can't be freely mixed with non-infinitesimals without "breaking" some pretty basic assumptions (if not axioms?) of number behavior. Perhaps this is simply a matter of our decimal representation system being inherently flawed/limited, and indeed the wording of the article does imply that perhaps this is the case (1 > 0.999... might hold true in "other number systems.") --Lode Runner

You feel that, because this is contrary to your intuition, it likely violates some basic axioms. But the fact is that it is the basic axioms that lead to 0.999...=1, and trying to invent a number system that agrees with your intuition would violate those axioms (or be inconsistent and thus invalid). It seems contrary to someone who first encounters this without the benefit of knowing all these axioms and how they function. But in mathematics, as in life, some counterintuitive things are in fact true. You likely also learned in kindergarten that the primary colors were red, yellow, and blue; that Columbus was the first white man to visit North America; and that distance is not a quantity that changes. The correctness of 0.999...=1 can be formally demonstrated to someone with enough education (or informally to someone with a basic knowledge). That's the purpose of the page. Look at the arguments of those who don't believe this, and I think you'll find they all boil down to, "It goes against my intuition, so it's wrong." Clearly, were it really wrong, there would be more rigorous arguments against it. Calbaer 03:03, 7 November 2006 (UTC)

Lode, I feel your pain. You gotta understand the people who wrote this article and so passionately defend this 0.999...=1, are doing so under the rules of "real analysis." To them, infinite summations equal their limits, the infinitesimal is equal to zero, and multiple numbers on the number line can all represent the same number. I completely disagree, and I wish they made it clearer that these assumptions aren't true for all of mathematics or reality. But then, it would take away the need for debate, and take away their fun and shock factor. --JohnLattier 01:24, 7 November 2006 (UTC)

What "assumptions" are there? The only assumptions I can think of are axioms and definitions, and mathematicians generally agree on what those are. What part of mathematics uses different assumptions that give these results inconsistent with the rest of mathematics, and which assumptions are different? -- Schapel 02:22, 7 November 2006 (UTC)

Whilst that "truncate and compare" system works perfectly well for the integers, it doesn't work here. Unfortunately, 0.999... is the only counterexample I'm aware of, and that's obviously not going to be very convincing to you. However, I can dispute your complaints that 0.999... and 1 aren't the same "kind" of number. It's true that 1 is much more elegant, written down, than 0.999..., but it doesn't mean that 0.999... is some kind of "flawed" number. It has an infinitely long decimal expansion, and you can't write it on paper. But being able to write a number on paper is certainly not a requirement for being a legitimate number. In fact, there are finite natural numbers that can't be written on paper, because they're just too big (as an example, see Graham's number). There are other real numbers that can't be written on paper, because they're infinitely long, just like 0.999.... No irrational number (such as e, or π, or sqrt(2)) can be written on paper as a decimal expansion. Does this mean they're "bad" numbers, or beyond the realm of human understanding? Not at all.

As a side note, since infinitesimals don't exist in the real numbers, if we admit them, we are no longer discussing the real numbers. Maelin 01:07, 26 October 2006 (UTC)

Quite right. And the notion of the infinitesimal, while it may seem strange is more intuitive than the notion that 0.999...=1, so perhaps we shouldn't be describing the real numbers. (idiocy by me reverted)

I would of course argue that while the numbers are bad, the entire concept of infinity is beyond (possibly just slightly) the realm of human understanding. If it were inside the realm of typical numbers then we would be able to define the complete set of operations as we can for other numbers, which we can conceptualise intuitively. Ansell 01:15, 26 October 2006 (UTC)

You said: "I haven't studied anything beyond precalc/trig, so I don't know if any of this can be formalized but I do believe that this is a very compelling semantic argument against 1 = 0.999... While I'm not sure..." ----- In this short intro, you mentioned that you "don't know", and you're "not sure", and you also mentioned that you "do believe". Please, this is not a guessing game or a matter of faith or belief. Math has rigorous rules, axioms, definitions, and so forth. In math, something is true, false, or unknown. This particular topic is a definition, which is not something that can be researched or questioned. It is simply a definition. 0.999... is the limit of 1-1/(10^n) as n goes to infinity. This limit converges, and is exactly equal to 1. Just as a=b and b=c allows us to conclude that a=c. So, it goes like this:

$0.999\dots =\lim _{n\to \infty }(1-{\frac {1}{10^{n}}})$ (by definition).

$\lim _{n\to \infty }(1-{\frac {1}{10^{n}}})=1$ (trivial result).

Hence, $0.999\dots =1$. End of story.-- tparameter Nov 6, 2006

A minor correction. The definition of $0.999\dots$ isn't actually $0.999\dots =\lim _{n\to \infty }(1-{\frac {1}{10^{n}}})$. It's $0.999\dots =\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}$. To reach the line you suggested, you would show that $\sum _{k=1}^{n}{\frac {9}{10^{k}}}=1-{\frac {1}{10^{n}}}$ and then use substitution. Dugwiki 20:28, 6 November 2006 (UTC)

You say "potayto", I say "potahto". Are you sure that you would classify this as a "correction", minor or otherwise? I guess I might ask, would you similarly correct:

$1-{\frac {1}{10}}$ to be: ${\frac {9}{10}}.$

Or, would you characterize them as identically equivalent? Tparameter 00:01, 7 November 2006 (UTC)

Yes, it's a correction. Let me put it this way, would you define the expression "0.29" as "the number which equals 1 - 0.71"? No, of course not. It's defined as ${\frac {2}{10}}+{\frac {9}{100}}$, or equivalently, as ${\frac {29}{100}}$. Likewise, more generally, a decimal number $0.a_{1}a_{2}a_{3}\dots$ is defined as $\sum _{k=1}^{\infty }{\frac {a_{k}}{10^{k}}}$, or equivalently as $\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {a_{k}}{10^{k}}}$.

So the earlier line that says "$0.999\dots =\lim _{n\to \infty }(1-{\frac {1}{10^{n}}})$ (by definition)" is incorrect. That's not the definition. It turns out to be numerically equal to the definition, but that requires an additional (fairly easy) arithmetic step to prove. Dugwiki 21:26, 6 November 2006 (UTC)

I hate doing this, and please don't take me too seriously, but let me be clear. Are you saying that, using your example, it is incorrect on the one hand to say:

0.29 := 1 - 0.71

And a suitable "correction" would be to instead say:

0.29 := 0.2 + 0.9

Am I right, or am I right? To me, these are equivalent. Nevertheless, in the original definition that I used that warranted your "correction", the term dealing with infinity was simplified as much as possible, thereby giving what I have seen as the most intuitive definition of this particular topic.Tparameter 00:01, 7 November 2006 (UTC)

I'm saying that what you claimed is the "definition" of how to calculate the symbol 0.999... was not actually the definition of that symbol. It was a seperate formula that happens to have the same numerical result as the definition of what the decimal expression 0.999... means. As an analogy, in the above example, while the equation 0.29 = 1 - 0.71 is true, it is not technically correct to say that "0.29 is defined as that number which is 1 - 0.71". The actual definition of what is meant by a decimal expression is that it is the sum of the decimal numeral components times their corresponding powers of the decimal base. Thus 0.29 is defined as "2/10 + 9/100", or alternatively as "29/100". Using that definition, you can then arithmetically prove that 29/100 = 1 - 71/100 (arithmetic step) = 1 - 0.71 (substitution using the definition of 0.71). And more generally, for infinite decimal expressions, it represents the infinite sum of those components.

To see why this might be technically important, consider if we were talking about the repeating decimal 0.123412341234... . In order to calculate a fractional equivalent, you would want to start by using the definition of what that symbol represents, namely that it is the infinite sum 1/10 + 2/10^2 + 3/10^3+ 4/10^4 + 1/10^5 + 2/10^6 +..., which is equivalent to 1234/10^4 + 1234/10^8 + 1234/10^12 +.... You wouldn't claim that "by definition" 0.123412341234... = 1 - 0.876587658765... , even though the equation is arithmetically true.

It's kind of a subtle point, but hopefully that explains what I'm getting at. Dugwiki 16:02, 7 November 2006 (UTC)

Thanks for clearing up what you meant. We will, however, have to agree to disagree on this matter. It would be fun to discuss this topic further, but this is not the right forum. As a (hopefully) final word, I think you created a straw man when you used the following phrase:

'what you claimed is the "definition" of how to calculate the symbol 0.999...'

As a matter of fact, I never mentioned calculating anything. Hence, we are discussing two completely different topics. This incorrect assumption is probably the source of your disagreement. Beyond that, we might discuss what constitutes a mathematical definition. In fact, this would be a great article!Tparameter 15:23, 8 November 2006 (UTC)

As the final, final (hopefully) word, the line I objected to was:

$0.999\dots =\lim _{n\to \infty }(1-{\frac {1}{10^{n}}})$ (by definition)

My objection lies that you appeared to be justifying that equation as true "by definition of how the symbol 0.999... is calculated". But the definition differs from the equation provided, and requires some slight arithmetic to match it. Now if you meant something else by the phrase "by definition", my mistake. But usually that phrase means exactly what it says - that line of the proof matches the definition of a term involved. So I'm not saying the equation is untrue. I'm saying you listed the wrong justification. Dugwiki 16:02, 8 November 2006 (UTC)

In my proof, 0.999... is not "calculated" at all, contrary to your assumption. Instead, we evaluate the limit. This is an important distinction that you seem to be missing. 0.999... was simply defined. Let me lay out my logic again in generic terms:

1. a = b (defining a)

2. evaluate b, yielding b = c.

3. since a = b by definition, and since b = c, we conclude that a = c.

With regard to the subject of what constitutes a mathematical definition, again, let's agree to disagree.Tparameter 21:32, 8 November 2006 (UTC)

you're just using equations to represent numbers. --JohnLattier 01:27, 7 November 2006 (UTC)

It's interesting, John, that given the choice between

• Overcoming your initial intuition, and accepting the equality of 0.999... = 1; an equality with virtually no real world application at all; or
• Denouncing the entire mathematical branch of real analysis; a branch that is criticially important to engineers and physicists everywhere, a branch that has deep and fundamental applications in the creation of computers, the internet, automobiles, and is crucial in our understanding of mechanics, electrodynamics, quantum mechanics; denouncing this as "not consistent with reality",

you choose the claim that every mathematician since the 17th century has been using a "phony" system that is "wrong". Real analysis is a cornerstone of modern physics and engineering, but you prefer to dismiss it, and claim you know better.

There is no point in discussing this with you, John. You don't want to listen. You don't want to learn. You don't answer our questions, you ignore our attempts to help you understand things. Instead of working to learn of the techniques involved, you accuse us of using some kind of "sheister mathematics" to pull the wool over peoples' eyes. If you don't want to understand, what are you even doing here? Maelin (Talk | Contribs) 03:12, 7 November 2006 (UTC)

No need to continue with the Real Analysis discussion. Plain and simple, the guy doesn't understand limits. I hereby prescribe one semester of precalc followed by one semester of calc I. If that doesn't work, there's no hope.Tparameter 06:14, 7 November 2006 (UTC)

Tell me, what reasoning is it that real analysis redefines the infinitesimal as finitely equal to 0? If it's because the only number when divided by 2 still equals itself is 0, and the infinitesimal divided by 2 is itself? That's just not going to cut it. The infinitesimal is a legitimate concept, whereas dividing an infinite fraction by 2 doesn't mean anything. We must first find, or create, a set of rules for manipulating the infinitesimal/infinite, and only then can we attempt to build an entire mathematics system upon it. I am afraid we will find that the founders and proponants of real analysis actually think the infinitesimal can be divided by 2.

Tpara if you understand limits so well, graph the simple function y=1/x and tell me what value of x returns y=0. Exactly. Just because the limit as x->infinity is 0, that doesn't mean 0 is a legitimate value for y. Although your pals would argue it is. Same goes for limits of infinite summations. --JohnLattier 06:23, 7 November 2006 (UTC)

You're talking apples and oranges my friend. Nevertheless, I will graph your function - but, I will graph it on the z-plane. Using your example of y=1/x, I can easily conclude that the value of x s.t. y = 0 is x=1/0=infinity. Since you didn't define the space to graph this result, I will choose to use a stereographic projection on the Riemann sphere. Hence, the point x is mapped to the North Pole when y is sent to 0. The Riemann Sphere is topologically equivalent to the xy-plane with infinity, which is represented on the sphere as the North Pole. That's right, there is a point infinity, and we use the Riemann sphere to show this in Complex Analysis. Every line in every possible direction on the xy-plane goes to infinity - which is one and the same point on the z-plane. Basically, the infinite xy-plane and any finite sphere map 1-1 and onto, a bijection. And guess what? The results from this analysis model reality. A Stream Function would be a great example. You need to understand that there are different ways to look at things in analysis. The fact that you don't understand the math topics that you're discussing is not evidence that they are not true or real. Instead, it is just evidence that you are not a mathematician. That's okay. But, you should be more humble on these topics IMO, because you really don't know what you're talking about. You need to know that the results we're discussing in analysis give EXACT answers, and these results model reality as precise as we wish, limited only by constraints on time. All of science is built on these facts. Instead of arguing on topics you don't understand, why don't you study math and find out why these things are true? Tparameter 14:29, 8 November 2006 (UTC)

My suggestion to you, John, is that you accept that these things are true in math. And then simply advocate that math doesn't model reality. Then you could find peace knowing that 0.999... is simply a mathematical construct.Tparameter 14:29, 8 November 2006 (UTC)

As usual, John, your ill-defined ideas result in misunderstandings. No finite value for x returns y = 0. This is mirrored in the fact that no finite number of nines in a 0.9999999 style number results in 1. This article is not discussing a number with a finite number of nines.

By the way, John, would you like to go to all the engineers who were responsible for your computer and tell them that the mathematical tools they used to do it are wrong? How about visiting the guys at NASA who landed a man on the moon, and telling them that the real analysis techniques they utilised are rubbish? Nonzero infinitesimals are a legitimate concept, in the mathematical systems in which they are properly defined, which do not include the real numbers. And yet the real numbers and their associated analysis techniques are exceptionally useful to engineers and physicists and heavily studied by mathematicians. You need to accept this, crying that they are "invalid" and "contradictory" is pointless and foolish. Maelin (Talk | Contribs) 06:38, 7 November 2006 (UTC)

Just to address JL's statement directly, you are correct that there does not exist a finite number x such that ${\frac {1}{x}}=0$, even though the $\lim _{x\rightarrow \infty }{\frac {1}{x}}=0$. However, normally, in the reals the symbol ${\frac {1}{\infty }}$ is defined to mean $\lim _{x\rightarrow \infty }{\frac {1}{x}}$ if that limit exists. Thus under that definition of that symbol you could correctly say that ${\frac {1}{\infty }}=0$. Now in a different number system, such as the hyperreals, the definition of that symbol might differ. But for purposes of the 0.999... article the standard real number definition would apply.

Likewise, under the reals, the symbol $\sum _{m=1}^{\infty }$ is normally defined as $\lim _{k\rightarrow \infty }\sum _{m=1}^{k}$ if it exists. Again, under something like the hyperreal number system, that definition might differ, but for purposes of the reals and this article it would apply. Dugwiki 16:26, 7 November 2006 (UTC)

your loophole

Moved from main talk page. --Brad Beattie ^(talk) 02:22, 26 October 2006 (UTC)

1/3 = 0.333... TRUE. 1/3 * 3 = 1 TRUE. 0.333... * 3 = 1 TRUE. 0.333... * 3 = 0.999... FALSE (the mistake in logic).

As far into the 0.333... as you want to go there will always be a bit more. That little bit more is what theoretically adds up thrice to produce the 1/infinity that is needed to push 0.999... to 1. However it can never be reached. Such is infinity. I'm worried people don't understand the concept of infinity. You can't just blindly add, subtract, divide, or multiply without consequence, and such is the outcome of any of these "simple" proofs.

Yes you can. These operations are justified most easily by the infinite series formalism. Vague statements like "little bit more" and "never be reached" don't fly in mathematics. Melchoir 22:07, 25 October 2006 (UTC)

0.999...=1 becuse each 9 brings the number closer to 1 each time, until it brings it infinitely close to 1, since the 9's go for infinity. Infinity is the largest possible number tht goes on forever. To put it this way, if you could count from 1 to infinity, it would take you one eternity to do so. It's the same concept with 0.00000...1=0 and 0.19999999999999...=0.2. I guess you could say there is an infinite number of numbers like this. AstroHurricane001 22:19, 25 October 2006 (UTC)

This "little bit more" isn't my responsibility to account for. It's the common trick used by champions of the 0.999...=1 camp. They change the number of significant digits which truncates a "little" piece that they never acknowledge. It's so simple to see through, I'm amazed people don't get it. Blizzard Entertainment, for instance, made the bold statement that if x = 0.9999... then 10x = 9.9999... But that's not true. They are not accounting for the infinitely last sig digit.

And there is no last digit. Read the article or get off the talk page. Melchoir 22:46, 25 October 2006 (UTC)

Read the article? I did, and it's chock full of holes and misunderstandings. Like this sentence (note the use of the word "students"): "Some students interpret "0.999…" (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity."" If you can't understand the concept of a "last digit" that is never reached, then you don't understand the concept of infinity. Those that say there is no 0.000...1 are denying the very zero dimensional unit by which our entire mathematical system is based. I'm sorry for them, even though it's not all that important in the end.

Yeah, that material is cited to articles in Mathematics Education Research Journal and Mathematics Teaching. The later treatment of last digits is due to books published by Dover and Oxford University Press. I trust your theories about who doesn't understand what are reflected in similar literature? Melchoir 23:10, 25 October 2006 (UTC)

Eh, I looked at the article, and it does absolutely nothing for the debate. Even still, you could find great, famous professors who are strong proponents of 0.999...=1, or the Big Bang being the beginning of space/time, or that fire has 1 syllable, or that classic tennis balls are green. That doesn't make it fact. Someday I'll sit down and research this, but in the mean time I have 2837492873 biology research articles to sift through and 0.000...1 days to do it.

Oh, if you wanted a debate, just enter the relevant search term into Google and you'll get all kinds of incoherent shouting. This is an encyclopedia article. Melchoir 23:42, 25 October 2006 (UTC)

If this illogical article is passed off as fact, it will confuse and misinform the masses. Is there a way to remove it or place it in a flawed logic category?

Yes, you can go find a reliable source that disputes a claim made in the article. When you bring that source to our attention, it will be integrated into the article in a manner consistent with WP:NPOV. If you have no sources, then you have no power here. This is an encyclopedia, not a court of truth. Melchoir 23:54, 25 October 2006 (UTC)

Of course, even if this was a court of truth, the article would stay, since it is true. It's inappropriate of you to project your own misunderstanding of mathematics in general, and decimal expansions in particular, onto others who happen to actually know what they are talking about. Read the article and this arguments page. All the answers are found within. -- Meni Rosenfeld (talk) 09:18, 26 October 2006 (UTC)

0.333...+0.666... isn't 0.999...

Moved from main talk page. --Brad Beattie ^(talk) 02:23, 26 October 2006 (UTC)

0.333...+0.666... is 1, not 0.999... There is always another, smaller 0.0...3 and a smaller 0.0...6 unaccounted for that, together, break towards 1/inf that ends up giving 1. - JohnLattier

"Always"? Melchoir 01:58, 26 October 2006 (UTC)

Yes. People throw the 0.333...+0.666... = 0.999... into the debate, or something of the same logic, all the time, but have no reason for doing so other than the flawed logic that if 0.3+0.6=0.9, it is the same for every value in the infinite set. But that is not the case, as number of approaches tends towards infinity, the amount unaccounted for approaches 1/infinity at the same rate. This 1/infinity cannot be truncated to give 0.999...

Um, I'm no math expert, but if I start with 0.3 and only add digits to the right of the three, I'm not approaching infinity; I'm approaching 0.4 - aren't I? --Badger151 02:05, 26 October 2006 (UTC)

Well, if you add infinite 9's you approach .4, but if you add infinite 3's you reach 1/3. Note the difference in "approach" versus "reach." Although, who has time to add infinite digits?

"Every value in the infinite set"? "Number of approaches tends towards infinity"? What are you talking about? Melchoir 02:08, 26 October 2006 (UTC)

Every value in the infinite set = every digit/value/number after the decimal, in the infinite/neverending set/array/collection of digits/values/numbers. Number of approaches tends toward infinity = as you move infinitely further from the decimal, infinitely closer to the infinitely small

And where, in the definition of the addition of two decimals, do those concepts arise? Melchoir 02:19, 26 October 2006 (UTC)

0.999... "approaching" 1

Just a little note to help clarify things for people who (frequently) try to say things like "0.999... approaches 1, but doesn't reach it", this is actually impossible. 0.999... is a single number, not a process, sequence, or function, and so you can't say anything about it "approaching" anything. What you may be thinking of is the sequence (0.9, 0.99, 0.999, 0.9999, ...), which certainly does have a limit, and so may be said to "approach" 1. Of course, it is also easy to see that it "approaches" 0.999..., and if you assume that any convergent sequence must have a unique limit (unique being the keyword), then the two things the sequence approaches must be equal. So remember, kids, 0.999... doesn't "approach" 1 because it isn't moving, and poor wording can be your downfall. Confusing Manifestation 02:33, 26 October 2006 (UTC)

Excellent! I love it. Very true. The laypeople see this exact value as a process, which I suppose shows that they have an intuitive sense for sequences - but, little formal knowledge. Nov 4, 2006

The point people are trying to make when they say "0.999... is approaching 1" is proving that 0.999 does not equal 1 in the case of limits. Take f(x)=(x-1)(x+3)/(x-1) for example. The graph looks like the function y=x+3, but there is a hole in the graph at x=1. The limit as x approaches 1 is 4, but x cannot equal 1. Coming from the left side, x can equal as infinitely close to 1 as it wants, but it cannot equal because the function is undefined there. 0.999... gets infinitely closer to 1, but in this case, does not and cannot equal 1. I know the number 0.999... itself is not a function and therefore does not have a limit, but in the case of limits, 0.999 cannot equal 1. Unsigned.

Wrong, 0.999... IS DEFINED as a limit. 0.999... = 1 - 1/(10^n) as n goes to infinity. This is a limit. This limit equals exactly 1. B Nov 5, 2006

0.333... < 1/3

0.333... is less than 1/3 because every time you add another 3, the decimal only gets 10 times closer, but never really hits the fraction of 1/3. The difference between the two numbers gets infinitely small, but the numbers are not the same.

You're thinking of a finite number of threes, with a final three at the end. The value 0.333... does not have an end, so it is exactly 1/3. Read the article; this misconception is explained. -- Schapel 03:01, 26 October 2006 (UTC)

Exactly?

Moved from Talk:0.999... Supadawg (talk • contribs) 02:54, 26 October 2006 (UTC)

It's not. Because 0.3rec DOESN'T equal 1/3. It's infinite, it does not equal it. You cannot write thirds or ninths in decimal form. They say "OMG DO LONG DIVISION!!!" but it's still infinite. Infinite numbers cannot have answers. Beelzebud 22:35, 25 October 2006 (UTC)

The number is .000000...01 off

this article is a failure

0.999... dose not equal 1 (except in pure maths) 0.999... Tends towards 1.

Any moron with any grasp of reality knows that 0.999... != 1 for the simple fact that 1 is an integer, and 0.999... is not. Thus, morons have a stronger grasp of reality than mathemeticians that claim "0.999... = 1" --207.72.2.150 23:19, 25 October 2006 (UTC)

And you have shown by example that at least one moron believes that 0.999... has a reality other than 1. Nov 4, 2006()

Well, that actually is one legitimate point of view—not all mathematicians consider "integer reals" to be identical to their corresponding integers. Looking at it that way, it's true that 0.999... is not equal to 1, but it's also true that 1.000... is not equal to 1, and it's still true that 0.999... equals 1.000... . --Trovatore 03:36, 26 October 2006 (UTC)

<sarcasm> any moron with any grasp of reality knows that 6 / 3 != 2, since 2 is an integer and 6/3 is a fraction.</sarcasm> -- Meni Rosenfeld (talk) 09:25, 26 October 2006 (UTC)

Difference

Moved from Talk:0.999... Supadawg (talk • contribs) 03:18, 26 October 2006 (UTC)

1 - 0.999... can only equal 0.000... which is clearly equal to 0. Therefore they're the same. —Ashley Y 01:01, 26 October 2006 (UTC)

1-0.9 = 0.1, 1-0.99 = 0.01, 1-0.999 = 0.001. 1-0.999... = 0.000...1 It's the same as 1/infinity. It's the infinitely small value. It's the number "next door" to zero on the number line as the number line is made of an infinite number of 1/inf units. But it's not zero. The day mankind defined the decimal system, mankind gave birth to 1/inf. It is not derived. It is the foundation.

Where are you getting this stuff? Melchoir 01:13, 26 October 2006 (UTC)

Though it is possible to construct a consistent number system (see non-standard analysis) which includes infinitely small values, this article is set in the context of the standard real numbers, where the only non-negative number smaller than all 10^-i for all i is 0. --Macrakis 01:28, 26 October 2006 (UTC)

This article actually sits in the Extended real number line as it relies on infinity to define the length of the number. Or is the concept of infinity defines outside of the number line concept? Ansell 01:35, 26 October 2006 (UTC)

An 'infinite sequence' is simply a sequence that is longer than any particular number. Limits are similarly defined without an object called 'infinity'. The infinity symbol in things like infinite sums and so on is a notational convenience, and does not commit you to an object called 'infinity'. That said, it is (as I mention above) possible to define number systems which do include infinities and infinitesimals. --Macrakis 01:48, 26 October 2006 (UTC)

OK, this is some kind of weird folk mathematics, but let's go with it. Suppose x = 1 - 0.999... = 0.000...1 = 1/inf. Then we have x * 10 = 0.000...10 = 0.000..1 = x. Therefore 10*x = x, therefore x = 0. —Ashley Y 01:44, 26 October 2006 (UTC)

x*10 would just have one less zero before the assumed 1 at the assumed end. 10x wouldn't equal x, it would equal 10x. You can't just throw around significant digits like that without accounting for them.

from above:"1-0.9 = 0.1, 1-0.99 = 0.01, 1-0.999 = 0.001. 1-0.999... = 0.000...1" - The first three examples establish a pattern that every time we add another nine to the right of the decimal point, we add another zero between the decimal point and the one in the answer. Why, then, in the fourth example is this pattern broken? Further, How is 0.0000...0001 the number "next to zero", given that I could just add another zero to the series, creating a number that's even closer to zero? --Badger151 02:24, 26 October 2006 (UTC)

the "..." is just used to mean an infinite repeat. i guess i could've said 1-0.999...=0.0000...1 but its the same thing. adding more zeroes to the infinite set of zeroes does nothing for the value of the 1/inf. BUT a lot of people use this fact in a trick. They add a significant digit, manipulate it without accounting for /inf's, then take the significant digit away. That's not mathematically sound.

But doesn't an elipsis merely represent what isn't written? When writing out an infinite decimal, we establish the pattern and then add the elipsis, enabling the brain to fill in what the elipsis represents - and by placing the elipsis at the trailing edge of the number, we understand that the established pattern continues without end. But in placing the elipsis in the middle of the number, I think that you imply a termination to the decimal: nothing may come after this digit. The broken pattern I'm seeing is moving the elipsis from the right side of the decimal to its middle. What is more, the numeral 1 in the answer is always found in the terminal decimal place - that slot farthest to the right. The location of that 1 is, in fact, determined by the location of the rightmost 9 in the original equation; as soon as we add another 9 to the decimal in the original expression, we push the 1 in the answer over one more place to the right to reflect the position of the rightmost 9 in the original equation. 0.999... has no rightmost 9, though, and so I think that the answer to 1 - 0.999... is not 0.000...0001, but is 0.000... - with no end and therefor no 1. --Badger151 03:26, 28 October 2006 (UTC)

Obviously, in the reals, 1 - 0.999... = 0.000... . However, the notation 0.000...1 is ambiguous, and can mean either of:

• a finite, unspecified number of zeros, and then 1;
• infinitely many zeroes, and then 1 in position ω.

Of course, the latter only has any chance to be meaningful in a structure completely different from the real numbers; and in such a structure, it just might be possible for 1 - 0.999... to be equal to it. -- Meni Rosenfeld (talk) 07:04, 28 October 2006 (UTC)

Failed Logic

Moved from Talk:0.999... Supadawg (talk • contribs) 03:18, 26 October 2006 (UTC)

1/3 does NOT exactly equal 0.33333....., just like 0.9999...does NOT equal one. Think about it- no one can prove that 0.333..... does. Also, in the second proof, the last digit would not equal 9 as multiplying any number by ten will end up with the last digit as zero. Also, I fail to see any proof that many textbooks support this idea. 0.9999999999..... of a meter does not equal one meter. You cannot prove otherwise, as to do so you must find its last digit, which is impossible.

Oy. 0.333… is easily proved equal to 1/3 by summing the implied geometric series. There are no last digits to multiply to zero. And all of the cited references support the article; that's why they're cited. There are plenty of textbooks among them. Melchoir 00:44, 26 October 2006 (UTC)

Step 2???

Moved from Talk:0.999... Supadawg (talk • contribs) 03:18, 26 October 2006 (UTC)

I'm no mathematician, but if you can't tell me that .9999-> of an apple is the same as 1 apple, then I'm not going to accept that .9999-> = 1.

.9999-> of an apple is the same as 1 apple. Melchoir 22:01, 25 October 2006 (UTC)

No!!

Moved from Talk:0.999... Supadawg (talk • contribs) 03:18, 26 October 2006 (UTC)

How can this article be a featured article when it is obviously so controversial. You could travel 90% of the way to a destination, and then travel 90% of the remaining distance. You could keep traveling 90% again and again, but you would never hit the exact point of the destination, which is infinitely small. This should have never been featured, especuially because I just proved it incorrect!!

I do this everyday. Everytime I take a step, there's a point where I take 90% of my step, 99% of my step, 99.9% and so on. Eventually I complete my step, and I've covered every point in between (there's infinite points in my path). "Infinitely small" for most purposes means 0. A point is infinitely small, and occupies no space. Falsedef 05:10, 26 October 2006 (UTC)

No. Argyrios 03:46, 26 October 2006 (UTC)

"...researchers have studied the reception of this equation among students, who often reject the equality. The students' reasoning is often based on an expectation... that 0.999… should have a last 9." 153.104.208.67 04:42, 26 October 2006 (UTC)

Question

It seems to me that all of the criticisms of this concept are coming from the fallacy that the number .999... actually corresponds to some physical quantity that can be measured.

Imagine cutting a pie into perfect thirds. You can pick up 1/3 of the pie, but does that really equal .333...? No, it doesn't. Because in the real world your fraction would inevitably end. And even if you did have three pieces that were each .333...%, wouldn't that still leave .000...1% of the pie left over? The same goes for driving 99.999...% of the way to a destination or having 1/999... of an apple. It doesn't make sense in the real world.

But, that said, try doing math without the concept of .333... being equal to 1/3. Without the ability to round a number off, math would be impossible. So this question is a bit like a Zen Koan. It's not meant to teach us a fact or produce an answer, but it's a gateway to thinking in a new way. If you always try to think of digits as representing real quantities, higher level math would be impossible. Even though it is impossible to define it in real-world terms, the math still makes sense. The real lesson, therefore, is for the math student to trust his paper work rather than try to relate it to an actual, physical object.

My question, for you mathematicians, is this: Is the above interpretation correct? Or has this problem driven me insane?Jboyler 04:03, 26 October 2006 (UTC)

If a pie has 3 trillion atoms, and I was able to subdivide it into 3 pieces of 1 trillion atoms, then yes I have 3 pieces of pie that are 0.333... the size of the full pie. If you have 3 apples, and I take 1 away, you'd have 2/3 apples left, or exactly 0.666... In the real world, repeating decimals don't magically end. Decimals are just representations of real numbers, and sometimes whole numbers. Mathematics is theory that can represent the real world.

I drive 99.999...% to a destination every time I take a ride in my car. There are infinite points inbetween you and your destination. At some point, you'll cross every percentage of that distance possible when you finally arrive at your destination. Your destination can be represented as 100% of the distance, 100.00...% or 99.999...%.Falsedef 05:24, 26 October 2006 (UTC)

"There are infinite points inbetween you and your destination."

Not unless you are driving to a location that is infinitely far away. Otherwise there is a finite number of points between Kansas City and New York City, just because you can't count that high, or we don't have a way of describing the quantity of those points, doesn't mean that there is an infinite value. I can assure you there are more points between New York City and Seattle that there are from NYC and KC. Neither have infinite points because there are extremely close in comparison to most points within the galaxy.

71.234.110.209 08:11, 26 October 2006 (UTC)

Any two points that are not equal will have points inbetween. This is a concept most people find hard to understand, and one you have a hard time understanding. A point does not take up space. It doesn't matter the distance between two points (greater than 0), whether it's 1 mile or the size of the Milky Way, they all have infinite points inbetween. That's how lines work. It doesnt' amtter if you can "measure" it, since I don't need to measure it to know I've passed it.

When I travel over 1 mile, I know that I've passed 50% of a mile, 0.5 of a mile, or half a mile. I know I've passed 0.1 miles, I know I've passed 0.2, 0.22, 0.3 -- whatever decimal you can think of less than or equal to 1, I will have passed. In fact, an infinite amount of decimals will have been passed. How do I know i've passed 0.333... of a mile ? Because 0.5 of a mile is longer. When I reach 1, i will have passed 0.9, 0.99, 0.99 and so on. But what happens if 0.999... goes on into infinity? It'll equal 1. 1 mile, 1 apple, whatever. Falsedef

"Mathematics is theory that can represent the real world." Well, this is kind of my point (but I don't think you mean it the same way I do). Writing down digits on paper is sort of a model. Like if you were looking at a model airplane, something that is meant to approximate the real thing. I'm struggling with the concept that (1/3=0.333) so (3/3=.999...) and also (3/3=1). The math on paper is clearly correct, but it looks illogical. I keep coming to the conclusion that this is a mathematical abstraction done for the sake of figuring it out on paper.

If I have three slices of pie that add up to 0.999... of a pie, then somewhere out there is .000...1 of a pie missing. This means I am missing an infinitely small piece of pie, and as you pointed out above "A point is infinitely small, and occupies no space." If it occupies no space, how can it exist? How could I add, subtract, or multiply something which does not exist? This leads me to my idea that the problem of 0.999... cannot be related to real, physical objects because you're doing math with objects which cannot exist.

Or did I just prove myself wrong?Jboyler 07:16, 26 October 2006 (UTC)

"Then somewhere out there is .000...1 of a pie missing."

If such a number of pie did exist (it's not a recognized in real numbers), it would be equal to 0. Infinity is a hard concept to grasp, but anything trailing an infinite amount of 0s would be swallowed up by the 0s and become nothing. Infinity doesn't stop. Finite values become inconsequential in comparison to infinity. It wouldn't matter if it was 0.000...90210. When something grows infinitely small, you can't say "here's a piece of an infinitely small pie", because that piece of a pie would have never stop shrinking for you to say that there was such a piece. Why is 1/3 exactly equal to 0.333...? Because once an infinite amount of trailing 3s are produced, it becomes exact. There's nothing left but perfect precision.

I say "never stops", but that might give the impression that infinity is an ongoing process -- once something has become infinite, that's not the case anymore. 0.333... is no longer a process of adding trailing threes, it's defined as an infinite trailing number of a threes, and has therefore already reached 1/3rd. Trying to describe infinity is difficult, a lot harder than ancient mathematicians trying to describe 0. Falsedef 09:20, 26 October 2006 (UTC)

Again, if you have a two infinitely huges pizza pies covered in slices of pepperoni and someone takes 100% or all of the pepperoni off one pizza and someone only removes 99.999... of the pepperoni off the other, there will still be at least 1 piece of pepperoni on the second pie. I don't know why its hard for people to understand things they cannot name, quantify or visualize. 71.234.110.209 08:11, 26 October 2006 (UTC)

Two pronged question - (and please let me say this article has kept me amused for a good hour - I need to get a life)For the higher maths people out there - does 1-0.999... = 0.000...? by definition? Also taking the assertion that 1/3 = 0.333... when multiplied by 3 = 1 again by circular logic, but does 1/3 + 1/3 + 1/3 = 1 or 0.999... or both?

Thanks for making me think over a dull lunchbreak

Wait, if the formula is true...

Moved from main talk page. --Brad Beattie ^(talk) 05:18, 26 October 2006 (UTC)

then doesn't that just say that infinity = 1?

...999 = c

.1c = ...999.9

.1c - c = .9

-.9c = .9

c = -1

...999 = -1

|...999| = |-1|

...999 = 1

It just feels like the whole 1 = 2 thing. --Zonis 04:08, 26 October 2006 (UTC)

Infinity is not a real number. ...999 is not a real number. 0.999... is a real number. 153.104.208.67 04:36, 26 October 2006 (UTC)

Really? In what way is 0.999... more real than ...999? 198.54.202.254 10:03, 26 October 2006 (UTC)

Where did you learn to use the absolute value operation like that? It doesn't just remove minus signs! Melchoir 04:50, 26 October 2006 (UTC)

Awww, what's the matter Melchoir - don't you like any evidence that what you are publishing is rubbish? Zonis's argument is just as good as your fake argument that 0.999... = 1. See, you just have to imagine that ...999 is infinite, then everything will be quite clear to you. 198.54.202.254 10:01, 26 October 2006 (UTC)

Did you even bother to read the last step, where Zonis basically says something like "|-4| = |4|, so -4 = 4"? That's in addition to using ...999, where by definition decimal expansions of real numbers (as opposed to 10-adics) start somewhere (are all zeros to the left of some point). -- Meni Rosenfeld (talk) 10:33, 26 October 2006 (UTC)

"Ma atta omer Meni? Atta yodea klum." Did you read Zonis's comment properly? The last step says ...999 = 1, not -...999 = ...999. Anyway, Zonis's argument is also false. He knows it but is using similar logic to your's to prove a point that what you are publishing is nonsense. 198.54.202.254 06:02, 27 October 2006 (UTC)

"The last step says ...999 = 1, not -...999 = ...999." And? The point is still the exact same: just because |x| = |y|, it does not mean x = y. - KingRaptor 07:23, 27 October 2006 (UTC)

Actually, since Zonis already concluded ...999=-1 (which even seems to be true in the 10-adics), claiming ...999=1 is the same as claiming -...999=...999 (and -1=1). --Huon 08:45, 27 October 2006 (UTC)

The problem with the above argument lies in the fact that "c" is not a finite number. I'm going to assume that when the author writes "...999" what he is talking about short-hand for "the limit of the sequence {9, 99, 999, ...}" That sequence has no finite limit, so c=...999 is not a finite number.

Thus the phrase ".1c - c" in the proof has a problem, because you can't simply subtract two two infinite terms. "Infinity minus infinity" is not well defined, and can't be claimed to be a specific finite number or infinity. As an example, consider set subtraction for sets A and B where "A-B" is the set of elements that exist in A but do not exist in B. For finite A and B, the number of elements in A-B is the number of elements in A minus the number of elements in B. If A and B are infinite, though, the number of elements in A-B is indeterminate. Some examples:

1. A=all integers >=0 and B = all integers >=3. Then A-B is {0,1,2} with 3 elements
2. A=all integers >=0 and B = all integers >=100. Then A-B has 100 elements.
3. A=all integers >=0 and B = all even integers. Then A-B is the set of odd integers and has an infinite number of elements.

So a statement such as "Infinity - Infinity" isn't well defined, and thus the line ".1c - c = .9" is a place where the above proof breaks down, since both .1c and c represent infinite values. Dugwiki 22:06, 27 October 2006 (UTC)

You should have specified that for the finite case, B must be a subset of A for the formula to work. Anyway, this example illustrates the idea, but it doesn't quite fit since the types of infinities sets have is different from the ones that relate to real numbers. -- Meni Rosenfeld (talk) 07:30, 28 October 2006 (UTC)

Correct, I forgot to say that B must be a subset of A above. My bad. As far as set infinities versus real numbers, I was making an analogy between infinite countable cardinality sets and the concept of "positive infinity" in the real numbers. For the heck of it, though, I'll give another similar example of why "subtracting infinities" is perilous using using real numbers.

Consider forumlas of the form f(x)-x and calculating the limits as x approaches infinity.

1. If f(x)=x+1, then the limit of f(x)-x as x approaches infinity equals 1
2. If f(x)=x+100, then the limit of f(x)-x as x approaches infinity equals 100
3. If f(x)=2x, then f(x)-x as x approaches infinity has no finite limit

So this is another similar example of problems associated with trying to subtract one infinite value from another infinite value. In the original post in this thread, in fact, you have a formula similar to .1x - x, which is negatively unbounded and has no finite limit as x approaches infinity. Hopefully this one fits a little better since it uses real number functions instead of sets. Dugwiki 19:08, 30 October 2006 (UTC)

Bank on it...

I tried telling my bank that $0.999... = $1

They said to stick that into my infinity...

Melchior said it best

"

It's true that the decimals 0.999… and 1.000…, left unevaluated, are equivalent but not equal."

Thanks man, you wrapped this up after way too much discussion. The are indeed "equal in value, measure, force, effect, significance, etc." but not "the same as" each other. Thank you dictionary.com and Melchiors for tag-teaming a great explanation.

Maybe you misunderstood. 0.999... and 1 are different in the same way that 2/2 and 1 are different. They are different sequences of symbols, they are different arrangements of pixels on your screen... But they represent the same number. -- Meni Rosenfeld (talk) 09:32, 26 October 2006 (UTC)

1.000... has the property where I can choose two integers, a and b, such that that the decimal expansion of a/b = 1.000.... However, no two integers exist, a and b, such that the decimal expansion of a/b results in 0.999... 71.72.135.142 05:17, 27 October 2006 (UTC)

Interesting point. Are you claiming this as a proof that 0.9999.... = 1 (since all recurring decimal expansions are rational, so 0.9999... must equal some a / b and so therefore some other decimal expansion), or as an argument that it is not equal to 1, since it is a difference between the two notations? The second would of course mean that you're saying that 0.999... is not rational. That'd be difficult to argue I think.

I'm not claiming it as a proof of anything. I just find it interesting that the decimal expansion 0.999... cannot be arrived at by the long division algorithm when using integer values that aren't already in the form x.999... (since 0.999... equals 1, then 0.999... is also an integer, even though it doesn't "look" like one). This makes such representations unique in that they can't be derived via division or multiplication algorithms unless one or more of the operands is itself a recurring decimal. In contrast, values such as 0.333... can be arrived at without using a recurring decimal operand.

Whoever said that, that's brilliant. /agreed. Edit: I'm referring to the comment "1.000... has the property.." Two numbers that don't even share the same characteristics can't have the same identity --JohnLattier 11:49, 27 October 2006 (UTC)

a = 1, b = 1. a/b = 1, which has 2 decimal representations, and the long division algorithm happens to give only one of them (the 1.000... one and not the 0.999... one). You're confusing a number with the notation used to represent it. -- Meni Rosenfeld (talk) 09:59, 27 October 2006 (UTC)

Wrong again Meni. A radix system is designed with unique representation. In your answer above, you assumed that 0.999... = 1 and then stated without precedent 1/1 = 0.999... - this is not mathematics. You will have to do much better than this - I suggest you continue to research the subject. 198.54.202.254 11:47, 27 October 2006 (UTC)

• Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Melchoir 17:37, 27 October 2006 (UTC)

Cut the stuff about students and novices out. I have a BS. I was a student a very long time ago - probably before you were born. 198.54.202.254 12:16, 29 October 2006 (UTC)

Questions about the article

What is the last number before 1.0 that is not equal to 1.0?

Is 11:59pm and 59.999... seconds: today or tomorrow?

If 0.999...=1 then why don't you do the same for other repeating decimals?

Do you mean write a wiki article on each of the infinite other equalities? "We" do do the same for all other decimals ending in a repeating 9. e.g. 1.99999... = 2. This is covered in the article under generalisations.

I already know the answers, but go ahead and amuse me.

Why don't you sign your comment so we know who you are? The Wiki team will respond that there is no previous or next number because as they claim the real number system contains no infinitesimals (not only do these not exist but they are not even well-defined. For a good laugh, see Wiki's definition of infinitesimal). Thus the real number system contains holes. Hmmm, this is kind of strange because in Calculus we talk about continuous functions. We further define continuity in terms of limits. I wonder what happens to continuity when we reach these holes? Does a function experience what is called a bump? See, the problem lies with the subject of real analysis - it is mostly illogical nonsense that contains serious contradictions. 198.54.202.254 06:15, 27 October 2006 (UTC)

Hard to know from this if you're a troll or genuine, since you seem to be claiming as evidence something that proves the exact opposite of what you're arguing. It's exactly because there are no holes that there are no infinitesimals. If there were infinitesimals there would be holes, for example half way between the infinitesimal and the next number. Since the real numbers are continuous and have no holes, there can be no infinitesimals. You seem to be saying that if the real numbers do not have infinitesimals, they would have holes. Can you point to any of them?

Anonymous: It would help if you read my comment carefully - I do not believe in 'infinitesimals'. Furthermore, I agree that reals do not have holes. I am saying that Wiki's arguments imply that holes exist and they use this as an excuse to support their argument which goes something to the effect that the underlying structure of the reals must have 0.999... = 1. This is false. And can I point to an example? Why of course: 0.999... - if this were a rational number, one would be able to express it as a ratio of two integers. This is impossible unless you start out by assuming the two are equal in which case you are defeating your argument through faulty logic. 198.54.202.254 11:40, 27 October 2006 (UTC)

Is anyone going to answer the above questions? Here's my signature. --JohnLattier 11:46, 27 October 2006 (UTC)

• Within the standard reals there is no last number before 1.0 that is not equal to 1.0: After every number x less than 1, there is always (x+1)/2, which is again less than 1, but greater than x.
• 11:59pm and 59.999... seconds is midnight; whether that's included in "today" or "tomorrow" is irrelevant.
• We do the same for all repeating decimals. 2.9999...=3, 0.3333...=1/3, 0.232323...=23/99.

Does this answer your questions? --Huon 14:14, 27 October 2006 (UTC)

Are you as good as Brittanica?

Your article states that second or third year university courses are required to understand the formal proof that 0.999... = 1. So why don't you include this proof in your article? The Brittanica includes controversial proofs regardless of education level required. Please publish a formal proof. The so called Mickey-Mouse proofs in your article say nothing. 198.54.202.254 12:00, 27 October 2006 (UTC)

I'm not quite sure why you felt the need to delete the last few comments when you wrote that... Quendus 12:16, 27 October 2006 (UTC)

??? Are we talking about the same thing? 198.54.202.254 14:48, 28 October 2006 (UTC)

How formal do you want it? A proof from the ZF axioms and first-order predicate logic on up might be a little more dense than is appropriate in an encyclopedia article. I seem to recall that there's a guy who is building a forest of machine-readable proofs of every well-known result in mathematics from that basis up, with an automatic proof verifier to demonstrate their logical consistency; can anyone here remember what it's called, and the URL of his website? -- The Anome 14:54, 28 October 2006 (UTC)

Formal is not similar to a 'limit'. You cannot get as formal as you like. Thus formal means exactly one thing - formal. So where is your formal proof then? 198.54.202.254 15:02, 28 October 2006 (UTC)

How much more formal do you want? More formal than ZF+logic+real number axioms? If so, could you tell us what you have in mind? See Tarski's axiomatization of the reals and Dedekind cut for more details. -- The Anome 15:04, 28 October 2006 (UTC)

The Anome, is Metamath [1] the Web site you're remembering? I'm not sure they have a proof that 0.999... =  1 specifically, but their proof that 2 + 2 = 4 is 22,607 steps long. I wonder if this is the level of formality that 198.54.202.254 had in mind. —Caesura^(t) 15:14, 28 October 2006 (UTC)

Yes, Metamath is exactly what I had in mind. Perhaps if 198.54.202.254 asks nicely, they could code one of the proofs below up for them? Then 198.54.202.254 could print it out and have the resulting fully rigorous telephone-book-length proof bound in hard covers for their reading convenience. -- The Anome 15:29, 28 October 2006 (UTC)

The infinite series and Cauchy sequence proofs are formal proofs. Another one is given by Meni Rosenfeld as a new proof for the skeptics. --Huon 15:18, 28 October 2006 (UTC)

Regarding a new proof for the skeptics: in spite of my disagreement with Meni on the exact wording of the intro, I have to say that Meni rocks. That's about as simple as it gets. -- The Anome 15:35, 28 October 2006 (UTC)

Thanks :-) The proof itself is not my own invention, obviously, but I think the request to pinpoint the exact contended step is just the right thing for such a debate.

And another thing... I didn't want to make this public, but I guess I should warn everyone: I have strong evidence that 198.54.202.254 has a long history of trolling with this article (though this is not the same IP he used before). Please do not feed him. -- Meni Rosenfeld (talk) 16:21, 28 October 2006 (UTC)

(198.54 too, huh?) What you see are the proofs that 0.999… = 1, including all the essential details. It takes a specialist course for mathematicians only to establish the foundations of these proofs, not the proofs themselves, which are just as simple as they appear. Melchoir 16:24, 28 October 2006 (UTC)

Half-Life

Could someone explain how the concept of the half-life and exponential decay of radioactive elements relates to this? I was always taught throughout school and college that an element never reaches "0," because of there being the inability to reach zero by dividing the same number in half, over and over again (1/1, 1/2, 1/4, 1/8, etc.).

Yet, the wiki article for half-life states that:

"Of particular note, atoms undergo radioactive decay in whole units, and so after enough half-lives the remaining original quantity becomes an actual zero rather than asymptotically approaching zero as with continuous systems."

Is this proving this .999... article any further? Helping it or damaging the concept at all? Do they even relate?

Thanks Punchinelli 12:55, 27 October 2006 (UTC)

Radioactivity is a process. This relates to the "infinite process" described in this article. The 0.999... is not a process that gives a number that keeps getting closer to 1. It is a definite number that has the value of exactly 1. The half-life article says that the quantity of radioactive atoms becomes zero eventually because there are a finite number of atoms in a material, and if you wait long enough each one of them will decay. -- Schapel 13:26, 27 October 2006 (UTC)

But isn't writing down the nines also a process? And note that radioactive decay is a stochastic process, in the sense that in principle that last uranium atom can linger around "forever", decaying at the same point where the "last" nine is (finally!) written down... ;-) mdf 13:43, 27 October 2006 (UTC)

Yes, if you wrote the nines down, it would be a process. However, 0.999... does not involve any process. It is a number. I understand that radioactivity is random, and the amount of time a given amount of uranium takes to decay is unbounded, but the amount of time a given amount of uranium takes to decay is finite. That last uranium atom has a zero probability of lingering literally forever. If you wrote down the nines in 0.999..., however, you would truly never finish no matter how long you took. -- Schapel 14:30, 27 October 2006 (UTC)

You can prove that 0.999... == 1. But is this also the proof that 0.999... is not a "process"? On its face, the very word "repeating" has connotations of mechanicity that are extremely difficult to overcome. But fortunately, there are many finite (and infinite) processes which map to numbers, given a suitable computational substrate. And this gets back to why people are finding it "hard to swallow" the proofs here ... math people are using one CPU, everyone else has another, and while the instruction sets are close, they aren't exactly the same. mdf 16:00, 27 October 2006 (UTC)

Yes, the proof that 0.999...=1 is a process, but it terminates with the definitive result that the equality holds. Writing the digits of 0.999... down is also a process, but it does not terminate, and never reaches the value of 1. Trying to prove something by writing 0.999... down would neither prove nor disporve anything, because you could never complete the proof. -- Schapel 17:49, 27 October 2006 (UTC)

[Edit conflict]An atom could stay around forever, but the probability of that is 0. Writing down the nines is probably a process, but we don't actually write down infinitely many nines - we use a (finite!) notation which specifies that there are infinitely many nines. A possible rigorous notation for that is:

$\{(n,9)|n\in \mathbb {Z} \land n>0\}\cup \{(n,0)|n\in \mathbb {Z} \land n\leq 0\}$

And no, I don't think decay has anything to do with 0.999..., since you don't have any concepts that deal with the physical "end of time". -- Meni Rosenfeld (talk) 14:33, 27 October 2006 (UTC)

Many infinite processes have finite representations. My own notation looks like this: printf("0.9"); for(;;) printf("9");, simply because I don't know of any TeX compiler. Anyways, it would be surprising to me if there were not a connection between a proof that 0.99999.... == 1 and the P(n_uranium_atoms > 0) == 0, "forever". Both statements are limits, the usual epsilon-delta arguments would apply. mdf 16:00, 27 October 2006 (UTC)

Maybe. But I guess it is subjective what to count as a "connection". -- Meni Rosenfeld (talk) 16:49, 27 October 2006 (UTC)

I think my main question it this: can you rewrite that statement, using similar concepts in the statement itself, to fit this .999... solution? I.E., instead of "Of particular note, atoms undergo radioactive decay in whole units, and so after enough half-lives the remaining original quantity becomes an actual zero rather than asymptotically approaching zero as with continuous systems." could you write, "Of particular note, the nines in .999... approach 1 in whole units, and so after enough repititions of 9, the remaining original quantity becomes an actual one rather than asymptotically approaching one as with continuous systems." I apologize for my lack of experience in this subject, but the two seem to relate to me. Punchinelli 18:09, 27 October 2006 (UTC)

No, that's not a parallel case. In the radioactive decay case, with probability 1, there will be some finite amount of time after which the number of atoms becomes an actual zero. There's no finite number of 9s that makes .999...9 equal 1. You have to take infinitely many 9s. --Trovatore 18:10, 27 October 2006 (UTC)

Thank you. I think I need to go back to my old teachers and tell them to stop telling students that the half-life of radioactive decay of an element will never reach zero.Punchinelli 18:17, 27 October 2006 (UTC)

First, be careful with your terms: The half-life is a characteristic of a particular reaction, which specifies the constant amount of time it takes for the sample to reach half its size. It doesn't change and it is definitely never zero.

What you meant is that the sample size reaches 0. Be careful here as well. There are mathematical models and there's physical reality. Unless you bring a sample of radioactive isotopes to your teachers and show them the moment the last atom decays, you don't have a real case. What you do have is a mathematical model which may be negligibly more accurate than your teachers'. And for all you know, tomorrow your model will be proven to be incorrect as well. -- Meni Rosenfeld (talk) 07:12, 28 October 2006 (UTC)

Sqaure root and square question

At first glance I could understand how one would be skeptical that these .999... and 1 are not equal BUT after having read the article and the Arguments (ad nauseam) I can accept that they are equivalent, however there are a couple of things I don’t understand perhaps due to my limited understanding of advanced mathematics.

I was taught that the square root of an integer would be equal to the square root of any function that equals the integer. Thus the equation √9 = √8+1, seems simple enough to me.

But I don’t see how the √1 = √.999… I would think that since the √1 = 1 either √.999…= 1 or √.999…= .999… but neither of these answers makes any sense either! Can anyone clarify where is the error here?

In addition, I have a similar complaint when I compare squares of these two numbers 1^2=1, can it also be stated then that .999…^2=1? Or in fact is .999…^2 = .999… which begs the question, how can that be so when any finite number of digits to the right of the decimal will at some point not equal zero? To phrase another way .9999*.9999=.99980001 so how can .999…*.999…=.999…?

thanxSteevnd 20:07, 27 October 2006 (UTC)steevnd

About the square roots, if .999... = 1, then you can interchange them as much as you like. It makes perfect sense that √.999... = √1, because that is the same as saying √1 = √1. It's the same thing with the squaring. .999...^2 = .999... is the same as saying 1^2 = 1.

slippered sleep 20:24, 27 October 2006 (UTC)

One way to answer the question is that when we say we've proven that ".999... equals 1" we have proven that they are, in fact, different ways of symbolizing the same number. They are totally interchangable with respect to functions, because they represent the exact same element in the set of real numbers. So since .999... = 1, you can use substitution to say that √.999... = √1.

Another way to look at the question would be to observe that the limit of √f(x) as x->infinity equals √(limit of f(x) as x->infinity), assuming that f(x)>=0 for all x and f(x) is a continuous function. So since .999... equals the limit as x approaches infinity of the sum of all terms 9*10^-k for k=1 to x, it follows that √.999... equals the sqaure root of that limit. Thus because that limit equals 1, we get that √.999... likewise equals √1, which is 1. The same sort of argument likewise works for raising to a square power. Dugwiki 20:39, 27 October 2006 (UTC)

I guess my confusion arises in trying to apply squares to infinitely repeating decimals. If we take .999…9*.999…9 a curious pattern develops where the result of the equation will always equal .9(n)80(n)1 where n=number of positions in the original numbers -1.

i.e.

.999999999999*.999999999999=.9999999999980000000000001 .9999*.9999=.99980001

Steevnd 20:56, 27 October 2006 (UTC) steevnd

The 8, 0s, and 1 result because the 9s end. If there are an infinite number of 9s, they never end, and you never reach the 8. -- Schapel 21:04, 27 October 2006 (UTC)

Thank you those of you who took the time to answer my question. I believe I could explain this concept now to my 11 year old son or if necessary his math teacher. I notice that this only applies in cases of the repeating decimal is one less than base. Meaning that .333... != .4 but .3999...=.4, maybe obvious to you mathematicians, but quite an insight for a layman like me. I assume this can be proven using the same proofs as for .999...=1. 15:47, 30 October 2006 (UTC)Steevnd

It's illogical to add, subtract, multiply, divide, square, etc. a number with infinite decimal digits. Although don't tell that to the author of this article or the people at Blizzard. Their whole understanding of existence would crumble. -John Lattier 841pm 10-27

Nonsense. The square root of 2 has infinite decimal digits. Square it and you have 2. -- Schapel 00:49, 28 October 2006 (UTC)

Sure the square root of 2 is irrational. What John is saying is that you cannot take the square of the square root of 2 unless you know it is the square root of 2. You have not established that 0.999... = 1. Where is your formal proof? I read the so-called proofs in your article and they are all false - even by your own admission. Can you publish a formal proof? You talk about inventing the real numbers. When calculus was invented, the real numbers were not even on the horizon. Who invented the real numbers? I don't think Archimedes ever referred to any numbers as real, did he? 198.54.202.254 14:55, 28 October 2006 (UTC)

The proofs are in the article and are correct. Do you see an error somewhere? If so, could you point it out? I never spoke of someone inventing the real numbers, but Georg Cantor gave the first rigorous definition of them. -- Schapel 18:54, 28 October 2006 (UTC)

Look at the section called 'A closer look at Meni's Proof'. 198.54.202.254 15:19, 29 October 2006 (UTC)

I see the section. Where is the proof it refers to? Can you simply state any problem with any proof in the article? -- Schapel 15:24, 29 October 2006 (UTC)

You saw the section. Now if you read it, you will see where it goes wrong. I have stated after each step of the proof what the problems are. 198.54.202.254 05:46, 30 October 2006 (UTC)

If that were true, we would never have invented the real numbers, because they'd be useless. What would be the point of taking our nice, arithmetic-friendly rational numbers, and mixing them up with those nasty irrational numbers with which arithmetic is "illogical"? Maelin 01:29, 28 October 2006 (UTC)

Schap--You're right, then. I should have said it is illogical to add, subtract, multiply, divide, square, etc. a number if in doing so you are increasing its significant digits. -John Lattier 10/27 1139pm

1.5 has two significant digits. 1.5 * 1.5 = 2.25, which has three. Doing arithmetic on real numbers is never illogical (except for division by zero, which is undefined). The real numbers are a field, so addition or multiplication of two numbers is always defined and always sensible, for any two real numbers, be they small, large, rational, irrational, whatever. Maelin 04:31, 28 October 2006 (UTC)

John is mixing up mathematics with notational conventions used in the natural sciences. Generally, in the latter, when one writes x = 1.5 he actually means 1.45 <= x <= 1.55. So 2.1025 <= x² <= 2.4025, which, losing some information and using the notation again, would be written as x² = 2. Of course, all of this has nothing to do with mathematics or the decimal representation. -- Meni Rosenfeld (talk) 07:22, 28 October 2006 (UTC)

But 1.5 isn't a number with infinite decimal digits. Ya I realize I didn't write it the second time. But anytime you manipulate an infinite decimal digits number using a method that "carries over" a value, it becomes illogical. Like 10*0.999... doesn't equal 9.999... because you are manipulating the infinitesimal. Same with 3*0.333... It's not 0.999...

In fact, 1.5 can be represented with infinitely many decimal digits (either as 1.5000... or as 1.4999..., whichever you prefer). Anyway, what's this rule against "manipulating the infinitesimal"? I've never heard of anything like that. On the contrary, multiplying infinite decimals works in more or less the same way as multiplying finite decimals. I'm sure you'd agree that 2 × 0.333... = 0.666..., so why not 3 × 0.333... = 0.999...? —Caesura^(t) 00:50, 29 October 2006 (UTC)

I agree that 0.333...+0.333...=0.666... because there is no conflict with the infinitesimal. But when you add another 0.333... think of it like every digit goes into error mode because it is awaiting the carry-over from the next digit. You'll never reach a conclusion (just like these debates never reach a conclusion lol). It would be convenient if these infinite decimals were expressable as a manipulatable number like a fraction. And they are: 1/3. Three one-thirds gives 1.0, not the entirely separate expression 0.999... --JohnLattier 02:24, 29 October 2006 (UTC)

This is correct John. In fact, this is laughable because they can't fo anything with 0.999... in any event. Whether anyone likes it or not, our arithmetic in any radix system always takes the form of an approximation. Any calculation involving pi or an irrational number can never be exact unless you write it in terms of pi or that irrational number in which case you do not know what the value is anyway - not even an approximation. To find the approximation, you evaluate the answer by substituting values for pi, e, etc. 198.54.202.254 05:28, 29 October 2006 (UTC)

JohnLattier, I'm not sure where you learned this "problem" with infinite decimals, but it's wrong. Infinite recurring decimals are rational, and more importantly, real numbers just like every other number. Because the real numbers are a field, then addition is always defined and sensible, and multiplication is always defined and sensible. The different ways we represent numbers have their respective advantages and disadvantages, but they're all representations for the same number and they are all allowed to be added and multiplied in the same ways. If we weren't allowed to add 0.333... to 0.666..., we wouldn't be allowed to add 1/3 to 2/3, since they are the same number. The fraction proof just shows that we can do the same arithmetic on two different representations for the same numbers, so the two different representations we get at the end must also be the same number. Maelin 07:50, 29 October 2006 (UTC)

What utter nonsense you write! The Greeks were able to add 1/3 and 2/3 long before the decimal system was invented. Adding fractions has nothing to do with radix systems. Furthermore, just because the real numbers are a 'field' does not mean that the four basic arithmetic operations are always defined. What is 1/0, 0/0? You are so arrogant that you don't realize how stupid you really are. Listen Aussie boy, get some education and then try to assert your knowledge. At this time, you are revealing to everyone what a fool you are. 198.54.202.254 12:12, 29 October 2006 (UTC)

Division by 0 is not defined because 0 does not have a multiplicative inverse in the real numbers. However, one-third plus two-thirds is always defined, no matter whether you write it 1/3+2/3=1 or 0.333...+0.666...=0.999.... -- Schapel 12:44, 29 October 2006 (UTC)

does this prove its not equal?

53.012>52.984 right, we can all agree on that Dispite 9>0 and 8>1 and 4>2 it does not come into play because "if the first non equal didgit of a positive number is larger than the corresponding didgit of another positive number then it is larger than that second number", so the 3>2 overides the fact that all the decimal places are smaller, as they come after the first non equal numbers so dont come into play when deciding whether number is larger or not,they only come into play when deciding by how much they are differant in size so 1.000>0.999 as 1>0, it does not matter that all the 9's are bigger than the 0's and that there is an infinate number of them,as 1>0 and so the diferance is infanatly small, and infanately small is not equal zero. now i need some sleep, yours senserarly sem, fingers crossed im beat maths, or i will have to delet my account and hang my head in shame, shame fulled shamey shame of the worst kind. I blame the flaws in the base system.

Where did that quote come from? Melchoir 01:07, 29 October 2006 (UTC)

The spelling and grammar are exceptional. Is this your son Melchoir? 198.54.202.254 06:57, 29 October 2006 (UTC)

"if the first non equal didgit of a positive number is larger than the corresponding didgit of another positive number then it is larger than that second number"

This isn't a rule in mathematics. It is an observation that generally holds true, without any obvious counterexamples. Analogously, a person learning integrals might see, eg. ∫(2x dx = x² + c and ∫2x^-3 dx = -x^-2 + c. They might conclude that ∫nx^n-1 = xⁿ + c. This pattern is also generally true, but can fail. Here it fails when the power is -1, while the larger number one fails when one of them ends in .999... -- kenb215 talk 18:45, 9 November 2006 (UTC)

Isn't that the step function? Step(0.999...) < Step(1)--JohnLattier 21:52, 31 October 2006 (UTC)

I'm not sure which specific function you're talking about ("Step function" can refer to a class of similar functions). But since 0.999... = 1 in the real number system, it follows that f(0.999...) = f(1) for all real functions f(x). Dugwiki 20:07, 9 November 2006 (UTC)

A close look at Meni's 'proof'

1. 0.999... is not greater than 1. In other words, 0.999... <= 1.

A reasonable assumption by the architecture of any radix system.

2. So 1 - 0.999... >= 0. Denote c = 1 - 0.999...

Problem really begins at this step. Why? Meni is using the properties of Metric spaces which define the real numbers to conclude that the distance between any two real numbers can only be greater than 0 if they are not the same numbers. So already Meni has assumed in his proof that 1 and 0.999... are not equal by his inequality 1 - 0.999... >= 0. If I were marking Meni's exam, then he would receive no marks at all starting from this step.

Okay, but let's give Meni a break and overlook this error. Perhaps the rest of his logic is in order if this step is correct.

3. Let n be any natural number.

  There exists a natural number m such that 10^m > n. 
  

A reasonable assumption.

4. 10^(-m) < 1 / n.

Somewhat problematic assumption since Meni does not prove this for all m. He erroneously assumes it is true.

5. 0.999... is greater than 1 - 10^(-m).

A reasonable assumption if previous step is true (alas it's not!)

6. Therefore, c < 10^(-m).

  So c < 1 / n. 

Logic in severe problems at this step. I cannot continue to mark Meni's work because it is fatally flawed.

All conclusions from here on are nonsense. 198.54.202.254 05:55, 29 October 2006 (UTC)

Just go away. Please. It's so tiresome having to trawl through all your rubbish trying to find people who are actually here because they want to understand the article. You are undermining everyone's hard work and wasting everyone's time. Please stop. Maelin 07:20, 29 October 2006 (UTC)

Excuse me?! The only rubbish is what you write. If you have a problem with what I write, then don't read it! Who do you think you are? 198.54.202.254 12:07, 29 October 2006 (UTC)

Even though I know it won't stop your trolling, and despite my better judgement, I'm going to point out why your claims are false in order to save others from being misled by them. I am also going to do it in very short steps so it is completely obvious how each result was derived and what properties of the real numbers were involved. It may be useful to remember that the real numbers are a field, since some of the steps in the proofs rely on the properties of fields.

2. So 1 - 0.999... >= 0. Denote c = 1 - 0.999...

Problem really begins at this step. Why? Meni is using the properties of Metric spaces which define the real numbers to conclude that the distance between any two real numbers can only be greater than 0 if they are not the same numbers. So already Meni has assumed in his proof that 1 and 0.999... are not equal by his inequality 1 - 0.999... >= 0. If I were marking Meni's exam, then he would receive no marks at all starting from this step.

Wrong. Meni hasn't assumed anything. This step is entirely valid, and here's a proof.

• We have 0.999... ≤ 1, by step 1 of the proof
• By multiplication properties of inequalities and multiplying by -1, we have -0.999... ≥ -1
• Now, by addition properties of inequalities, we have (-0.999...) + 1 ≥ (-1) + 1
• By commutativity of addition, we have 1 - 0.999... ≥ (-1) + 1
• By additive inverses, we have 1 - 0.999... ≥ 0 as required

4. 10^(-m) < 1 / n.

Somewhat problematic assumption since Meni does not prove this for all m. He erroneously assumes it is true.

It is true. The proof relies strongly upon one of the multiplication properties of inequalities, specifically, the fact that For any real numbers, a, b, c: If c is positive and a > b, then a × c > b × c and a / c > b / c. Here is the proof:

• We have 10^m > n, by step 3 of the original proof
• Since n is positive and 10^m > n, then (10^m) / n > n / n, by the above property of inequalities
• By algebra and since n ≠ 0, we obviously have n / n = 1, and so (10^m) / n > 1
• Now, since 10^m is positive and (10^m) / n > 1, then (10^m) / (n × 10^m) > 1 / (10^m), by the above property of inequalities
• By algebra and since 10^m ≠ 0, we obviously have 10^m / 10^m = 1 and so 1 / n > 1 / (10^m)
• By the reversal property of inequalities, we have 1 / (10^m) < 1 / n
• By negative integer exponentiation, we have 1 / (10^m) = 10^(-m)
• So we have the result that 1 / n > 10^(-m), exactly as in Meni's proof.

6. Therefore, c < 10^(-m).

So c < 1 / n.

Logic in severe problems at this step. I cannot continue to mark Meni's work because it is fatally flawed.

Wrong, again. This step is entirely valid. It relies strongly on properties of inequalities again. Here is a proof.

• We have 0.999... > 1 - 10^(-m), from step 5 of the original proof
• Then, by subtraction properties of inequalities, we have (0.999...) - 0.999... > (1 - 10^(-m)) - 0.999...
• By additive inverses and commutativity of addition, this implies 0 > 1 - 0.999... - 10^(-m)
• Since we defined c = 1 - 0.999..., we have 0 > c - 10^(-m), by properties of equality
• By addition properties of inequalities again, we have 0 + 10^(-m) > c - 10^(-m) + 10^(-m)
• By additive inverses and the definition of zero as the additive identity, this means 10^(-m) > c
• By reversal of inequalities, we have c < 10^(-m) which is the first result we needed
• Now, since c < 10^(-m) and 10^(-m) < 1 / n as proved before, and by transitivity of inequalities, we have c < 1 / n as required.

If you have more problems with Meni's proof, please post them below and I will prove each step slowly. Maelin (Talk | Contribs) 07:10, 30 October 2006 (UTC)

How is it wrong? Meni is assuming that 0.999... < 1. However, what he wishes to prove is that 0.999... == 1. You are wrong in stating that he can start the proof with this assumption. It is an assumption after all, because he uses this inequality to reach the next step as you correctly pointed out. Can you see why it's wrong now? In plain words, if 0.999... is not less than 1, then you cannot use any of the properties you mentioned to arrive at -0.999... > 1. Does this make sense to you now? It's like saying in order to prove the sky is blue, let's begin by saying that it is either blue or gray. Once you understand this, I'll proceed to explain the rest to you on one condition: you watch your attitude!198.54.202.254 08:55, 30 October 2006 (UTC)

Meni is not assuming that 0.9999... < 1. What Meni does assume is 0.9999... <= 1, which you yourself called a "reasonable assumption by the architecture of any radix system". --Huon 09:39, 30 October 2006 (UTC)

So are you telling me that 0.999... <= 1 does not imply that 0.999... < 1? This is part of Meni's proof. Of course it's reasonable except when you attempt to use it to prove something else. Each proof statement must be valid. 198.54.202.254 14:08, 30 October 2006 (UTC)

What are you talking about? 0.999... ≤ 1 implies that either 0.999... < 1 or 0.999... = 1; this is, in fact, the very definition of "less than or equal to". Meni's proof does not do anything of the sort. His proof basically takes the assumption that 0.999... ≤ 1, thus one of {0.999... < 1, 0.999... = 1} is true, and he shows that only one of these is possible, due to properties of the real numbers. Maelin (Talk | Contribs) 14:17, 30 October 2006 (UTC)

Exactly my point. I said Meni has made an assumption that 0.999... < 1. What do you find so hard to understand about this? The following steps build on this assumption (not the assumption that 0.999... might be equal to 1). His proof is thus invalid. 198.54.202.254 19:20, 30 October 2006 (UTC)

Do you really not understand what "less than or equal" means? --jpgordon^∇∆∇∆ 19:39, 30 October 2006 (UTC)

This is the structure of Meni's proof: First we agree that 0.999... is, at most, 1. Then we know there is some non-negative number c = 1 - 0.999.... We then show that, due to the properties of the real numbers, c can only be equal to zero. Thus, 0.999... is equal to 1, since their difference is zero. There is no assumption that 0.999... < 1, there is only an assumption that 0.999... ≤ 1. If you disagree, please describe, in detail, in which step of the proof the erroneous assumption is made, using the numbering of the steps that Meni has used. Maelin (Talk | Contribs) 01:07, 31 October 2006 (UTC)

The structure of Meni's proof is nothing but nonsense. Look, it doesn't matter whether you assume 0.999... < 1 or whether you assume 0.999... <= 1 - the fact is that you are still assuming either way that 0.999... < 1. You cannot begin a proof by assuming the opposite is true unless it's a proof by contradiction which Meni's proof is not. In fact, Meni's proof is so amateurish, he should be ashamed of stating it. Look, Meni's proof is prehistoric next to the proof provided by Rasmus. Rasmus's proof is far more convincing than Meni's even though both proofs are completely wrong. Rasmus was refuted. I will not even waste my time with Meni. Go back and read your archives - you have been given plenty of solid mathematical proof which you have ignored because you believe whatever you want to. 198.54.202.254 11:07, 4 November 2006 (UTC)

Look, it doesn't matter whether you assume 0.999... < 1 or whether you assume 0.999... <= 1 - the fact is that you are still assuming either way that 0.999... < 1. That's just wrong. Utterly wrong. Blitheringly wrong. (You supply the adverb) wrong. Assuming that X <= Y is exactly that: assuming that either X < Y or X = Y. Do you really not understand what "less than or equal" means? --jpgordon^∇∆∇∆ 15:16, 4 November 2006 (UTC)

Yes, X <= Y means X < Y or X = Y. Therefore 2 <= 2, even though 2 is not less than 2. -- Schapel 16:54, 4 November 2006 (UTC)

Assuming 0.999... ≤ 1 is no different to saying, "Let's assume that it's not the case that 0.999... > 1". It means when we define c as 1 - 0.999..., we don't have to worry about c being negative. Nothing else is assumed. Stop claiming otherwise. Maelin (Talk | Contribs) 05:14, 5 November 2006 (UTC)

For comments such as "You are so arrogant that you don't realize how stupid you really are. Listen Aussie boy, get some education and then try to assert your knowledge", I have initiated a template warning process on 198.54.202.254's talk page. Please continue adding template messages {{subst:npa3}} and {{subst:npa4}} as appropriate. At {{subst:npa5}}, the user will be blocked by an administrator. Argyrios 18:11, 30 October 2006 (UTC)

Isn't the assumption 0.999...<1 using to prove 0.999...=1 using something called proof by contradiction or similar? x42bn6 Talk 20:36, 30 October 2006 (UTC)

If that assumption were made, it would indeed lead to a proof by contradiction. But it isn't made - a <= b does not imply a < b. --Huon 20:43, 30 October 2006 (UTC)

Using the assumption that c=0 is circular reasoning. That's your mistake. c only equals zero to someone who is convinced that 0.999... = 1. --68.211.195.82 13:51, 1 November 2006 (UTC)

If that assumption were used, it would indeed be circular reasoning. Luckily, it's not. The assumption made, in terms of c, is that c is greater than or equal to 0, that is, that c is not negative. Are you suggesting the opposite to be true and 0.9999... to be greater than 1?

After that assumption is made, Meni shows that c is not greater than 0. That part of the proof is, in fact, independent of the assumption. --Huon 17:12, 1 November 2006 (UTC)

The algebraic proof is wrong

the argumentation used has a fault in from line 2 to line 3.

"c = 0.999… 10c = 9.999… 10c − c = 9.999… − 0.999… 9c = 9 c = 1" Because this is an equasion, in this step the same value has to be subtracted form both sides of the equasion for it to continue beeing true. The author does not consider this. Actually c would equal 0.999…, (which is the inital statement) if the equasion continued valid.

c =0.999… 10c =9.999… 10*0.999…-0.999…=9.999…-0.999… 9=9 this is the mathematically correct version of the equasion, which actually has no meaning at all.

In the third step, c is subtracted from both sides. On the left side, it is written as c, and on the right side it is written as 0.999... The first step says these are the same. -- Schapel 12:50, 29 October 2006 (UTC)

I don't understand what's being argued in this section... and I don't think it's because of the usual reason (i.e. my poor maths). Could you please restate the objection as clearly as possible. --Dweller 11:40, 2 November 2006 (UTC)

0.000...1

What's the struggle about? It seems like people understand 0.999... but not its inverse, 1.000...1.

Whoa right there. 0.000...1 is utterly meaningless. What you've essentially said (laymans' terms) is "I have a 0, with a decimal point, followed by an INFINITE number of 0s. AFTER this infinite number of 0s ends, I put a 1, and therefore have a number a little bit bigger than 0" This is patently absurd - an infinite number of 0s, by definition, has NO end. Therefore, there is no "end" on which to "tack" a 1 - the 1 simply doesn't exist. What you have is 0, followed by a decimal point, followed by an infinite number of 0s - that's zero. The infinite sequence of 0s after the decimal point, by definition has no end - therefore any argument based upon something that "follows" them is meaningless. Srpnor 07:55, 9 November 2006 (UTC)

So tell me, if you are one of these struggling individuals, what makes 0.999... any more applicable than 1.000...1 or any other number with the ellipses inclusive? Perhaps you are struggling with understanding infinity? --JohnLattier 02:44, 29 October 2006 (UTC)

So if 0.999... is the reciprocal of 1.000...1, then just for fun, what's the reciprocal of 1.000...2 ? --Trovatore 02:56, 29 October 2006 (UTC)

I don't understand what the fuss is about having an inverse for .999... Tell me Trovatore, do you know the inverse of pi, e or any other irrational number? I suppose you are going to tell me 1/pi, 1/e and 1/a where a is an irrational number. Yet in all these cases you do not know the full extent of any of these numbers but you are confident that you can find the inverses. There is no debate that the limit of the series 9/10+9/100+9/1000+... is 1. However, 0.999... is strictly less than 1. 198.54.202.254 05:30, 30 October 2006 (UTC)

0.999... is strictly less than 1? How much less than 1 is it? What's the decimal expansion of 1-0.999...? -- Schapel 12:43, 30 October 2006 (UTC)

You say "There is no debate that the limit of the series 9/10+9/100+9/1000+... is 1" (I guess you meant sum of the series). Agreed. Then you say "0.999... is strictly less than 1". The notation 0.999... is defined to be the sum of the series you gave. Therefore you contradict yourself. --Kprateek88(Talk | Contribs) 13:25, 30 October 2006 (UTC)

People understand 0.999... because there are an infinite number of nines. People do not understand 1.000...1 because there cannot be an infinite number of zeros and then a one. If there are an infinite number of zeros, the zeros never end and there can be no last digit at all. The inverse of 0.999 is 1 because 0.999 is 1 and is its own inverse. -- Schapel 22:18, 29 October 2006 (UTC)

1.000...1 is not a defined number in the reals - the notation makes no sense. We define 0.999... as the limit of the sequence {0.9, 0.99, 0.999, ...} which can be succinctly expressed as $0.999...=\lim _{N\to \infty }\left(\sum _{n=1}^{N}{9 \over 10^{n}}\right)=\sum _{n=1}^{\infty }{9 \over 10^{n}}$. Note that we are not actually making reference to any "infinitieth decimal place" here, we are just using infinity in neat mathematical notation as a way of saying, "Keep putting on more nines and don't stop". An "infinitieth decimal place" does not exist, which is why 1.000...1 does not exist. We can't say, "Go to the end of this endless string of zeros and place a 1," because that makes no sense. It's important to realise of course that we are using the ellipsis (...) notation in a specific way here, meaning "this simple pattern continues forever". Sometimes the ... is used to mean "This simple pattern continues for a finite length that is obvious from context, but it would be tedious writing the whole thing out," but that is not what it means in the context of this article. Maelin 00:37, 30 October 2006 (UTC)

Response to Schapel -- Hm? Certainly there can be an infinite number of zeroes, followed by a one—as a formal string of characters. No problem with that at all. It has order-type ω+1.

Whether there's any sensible sort of arithmetic that can be performed on these strings, and whether it has anything to do with the real numbers, is quite another question. I'm trying to get JohnLattier to explore why the reciprocal of 0.999... should be that particular string, rather than say 1.000...2, or 1.000...314159. --Trovatore 00:44, 30 October 2006 (UTC)

It's about as sensible as trying to perform any reasonable arithmetic with 0.999... 198.54.202.254 05:33, 30 October 2006 (UTC)

As you point out yourself, such a "formal string of characters" you're referring to is not allowed as a decimal representation of a real number. As such, you cannot have an infinite repeat of any digits, followed by something else. -- Schapel 01:03, 30 October 2006 (UTC)

As a decimal representation, no. At least not as a decimal representation of a real number (a decimal representation of something else--well, maybe, depends on what you mean).

But what you seemed to be saying was that it doesn't make sense to speak of doing infinitely many things, and then something else. That's not true. That sort of thing is a set theorist's stock in trade. --Trovatore 01:08, 30 October 2006 (UTC)

We never do infinitely many things with any arithmetic or calculus. The fact that we can evaluate integrals is because these are in fact limits of infinite average sums that are telescopic in nature and hence can be evaluated finitely. 0.999... is not an average sum. It is an infinite sum. There is no discrepancy. No mathematics falls apart by not having 0.999... = 1. Not the Archimedean principles or any other properties of the reals. The only thing that falls apart is your logic when you try to evaluate infinite representations of numbers using an arithmetic that is defined only for finite representations. Think about it: the first arithmetic operations took place only with finitely represented quantities. Example 1/3 + 2/3 = 1. Not 0.333... + 0.666... = 1. 198.54.202.254 05:40, 30 October 2006 (UTC)

Pi is an infinite sum. e is an infinite sum. The square root of 2 is an infinite sum. Every irrational number is an infinite sum. Yet we can do arithmetic with them. In fact, since the real numbers is the union of the rationals and the irrationals, and since the real numbers form a field, we can do addition and multiplication with all of the reals. We can even do division with all of them except for zero, since division by a number is actually multiplication by its multiplicative inverse, and all reals except for zero have multiplicative inverses. Please stop propagating this misconception that arithmetic is not possible with rational numbers, or irrational numbers, or infinite sums, or recurring decimals, or nonterminating decimals, because it is wrong. Arithmetic is possible with EVERY REAL NUMBER, with exactly one exception being division by zero. Maelin (Talk | Contribs) 10:20, 30 October 2006 (UTC)

No, you are completely wrong. Pi, e and every real number is not an infinite sum as you claim. You cannot do any sort of arithmetic with these numbers besides approximation unless you use the actual symbols, i.e. pi, e, etc in which case you still have only an approximate idea of what the value is in your mind. You keep stating that the reals are a field. I suppose you are busy learning about groups and fields (Abstract Algebra) in your college courses and now you think you know it all eh? Am I right in assuming that you are so wet behind the ears that perhaps I am wasting my time with you? Read carefully now because if you respond stupidly to my posts one more time, then I will unfortunately ignore you. Hey, I just don't have the energy to converse with someone who is purely argumentative. Please read what I post carefully and remember I am a graduate. I have passed all these courses with good marks a very long time ago. 198.54.202.254 14:16, 30 October 2006 (UTC)

Please don't make inflammatory comments like this again. This is mathematics, where credentials are meaningless. I don't care if you're the Dean of Mathematics at Oxford University and three-time winner of the Fields' Medal. I'm here to discuss mathematics, not be insulted. If you want to talk maths, talk maths. You're not going to impress anybody with your "I'm a graduate, so respect me, whippersnapper" rubbish. Maelin (Talk | Contribs) 14:44, 30 October 2006 (UTC)

I am not going to discuss this with the IP troll. JohnLattier, I am still interested in your thoughts. --Trovatore 05:55, 30 October 2006 (UTC)

Well, if you look upon 1/.999 and 1/.9999 and 1/.9999999 you see a pattern. This pattern looks to go off towards 0.000...10...10...10... and the number of ... goes on forever. I don't know if it is sound or not to have a 2 somewhere in there, but by the looks of the pattern, a 1 is more suitable. Also, Maelin said, "We define 0.999... as the limit of the sequence {0.9, 0.99, 0.999, ...}" No, we don't define it that way. The limit is clearly 1. A limit is the value that is being approached. 0.999... is not being approached, it is that value. 1 is being approached. --JohnLattier 12:26, 30 October 2006 (UTC)

If 0.999... is not the limit of the sequence, then what is it? Obviously, it isn't greater than 1 (surely everyone agrees with that), so it must therefore be equal to or less than 1. If it's equal to, then it is the limit of the sequence, just as 1 is. If it's less than 1, then some specific value in that sequence must be closer to 1 than 0.999..., and we can calculate which value that is. Every single value in that sequence has only a finite number of 9s, so this implies that 0.999..9 (with some calculable, finite number of nines) is greater than 0.999... (with infinitely many nines). This follows directly from the epsilon-N definition of a limit of a sequence, which I won't describe here in detail. If you claim that 0.999... is not the limit of the sequence, you need to be able to point out how many nines are needed to get closer to 1 than 0.999.... Maelin (Talk | Contribs) 13:28, 30 October 2006 (UTC)

"No, we don't define it that way." Well, yes, if you define it some other way, you are bound to get different results. Perhaps I define the formal symbol 0.999... to mean "the current age of my dog Sparky". This is a perfectly valid choice, but it happens to be inconsistent with the one the mathematical community has agreed upon. The mathematical community has, as a matter of convention (and that's all notation is—convention), agreed to use the symbol 0.999... to denote the limit of the sequence {0.9, 0.99, 0.999, ...}, as Maelin suggested, just as it has agreed to use the symbol 2 to denote the successor of 1 in the integers. You are, of course, free to use these symbols differently, but then you might as well walk into a chemistry classroom and use Xe to denote oxygen and O to denote xenon, talking about the copiousness of Xe₂ molecules on Earth. You have to agree to the notational conventions if you want to be understood. —Caesura^(t) 14:55, 30 October 2006 (UTC)

John, you're right, there is a pattern in the sequence 1/0.9, 1/0.99, 1/0.999, etc. The correct way to generalize such patterns to the infinite is not always obvious, but your conjecture is sort of a reasonable one on its face (at least, not absurd on its face).

But then let's try a different pattern and see how it works out: Let's look at the sequence 1/1.2, 1/1.02, 1/1.002, etc. By your reasoning, the pattern here should tell us the reciprocal of 1.000...2, should it not?

Well, this sequence goes 0.833..., 0.980392..., 0.998003992..., 0.999800039992..., and so on.

So what should be the reciprocal of 1.000...2? Perhaps 0.999...8 ? Or 0.999...8000...3999...2 ? Pick the one that seems most natural to you and let me know. I have another point to make once you do. --Trovatore 16:36, 30 October 2006 (UTC)

Hang on, friend, I'm not sure about "there's a pattern" here and I'm not sure where you're going with this line of thinking, but I'm pretty sure you're math is off.

1/1.2 = .8333...

1/1.02 = 0.9803921568627450 9803921568627450... (space is for easily locating repeating portion)

1/1.002 = 0.99800399201596806387225548902196? (i gave up after this far don't know where it starts to repeat, if at all) Steevnd 17:42, 30 October 2006 (UTC)steevnd

• I think its period is something like 166. --jpgordon^∇∆∇∆ 17:46, 30 October 2006 (UTC)
  • The "dot dot dot" in what I wrote above doesn't mean "repeating"; it just means that I've left stuff out. Now, will everyone please let John answer the question? --Trovatore 18:08, 30 October 2006 (UTC)

I'd say the concept seems impossible to write down on paper. It does start with 0.999..., approaches equivalency with both 0.999... and 1.0, never reaching either of course. I don't have a strong calculator, but it appears to be reaching towards 0.9...80...39...20...159 or something along those lines. I don't know how to make sense of it. Please enlighten me. --JohnLattier 21:29, 31 October 2006 (UTC)

Good enough; we're really only interested in the second infinite block anyway, which you appear to agree would be 8000... according to your reasoning. So the reciprocal would start with 0.999...8000... and go on from there.

Next question: Is this number 0.999...8000...whatever... greater than 0.999...? It ought to be, if 1.000...1 is greater than 1.000..., am I correct? --Trovatore 21:47, 31 October 2006 (UTC)

I was thinking 0.999...8- is less than 0.999... because 1.000...2 is greater than 1.000...1 --JohnLattier 22:26, 31 October 2006 (UTC)

But how can it be? You start with the infinitely many 9s, and add an 8. Hasn't that got to make it bigger, the same way that adding a 1 after the infinitely many 0s in 1.000...1 makes it bigger?

This is where I was going all along; this is a contradiction in your analysis. Unless you can fix it somehow, your claim that the reciprocal of 0.999... is 1.000...1 doesn't seem to fit into any larger scheme, so it's pretty much doomed from a mathematician's point of view. --Trovatore 22:31, 31 October 2006 (UTC)

You aren't adding an 8. You are substituting a 9 for an 8 at an infinitesimal position in the number. Like how 0.999...9 > 0.999...8.--JohnLattier 22:37, 31 October 2006 (UTC)

So are you saying that 0.999...9 is the *same* as 0.999... ? I would have thought 0.999... was the same as 0.999...0, yes? and then you're substituting the 8 for a 0, not for a 9. --Trovatore 22:42, 31 October 2006 (UTC)

I see 0.999... as 0.999...999...9...9...999...99999999... with an infinite number of ...'s --68.211.195.82 22:47, 31 October 2006 (UTC)

Then why would its reciprocal have a 1 after only the first ...? --Trovatore 22:48, 31 October 2006 (UTC)

A great way to represent 0.999...0 is 10*0.999...minus 9 And this is a trick used by Blizzard Entertainment (probably unknowingly) to truncate the infinitesimal. http://www.blizzard.com/press/040401.shtml (not to mention improperly labeling the the limit as 0.999... when it is in fact 1.

You ask why would its reciprocal have a 1 only after the first ...? Dealing with infinity is, literally, impossible. Like, when you see 1.000...1 you are in essence seeing 1.000...00...000...0000000...1; the 1 never actually is reached. Such is infinity. Existence is so interesting, huh? --68.211.195.82 22:54, 31 October 2006 (UTC)

If it's so impossible to deal with infinity, then why did you start pontificating to us about just how to do it?

Actually, though, it's not impossible at all. It's what set theorists do for a living. It can be tricky in places, but it's quite doable.

What I had thought you were attempting to give us was some new way of thinking about infinity, not just throwing up your hands and saying "it's a mystery". Offering new approaches to mathematical concepts, including infinity, is perfectly licit, even encouraged. But you have to be able to answer legitimate questions. Thus far you haven't thought your approach through well enough to be able to do so. --Trovatore 23:01, 31 October 2006 (UTC)

I'm just trying to point people in the right direction. Dealing with the advancement towards infinity is one thing. Actually dealing with infinity is another. It's the people that think you can actually deal directly with infinity or the infinitesimal that will conclude that the infinitesimal is exactly equal to zero. A conclusion to this debate is about as close as the last digit of 0.999..., but at the same time it's no reason to stop thinking about it. In any event, I'm not sure what I did to let you down so suddenly. --68.211.195.82 23:10, 31 October 2006 (UTC)

Numberline concept

Our mathematical system defines a numberline as a one-dimensional string of infinite points. Perhaps to look at the concept of dimensions would help people understand this debate.

0-dimensional = a point. (space-type)

1-dimensional = line: an infinite series of points. (line-type)

2-dimensional = plane: an infinite series of lines. (plane-type)

3-dimensional = space: an infinite series of planes. (space-type)

4-dimensional = time: an infinite series of spaces. (line-type)

5-dimensional = ?: an infinite series of times? we cannot comprehend this (plane-type)

-1-dimensional = that which, when strung together, makes a point. ditto. (plane-type)

A numberline is a 1-dimensional string of infinite points. There are an infinite number of points on the whole line and even between any two points. 0.999..., 1.0, and 1.000...1 represent three consecutive 0-dimensional points on a numberline such that if you say, "gotcha, there are an infinite number of points even between two points," you are only just beginning to understand the value of 0.999... Gotta run, discuss-- --JohnLattier 12:37, 30 October 2006 (UTC)

Points on the real number line are not consecutive; they form a continuum. Between any two distinct points, there are infinitely many points between them. Between any two distinct points, there is a non-zero distance between them. -- Schapel 12:50, 30 October 2006 (UTC)

This is exactly why Metric space theory cannot be applied to real numbers. The main idea in a real space which is metric is that two numbers/points are the same if the distance between them is zero. In real life, this is untrue and is the real reason behind all misunderstandings regarding this very issue. You can define a point as a location but it is not very helpful assigning real numbers to points on a number line. Real numbers are finite and have finite representations. This is how arithmetic was invented - with finite quantities. This characteristic gives real numbers their realness. The limit concept is derived fro the upper bound principle. The reason we talk about 1 being the limit of the series 9/10+9/100+9/1000+... is because we don't know what the sum of this series is nor can we ever hope to find out. To blatantly define the limit of an infinite sum as the number illustrates a human tendency to resolve the uncertainty. 198.54.202.254 05:59, 31 October 2006 (UTC)

Ahhh, finally, the core misunderstanding surfaces. Firstly, the real numbers ARE a metric space. They are that, by their very construction, and the metric is defined by d(x,y) = |x - y|. Raising issues of "real life" reveals a failure to appreciate the nature of mathematics, and is characteristic of folk mathematics. In maths, we have created the real numbers as a model for quantities we observe in real life - but that model is a wholly-developed mathematical number system, all the same. It is formally defined, and it is subject to all the rigourous methods of axiomatic proof, just as any other mathematical number system is. If you want a number system in which every number is "finitely representable", then you want the rational numbers, because there are uncountably many numbers in the reals that are NOT finitely representable. Maelin (Talk | Contribs) 06:11, 31 October 2006 (UTC)

To 198.54.202.254: You claim that real numbers have finite representations. I guess you had a finite alphabet in mind. All possible finite strings over a finite alphabet (and in fact, also over a countably infinite alphabet) form a countable set, while the set of real numbers is uncountable. So the reals cannot have a finite representation over an at most countable alphabet. --Kprateek88(Talk | Contribs) 11:47, 31 October 2006 (UTC)

1.000...1 is without meaning; it requires a lack of understanding of the concept of "infinite".

--jpgordon^∇∆∇∆ 14:50, 30 October 2006 (UTC)

While 1.000...1 is not a formally defined term, I think I have a mathematical definition of the concept it probably is meant to convey. Namely, define 1.000...1 as the limit of the sequence {1.1, 1.01, 1.001, 1.0001, ...}. Or, arithmetically, it would be the limit as x approaches infinity of f(x)=1 + 10^-x. You'll find when you try and calculate this number that, not surprisingly, the limit equals 1. The proofs are similar to the proofs that the limit of {0.9, 0.99, 0.999, ...} equals 1. Dugwiki 18:29, 30 October 2006 (UTC)

Corrected your equation. mdf 19:02, 30 October 2006 (UTC)

Thanks. :) Also, another interesting thing I just noticed is that another poster had tried to ask about what 1.000...2 would represent. Extending this limit idea, 1.000...2 would be the limit of the sequence {1.2, 1.02, 1.002 ...}, which is the limit as x approaches infinity of f(x)=1 + 2*10^-x. Again, that limit equals 1. In fact, you can show that for all postive finite integers k, "1.000...k" equals 1 when 1.000...k is defined as the limit of {1.k, 1.0k, 1.00k, 1.000k, ...}. In other words, 1.000...1 = 1.000...2 = 1.000...k = 1 Dugwiki 20:21, 30 October 2006 (UTC)

Yes, this is obvious. But it's beside the point. JohnLattier was obviously not defining the meaning of these strings according to their limits in the usual sense. I was asking him what the reciprocal of 1.000...2 would be, according to his reasoning. --Trovatore 20:24, 30 October 2006 (UTC)

The reason I posted in this thread and not the thread you're talking about is that it is a footnote for my post. I never said it was directly related to the other JohnLattier thread. So it's not "beside the point" - it's just an observation that in fact all such uses of that notation will end up having the same value. And judging by some of the earlier posts in this talk page, that fact apparently is not obvious to all the readers here; it's only obvious to people who are used to dealing with limits. Dugwiki 20:41, 30 October 2006 (UTC)

Definition of "never"

Never (nĕv'ər; adverb; from Old English nǽfre): when $\sum _{n=1}^{\infty }{9 \over 10^{n}}$ actually arrives at 1 --JohnLattier 21:39, 31 October 2006 (UTC)

If you're thinking of the process of adding up each of the terms in sequence of operations, you're correct in that the process would never end and would never reach 1, because you never account for all the terms. However, if we simply realize that the expression is a geometric series, we can quickly account for all the terms and compute that the answer is 1. This confusion is already explained in the article in the "Skepticism in education" section. -- Schapel 21:54, 31 October 2006 (UTC)

What do you mean by "when"? - the expression you give is a single fixed value. It doesn't change or become equal to 1 at some particular point, it simply is equal to 1. Mdwh 22:04, 31 October 2006 (UTC)

"all the terms" Schapel, what do you mean by this?--JohnLattier 22:09, 31 October 2006 (UTC)

"All the terms" means all the amounts each of the nines contributes to the total sum. -- Schapel 22:27, 31 October 2006 (UTC)

I think this gets to the root of the disagreement. You believe it's possible to account for all the 9's in 0.999... am I correct? I say you cannot. Also I say if all the 9's were finally accounted for, you'd get infinite carry-over at each position which would terminate at the ones digit. However, I see it as by definition never happening. That's what the ... means to me, and that's what I think it should mean to anyone who interprets it. Dancing around pretty formulas and tricky "proofs" that fiddle with the infinitesimal is something we should be careful with. --JohnLattier 22:35, 31 October 2006 (UTC)

Of course you can account for all the 9's in 0.999...; there's an expression in the geometric series article that does just that. We're not "dancing around" anything. Careful analysis shows all the proofs presented in the article are correct according to commonly accepted axioms of mathematics. The ... really does mean an infinite number of nines. It all has been done very carefully already, as can be shown by the numerous references. On the other hand, I've seen you make many careless statements that are clearly incorrect. If anyone should be careful, it's you. -- Schapel 00:29, 1 November 2006 (UTC)

The definition of an infinite sum is the limit, if it exists, of the related finite sum sequence, that is the limit as k grows to infinity of $\sum _{n=1}^{k}{9 \over 10^{n}}$. You are correct that none of the corresponding finite sums equal 1, but even so the limit of that sequence of sums is 1. Dugwiki 22:59, 31 October 2006 (UTC)

Think of the limit as the asymptote that is being approached as the sequence progresses infinitely. Sadly, it is never reached; hence the definition of the word "never." This applies specifically to an infinite sum that is awaiting "carry over," a.k.a. all 9's, it goes into error. Fortunately there is no carry over error in a sum like, say, 0.333... --68.211.195.82 23:18, 31 October 2006 (UTC)

Let me remind you that "0.999..." represents one element in the set of real numbers (and some other ways to represent that element are "1", "2-1", "2*0.5"), it is not some "process" that is "approaching" or "progressing" and is "awaiting" anything, or something that can "go into error". Please use precise mathematical terms. I think one cause of confusion might be the misconception that the decimal expansions and our system of writing numbers is something more fundamental than the set of real numbers itself. --Kprateek88(Talk | Contribs) 10:39, 1 November 2006 (UTC)

Simply because there are multiple equations that give 1 doesn't mean there are multiple decimal representations. People should not even make that appeal. --68.211.195.82 13:48, 1 November 2006 (UTC)

$\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}\equiv \lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}=1$ is also an equation that gives 1. It is also the definition of the "0.999..." . --Kprateek88(Talk | Contribs) 15:54, 1 November 2006 (UTC)

82, that wasn't an argument intended to win a debate. He's trying to help you achieve enlightenment on this subject. There is no debate, because no one has produced any reputable secondary sources in disagreement with the article, and in Wikipedia secondary and tertiary sources are the only things that matter. (If you don't like that, consider a different wiki encyclopedia.) This talk page has morphed into a help desk discussion, but I think many of those in disagreement with the article are in denial about this. Or perhaps this is a hobby for them, like circle-squaring is for some other people. (The same people?) --dreish~talk 15:57, 1 November 2006 (UTC)

68.211, please prove this claim that you seem to be making that "every real number has at most 1 decimal expansion". A formal, rigourous proof, please, not just hand waving. You don't have to do it in excruciating detail, but do be prepared to back up each step with a formal mathematical argument if we should choose to challenge it, just as we are prepared to do for the proofs on this article. So far many people have claimed that real numbers can only have one decimal expansion, but nobody has proven it. Maelin (Talk | Contribs) 15:59, 1 November 2006 (UTC)

The heart of the debate is going to fall on the definition of what the symbol "0.999..." represents. As stated above, the mathematical definition of an infinite decimal expansion is typically that its value is equal to the limit of the sums of the corresponding finite decimal approximations. While calculating the value of a limit typically involves a process, it does have a definite real value, often times a value that is never actually reached by the corresponding approximations.

So the main question boils down to this - if you are willing to accept that an infinite decimal expansion is defined by the corresponding limit of the finite decimal approximations, then since you can prove that this limit equals 1 you must conclude that likewise 0.999... = 1. However, if you have a problem with the limit definition, and are proposing an alternative mathematical definition for the symbol "0.999...", then the question becomes what that new definition is, why and when it should be used and what value that definition gives you, if any.

As an interesting aside, I would bet that using the limit definition above you can prove that all real numbers have at most one equivalent infinite decimal expansion, and also either one or zero finite decimal expansions. For example, using the limit definition 0.25 = 0.24999... , and there are I think no other decimal expressions for that number. Likewise, 0.333... has no finite decimal expansion equivalent and no other infinite decimal expansion equals the same value. I'll leave proofs or counterexamples to the reader, though, since it's kind of outside the scope of this discussion. Dugwiki 16:50, 1 November 2006 (UTC)

"if you are willing to accept that an infinite decimal expansion is defined by the corresponding limit of the finite decimal approximations," I disagree with this. The limit is the value that the infinite progression approaches as the number of summations increases. But the expression never literally reaches the limit. That's based on the definition of limit to a convergence number, and that's the whole reason why we use limits in the first place. If the limit equaled the expression, there would be no use for limits. This does not detract from the fact that a limit has many highly important uses. Still, they are not exactly identical. If you think the summation progression finally reaches the limit, and says "Whew. We made it." then you are dead wrong.

You want a formal proof? I have no clue what a formal proof would look like. But I'll give it a shot.

I. If $A=B$, then $A-B=0$

II. Set $A=1$

III. Set $B=0.999...=\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$

IV. $A-B=1-\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$

V. $1=\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}$

VI. $A-B=\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}-\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$

VII. $\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}\neq \sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$

VIII. $\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}-\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}\neq 0$

IX. $1-\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}\neq 0$

X. $1\neq 0.999...$

--JohnLattier 18:07, 1 November 2006 (UTC)

You're restating your assertion at step VII without proof. The rest is just window dressing. --dreish~talk 18:48, 1 November 2006 (UTC)

VII is a direct contradiction of the definition of the sum of an infinite series. For your convenience, here is a list of the things that you can do with a defined term (or symbol) you don't like:

• Ask whether it is well-defined.
• Claim that it is hard to apply.
• Claim that it does not lead to enough interesting results to justify its existence.
• Claim that it is made obsolete by a more powerful and useful concept.
• Seek formally equivalent but simpler definitions.
• Complain that it is educationally inappropriate.
• Complain that its choice of wording is intuitively misleading.
• Set up competing definitions of new, different terms.

Here is a list of things that you cannot do with a defined term you don't like:

• Set up competing definitions of the same term. (If a term means more than one thing, it is useless.)
• Claim that the definition is wrong. (Definitions are always right, duh, by definition.)
• Refuse to use it when interpreting the words of mathematicians who do use it. (This is just misquoting people in an effort to hear what you want to hear; as a result, you can't comprehend what they're actually saying. At best this makes you a poor communicator; at worst, a sociopath.)

Now, it sounds to me as if the option closest to you is "Complain that its choice of wording is intuitively misleading." What we call a sum is not what you think. Well, too bad. Welcome to mathematics. Melchoir 19:09, 1 November 2006 (UTC)

If the geometric series does not account for all the 9's in 0.999..., which of the 9s is it missing? — Loadmaster 23:42, 1 November 2006 (UTC)

an infinite number of them --JohnLattier 00:36, 2 November 2006 (UTC)

• And which are those? --jpgordon^∇∆∇∆ 00:40, 2 November 2006 (UTC)

Perhaps I misunderstood your question. What are you asking? --JohnLattier 00:50, 2 November 2006 (UTC)

• • Same thing Loadmaster asked. Which of the 9s is it missing? Identify one for us. Describe it mathematically. --jpgordon^∇∆∇∆ 01:28, 2 November 2006 (UTC)

An infinite extension does not have the luxury of having an address for every digit. -JohnLattier --68.211.104.154 02:34, 2 November 2006 (UTC)

The set of '9' digits in the decimal expansion of 0.999... can be put in a one-to-one correspondence with the set of natural numbers. For any particular nine in the decimal expansion, it has a unique natural number giving its position in the decimal expansion, which is exactly how many positions after the decimal point that digit is. Thus, every nine has exactly one unique address, and every address contains exactly one nine. If you don't follow this, John, try reading about Hilbert's Hotel. It might help your understanding of countable infinities. If you still disagree, then please describe, in detail, the set of digits in the decimal expansion that do not have addresses. Maelin (Talk | Contribs) 05:54, 2 November 2006 (UTC)

More to the point, you (JohnLattier) have given a unique address to every 9 digit: it's the n in all of the summation equations your listed above. Are you saying that there are some 9s that don't correspond to any n? — Loadmaster 22:28, 6 November 2006 (UTC)

To Lattier: You say "if you are willing to accept that an infinite decimal expansion is defined by the corresponding limit of the finite decimal approximations," I disagree with this. The limit is the value that the infinite progression approaches as the number of summations increases. What you don't agree with, and what you say seem to be the same thing. I would recommend reading limit of a sequence and infinite series, and particularly understanding the definition of the latter. --Kprateek88(Talk | Contribs) 10:12, 2 November 2006 (UTC)

Kprateek, this may be the source of your misunderstanding, then. An infinite series that converges towards a limit is only approximated by that limit. What about that do you not understand? --68.158.245.160 09:30, 5 November 2006 (UTC)

An infinite series itself is a pair of sequences, one being the terms itself, and another being the partial sum sequence of the first sequence (of course either can be determined from the other). A sequence itself in this context is a function from $\mathbb {N}$ to $\mathbb {R}$. The limit (of the partial sum sequence) is a real number. So what do you mean by the series being approximated by the limit? Any two real numbers are equal, xor unequal. There is no question of one real number being "approximately equal" to another. Many people here somehow don't seem to like the definition of the sum of an infinite series (and instead of saying so plainly they go on to say that the definition is wrong), but not one has come up with a precise rigorous alternate definition, let alone with reasons as to why his/her definition would be more suitable. --Kprateek88(Talk | Contribs) 10:18, 5 November 2006 (UTC)

From infinite series "We say that this series converges towards S, or that its value is S" Great line. Pick one or the other. Converging towards something and equaling something are two completely separate identities. Is this the false propoganda that is being spread? Something needs to be done about this. --JohnLattier 11:25, 2 November 2006 (UTC)

Yes, something needs to be done: you need to understand that the sum of the infinite series is defined (I really wanted to put that in all caps) to be equal to the limit of the partial sum sequence. As Melchoir pointed out above, you can't just say "this definition is wrong". --Kprateek88(Talk | Contribs) 12:31, 2 November 2006 (UTC)

Perhaps you mean, "the sum of the infinite series is defined (I really wanted to put that in all caps) to be ^^Approximated by^^ the limit of the partial sum sequence. Or are you suggesting we rewrite the dictionary? --131.96.2.208 19:42, 2 November 2006 (UTC)

I mean what I say. This is the standard definition of the sum of an infinite series. You can consult any elementary real analysis book. --Kprateek88(Talk | Contribs) 10:54, 3 November 2006 (UTC)

Limit 3. Mathematics. a. a number such that the value of a given function remains arbitrarily close to this number when the independent variable is sufficiently close to a specified point or is sufficiently large. The limit of 1/x is zero as x approaches infinity; the limit of (x − 1)2 is zero as x approaches 1. http://dictionary.reference.com/browse/limit

Next, I suspect you shall want to retitle this article "close = equal" and pass it off as fact? --131.96.2.208 19:47, 2 November 2006 (UTC)

The limits mentioned here are limits of sequences, not of functions. Where would the function be in 0.9999...? In the words of your dictionary:

b. a number such that the absolute value of the difference between terms of a given sequence and the number approaches zero as the index of the terms increases to infinity.

The sequence is the sequence of partial sums (0.9, 0.99, 0.999, ...). --Huon 19:59, 2 November 2006 (UTC)

"the number approaches zero" /applaud --JohnLattier 02:05, 3 November 2006 (UTC)

I am going to try this one last time, and if it doesn't work I shall dismiss you as either irrational or a troll. 0.999... is a number. It is an element of the set of reals. It does not go anywhere. It does not approach things. It does not move. It is a single value, which happens to have a recurring decimal expansion, just like 1/3 or 5/7. Do you understand this? Maelin (Talk | Contribs) 02:33, 3 November 2006 (UTC)

You should not be so quick to dismiss everyone who differs from your solidified opinion. 0.999... is a number. It is a fixed point. All points on the numberline are of infinitesimal length on the numberline and said to be zero dimensional. Zero in this sense meaning it is extended in neither the x,y,z or any other axis (not that it equals 0 length). If points were zero length, there would be no numberline formed by stringing infinitesimals. 0.999... is considered to be the point just before 1 on the number line. Since manipulation of infinitesimals is valid only when dealing with finites, trying to find the next lower number on the numberline is impossible. But from any finite number, one can identify at least conceptually an infinitesimally larger or smaller number. Of course, being a fixed point, it does not have any kind of "movement" properties itself. But, as you already know and are just making a case of semantics, the number can be represented by an infinite sum. It goes without saying that as the number of summations tends towards infinity, the number produced by this representative sum goes to 0.999... and approaches the limit 1. Why should I even have to explain this simple concept to you? --68.158.245.160 09:47, 5 November 2006 (UTC)

Just want to make clear, 0.999... is a fixed point, but does not have a fixed value. It's location is fixed because 1 is fixed, and its location is defined in relation to 1. It's value it not fixed / not defined. Make sense? --68.158.245.160 10:08, 5 November 2006 (UTC)

All of modern mathematics is founded on axiomatic set theory. Also in modern mathematics, the real numbers are defined to be a set. If you have a specific member of a set, it has a fixed value, in that each member is a specific value of the set. There is no "fixed point" on the real number line that does not have a "fixed value". How can one fixed element that is a member of a set not be a specific element of that set? -- Schapel 16:53, 5 November 2006 (UTC)

[0,1) vs. [0,1]

Interesting point brought up on the article discussion page.

I. Interval [0,1) is defined as the interval from 0 up to but not including 1.

II. Interval [0,1) upper bound is defined as 0.999...

III. As an upper bound, 0.999... is a member of [0,1).

IV. By definition (I.), 1 is not a member of [0,1)

V. Therefore $0.999...\neq 1$

--JohnLattier 00:34, 2 November 2006 (UTC)

You just made (II) up. And it's wrong on so many levels. Melchoir 00:52, 2 November 2006 (UTC)

I think I just won the debate. gg --JohnLattier 00:58, 2 November 2006 (UTC)

Where the upper bound of [0,1) defined as 0.999...? --jpgordon^∇∆∇∆ 01:32, 2 November 2006 (UTC)

...and, if it is, then what is $(1+0.999...)/2$? Clearly it must lie in (0.999...,1). But if 0.999... is the upper bound of [0,1), then (0.999...,1) is empty! Have you found a violation of closure over the real field? If not, how do you explain this? (By the way, there is a point at which we're going to have to decide whether you truly don't believe than 0.999...=1 or whether you're an Internet troll — If you could answer our queries rather than issue broad proclaimations with neither academic nor rigorous mathematical support, it would help us decide that you're not one....) Calbaer 01:40, 2 November 2006 (UTC)

JohnLattier, you're not winning anything by simply making up statements that are clearly incorrect. The interval [0,1) does not have an upper bound. For any number that x that lies in that interval, there are infinitely many values that are greater than x and less than 1, and therefore no value in that interval can be an upper bound. -- Schapel 01:56, 2 November 2006 (UTC)

"For any number that x that lies in that interval, there are infinitely many values that are greater than x and less than 1," Exactly, Schapel, that's what the ... in 0.999... represents. You are starting to understand infinity, perhaps?

Calbear, I don't appreciate you trying to brand me a so-called troll, nor do I appreciate the fact that you took it upon yourself to cut & paste my article suggestions. Please stick to legitimate mathematics discussion. --JohnLattier 02:04, 2 November 2006 (UTC)

You are a troll. The benefit of doubt has long passed. There is no point to your participation here. You have no chance of accomplishing anything. Just as your effort here is wasted, it has become apparent that our effort spent on you is wasted as well. I will not address you again, and I strongly urge other Wikipedians to follow suit. Melchoir 02:26, 2 November 2006 (UTC)

I was merely putting forth the possibility that he actually wasn't convinced 0.999...!=1, not saying he actually is a troll. In fact, he might be convinced of 0.999...!=1, but, if so, he's too convinced, so much so that he will not consider arguments contrary to his way of thinking. Which means that, although not a troll (and I'm sorry if he thought I was saying that I knew he was a troll), he, I suppose should be treated similarly. Calbaer 02:35, 2 November 2006 (UTC)

I suppose we can never be absolutely sure, but I'm satisfied that he's a troll, and as long as he acts like one, we should treat him accordingly. Melchoir 03:56, 2 November 2006 (UTC)

I agree that it's pointless to discuss mathematics with JohnLattier. He continues to make obviously incorrect statements, while at the same time acting as if he knows better than those trying to explain mathematics to him. Whether he's a troll wishing to prolong the arguments as long as possible, or so convinced he's correct that he's unwilling to listen to anyone else, it doesn't matter. The conversations get no where, so there's no point in continuing them. -- Schapel 03:08, 2 November 2006 (UTC)

There is no space in any science for people who refuse to entertain the notion that they might be wrong. People who, when presented with a counterargument that they can't rebut, don't say, "Hmm, I see," and think about it, but rather ignore it entirely and carry on arguing their same erroneous points in a different place. In this way, people convinced of a thing beyond the point of reason are no different to trolls arguing a point they don't believe for their own fun. JohnLattier, if you really want to continue this discussion, you're going to have to acknowledge that you might just be wrong, and start using this debate as a way of finding the truth, not as a way of proving that everyone else is wrong. Maelin (Talk | Contribs) 02:55, 2 November 2006 (UTC)

Of course I can be wrong. I can be wrong, you can be wrong, a professor can be wrong, an entire system of mathematics as we know it can be wrong. I don't claim to have figured it all out. But I do feel like I can see very clearly both my own arguments and your arguments, meanwhile I feel that there are about 6 people that roam this particular board that are on automatic defensive of this article as if it were there own (it might be, who knows). I've seen right through every so-called proof. Down to the root of it all, which is a misunderstanding of the infinitesimal, a misunderstanding of limits, a misunderstanding of infinity. These all tie in together to a general misunderstanding of convergence. I feel bad to burst your bubble. I know you guys probably get a real kick out of strutting these mathematical fluffs in front of lesser educated people. Seeing the look in their eyes when you say 0.999...=1, practically contradicting the entire universe they have grown to understand. How can two separate numbers with two completely unique properties, with two completely different definitions, be the same, they ask. How can one number, which is defined clearly as the largest possible number less than one, be not less than one but equal to one, they ask. Then you beautify it with one, two, maybe three proofs. Each proof accomplishes the same set of strategies: truncate an infinitesimal, assume an infinitesimal is zero, assume a limit is exactly identical to the number converging toward it. It's pretty, I have to admit. I can follow your proofs and really believe them, until I look more closely at them. I'm a biology senior, about to move on to graduate school. I am by no means a math expert. But at one time, I was a physics major at the top of my class in calculus and differential equations (don't ask me to site diff eq, that was about 10 years ago). I discovered calculus on my own, in my bedroom, before I even knew what it was. I discovered a formula for the Fibonacci sequence. I discovered a formula for pi. These, on my own, without any guidance from math books. I've enjoyed math in the past. But now I'm a biology major, with more practical goals in mind. This debate intrigues me. I do think the conclusion is obvious, but I don't think the means to that end is resolved yet, obviously. I'm afraid this entire debate might even come down to a battle of definitions. Definitions are not easily debated, as they are the foundation of proofs. If people disagree on a definition, it will get nowhere. Another thing that will get nowhere is calling people names like troll or crank, or editing, deleting, or moving people's posts, and even the "you have an extreme problem with grasping..." kind of comments. I may have made some like this, if so, sorry. I feel like a few of you are doing a great disservice to the uneducated but rising mathematicians by warping their concept of what a point is. Just like in astronomy where the vast majority of the highly educated truly believe, and teach it like it's fact, that the Big Bang was the beginning of time and space rather than just a local occurrence. If you argue it, you are entirely shunned in the astronomy community. It's a sad fact. I left astrophysics partly because of this. It was too overwhelming. And I feel the same atmosphere here. People who buck the system are probably wrong, usually wrong, but could be right, and should never be made to feel ashamed for what they see as right. If anything they should be encouraged to proceed in their path of knowledge. Not that I am a minority in this debate. It appears, other than this board, I am a majority. So I encourage you couple of noble article defenders, keep thinking, don't let anything discourage you. Hopefully there is a truth and it can be figured out by someone. --JohnLattier 03:02, 5 November 2006 (UTC)

(1+0.999...)/2 doesn't equal 0.999... but is a more accurate representation of 0.999... You're getting there, keep trying. --JohnLattier 02:08, 2 November 2006 (UTC)

Heh — (1+0.999...)/2 is more accurate at representing 0.999... than 0.999... itself! That's like more human than human, more perfect than perfect, or more godly than God Himself! Anyway, I take it from this that you'd say that 0.000...1/2 is more accurately 0.000...1 than 0.000...1, etc. Saying something is an accurate representation yet not equal certainly precludes any algebraic argument. Or logical argument. It is clear that you are too convinced of your own opinion to be moved by any arguments or explanations, whether they appeal to intuition, logic, mathematics, authority, published works, or anything else. Calbaer 02:25, 2 November 2006 (UTC)

The closest possible number less than 1 but not equal to 1 is clearly a legitimate concept. Universally, it is understood as the value 0.999... If you try to devise ways to make your understanding of 0.999... closer to 1, you are only getting closer. You will never reach the true value of 0.999... by definition because it is infintiely repeating (hence you also won't reach 1! debate over) You can shrink the infinitesimal gap between 0.999... and 1 all you want, by (1+0.999...)/2 and you will only get closer. Another way to possibly think of it is, if you perform 1.999.../2 you are actually substituting a ...4999... at an infinitesimal position in the fraction. In fact it is at a position that is infinitesimally smaller than the infinitesimal. Even so, having a 4 somewhere in the digit would seemingly make the 0.999... smaller. The final conclusion is, of course, (1+0.999...)/2 has absolutely no logical answer because you cannot divide "..." It's ironic you should say I am too convinced of my own opinion. I'd say the same of someone who thinks "closest to 1 without reaching it" reaches it. And "up to but not including 1" includes 1.--JohnLattier 11:04, 2 November 2006 (UTC)

In the real numbers, there are no such entities as numbers that are closest to a value without equalling it. They do not exist, because the real numbers form a dense set. For any real numbers x and y, EITHER there are infinitely many numbers that lie strictly between them (infinitely many distinct values for c such that x < c < y or y < c < x), OR they are equal (x = y). Sorry, but this is a fact of the real numbers and there is simply no avoiding it. It's not a 'lack of understanding of infinity', it's a direct consequence of the construction of the real numbers. There can be no debate about this, any more than there can be debate that 2+2=4. If you don't like this property, you need to invent your own number system, but you can't claim to be working in the reals if you continue to make claims about mythical beasts like "the closest number that isn't equal". Maelin (Talk | Contribs) 12:14, 2 November 2006 (UTC)

"The closest possible number less than 1 but not equal to 1 is clearly a legitimate concept." This is exactly why those who have trouble with this concept should not be hostages to their intuition. In real numbers this is not a legitimate concept. A study of real analysis can help one understand this rigorously, although most people interested in mathematics can come to a casual understanding of this.

Since Maelin decided that one more response wouldn't hurt, I'll allow one of myself, too. It is always a mystery to me what the true intentions of a seemingly irrational person is. Is he trying to get attention? Does he believe himself too correct to bother to check reputable sources? (In this case, I suspect so, since "universally, it is understood" is used to describe something that is universally though of as wrong among mathematicians.) Or does he think that he is correct and established knowledge wrong, fancying himself a Copernicus or Einstein without the associated apprenticeships and doctorates? Alas, it is part of the human condition to resist all types of evidence to prove an earlier belief wrong. Some people just have that condition more than others.

Before I saw this article a few days back, I'd not thought much about 0.999..., but my mathematical intuition (and mental epsilon-delta analysis) showed me that it was indeed equal to 1, just like how -0 is equal to +0. While 0.999...=1 may not be terribly interesting to someone with a solid mathematical foundation, the best way to teach 0.999...=1 — through reason and patience rather than dogma — to others is. Calbaer 20:42, 2 November 2006 (UTC)

The dense set is a method of approximation, by definition. Not equality. Further, a number is a 0-dimensional point. --68.219.44.220 12:22, 2 November 2006 (UTC)

What does this even mean? How can a set be a method of approximation? A set is a collection of objects, and a method of approximation is a technique. How can a collection equal a technique? —Caesura^(t) 02:15, 3 November 2006 (UTC)

dense set "X is called dense (in X) if, intuitively, any point in X can be "well-approximated" by points in A" okay it is not a method, it is a set of numbers as you said, but that's beside the point. The point is it is an approximation, not equality. --JohnLattier 02:44, 3 November 2006 (UTC)

To JohnLattier: above you say "The final conclusion is, of course, (1+0.999...)/2 has absolutely no logical answer because you cannot divide "..."". Real numbers form a field with the operations of addition and multiplication. $1\in {\mathbb {R} },0.999...\in {\mathbb {R} }$. Therefore by the field axioms, $(1+0.999...)\in {\mathbb {R} }$. As $2\neq 0$, 2 has a multiplicative inverse, by the field axioms. This is (1/2). Therefore $(1+0.999...)/2\in {\mathbb {R} }$, again by the field axioms. Please avoid vague statements like "you cannot divide "..." ". --Kprateek88(Talk | Contribs) 15:57, 3 November 2006 (UTC)

Adding 1 to 0.999... forms a definite answer, 1.999... However dividing that by 2 does not give any definite answer. It seems if you manipulate the infinitesimal, it is not valid mathematics. 0.999...^2 would have no meaningful answer, while 0.999.../0.999... would. 0.999.../2 would have no meaningful answer, while 0.999...-1 would. There should be a set of rules for manipulating the "..." But I cannot find any anywhere. Explain to me why 0.999... is considered a real number. Perhaps that should be questioned? --JohnLattier 03:09, 5 November 2006 (UTC)

"It seems if you manipulate the infinitesimal, it is not valid mathematics." LOL! Pot, calling kettle! Come in, kettle! Holy moly. Where'd you all find this guy?! B, Nov 5, 2006

Wow, that's quite a wild reaction you had there, B. The infinitesimal is a valid concept, but when one goes about trying to manipulate it, it might or might not be a valid function. Like I said, squaring an infinitesimal doesn't yield a meaningful result. But adding a finite number does. I'd like to see someone come up with a system for computing whether a manipulation of the infinitesimal is valid or not. In the meantime, I'd also like to see both a decrease in the number of rude or arrogant comments and an increase in the number of people who understand the concepts of infinity and infinitesimal. --68.158.245.160 09:26, 5 November 2006 (UTC)

Replying to John's comment above: We are all, of course, aware of the fact that anyone can be wrong, including proffesors. That doesn't change the fact that experts in a field know much more about what they are talking about than laymen, even smart laymen. People aren't born with knowledge, they learn it - and can be quite wrong if they have not learned about something. If mathematicians (and here I mean every single mathematician) say that people are wrong to believe that 0.999... is not 1, then it is indeed the amateur people who are wrong, not the mathematicians (at least, with extremely high probability). The amateurs just don't know any better. Likewise, when astrophysicist say that the big bang theory is about the big bang being the beginning of space and time, then this is indeed what the theory is about, no matter what common people think they know about it. People find it easier to imagine that the big bang is just some local event - that agrees with everyday intuition, and sadly this is what is commonly taught at schools. But people who have actually studied (as in "learn" and as in "research") astrophysics understand that this is not a satisfactory description, and it can only be thought of as the beginning of spacetime. Of course, this is just a theory, and theories can be wrong - but at least the experts know what this particular theory says.

Every proof in the article does the same thing: Use the properties of the real numbers. If you think that the result 0.999... = 1 is wrong within the real numbers, then come back after you've learned a thing or two. If you are simply saying that the system of real numbers is inadequate for one purpose or another, then keep in mind that this is counterintuitive and contrary to current knowledge (mathematical and physical), and that you have not yet given a single valid reason why this is so. -- Meni Rosenfeld (talk) 18:17, 6 November 2006 (UTC)

Error in logic in calculus and analysis section

The author of the article refers to 0.999... as $1-\lim _{n\to \infty }1/10^{n}$ but that's circular reasoning. The LIMIT portion is obviously 0, as limits are defined as the number that is being approached.--JohnLattier 03:32, 29 October 2006 (UTC)

It's not circular reasoning; it follows directly from the limit definition of the real number represented by a decimal. Melchoir 03:53, 29 October 2006 (UTC)

No, if the author is going to define that limit to be BOTH 0 and 0.000...1 he is using his own conclusion to prove his proof. That limit is 0. That limit is not 0.000...1. A limit is defined as the value you approach as you go to infinity, not the infinitely repeating decimal number before it. The author's brain is starting with the assumption they are the same number, which is circular thinking.--JohnLattier 08:57, 29 October 2006 (UTC)

Who is this "author" you speak of? I kinda think we should just discard any comment that uses the meaningless notation 0.000...1 -- there's no reason to consider it. --jpgordon^∇∆∇∆ 14:13, 29 October 2006 (UTC)

There is no such thing as 0.000...1, if "..." represent infinitely many repetitions. But 0.999... is the limit of the sequence 0.9, 0.99, 0.999, ..., which is the same as the limit of 1-0.1, 1-0.01, 1-0.001, ..., and hence the same as 1 minus the limit of 0.1, 0.01, 0.001, ... There is no circularity in this, only an application of the limit of a sequence to define the value of a non-terminating decimal, and some uncontroversial manipulations of such sequences and limits.

By the way, people who have difficulties accepting 0.999...=1 might be glad to know that no serious difficulties would arise if mathematicians instead had decided to declare all decimals ending in 999... to be undefined, thereby creating a one-to-one correcpondence between decimal notation and real numbers (except for such possibilities as appending any number of zeros to a terminating decimal). I think that is a less weird and more understandable way of avoiding 0.999...=1 than the non-standard number systems.--Niels Ø 09:50, 29 October 2006 (UTC)

You need nonterminating decimals, otherwise you don't even get all the rationals, e.g., 1/3. No mathematician will accept the "undefining" of a nonterminating decimal just because it is comprised of 9's - that's an utterly ad hoc move to try to satisfy people who don't understand what you're doing in the first place. The problem lies not in the 9's, but in failure to appreciate "nonterminating". Talk of "0.000...1" is irredeemable. Three contradictions: 1) Every symbol in a decimal expansion has a position indexed by a natural number (a positive integer) and an associated power of 1/10. What position could the 1 occupy in "0.000...1"? What power of 1/10 would the 1 correspond to? The infinity'ith? Impossible! 2) Every decimal is ultimately shorthand; e.g., 1/4 = 0.25 = 0.250000... with recurring zeros. What could "0.000...1000..." mean? Nonsense! 3) What symbol would come just before the 1 in "0.000...1"? Is it a zero? In which case, why do the recurring zeros stop recurring? If it's not a zero, why is the symbol ignored? Inconcievable! So: "0.000...1" is not a legitimate number; sadly, any post which attempts to make use of "0.000...1" instantly disowns any mathematical viability it might have aspired to. Harsh, but true.

Yes of course we need non-terminating decimals. What I am saying is that some of the flatworlders coming this way might find it easier to understand the logical status of the "0.999...=1" issue if we told them that math could have dealt with it by (i) leaving 0.999... and all other decimals ending in 999... undefined, thereby (ii) creating a system with one-to-one correspondence between the reals and decimal notation. Here, (i) is an ugly complication when we at the same time do define e.g. 0.333..., but (ii) is in itself quite satisfying. History, tradition, convention among mathematicians happens to have chosen another, equally valid, solution: Avoiding the complication (i), we accept that some numbers do not have a unique decimal representation, which, however, in itself is a rather ugly complication.--Niels Ø 07:52, 1 November 2006 (UTC)

Can someone please explain why a person by the name "AzaToth" removed my message and noted "vandalism?" I signed the message. Anyway here it is, again. I'd appreciate it if you refrain from deleting others' posts. -JohnLattier 1047p 10/29 1.000...1 is just as legitimate as 0.999... -JohnLattier 916p 10/29

• Perhaps "vandalism" was too harsh a word, but it's kinda hard to deal with people whose are simultaneously stubborn and utterly wrong. --jpgordon^∇∆∇∆ 04:05, 30 October 2006 (UTC)

It depends on what you mean by .... If you mean an infinite number of zeros followed by a one, that isn't the decimal representation of a real number. On the other hand, 0.999... is a decimal representation of a real number, even though there are infinitely many nines. If you mean there are a finite number of zeros followed by a one, yes that number is legitimate, but that number would be greater than one by a finite amount. On the other hand, by 0.999... we mean that there are truly an infinite number of nines, and that number is not less than 1 by any amount at all.

If you disagree, look at it this way: in the decimal representation of a number, each digit is at a specified position, and that position is given by an integer. For example, in 0.456, the 4 is in the position 1, the 5 is in position 2, and the 6 is in position 3. In 0.999..., there is a 9 in position 1, and in position 2, and in position 3, and so on for all the integers. In 1.000...1 with an infinite number of zeros, what integer position is the last 1 at? -- Schapel 04:01, 30 October 2006 (UTC)

"in the decimal representation of a number, each digit is at a specified position" -I disagree. Tell me all the specified locations of the digit 8 in the number $\pi$

As such, the value of the infinitesimal 1 in 1.000...1 ( aka 1/0.999...) has no specified non-infinite value. --JohnLattier 04:57, 30 October 2006 (UTC)

That's a cheap way out. There are infinitely many 8s in the decimal expansion of pi, so obviously we can't list them all, and anyway, asking for such a thing demonstrates having missed the point. What we want to do is talk about specific digits. In the decimal expansion of pi, the second digit 8 occurs in the 18th decimal place. Where does the second digit that is a 1 occur in 1.000...1? If you can't specify that, you fail the criterion that every specific digit in a decimal expansion of a number occurs at a specific and well-defined position, given by an integer value. Maelin (Talk | Contribs) 11:31, 30 October 2006 (UTC)

I don't know where you get this "criterion," but whatever your source is, it's wrong. 0.999... is approaching 1 from opposite direction that 1.000...1 approaches 1. If you can understand 0.999... you can understand 1.000...1. Listen, if you can list for me the specific locations of 9 in 0.999..., then I will tell you where the 1 is located. "Every" or "so on for all the integers" doesn't cut it. I want an exact list. If you can't, your criterion rules out both 0.999... and 1.000...1 as mathematical concepts, which tells you your so-called criterion needs to be questioned. --JohnLattier 18:06, 30 October 2006 (UTC)

Mr. Lattier: You're talking about an infinite sequence:

0.1

0.01

0.001

0.0001

In each term in the sequence separately, the "1" is an a specific position. The limit of the sequence has a decimal expansion in which every digit is 0. It does not have any decimal expansion with any "1" in it. Michael Hardy 18:13, 30 October 2006 (UTC)

How does not being able to enumerate where every nine is in 0.999... make it not a mathematical concept? If I ask for a list of all the integers, you can't write it down, either. Does that make integers not a mathematical concept? -- Schapel 18:15, 30 October 2006 (UTC)

Your statement that "0.999... is approaching 1" is incorrect, since 0.999... is a single fixed real number, and a single fixed real number cannot approach anything. As for the criterion, try looking at it like this: If you choose any particular 9 in the expression 0.999..., then I can tell you which power of 10 it is multiplied by, but you cannot tell me which power of ten is associated with the second 1 in the (ficticious!) number 1.000...1 - can you? Madmath789 18:17, 30 October 2006 (UTC)

Dispite the fact that JohnLattier wants us to list infinitely many positions (of 8's in the decimal expansion of π) as opposed to the *single* position asked of him (for the other 1 in "1.000...1"), we can still beat him at this game. There exist algorithms for calculating π to any accuracy - check History of numerical approximations of π. We can therefore build a Turing machine P that accepts as input a natural number i, calculates π to an accuracy whereby the i'th digit is known for certain, and then outputs the i'th decimal digit of π. We can then build another machine (not quite Turing) which is set on an infinite loop and calculates P(i) for each natural number i and prints only those i for which P(i)=8. Done.

Next: the following set S is an enumeration of all the decimal positions in .999... which are 9's: S = {i : is a natural number}. S is an exact list. This should be obvious from the definition: 0.999 = \sum_{i=1}^{\infty} 9 * 10^{-i}. So it's an infinite list. There are infinitely many decimal positions. It's an enumeration. Deal with it.

The game can also be played the other way. Given a natural number i, what is the digit occupying the i'th position in a decimal? For 0.999... it's a 9; for π it's P(i). What could it be for "1.000...1"? I shall expand the list of challenges to this "1.000...1" nonsense to five, since John has so far only presented one (utterly inadequate) response to my original three:

1/ What could the position of the other 1 be in "1.000...1"?

2/ What could "1.000...1000..." possibly mean?

3/ What digit could come before the other 1 in "1.000...1"?

4/ For a given i, what digit would sit in the i'th dcimal position of "1.000...1"?

5/ How could you find a series b_{i} so that "1.000...1" appears in the form \sum_{i=0}^{\infty} b_{i} * 10^{-i} ?

Challenge 5 is the money-question; it contains the very definition of the infinite decimal. If you can't do 5, you don't have a real decimal. Various other challenges are obstructions to 5.

As a demonstartion of what is required, let's go through these challenges as applied to 0.999... (some repitition involved)

1/ The positions of all the 9's in 0.999... are enumerated by S = {i : is a natural number}

2/ Not applicable - 0.999... suffers from none of this final-digit-following-recurring-digits ass. It's 9's all the way down.

3/ For every 9 in .999..., a 9 comes before the 9, unless it's the first (leftmost) 9, in which case the digit before it is a 0 on the other side of the decimal point.

4/ For any given i, a 9 sits at the i'th position in the decimal .999...

5/ Put b_{0} = 0, b_{i} = 9 for i>0. Then 0.999... = \sum_{i=0}^{\infty} b_{i} * 10^{-i}

Just to humor JohnLattier, let's start him off: put b_{0} = 1, b_{1} = 0, b_{2} = 0, b_{3} = 0. That gets you the 1.000 bit. Now continue. Notice that the definition of the infinite decimal \sum_{i=0}^{\infty} b_{i} * 10^{-i} leaves *no room* for non-specified positions.

To humor you,

1. It is not a requirement that every number have a position. To demand such shows a lack of understanding of infinity. Saying "all the 9's" doesn't cut it.

2. 1.000...1000... is one way to represent the approachment of 1 from the direction furthest from 0. "ass. It's 9's all the way down" again, doesn't cut it.

3. "For every 9" I'm sorry, did you just account for every 9? Think about infinity again.

4. This is a good one. Any value "i" you pick would have a 0. But if you grasped infinity you would not be looking for an i'th position. You cannot reach the 1. That does not take away from the concept of a number as near to one as possible without reaching it.

5. $1.1+\sum _{i=0}^{\infty }(-.09*10^{-i})$ --JohnLattier 04:13, 1 November 2006 (UTC)

John, what is 1.000...1 - 1? Maelin (Talk | Contribs) 04:16, 1 November 2006 (UTC)

Saying "it is not a requirement that every number have a position" shows a lack of understanding of arithmetic. Of course each digit needs to have a well-defined position, otherwise you would not know the value of the digit in the decimal representation. Saying every digit is nine, on the other hand, gives the exact position and value of each nine. What's the problem with saying that? Just saying it "shows a lack of understanding of infinity" makes no sense. Each nine is in a definite location a finite distance away from the decimal point. -- Schapel 04:44, 1 November 2006 (UTC)

Maelin: 0.000...1

Schapel: Okay, then list for me every position of 9 in 0.999... And don't say "all" or "every." I am giving you that in the instruction. List each position in terms of units to the right of the decimal. Ready...go! --JohnLattier 05:04, 1 November 2006 (UTC)

Let me digress a bit first. Would you agree that f(x) = 1 is a well defined function over the real numbers? If not, what is the problem with it? -- Schapel 05:06, 1 November 2006 (UTC)

Explain your idea of "well defined" so we are on the same page--JohnLattier 05:09, 1 November 2006 (UTC)

That a function is "well defined" means that I have unambiguously specified the output of the function given the input to the function. If I give two people that definition of the function, and I ask them both what f(x) is for a given x, they should always agree on the answer. In computer programming parlance, I have stated an algorithm for computing f(x). -- Schapel 05:13, 1 November 2006 (UTC)

ok, f(x)=1 is well defined, no matter what you plug in for x, you will output a 1. right? example: x=6,000,000,000; f(x)=1 --JohnLattier 05:27, 1 November 2006 (UTC)

Exactly. How about a(i) = 9 for any positive integer i? Is it well defined? If not, explain why not. -- Schapel 05:31, 1 November 2006 (UTC)

a(i) = 9 is fine, no less well-defined than f(x) = 1; i said "any" to mean if I were to pick a valid finite number, yes. --John Lattier --68.211.195.82 07:07, 1 November 2006 (UTC)

All integers are finite. What do you mean by a valid number? Are some numbers invalid? If so, which ones? Is a(i) not well defined for some values of i, where i is a positive integer? If so, which values of i? -- Schapel 12:03, 1 November 2006 (UTC)

Just wondering where you're going with it. Didn't want to see it lead to you throwing an infinite or infinitesimal number in a function as the independent variable; it may or may not be sound. You're fine, just keep going. -John Lattier --68.211.195.82 14:17, 1 November 2006 (UTC)

I'm going to define a number n whose decimal representation according to the definition and notation in that article are given by $a_{0}=0$ and $a_{i}$ = a(i), where a(i) is the function I defined above, a(i) = 9. Is that decimal representation well defined, in the sense that each digit in the decimal representation is well defined? If not, why is it not well defined? Have I not unambiguously defined what each digit in the decimal representation is? -- Schapel 14:59, 1 November 2006 (UTC)

While a(i)=9 is well defined, "i" is not. --JohnLattier 11:16, 2 November 2006 (UTC)

As I said before, i is just an integer. -- Schapel 12:39, 2 November 2006 (UTC)

Wow, John, you completely missed the boat on the "0.000...1000..." challenge. The idea was, how on earth would you present some sensible b(i)'s for this "0.000...1000..." fiction? Decimals are defined as \sum_{i=0}^{\infty} b_{i} * 10^{-i} - you only specify b(i)'s. John has three problems: 1/ you don't get to throw in other operations as in 1-\sum_{i=0}^{\infty} a_{i} * 10^{-i}, 2/ you don't get to add a "last" term on the end - that would be (\sum_{i=0}^{\infty} c_{i} * 10^{-i} ) + c_{\omega}. There isn't a "last" term in a limit to infinity by definition. Have you read Hilbert's paradox of the Grand Hotel yet, John? Specify the b(i)'s, all the b(i)'s, and only the b(i)'s, then you have a decimal. Presently, John does not. 3/ John requests a list of all the 9's in 0.999... but appears to have overlooked the fact that he will have *identical* difficulties in locating all the zeros in his "0.000...1" fantasy. John's position is not even internally consistent. On the upside, it would make a superb Uncyclopedia article.

Sorry I didn't play along with his silly challenge. "John requests a list of all the 9's in 0.999... but appears to have overlooked the fact that he will have *identical* difficulties in locating all the zeros in his "0.000...1" fantasy." That's exactly my point. The two numbers both share any kind of infinity-based error you throw at them. It's easy to see through his trickery: (i) is not defined, so neither 0.999... nor 0.000...1 has a fixed value, but using semantics, namely the unmathematical "all the 9's" can make 0.999... appear to have a fixed finite value. I feel sorry for the readers that this 0.999...=1 garbage is on a place like wikipedia, a site I respect.--JohnLattier 02:21, 3 November 2006 (UTC)

As I've repeated before, i is clearly defined as a positive integer in the function a(i)=9, and this function gives the decimal expansion 0.999.... However, there is no function b(i) that gives the decimal expansion 1.000...1, and therefore it is the only one with an "infinity-based error". Go ahead and ask any math professor. -- Schapel 02:50, 3 November 2006 (UTC)

Then go ahead and clearly define "i" --JohnLattier 03:14, 3 November 2006 (UTC)

Wow. If you're going to argue about how i is defined in this context, then it is established that you do not accept any definition, no matter how simple. Nov 6 06()

i is any member of the set of positive integers, as I have said many times before. What a positive integer is is clearly defined. If I ask if 1, one million, or one googol is a member of the set, any mathematician will tell you they are. If I ask if 0, one-half, or infinity is a member of the set, any mathematician will tell you they are not. You seem to be having extreme difficulty with elementary mathematical concepts such as sets, integers, and arithmetic, much less anything involving infinite sums. -- Schapel 03:23, 3 November 2006 (UTC)

Proofs need to be reexamined

$3\times 0.333\dots {\mbox{equals}}0.999\dots$

--When performing multiplication that leads to infinite carry over, each fractional digit is suspended in an error state awaiting carry over. The answer to this multiplication cannot be attained unless it can be expressed in some sort of non-infinite form such as 1/3. So the answer is actually 1. Not 0.999...

$10\times 0.999\dots {\mbox{equals}}9.999\dots$

--Also not a viable calculation. 10 X 0.999... is better expressed by the theoretical 9.999...0. The author of this article, as well as Blizzard, use this truncation to "forget about" the infinitesimal, and use the subtraction to return to 0.999... It is not sound.

from "Geometric progression" "one can apply the powerful convergence theorem concerning infinite geometric series"

--The author does not understand that convergence does not mean equality.

$0.999\ldots =\lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}=\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)=1-\lim _{n\to \infty }{\frac {1}{10^{n}}}=1.$

--The author does not understand that the limit represents the asymptote which is being progressed towards. This asymptote is not reached, as the author would like you to think by the bold declaration of =1. The limit equals 1, of course, because a limit is defined to be that which is being approached. The expression, on the other hand, does not.

0.999… is then the unique real number that lies in all of the intervals [0, 1], [0.9, 1], [0.99, 1], and [0.99…9, 1] for every finite string of 9s. Since 1 is an element of each of these intervals, 0.999… = 1

--Several mistakes. First, if 0.999... really equaled 1, it would also lie in the interval [1,2] which it clearly does not. Second, it the statement that 0.999... and 1 are an element in "EACH" of these intervals shows and ignorance towards infinity. You can not account for each interval. Third, the use of the word interval shows that 1-dimensional distance (numberline) is being referred to, and no matter how small a distance between two points you measure, it will always be an interval.

Students of mathematics often reject the equality of 0.999… and 1

--Bias. This is not a professional statement. This is in essence saying that only the lesser educated (students versus professors) disagree with your dogma. The source that the author quotes doesn't even proclaim these nose to the air attitudes or conclusion.

Every element of 0.999… is less than 1

--Yes but you cannot draw from this that 1 is equal to 0.999...

$\lim _{n\rightarrow \infty }{\frac {1}{10^{n}}}=0$

--Certainly, it equals 0. The limit does. The expression does not! Author repeatedly makes this same mistake. The expression equals the infinitesimal. Best represented in decimal form by 0.000...1. The 1 is never reached.

this idea fails because there is no "last 9" in 0.999…

--This idea actually doesn't fail. While there is no last 9 in 0.999... and no last 1 in 0.000...1, there is no requirement that the theoretical last numbers must actually ever be reachable. That is not so subtly noted by the ...

From "P-adic number"

${\begin{array}{rl}9&=-1+10\\99&=-1+10^{2}\\999&=-1+10^{3}\\9999&=-1+10^{4}\end{array}}$

and taking this sequence to its limit, we can say (informally) that the 10-adic expansion of -1 is $\dots 9999=-1$

--Okay, can someone explain how $-1+10^{\infty }$ goes to -1, please. That's a positive infinity there. --JohnLattier 05:01, 1 November 2006 (UTC)

The majority of your arguments go to the equality itself and not the content of the article. I would invite you to follow the directive at the top of this page and move them to the Arguments page. As for:

Bias. This is not a professional statement. This is in essence saying that only the lesser educated (students versus professors) disagree with your dogma. The source that the author quotes doesn't even proclaim these nose to the air attitudes or conclusion.

The article provides several citations of studies showing that students do indeed often reject this equality. The reason the article does not provide sources showing mathematics professors rejecting the equality is not bias, but merely an unavoidable result of the fact that there are none.--Trystan 05:35, 1 November 2006 (UTC)

You claim that in the limit of 1 / 10^n, the limit is equal to zero but the expression is not. What expression are you talking about? Maelin (Talk | Contribs) 05:40, 1 November 2006 (UTC)

It would be a shame if any post that challenges the dogma is moved to arguments. No argument here. Just need to reexamine these proofs. I have plenty of posts on the arguments board if anyone wants to debate anything stated in this post. This post is merely an article-based comment. I don't have plans to add to it or respond to comments here.--JohnLattier 05:46, 1 November 2006 (UTC)

It's not dogma, it's a mathematically proven theorem. What your intuition says has no bearing on it. The fact that 0.999... = 1 is a direct consequence of the way we have constructed the real numbers. We don't expect people to take anything on faith in this article - there are several different proofs, including a couple of rigourous ones. Every result in them can be proven from earlier results, and so on, until you get down to the axioms. The fact is, there is no dispute here among professional mathematicians, because they know a valid proof cannot be disputed. Maelin (Talk | Contribs) 05:53, 1 November 2006 (UTC)

This is not something that's controversial, something that may or may not be true. It's something that is true, like the Earth not being flat, the Earth not being the center of the solar system, or George W. Bush being President of the United States. Many people have insisted on denying these facts and many more still believe one or more to be counterintuitive. But something can be counterintuitive and still be true. They, like $0.999\dots =1$, are Wikipedia:Verifiable. Intuition to the contrary has the weight of Wikipedia:Original research.

As I wrote in the FAQ, $0.000\dots 1$ 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 $d\cdot 10^{-k}$ toward the value of the number represented. It may help to ask yourself (JohnLattier), "How many places past the decimal point is the '1'?" It cannot be an infinite number of places, because all places must be finite. So what is your belief about $0.000\dots 1$? As you believe, is it or is it not a real number? What position is the "1" in? Do you believe that there is a real number that cannot be expressed in decimal notation, that is, as a limit of terminating decimals? What is ${\frac {0.000\dots 1}{10}}$? Is it $0.000\dots 1$? If it is, do we not have, for $c=0.000\dots 1$, $10c=c$, leading, through algebra, to $c=0$? Similarly, what is $(0.000\dots 1)(0.000\dots 1)$? Thinking seriously about these questions might help you challenge your current notions and hopefully come to a better understanding. Or answering them might enable people to isolate the flaw in your reasoning and help explain what is wrong with your thinkng. Which (hopefully) will in turn lead to a better article. (Not answering them will tell us that we should ignore you — not that I expect that.)

Similarly, do you believe that the article is wrong while there are mathematical textbooks and/or papers that are correct? (If so, please provide a reference; there are dozens of references on the page in support of 0.999...=1.) Or do you believe that all mathematicians writing on the subject are wrong and that you are correct? It seems as though you see this as a dogma that we all unthinkingly believe and that you fully understand the flaws of. You must realize that that isn't the case. In fact, judging from your willingness to dismiss mathematical fact as "not viable" as those who present them as ones who do "not understand," I dare say that your prose is more dogmatic than that of those who are trying to explain this to you. I hope that you, however, are not, and can think about the questions I've posed. Also, please don't be surprised if this moves to the Arguments page, since these in-depth discussions have little to do with the presentation and everything to do with arguments against 0.999...=1. (By the way, your question about how ...999 = -1 in the p-adic number system is explained by the fact that this is not the real number field and so doesn't have to behave by similar logic. But that has little to do with this article, since the segment you found objectionable is from P-adic number, not 0.999....) Calbaer (various times), 1 November 2006 (UTC)

All of these mathematical "proofs" require a cetain, non-intuitive definition of .999..., the definition being that that equals one. The intuitive way to think about it is that it is the largest number smaller than 1, which obviously can't be a real number. And how can you say that .999... is a number, but .000...1 isn't? The real problem is that there *is* controversy, and you won't accept it. Fresheneesz 23:10, 1 November 2006 (UTC)

No, they require the use of the real numbers as defined by mathematicians. The intuitive definition that some prefer is not mathematically rigorous and cannot be made so, so in that "intuitive" system 0.999... doesn't mean anything, so it's impossible to write an article about it. If there's controversy between mathematicians and people who don't know the subject, well, that's not much of a controversy. -- SCZenz 23:25, 1 November 2006 (UTC)

The "largest number smaller than 1" leads to immediate problems, first among them being the number (1+0.9999...)/2. What does your intuition say to that? Does it tell you there are numbers that cannot be divided by two? Concerning 0.000...1, that's not a decimal representation of anything, and you will have difficulties to define what it is except being a string of symbols. Can it be divided by 10? Can you square it? Sure, you could probably invent some strange number system where the string of symbols "0.000...1" does represent a number - but there is no standard system, while 0.9999... is well-defined in the mathematical canon. Finally, as for the controversy: There is none as far as mathematicians are concerned (if you doubt that, provide a reference). And the non-mathematicians who doubt 0.9999...=1 probably couldn't agree on a single alternative. --Huon 23:28, 1 November 2006 (UTC)

It's like asking what is the size of the open interval [0,1). It's exactly 1, even though the interval does not include 1 itself. — Loadmaster 00:02, 2 November 2006 (UTC)

Fresheneesz:A decimal expansion is taken to have an integer part and a function from $\mathbb {N}$ to {0,1,2,3,4,5,6,7,8,9} part. The properties of the real numbers that expansions represent have been studied. 0.999... has 0 as the integer part and the function part returns 9 for every natural number. What does 0.000...1 represent? It's definitely not a decimal expansion if the number of 0's is to be infinite - the integer is 0 and the function returns 0 for every natural number. That's it. The "1" at the end is gibberish. And, as we've said before, the definition of 0.999... comes directly from the general definition of every decimal expansion, and its equality to 1 can be established from their general properties.

Loadmaster: Not really sure about that, so the argument below is rather off topic. -- Meni Rosenfeld (talk) 21:10, 2 November 2006 (UTC)

What is the size of the open interval? [0,1)

It's like asking what is the size of the open interval [0,1). It's exactly 1, even though the interval does not include 1 itself.

Now this leads to some bizarre situations. (I hope I get the symbols right, correct me if I don't.)

• [0,1) is size one.
• [0,1] is size one.
• Therefore the two sets are equal, despite one being a superset of the other, containing an element that the other does not.
• Alternately, [0,1] is larger then [0,1), so 1 does not equal 1. Yup. Makes perfect sense to me! Algr

Actually, the first two bullets just imply that the interval [1, 1], better known as the singleton {1}, is "size" zero. Melchoir 04:26, 2 November 2006 (UTC)

True, but that is beside the point. If you know that a set is size one, and it's lowest element is zero, does it include one? Algr 04:52, 2 November 2006 (UTC)

Not necessarily; you've just provided the counterexample. Melchoir 05:02, 2 November 2006 (UTC)

Is bullet 3 true? How about 4? Algr

The sets aren't equal. They just have an equal Lebesgue measure. The Lebesgue measure is not a count of how many things there are in the set (in both intervals, this number is uncountable), it's more of a measure of how much of Euclidean n-space is occupied by the set. Maelin (Talk | Contribs) 05:31, 2 November 2006 (UTC)

Right, the sets are not equal (because they do not contain the same elements), but they have the same cardinality. — Loadmaster 22:35, 6 November 2006 (UTC)

Algr, your third bullet seems to assume that if two sets are the same size, then they are equal. Could you clarify this assumption? Clearly, {1, 2, 3} and {4, 5, 6} have the same size (measure 0 and cardinality 3) but are not identical. —Caesura^(t) 13:47, 2 November 2006 (UTC)

Here's another counterexample: [0, 1/2] ∪ [3/2, 2] is a set of measure 1 and minimum 0 that does not contain 1. —Caesura^(t) 14:02, 2 November 2006 (UTC)

Not great examples, as Algr has restricted to subsets of [0, 1] (explicitly at one part, implicitly at the other). A better example is $\mathbb {N}$ and $\mathbb {Z}$; same cardinality, yet clearly a proper subset. Generally, a proper subset can have the same size (cardinality, measure, etc...) as the superset. -- Meni Rosenfeld (talk) 21:10, 2 November 2006 (UTC)

This debate is a joke

I was blocked by the Wiki team just two days ago. The violation was supposedly 'vandalism'. I have no idea what was vandalized. To all those who waste their time and energy on this discussion I only have one thing to say: you are probably better off setting up another website that shows the falacies of all these so-called proofs on Wiki. Real Analysis is a sore joke - I do not care if most of the mathematics community think it is rigorous and beyond debate. Nothing is beyond debate - ever. Wiki talks about original research - this too is a joke. Everything once started out as original research. This very article is original research. Not all mathematicians agree with it. Someone asked why there is no list of mathematicians who disagree with it. The reason is simple: It's unimportant and nothing falls apart in mathematics. I started out by showing how false Meni Rosenberg's proof is and was shouted down by the Wiki trolls (sysops/administrators, etc). Someone called Trovatore ventured as far as saying that I am an IP troll. Well, need I say more? I will not waste my time with you anymore. 198.54.202.254 08:09, 2 November 2006 (UTC)

Your talk page shows that you (or someone using your IP) vandalized many pages, not this one. You did however get a warning (rightfully so I might add) for violating WP:NPA on this page. As for fallacies and false proof, your counterarguments have been rebutted, and referring to it as being "shouted down" does not change that fact. You clearly understand neither the purpose nor the workings of WP:NOR, and you certainly seemed to spend an awful amount of time "debating" (I use the term loosely) a subject that you deem unimportant.

Just my observations, is all. On a side note, I failed to find any block messages on either your talk page or the block log. - KingRaptor 12:44, 2 November 2006 (UTC)

You were blocked for edits like this one (there are many more where that came from), not for trolling here. If that edit was not yours (but instead someone else with your IP address), perhaps you should create an account to prevent this from happening again. —Caesura^(t) 13:42, 2 November 2006 (UTC)

This article is obviously not original research; there are many reliable sources cited. Saying that 0.999...=1 is unimportant and "nothing falls apart in mathematics" is completely incorrect. If 0.999... did not exactly equal 1, we would have a contradiction, and all of modern mathematics would fall apart because it would be able to prove any theorem, and thus all proofs would immediately be meaningless. -- Schapel 16:08, 2 November 2006 (UTC)

I am coming to your conclusion as well. Anything that challenges their false dogma is either edited out or placed in this obscure arguments section. If someone wants some real discussion about this topic, scroll down on this page and be tread very, very carefully with an open, skeptical mind. --JohnLattier 11:33, 2 November 2006 (UTC)

This page is linked to at the very top of the main talk page, and has grown enough to be archived - twice (and it needs a third one at this point). It is not obscure by any stretch of the definition. Also, you should substantiate your claim that "anything that challenges their false dogma is (either) edited out" with links to edits where such content was removed. - KingRaptor 12:44, 2 November 2006 (UTC)

I think he means "edited out of the article". This is true. We will not admit exceptional claims into an already high-quality article without exceptional proof. JohnLattier has been invited many times to provide reliable sources supporting his position. —Caesura^(t) 13:42, 2 November 2006 (UTC)

Feel free to post a rigourous, formal disproof, or a reference to a reliable source that challenges the view, at any time, guys. Maelin (Talk | Contribs) 14:10, 2 November 2006 (UTC)

John, I have memories of you saying you studdied maths. Why not print out a few of the arguments here from each side, take them to a mathematics professor at your university, and request their opinion? From my experience, mathematicians are happy to oblige when someone shows an interest in mathematics.

That's not going to happen. He ignores the more difficult questions we ask and gives wishy-washy answers to those he believes he understands. To think he would actually go to a mathematics professor — and that the two would have the patience for one another — is wishful thinking. He ignores our arguments; I think it's about time for us to ignore his and concentrate on those who have difficulty with the concept but come here with an open mind, which is clearly not the case of the person who began this section. (One last thing though: to call this section obscure is a laugh; it's linked several places and is no less active than the actual talk page!) Calbaer 21:38, 2 November 2006 (UTC)

Sure I can debate it with a mathematics professor. I'll probably annoy the crap out of him, but sure. Hopefully he will come up with something that doesn't have as many blatant errors as the [0.999...]] article. I definitely don't ignore your arguments. I've not only shown the holes in the article itself, but I've responded and refuted those who repeatedly misdefine terms, and provided 2 logical mathematical proofs for why the equality is wrong. The personal attacks are a laugh though; they really place my mind at ease about the quality of brains I'm engaging in e-conversation with. --JohnLattier 02:28, 3 November 2006 (UTC)

I hope your next reply, if any, will be to let us know what the professor says. Notes for improvement are always appreciated, as are serious answers to our questions about where the arguments provided lose people. However, ignoring the questions and arguments, calling them silly, calling everyone here except you dogmatic pea-brains, and generally saying that the entire field of real analysis is wrong (and you right) is not appreciated. If you come at the professor with a similar attitude, do not be surprised if your conversation is brief. And don't be surprised if any further posts are ignored. And it is not because you "won," unless your goal is convincing us that responding to you will have no effect on either your mind or the quality of the article. If that is your goal, however, please congratulate yourself. Calbaer 02:50, 3 November 2006 (UTC)

If you really think you do have proofs that 0.999... is not exactly equal to 1, please do get them published. I'm sure you'll easily be able to find a publication that will accept them, as long as they are valid proofs. Then you can gloat all you want about proving us wrong. -- Schapel 02:57, 3 November 2006 (UTC)

How would one go about proving, for instance, that $1\neq \ 2$? --JohnLattier 03:42, 3 November 2006 (UTC)

I'll assume you mean the natural numbers 1 and 2, and use Peano's axiomatization:

1. Suppose otherwise. That is 1 = 2.
2. By definition, 1 = S0, and 2 = S1, both natural numbers (Sa is the successor of a. The definition itself is not a part of the proof, and can be easily checked to be valid by axioms 1 and 2).
3. By steps 1 and 2, S0 = S1.
4. By step 3 and axiom 4, 0 = 1.
5. By steps 2 and 4, 0 = S0.
6. By step 5, there exists a natural number (namely, 0) which has 0 as its successor.
7. By axiom 3, there is no natural number which has 0 as its successor.
8. By steps 6 and 7, a contradiction. Our initial assumptions is therefore false. Namely, $1\neq \ 2$.

QED

Are you now going to project your lack of understanding of mathematics and\or peano axioms onto the valifity of this proof? -- Meni Rosenfeld (talk) 07:31, 3 November 2006 (UTC)

Is this your idea of a proof? Step 2 is already wrong. If 1=2, and 1=S0 then 2=S0 not S1. You assume they are different numbers to finish the proof. Just like you have to assume the infinitesimal is 0 to finish the article proofs. "project your lack of understanding of" you folks sure do a terrible job of trying to convey your thoughts without lashing out.

How would a step function work for a proof that $1>0.999...$?

1. If Step[A]>Step[B], then A>B

2. A=1

3. B=0.999...

4. Step[1]=1

5. Step[0.999...]=0

6. Since 1>0, 1>0.999...

simple, to the point, makes sense. --JohnLattier 02:26, 5 November 2006 (UTC)

• Why is 5 true? How is your step function defined? --jpgordon^∇∆∇∆ 02:53, 5 November 2006 (UTC)

step returns the value in the 1's digit. --JohnLattier 03:13, 5 November 2006 (UTC)

• Ah. OK, why is 1 true? --jpgordon^∇∆∇∆ 06:29, 5 November 2006 (UTC)

As I had pointed out somewhere else on this page, one of the causes for confusion is the misconception that the decimal system of writing is something very fundamental, something sacrosanct. The real numbers can be constructed rigorously using Dedekind cuts or equivalence classes of Cauchy sequences. Decimal expansions are only a representation, which is not unique. As is commonly done in mathematics, a function is defined using some representation of the elements in the domain, and for the function to be well-defined, it is shown that different representations of the same element in the domain, lead to the same element in the range. The 'step', if it is intended to be a function defined on real numbers, is not well-defined, as different representations of the same real number may have different digits in the 1's place. --Kprateek88(Talk | Contribs) 06:22, 5 November 2006 (UTC)

John, how can you blame us for assuming you are acting on bad faith, when you do everything to make us believe this is the case? Do you really not understand the difference between my proof and your "proof"? I have stated explicitly about what objects I am talking and which axiomatization I'm using. I've specified for every state the exact justification for it (which is usually some axiom). You haven't stated what system you are describing (presumably the real numbers), which construction of it you are using, and what led you to believe in the correctness of your steps 1 and 5 (is it because you "feel like it"? Sorry, not a valid argument in a rigorous mathematical proof). The definition given in the linked article doesn't agree with 5, and the definition you give isn't a definition at all becuase it is ambiguous. Again: decimal expansions are a derived, rather than a primary, notion.

Do you really not understand that in my step 2 I have done nothing more than quote a definition? Here, let me spell it out:

Definition: 1 is the unique successor (the existence and uniqueness of which are guaranteed by axioms 2 and 4, and which is denoted as S0) of the natural number 0 (which is guaranteed to exist by axiom 1).

Definition: 2 is the unique successor (the existence and uniqueness of which are guaranteed by axioms 2 and 4, and which is denoted as S1) of the natural number 1 (defined in the last step).

Do you, perhaps, not understand how a reductio ad absurdum proof work? Then please say so, and I will willingly repeat the proof without this approach.

And yes, this is my idea of a proof. If you disagree, it means you do not know what a mathematical proof is. Sure, I could have made things more formal by writing formulas in the language of logic; and of course, I have gone into much more detail than would appear in a serious mathematical work (since any one of my steps above would be completely obvious to mathematicians). But this is more or less what a proof should look like. Get used to it.

Frankly, it seems that you are not making any sincere effort to understand what people try to tell you. And you should not be surprised that people act accordingly. -- Meni Rosenfeld (talk) 17:55, 6 November 2006 (UTC)

The real debate--->>>

Okay. After reading everyone's take on this debate, I believe I have found the root cause of the debate:

$\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}\neq \ \sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$ "The limit is an approximation of the sequence as it progresses towards infinity."

OR

$\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}=\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$ "The limit is an exact equality to the sequence as it progresses towards infinity."

Resolve this and you resolve this entire debate instantaneously. But don't answer instantaneously. Think about it. The answer obviously lies in the meaning of "limit." Is it an approximation or a statement of equality? --JohnLattier 03:39, 3 November 2006 (UTC)

If you really want to know the answer, just read the Infinite geometric series section in the geometric progression article. -- Schapel 04:12, 3 November 2006 (UTC)

"Such a series converges" http://en.wikipedia.org/wiki/Geometric_progression

"Converge: (Mathematics) To approach a limit." http://education.yahoo.com/reference/dictionary/entry/converge;_ylt=Av8erCAA6Zxdb3JUlxs9SU.sgMMF --JohnLattier 04:38, 3 November 2006 (UTC)

I was referring to this part:

$\sum _{k=0}^{\infty }ar^{k}=\lim _{n\to \infty }{\sum _{k=0}^{n}ar^{k}}$

Now, according to you, the entire debate is resolved instantaneously. -- Schapel 14:16, 3 November 2006 (UTC)

Oh, I see you did notice it but still don't believe it. Perhaps this article on the sum of a geometric series will help. Note the sentence It is this limit which we call the "value" of the infinite sum. Perhaps you can understand it; it is written at the level of high school mathematics. -- Schapel 15:18, 3 November 2006 (UTC)

That article assumes that the reader realizes that the limit of the finite sums is the infinite sum, that is, the sequence $\sum _{n=1}^{N}{\frac {9}{10^{n}}}$ converges to the limit $\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}$, which is equal to $\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}$. For those who have trouble believing this, pick up any real analysis book from the library (which would be more authoritative than Wikipedia anyway; those who doubt the veracity of this article would have no reason to believe geometric progression). Calbaer 05:07, 3 November 2006 (UTC)

Definition of limit by Dr. Patrick Speissegger: "If (Sn) converges to the real number "s" we call "s" the limit of (Sn) and write s=lim Sn."

Does that satisfy you guys?

I suggest you consider:

$\sum _{k=0}^{\infty }ar^{k}\approx \lim _{n\to \infty }{\sum _{k=0}^{n}ar^{k}}$

for approximation, in any of your future proofs. Convergence and equality are mutually exclusive by definition. Thank you for your attention, and I hope you enlighten new young minds with what you have learned here today. --JohnLattier 06:01, 3 November 2006 (UTC)

Approximations cannot be used for formal proofs. And I would be amused to see what Professor Speissegger would say about your analysis. His email address is speisseg at math dot mcmaster dot ca. Good luck Calbaer 06:35, 3 November 2006 (UTC)

"Converges and equality are mutually exclusive"? The sequence a_n = 1 converges to the limit 1, and every element of the sequence happens to be equal to that limit.

Let's see what we have here. I assume we both agree on the following:

$\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}=1$

$\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {1}{3^{n}}}={\frac {1}{2}}$

$\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}$

And you disagree with the following:

$\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}=1$

$\sum _{n=1}^{\infty }{\frac {1}{3^{n}}}={\frac {1}{2}}$

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}$

Would you mind explaining what the value each of these summations is? Are they perhaps undefined? If so, why would you mind extending the definitions by setting them equal to the above limits? -- Meni Rosenfeld (talk) 07:42, 3 November 2006 (UTC)

Lim (N->infinity) Sum [N=1 to N] 9/10^N = 1 [A] is not the same as Sum [N=1 to Infinity] 9/10^N = 1 [B] So you want to define the limit [A] as the sum to infinity [B]? Let's assume that [A] and [B] mean the same thing. If this is true for all numbers, then what is the limit of the pi series? In fact, what is the limit of the series for any irrational number? Contrary to every statement made about duplicate representation, the radix system was designed with unique representation in mind. Mathematics is about order, not chaos. Can all numbers be represented in decimal? Absolutely not! In fact most numbers can only be approximated. You rely so much on real analysis which states that every real number can be represented in decimal - this is obviously false. Real analysis is the reason your article is false. 198.54.202.254 10:51, 4 November 2006 (UTC)

I think 198.54.202.254 might be onto something. Real analysis itself might be what's at fault. As for the above comment about limits that are infinite convergent sums, I do agree with your limits (I tried calculating them, and I got kinda close, I'll take your word they converge to those numbers). The summation itself, I don't agree with. There is no value for for N that causes the sum to identically equal the limit. --JohnLattier 01:59, 5 November 2006 (UTC)

You have not given any argument against the definition of the sum of an infinite series. What do you mean by "the pi series"? If it is some series whose sum is pi, then the sum of that series is pi (well, duh). And no, the radix system does not ensure a unique representation. It was not as if some great mathematician sat down and thought "ok, so now we have defined real numbers starting from Peano's axioms for natural numbers. Now let us try to come up with a good way to represnt them on paper." The notation evolved much before reals were formally defined. Also, your statement "Can all numbers be represented in decimal?" is quite vague. If you mean, "can all real numbers be represented as finite strings over a finite or countable alphabet?", the answer is no. But, if one says "can all real numbers be represented as countably infinite strings over a finite alphabet?", the answer is yes (an alphabet of size 2 will do the job). And yes, we rely on real analysis. We rely on axioms, definitions, proofs, and precise statements. --Kprateek88(Talk | Contribs) 12:08, 4 November 2006 (UTC)

A real valued sequence $a$ is a function from the set of natural numbers to the set of real numbers. Instead of using the notation $a(i)$ we will use $a_{i}$. We say that the limit of this sequence exists and is equal to $L$ (where $L$ is a real number), iff for every real number $\epsilon >0$ there exists a natural number $N_{0}$ such that for all $n\geq N_{0}$, $|a_{n}-L|<\epsilon$. It can be proven easily that any real valued sequence cannot have two distinct limits. We say $\lim _{n\to \infty }a_{n}=L$. Do you agree upto here? Now consider something like $\sum _{k=1}^{\infty }b_{k}$. What meaning, if any, are we to give to this? We know the meaning of $\sum _{k=1}^{n}b_{k}$. Consider $c_{n}=\sum _{k=1}^{n}b_{k}$. This is a real-valued sequence (and is called the partial sum sequence of the sequence $b$), and may have a limit. If it has a limit, then it is reasonable to define $\sum _{k=1}^{\infty }b_{k}:=\lim _{n\to \infty }c_{n}$. This is the accepted definition of the sum of an infinite series. If you don't agree, please give an alternate definition, or say whether you choose not to define the sum of an infinite series at all. In the latter case, as Meni said above, what's the harm in extending the definition to the limit of the partial sum sequence? --Kprateek88(Talk | Contribs) 11:50, 3 November 2006 (UTC)

0.999...!=1, a new religion?

This article makes people react like if the article is a blasphemy against their gods. The philosophy that gets entwined into the subject makes it close to a religion (in the sense of lifestyle). Will these people go to the grave believing their arguments? Do they go through steps of emotions like terminally ill patients, until they finally reach acceptance? The mythos of 0.999... seems much more interesting and shocking to me than the original problem ever did.

I think Will Hunting should be the new Messiah. "Humble carpenter janitor proves 0.999...=1 wrong and defies the mathematical empire. The empire strikes back, seizing the commandments of Will, and nailing a counterproof upon the board for everyone to see. Will Will Hunting's proof rise again in the form of a rebuttal?" falsedef 13:24, 3 November 2006 (UTC)

There's no "philosophy". There are a set of widely accepted axioms and definitions. According to those axioms and definitions, 0.999... is exactly equal to 1. Anyone who claims otherwise is by definition wrong. If you want to make up your own axioms and definitions such that 0.999... is not equal to 1, that would be equally valid. Emotions are completely irrelevant. The proof can be done by computer. -- Schapel 14:13, 3 November 2006 (UTC)

As a further answer to your questions, I suggest reading about cranks and trolls. -- Schapel 14:45, 3 November 2006 (UTC)

Just in case people don't want to follow the link:

"Crank" is a pejorative term for a person who

1. holds some belief which the vast majority of his contemporaries would consider counterfactual,
2. clings to this belief in the face of all counterarguments or evidence presented to him.

The term implies that

1. a "cranky" belief is so wildly at variance with some commonly accepted truth as to be ludicrous,
2. arguing with the crank is useless, because he will invariably dismiss all evidence or arguments which contradict his cranky belief.

And there we are. Arguing with cranks is useless, so we have to be careful to distinguish the cranks from those genuinely trying to understand the field of mathematics. Calbaer 17:36, 3 November 2006 (UTC)

You take yourself way too seriously, sometimes. falsedef 21:59, 3 November 2006 (UTC)

Responding to Schapel here: There's a temptation, faced with arguments like this, to retreat into formalism and just point at axioms and definitions, but I think it's an error. It makes it appear that those definitions and axioms are arbitrary, which they definitely are not. They are what they are for reasons, and those reasons (though good reasons) are subject to legitimate question. While there's a trivial sense in which a definition "cannot be wrong" as another poster said, there are more important senses in which it can be wrong; for example, if it fails to correctly capture the concept for which it was formulated.

The question "why are the standard definitions of the reals the 'right' ones?" leads to a much more interesting discussion than any that has appeared on this page. Unfortunately it's not a discussion that it's really possible to have with those who are unable or unwilling even to follow the trivial arguments deducing easy consequences of the standard setup. --Trovatore 07:29, 4 November 2006 (UTC)

Although I agree with what you say, I would say that this page is not the appropriate place for that discussion. First of all, the article has nothing to do with "why are the standard definitions of the reals the 'right' ones?" and second of all it will simply confuse people genuinely trying to understand why 0.999...=1. -- Schapel 16:57, 4 November 2006 (UTC)

I'm not sure what the comment about taking things too seriously is about. I wanted to give the definition of a crank so we could all mull it over. A good example of a crank is someone who's never bothered to learn the definition of real number or take a class on or read a book on real analysis — a mathematical field that's yielded some very interesting real-world results — and yet implies that he's found its fatal flaw and won't entertain the possibility that he is wrong. Sure, there's the occasional Copernicus and Einstein who shakes things up, but they generally have credentials above sophomore-level math and they impact science, not math. Calbaer 04:22, 5 November 2006 (UTC)

I was referring to Schapel's post; however, you're also taking things way too seriously.falsedef 09:12, 6 November 2006 (UTC)

[1,2]

Is 0.999... contained in the interval [1,2]? Discuss. --JohnLattier 02:04, 5 November 2006 (UTC)

Of course it is; 0.999... is strictly equal to 1. What is the point of this question? --jpgordon^∇∆∇∆ 02:48, 5 November 2006 (UTC)

Absurd--68.158.245.160 09:16, 5 November 2006 (UTC)

Also note that 0.999… is not contained in the interval [0,1), because that interval does not contain 1. Likewise, the set [0,1) ∪ (1,2] excludes exactly one real, which is 0.999…, a.k.a. 1.

— Loadmaster 22:41, 6 November 2006 (UTC)

[0,1) is defined as having 0 as its lower bound and 0.999... as its upper bound. The ")" means up to but not including 1. Do you have an alternate upper bound candidate? --JohnLattier 18:33, 8 November 2006 (UTC)

• You're wrong. [0,1) is defined as having 0 as its lower bound and 1 as its upper bound. An open interval does not contain its upper bound. --jpgordon^∇∆∇∆ 18:58, 8 November 2006 (UTC)

Anyone who thinks a set need contain its upper bound should read the supremum article. It seems some people (and/or sockpuppets) stick to critiquing this article in particular when so many mathematical concepts have identical underlying axioms. Calbaer 18:52, 8 November 2006 (UTC)

You are right about it not containing 1. The ")" specifically signifies that 1 is not included. 0.999... is therefore the upper bound, as it is the largest number less than 1. --JohnLattier 00:24, 9 November 2006 (UTC)

• No, the upper bound is 1, and it's not included in the interval. This approach won't work for you either, since we're talking about intervals within the reals, and in the reals, there is not and cannot be a "largest number less than one". Oh, by the way, [0,1) and [0,0.999...) describe the exact same interval; $\not \exists x\in \Re :x\in [0,1)\land x\not \in [0,0.999\ldots )$(did I get that right?) --jpgordon^∇∆∇∆ 01:39, 9 November 2006 (UTC)

Of course there can be a largest number less than 1. It might not be a number we can write on paper. But any finite number can be a reference point from which we can find an infinite decimal one point away. Whether it be 0.00029->0.00028999... or 1349->1348.999... or 0 and 0.000...1. Real analysis just keeps on coming up with some incorrect assumptions. [0,1) includes 0.999... but [0,0.999...) does not. Try to calculate the length of the interval [0,0.999...) and you can't because we cannot move from one infinite decimal to another, only from a finite number to an infinite decimal, without having to incorporate more "...'s" --JohnLattier 13:01, 9 November 2006 (UTC)

You're wrong, and if you're just going to make things up, we can't discuss this any more. Go get published, then come back here. Maelin (Talk | Contribs) 13:14, 9 November 2006 (UTC)

Manipulating the infinitesimal

1. Let's discuss the infinitesimal. Does it equal zero?
2. What is a legitimate numerical representation for it?
3. What kind of mathematical operations can lead to it?
4. What kind of mathematical operations can be performed on it?
5. How does the infinitesimal relate to practical situations?
6. How about in theoretical situations?
7. If a mathematical system does not acknowledge the infinitesimal, is it faulty?
8. Is the infinitesimal a number or just a concept?
9. If the infinitesimal is zero, why is it even distinguished by a separate definition and name?

I'll post thoughts later. --JohnLattier 02:09, 5 November 2006 (UTC)

• There are no non-zero infinitesimals in the real numbers, which is the subject matter under discussion here. All mathematical systems are faulty. --jpgordon^∇∆∇∆ 02:46, 5 November 2006 (UTC)

So in real analysis, the infinitesimal = 0? There's the problem, then. --JohnLattier 03:12, 5 November 2006 (UTC)

No, it's not a problem (see the FAQ); it's just how the real number system works. I'm sure there are infinitesimals in other number systems, but 0.999... equals 1 in the reals. Supadawg (talk • contribs) 03:20, 5 November 2006 (UTC)

Hmmm... While we have seen that discussion with John is fruitless, he has now raised some interesting questions, to which I will reply. I have also taken the liberty to attach numbers to them, to facilitate the process.

1. There's no "the infinitesimal". An infinitesimal is a number which is lesser in magnitude than any number of the form 1/n, where n is a positive integer (the usual definition is slightly different, but mostly equivalent). In the system of real numbers, there is exactly one infinitesimal, 0. In other systems (e.g. Hyperreal numbers), there can be many infinitesimals, usually infinitely many.
2. That depends on the system we are discussing. In the real numbers, there is only one infinitesimal, which can be properly represented as 0 or 0.000... or whatever. In hyperreals, they would usually be represented as a particular sequence from their equivalence class. An example would be the infinitisemial equal to the equivalence class of the sequence a_n = 1 / n (it is also commonly denoted ε).
3. In the case of hyperreals, taking the equivalence class of any sequence which converges to 0 in the sense of real numbers.
4. In the case of hyperreals, pretty much everything, including arithmetical operations such as addition, multiplication, division etc., and also the new hyperreal function of taking the standard part.
5. It possibly doesn't - it has not yet been proven that the physical universe contains arbitrarily small quantities - let alone infinitely small ones. However, there are cases where it is useful as an approximation, when using non-standard analysis approach to finding differential equations which are supposed to describe the universe.
6. Depends on the context, but the article non-standard analysis should provide some information.
7. Depends on what you mean by "faulty", but no. Some systems have non-zero infinitesimals (hyperreals, surreals, ...) and some do not (real numbers, integers, ...). None is more "faulty" than the other.
8. It is first of all a concept - any quantity which satsisfies my above definition. Within a particular setting, it can be a property of numbers (so, ε, 2ε and ε² are infinitesimal numbers, where 1, π and -7/2 are not).
9. First, because the fact that the only infinitesimal real number is 0 is a theorem, so having a separate definition with which we can formulate the theorem is useful. Second, this theorem does not generalize to every number system, and there are certainly those where there are other infinitesimals.

I'll also say that while there are certainly many systems which have non-zero infinitesimals, there are none that I know of where there is "the smallest positive number" (other than trivial ones such as the integers). It just doesn't make any sense. -- Meni Rosenfeld (talk) 08:14, 5 November 2006 (UTC)

Back up a step, John. What is it you want with infintesimals? Do you have an infinitely small orange you want to measure? I fear you want infinitesimals only to save yourself from a counterintuitive conclusion. Alas, infinitesimals will not help you separate 0.999... from 1. Infintesimals are typically introduced in an utterly non-rigerous manner to draw analogies between derivatives and fractions to aid teaching calculus to students. dy/dx is not a fraction involving infintesimals, but a derivative - a well defined limit of other quantities which, under certain manipulations, superficially behaves like a fraction. For instance df/du * du/dt = df/dt is not fraction multiplication but a highly non-trivial relation between the limits of composed functions. It's initially easier to pretend that "dy" and "dx" exist independently and that dy/dx is a fraction, but that is neither truth nor justification for dy/dx. Similarly, \int_{a}^{b} f(u) du/dx dx = \int_{a}^{b} f(x) dx is not a manipulation of fractions but merely a practical shorthand, a similar notation for an action that students don't get taught the truth of for a year or four later due to the complexities it summarises. Any problem which appears to require infintesimals can be more rigorously formulated with only ratios and limits; even if you did need infintesimals, they would not appear between 0.999... and 1, for the same reason it's hard to shoot a goal between only one post.

A few things that everybody needs to accept

We must all accept the following if what I shall charitably call this debate is to get anywhere.

1. All of the mathematics that this article deals with is done in the real numbers.
2. The real numbers are a formally defined set. All of their properties follow from their construction.
3. In all of mathematics, rigour and formality are essential. This guarantees that, unless your system is inconsistent, every single one of your proven theorems will be true, always, for every object on which it is applicable, under the axioms you used to prove it.
4. The real numbers form a field (mathematics).
5. The real numbers have the Archimedean property.
6. Claiming that parts of mathematics are "wrong" is meaningless. In mathematical logic, every statement falls under one of three possible descriptions: true under the axioms used; false under the axioms used; and undecidable under the axioms used. One may claim that a particular set of axioms gives results that do not properly reflect the real world. This does not mean that true statements are false. It means the axioms used are not a perfect model for how we intuitively perceive the physical world. To rectify this situation, use a different set of axioms.

Anybody who rejects any of the above statements needs to realise that there is then no point in discussing this article, any more than there is in discussing celestial mechanics with someone who flatly rejects the notion of gravity. Maelin (Talk | Contribs) 04:37, 5 November 2006 (UTC)

Absolutely superb post. Perfect. Every denier, PLEASE READ THE POST ABOVE THIS LINE! The fact is that people are arguing about a definition, which is defined in a rigorous system. If you want to create your own mathematical system - do so. However, arguing about a definition in math is ridiculous. B Nov 5, 2006

So that's what this is all about? 0.999...=1 in an alternate math system where the infinitesimal is redefined as zero and infinite sums are redefined as their limits? What a boring conclusion that would be. Well if real numbers don't represent reality, maybe someone should seriously consider changing its name. So, is there a name for the normal, defined number system with normal definitions like infinitesimal = smallest number greater than 1? The article should start out, "In the alternate mathematical system real analysis, using the elimination of the concept of the infinitesimal through the Archimedian principle,..." or something along those lines. You guys are just fooling people into thinking your article applies universally. You probably like that, gives you people to argue with? zzzz JohnLattier --68.158.245.160 05:21, 5 November 2006 (UTC)

I really wish I could understand why you so violently reject the idea that 0.999... = 1. Nonetheless, no mathematical system has been found where 0.999... and 1 are well-defined but unequal that is useful in any sense. And despite the mildly counterintuitive result in the real numbers, as anathemic as it inexplicably seems to you, the reals are used almost universally in the physical sciences for measuring continuous quantities. Perhaps it is you who needs to change, not the number system that has virtually universal acceptance in science. Also, since the article very clearly states that we're talking about 0.999..., the real number, your complaints about "fooling" people are nonsense. Maelin (Talk | Contribs) 05:39, 5 November 2006 (UTC)

If real analysis fails to properly describe the phenomenon, it doesn't matter how well the system works, nor how practical its uses are. It doesn't redefine the infinitesimal outside of its little system. Although that's the broad leap of faith a reader takes when he reads the article. The article pulls of the magic trick of applying this redefinition process to all of mathematics, as the first phrase causes a reader to infer, then the article goes on to take a very lopsided stance. What's the debate? In real analysis everything gets a new meaning, so there's no argument. The argument lies in the magician's trick of pulling the same curtain over the rest of mathematics (aka the real real analysis. Maelin, can you and your buddies put an end to the comments like "violently reject," "crank," etc. This is mathematics, not war. No one is getting violent on you, so there is no need for you to be so defensive. I'd appreciate it if all further comments are on a scientific/factual/Wikipedia-like adult level. --68.158.245.160 06:21, 5 November 2006 (UTC)

Real Analysis is a huge branch of mathematics that is consistent with the whole. It doesn't "fail" to define anything. It defines precisely everything that its conclusions say it does. Furthermore, the fact that you want the real numbers to be infinitely enumerable has nothing to do with reality. If you so disagree with the mathematics that has ironically allowed for the construction of, and logical operations of your computer - then go ahead and philosophically reject mathematics and create your own system. But, to argue about the rigorous logical conclusions of math, based on the axioms of math and its definitions - well, it just shows that you're coming in 2000 years late, ignoring all the previous work, and trying to fight over a defintion residing INSIDE of the system of mathematics. It's pointless. Nevertheless, I encourage you to formally study math - especially limits and formal logic - and then you will see that math is precise. Derivatives, which are limits by definition, are EXACT. They are not rounding - they are not approximations. And they model reality conceptually in a way that is infinitely more precise than we can possibly ever measure; which is really the crux of this article. B Nov 5, 2006.

Ignoring JL here and responding to Maelin's first comments in this section--I renew my point (made first to Schapel) that it's an error to retreat into formalism. Axioms do not make things true or false, only provable or refutable. True and false are semantic notions, whereas provable and refutable are syntactic ones, and it's important not to confuse the two.

The fact that the objectors show no sign of being able to actually follow a formal proof makes it tempting to try to "win" on that level alone, but if you go that way, people might think proofs are what math is actually about, which it isn't. Math is about the objects it describes. Proofs are a tool to find things out about those objects. --Trovatore 05:44, 5 November 2006 (UTC)

People "violently reject" these concepts because of various reasons, which are often a combination of (1) not being able to understand them (2) being stubborn and (3) knowing that in everyday experience, as well as in physics itself (due to quanta and the apparent lack of infinite quantities), things don't behave in the same manner as the axiomatically pure mathematics of real analysis. However, in spite of this, real analysis is useful while the alternatives are nearly useless, so refusing to learn the former due to aesthetic objections is the height of (2) on my list. Which is why the term "crank" is being thrown around; it is difficult to see any part of the definition of the word that fails to apply to the targets of the epithet.

Of course, a crank might say that something failing to accurate describe a real-world phenomenon must be flawed. If I have one apple and add another apple, surely I have two apples, don't I? Positive integers seem to work in a way that real numbers don't. Except that's not true. If I say I have two crates each of which holds 20 apples, you'd say I had 40 apples. But if I said I had a billion crates each of which holds a billion apples, you'd call me a liar, realizing that there aren't that many apples in the world. Do the integers break down when I run out of apples? No, of course not. Instead, they are a consistent system that helps describe certain phenomena, but not every number need correspond to a specific physical phenomenon. In the same manner, although infinite sums have no direct physical counterpart in a quantized world, 0.999... still equals 1. Calbaer 06:50, 5 November 2006 (UTC)

With all due respect, you may be creating more confusion. First, numbers are conceptual, and you might want to start there. Anyway, as a matter of fact, integers ARE real numbers. So, they can't possibly work in a way that real numbers do not. Second, the fact that we can't yet measure a distance, for example, below the quantum level - well, that's quite possibly just a matter of technology. Nevertheless, we can show conceptually that those distances exist; in which case, and which is most always the case, math is MORE ACCURATE than "reality". Last, "infinite sums" DO HAVE a direct physical counterpart in many cases, as in the case of 1. (Although 0.999... is not an "infinite sum", but I'm using your wording.) Also, as another example of "reality" and infinity - allow me: How many paths exist whereby an airplane can fly from NY to LA? Infinity exists. B Nov 5, 2006.

Positive integers as a system do work in a way that real numbers as system do not. Specifically, if the limit of a sequence of positive integers exists, then the limit is in the sequence (an infinite number of times, in fact). In the real numbers, that's no longer true. I'm merely trying to illustrate that not all mathematical expressions correspond to a specific physical reality. That doesn't mean the math is wrong.

And even when the math doesn't perfectly represent a physical phenomenon, is can still be useful. For example, integration has no place in infinite sums. But infinite sums can be approximated best using integrations, as in [2]. If we threw out calculus because we didn't like the underlying real analysis, we wouldn't be able to do many complex and necessary calculations, calculations for having the exact answer isn't as important as having a good answer.

As to the usefulness of real analysis: no, it isn't useful in, say, counting apples. However, it provides the basis for calculus, which has proven useful in almost every field of engineering and technology. And real analysis itself has proven useful in modeling control systems, financial markets, betting systems, communication systems (e.g., the Internet) and other complex phenomena. Just because a doubter doesn't bother to learn that for himself doesn't make it any less true.

That said, I must admit I'm a bit rusty with quantum physics, so perhaps some of my statements about quanta and infinity weren't correctly phrased. Like I said, I just wanted to illustrate that math is useful even when it's only an approximation of reality and it's still true even when they're no corresponding physical system. I'm sorry if my inexperience with physics caused any confusion. I only took it at a college sophomore level and thus should not be taken to speak with authority about it. Calbaer 20:54, 5 November 2006 (UTC)

I may disagree with Trovatore on several concepts, but he does the best job of making sound arguments. On the other hand, I request Calbaer get back on topic. This isn't a forum for personal jabs or psychological analysis.

Real analysis has shown me nothing to prove that it is worthy of being learned. Redefining common concepts and twisting them into a new, contradictory system, doesn't appeal at all. Just because a system produces results doesn't mean it describes reality. Look at Einstein/Hawking and their theories about bending of space/time. We all know space/time is a dimension set, not some physical entity that can be manipulated. I'd be perfectly content to leave you all to your fantasy and let you keep believing two separate points overlap, as you all seem firmly set in your unwavering faith. I just feel like there might be a young mind that stumbles upon this forum and gets swept up into the misconceptions. --68.158.245.160 10:04, 5 November 2006 (UTC)

"Real analysis has shown me nothing to prove that it is worthy of being learned." Equally, you have shown us nothing to make us suspect that you have the capacity to learn anyway.

Okay, so I guess we've established that you don't really disagree with the mathematical arguments that prove that 0.999... = 1 in the real numbers. Instead, you feel that the real numbers are 'flawed' for having this result. Unlike flat rejections of mathematical proofs, this idea ("the system is bad, because valid logical reasoning within it creates these bad results") has some merit, so let's look at that, now. Think about why you find the idea that 0.999... = 1 is abhorrent, why you feel that any mathematical system that asserts it is wrong. What aspect of your understanding of how numbers work "in the real world" leads you to feel that 0.999... shouldn't be equal to 1 in any "proper" number system? Why is that intuitive grasp of numbers above your suspicions, such that you prefer to reject the number system rather than your first impressions? Please consider this seriously, and then reply with your thoughts. Maelin (Talk | Contribs) 10:35, 5 November 2006 (UTC)

Your usage of the word "redefining" is interesting. Can you tell us what the "original" definition of the "common concepts" is? I guess that you cannot - the system of real numbers is so far the best attempt at rigorously defining something which resembles the physical universe. If you have another definition in mind, it is probably not rigorous and has little to do with mathematics (but we would love to hear it anyway).

Your use of "contradictory" is just as interesting. A contradiction is not the same as a counterintuitive result - it is the ability to prove (rigorously) both a statement and its negation within some system. Can you point out a single contradiction in real analysis? I won't go into the subtleties of consistency here, but unless you actually find a contradiction, you have no right to describe anything as "contradictory".

And... this has been said so many times that it hurts but... you place too much confidence in the power of decimal expansions. Decimal expansions are just a convenient notation for representing real numbers (and there are other notations). They have no importance on their own, and they happen to have a few quirks. One of those is the fact that some real numbers have 2 different representations, such as 1 which is represented by both 1.000... and 0.999... (the same way that in ternary notation, 1/3 has 2 representations : 0.1 and 0.0222..., and the same way that 2 can be represented as 2/1, 4/2 or other possibilities). Just because the decimal notation has this issue, doesn't say anything about the numbers themselves, and definitely is no reason to invent new numbers which break the structure of the real numbers.

To repeat: There are no two separate points which overlap. There is a number (or point, or whatever) which has more than one way to represent it. -- Meni Rosenfeld (talk) 16:56, 5 November 2006 (UTC)

By the way, saying that it was difficult to see any part of the definition of the word "crank" that wasn't being met wasn't supposed to be "a jab." One contributor called another a crank, and the alleged crank took offense. I wondered which part of the definition of crank was not satisfied. The alleged crank has not responded to that question, just as he has not responded to many of the more in-depth mathematical questions asked of him. The unwillingness to answer this speaks to the point of whether or not it is worth the time to answer him, a question which does have a place on this forum. Calbaer 20:54, 5 November 2006 (UTC)

Here's a simple experiment for 68.158.245.160. You will need one friend, two bycicles, and a road that runs north-south. Start a few hundred meters apart. Ride your bycicle towards your friend; do so on the eastmost part of the west side of the road. Have your friend ride their bycicle towards you on the westmost part of the east side of the road. Go hard - you won't hit each other when you pass, since we've *obviously* described different positions on the road. Get back to us, let us know how it turns out. Might like to make sure you health insurance is paid first though.

Need a convincing way of explaining that a step is correct

I 100% understand that 0.999... = 1. It doesn't freak me out or anything. I've led many of my friends to believe it too.

One of my friends, however, has a problem with this proof:

x = 0.999... 10x = 9.999... 10x-x = 9 x = 1

I think it's a great old proof. He has a problem with the second step though. He says when multiplying a number by ten it needs to end with a zero, I say you just shift the decimal point over one place. Anyone got some solid mathematical simple evidence to prove him wrong? I know he's wrong, he just refuses to accept my explanations, I need something more solid. Thanks.

83.70.213.7 22:24, 7 November 2006 (UTC)

I agree with your enlightened friend. This is the incorrect step in the process. Manipulating the infinitesimal is something people should be careful with. 10*0.999... has no definitive value. --JohnLattier 18:13, 8 November 2006 (UTC)

I suspect that that might be another experiential dogma that might be difficult to shake. What does your friend think 10/3 is? Or $10\pi$? Or $(10)(0.1)$? (For that matter, how does he write $(10)(0.999...)$?) Calbaer 22:43, 7 November 2006 (UTC)

The trick is that there is no place where the zero could go, no last digit which it could follow. Maybe a variation of Hilbert's Hotel is a good analogy: Consider a hotel with infinitely many rooms, numbered with natural numbers, all of which are taken. Now the occupant of room 1 leaves. Then the guest from room 2 can move up to room 1, the one from room 3 to room 2, and so on. In the end all rooms will still be occupied, even though one guest has left. The rooms correspond to the digits of 0.9999..., and the "room shift" is multiplication by 10: We get the one nine before the decimal separator, but still all digits behind it are nines, without a zero appearing. If your friend wants a truly rigorous proof of that step, you would have to use the definition of 0.9999... as the sum of a geometric series (see the article) and show that multiplying the series by 10 yields the same series plus nine, but that's more technical. --Huon 23:01, 7 November 2006 (UTC)

You can try a more formal version of the proof using the infinite sum definition of 0.999.

1. $y=0.999\dots =\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}$ (definition of 0.999...)
2. $10y=10\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}$ (multiply both sides by 10)
3. $=10\times ({\frac {9}{10}}+\sum _{k=2}^{\infty }{\frac {9}{10^{k}}})$ (seperating out first summand)
4. $=9+10\sum _{k=2}^{\infty }{\frac {9}{10^{k}}}$ (distribution)
5. $=9+\sum _{k=2}^{\infty }{\frac {10\times 9}{10^{k}}}$
6. $=9+\sum _{k=2}^{\infty }{\frac {9}{10^{k-1}}}$
7. $=9+\sum _{m=1}^{\infty }{\frac {9}{10^{m}}}$ (letting m = k-1)
8. $=9+y$ (by definition for y)

1. Thus $10y=9+y$, so
2. $10y-y=9$ and
3. $9y=9$
4. $y=1$

The above is essentially the same proof, pretty much, but is a little more formalized. Hope that helps. Dugwiki 23:11, 7 November 2006 (UTC)

Problem with your proof is that you let m=k-1 and then you assume that #$\sum _{m=1}^{\infty }{\frac {9}{10^{m}}}=\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}$ but the problem is that m is not equal to k. The analogy of Hilbert's hotel is completely irrelevant. I better post this before I get blocked again. It's amazing how many of my comments have been deleted by the Wiki team. John Lattier, I suggest you save your breath. 41.243.17.117 08:01, 8 November 2006 (UTC)

Ugh. You are just reaching at this point. Check the numbers, there's a minus one in the term and a 2 in the starting index. it's like this.

$\sum _{k=2}^{\infty }{9 \over 10^{(k-1)}}=\sum _{(k-1)=2-1}^{\infty }{9 \over 10^{(k-1)}}=\sum _{m=1}^{\infty }{9 \over 10^{m}}{\text{ since }}m=k-1$

If you're going to try to confuse matters like this, at least try to find something that requires more than a one-line proof and basic arithmetic. Maelin (Talk | Contribs) 08:24, 8 November 2006 (UTC)

Exactly my point: Step 8 assumes that m=k but this is untrue because m=k-1. Boy, you are so inexperienced. Go back and reread my post slowly. Think about it at least 5 times before you respond stupidly yet again. 41.243.17.117 09:36, 8 November 2006 (UTC)

If I read your post correctly, you are claiming that $\sum _{m=1}^{\infty }{\frac {9}{10^{m}}}\neq \sum _{k=1}^{\infty }{\frac {9}{10^{k}}}$ because of some perceived difference of summation indices? What's the difference supposed to be? Which of those terms is supposed to be larger? Fact is, each and every summand in one of the series also appears in the other one, and in the same order (which doesn't even matter, since the series converge absolutely). Thus, they are equal, no matter what we call the summation index. By the way, you just earned a WP:NPA warning. --Huon 10:01, 8 November 2006 (UTC)

Correct. The two sums are unequal because m is not equal to k. In fact, you cannot even begin to compare the two unless you express m in terms of k which is how you derived m to begin with. You ask which is larger - what do you think? Of course the summation involving k is larger since m starts at k=2 and not k = 1. This kind of mathematics demonstrates conclusively that you are all amateurs. Your statement that each and every summand in one series also appears in the other is false. Now unless you admit to your mistakes, I have nothing further to add. Finally, I could care less about your warnings or violations. You have already blocked me several times. I can simply come back under another IP at any time I wish. Is it not true that you are all fools? So what is this going to mean? Another violation because I am the only one more intelligent than all of you nincompoops. I have nothing but the utmost contempt and disdain for all of you starting with the chief idiot Prof. Hardy, followed by the pompous ignoramus who originally wrote this article (Ksmrq), not to mention your little nitwits such as Jitse Niesen (or whatever he is called) and Melchoir. Eat shit and die fools! 41.243.17.117 12:51, 8 November 2006 (UTC)

I agree with you 41. You have a very precise understanding of infinity, infinitesimal, and limits. In this particular proof, step 7 is incorrect. It should be from m=1 to m=infinity-1. But infinity-1 has no particular meaning, so step 7 is thrown out. Truncation of the infinitesimal. Very familiar theme. There would be no 0.999...=1 arguers without it. --JohnLattier 18:24, 8 November 2006 (UTC)

At least you're being consistent, in agreeing with yourself! However, you're incorrect. Infinity minus one is still infinity. Read about Hilbert's Hotel to see why. -- Schapel 18:42, 8 November 2006 (UTC)

While Hilbert's Hotel is not even worth mentioning, the infinity "truth" page is very much what I've wanted to see. The rules they post are very intriguing, but many of their answers that contain a manipulated infinity should be an error message, or a "different value of infinity." If you're suggesting 41.243- is me, no. We do have one thing in common, which is that it seems we are both aggressively censored by several of the 0.999...=1 proponants. I'm glad 41.243- is posting here; it's refreshing seeing that not everyone is tricked by the proofs in this article. --JohnLattier 19:00, 8 November 2006 (UTC)

The summation involving k is not larger. Both summations have exactly the same number of elements. There are exactly as many integers greater than zero as there as integers greater than one. There are exactly as many even integers as there are integers. That precise number is aleph null. Understanding this basic concept of infinity is essential to understanding the logically and mathematically sound arguments here. We are not defending the arguments, but attempting to explain them. If you do not wish to attempt to understand, you are free to leave the discussion. -- Schapel 13:11, 8 November 2006 (UTC)

For even more rigor, you would have to prove step 5, "infinite distribution". That would probably require explicit use of some arbitrary positive epsilon and an N such that the sum of the first N terms of the series is closer than epsilon/10 to the sum of the series. --Huon 23:32, 7 November 2006 (UTC)

I don't think an epsilon proof would be necessary, unless you feel the additional need to prove that for any constant c, if $\lim _{x\rightarrow \infty }f(x)$ exists, then $c\times \lim _{x\rightarrow \infty }f(x)=\lim _{x\rightarrow \infty }c\times f(x)$. But if you want to prove that explicitly, you might need to do an epsilon-style limit proof. Dugwiki 00:01, 8 November 2006 (UTC)

{Perhaps this is easier to read? _→AzaToth 23:35, 7 November 2006 (UTC) ${\begin{aligned}y&=0.999\ldots =\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}&\quad &{\text{definition of}}\;0.999\ldots \\10y&=10\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}&&{\text{multiply both sides by 10}}\\10y&=10\times ({\frac {9}{10}}+\sum _{k=2}^{\infty }{\frac {9}{10^{k}}})&&{\text{separating out first summand}}\\10y&=9+10\sum _{k=2}^{\infty }{\frac {9}{10^{k}}}&&{\text{distribution}}\\10y&=9+\sum _{k=2}^{\infty }{\frac {10\times 9}{10^{k}}}&&\\10y&=9+\sum _{k=2}^{\infty }{\frac {9}{10^{k-1}}}&&\\10y&=9+\sum _{m=1}^{\infty }{\frac {9}{10^{m}}}&&{\text{letting}}\;m=k-1\\10y&=9+y&&{\text{by definition for}}\;y\\10y-y&=9&&{\text{and}}\\9y&=9&&\\y&=1&&\end{aligned}}$

Above someone objected to the lines

• $10y=9+\sum _{k=2}^{\infty }{\frac {9}{10^{k-1}}}$
• $10y=9+\sum _{m=1}^{\infty }{\frac {9}{10^{m}}}{\text{letting}}\;m=k-1$

This follows by the simple corrolary that for all functions f(x) on the reals:

• $\sum _{k=2}^{\infty }f(k)=\sum _{m=1}^{\infty }f(m+1)$

One proof of this corrolary can be done by induction. First, we'll prove that for all natural numbers N>=2:

• $\sum _{k=2}^{N}f(k)=\sum _{m=1}^{N-1}f(m+1)$

(base case N=2)

• $\sum _{k=2}^{2}f(k)=f(2)=f(1+1)=\sum _{m=1}^{2-1}f(m+1)$

(induction on N) Assume that for a given natural number N>=2 that

• $\sum _{k=2}^{N}f(k)=\sum _{m=1}^{N-1}f(m+1)$

Then

• $\sum _{k=2}^{N+1}f(k)=f(N+1)+\sum _{k=2}^{N}f(k)$
• $=f(N+1)+\sum _{m=1}^{N-1}f(m+1){\text{ (by inductive assumption above)}}$
• $=\sum _{m=1}^{N}f(m+1)$

So by induction $\sum _{k=2}^{N}f(k)=\sum _{m=1}^{N-1}f(m+1)$ for all natural numbers N>=2.

Therefore

• $\sum _{k=2}^{\infty }f(k)=\lim _{N\rightarrow \infty }\sum _{k=2}^{N}f(k)$
• $=\lim _{N\rightarrow \infty }\sum _{m=1}^{N-1}f(m+1)$
• $=\sum _{m=1}^{\infty }f(m+1)$

Thus for interested readers the above is a more explicit proof of the lines in question. Dugwiki 16:52, 8 November 2006 (UTC)

The corollary you mentioned is false. Your induction statements are also false because m is defined in terms of k. Furthermore, if you accept such induction, then you ought to accept the induction that proves 0.999... < 1:

0.9 < 1;    0.9...k   < 1   =>   0.9....(k+1) < 1

How does Wikipedia hire people like you? Do they even pay you? The real debate here is not about 0.999... < 1. It's about the stupidity of the topic called real analysis. If you want to define an infinite sum as its limit, then you need to be able to do this for all real numbers. You are hell-bent on claiming that 0.999... < 1 but you cannot state the same facts for most real numbers. If you define real numbers in the way real analysis does (i.e. as the limit of a sequence or limit of an infinite sum), then you had better be able to do it for all real numbers otherwise you are being inconsistent and mathematics is about being consistent. So now if you can show me what the limit of 3.14... is, then I will accept 0.999... = 1. Don't tell me it's pi or state one of the many pi series - this is not an acceptable answer. Go to your math gods (Hardy et al) and ask them to help to you.41.243.17.117 06:33, 9 November 2006 (UTC)

0.9 < 1; 0.9...k < 1 => 0.9....(k+1) < 1"

If you reread the proof you'll see that the induction steps were specifically for finite N. Your induction above is likewise correct for finite numbers k. The remainder of the proof used what was shown by induction for finite N and extended it using limits. (P.S. I'm just a user, not an employee, same as you.) Dugwiki 16:31, 9 November 2006 (UTC)

What do you mean by 3.14...? Does the 14 repeat? How are you defining the sequence in the first place? -- SCZenz 06:45, 9 November 2006 (UTC)

Why bother? 41.243.17.117 just likes to be abusive, while asserting that real analysis is bunk and we're bad Wikipedia employees. Calbaer 06:56, 9 November 2006 (UTC)

You're right. If he replies abusively again, I'll certainly not feed him. -- SCZenz 06:59, 9 November 2006 (UTC)

The flaw in your induction trying to prove that 0.999... < 1 is that induction can only prove that something is true for any integer k. It's trivially true that for any finite k 9s after a decimal 0.99...9 < 1. When k is infinity, induction offers no insight. I'll leave the rest for other to dissect.--Trystan 06:50, 9 November 2006 (UTC)

"show me what the limit of 3.14... is" - this doesn't even make sense, since pi is not a sequence, it is a number. I guess you mean the limit of the sequence {a_n}, where a_i is the decimal expansion of pi truncated after i digits. In this case the limit is obviously pi. If you meant something different, you should try to be more mathematically explicit in what you are asking for. If you name any real number, we can find a sequence which has that number as its limit. By the way, Wikipedia editors are volunteers. Maelin (Talk | Contribs) 07:08, 9 November 2006 (UTC)

41.243- is right again. When you can locate the final digit in pi, I'll lend an ear about how 0.999... might equal 1. Until then you are chasing your tail. --JohnLattier 12:54, 9 November 2006 (UTC)