# Talk:0.999...

(Redirected from Talk:0.999)
Frequently Asked Questions (FAQ) edit Q: Are you positive that 0.999... equals 1 exactly, not approximately? A: In the set of real numbers, yes. This is covered in the article. If you still have doubts, you can discuss it at Talk:0.999.../Arguments. However, please note that original research should never be added to a Wikipedia article, and original arguments and research in the talk pages will not change the content of the article—only reputable secondary and tertiary sources can do so. Q: Can't "1 - 0.999..." be expressed as "0.000...1"? A: No. The string "0.000...1" is not a meaningful real decimal because, although a decimal representation of a real number has a potentially infinite number of decimal places, each of the decimal places is a finite distance from the decimal point; the meaning of digit d being k places past the decimal point is that the digit contributes d · 10-k toward the value of the number represented. It may help to ask yourself how many places past the decimal point the "1" is. It cannot be an infinite number of real decimal places, because all real places must be finite. Also ask yourself what would be the value of $\frac{0.000\dots1}{10}$. If a real number divided by 10 is itself, then that number must be 0. Q: 0.9 < 1, 0.99 < 1, and so forth. Therefore it's obvious that 0.999... < 1. A: No. By this logic, 0.9<0.999...; 0.99<0.999... and so forth. Therefore 0.999...<0.999..., which is absurd. Something that holds for various values need not hold for the limit of those values. For example, f (x)=x 3/x is positive (>0) for all values in its implied domain (x ≠ 0). However, the limit as x goes to 0 is 0, which is not positive. This is an important consideration in proving inequalities based on limits. Moreover, although you may have been taught that $0.x_1x_2x_3...$ must be less than $1.y_1y_2y_3...$ for any values, this is not an axiom of decimal representation, but rather a property for terminating decimals that can be derived from the definition of decimals and the axioms of the real numbers. Systems of numbers have axioms; representations of numbers do not. To emphasize: Decimal representation, being only a representation, has no associated axioms or other special significance over any other numerical representation. Q: 0.999... is written differently from 1, so it can't be equal. A: 1 can be written many ways: 1/1, 2/2, cos 0, ln e, i 4, 2 - 1, 1e0, 12, and so forth. Another way of writing it is 0.999...; contrary to the intuition of many people, decimal notation is not a bijection from decimal representations to real numbers. Q: Is it possible to create a new number system other than the reals in which 0.999... < 1, the difference being an infinitesimal amount? A: Yes, although such systems are neither as used nor as useful as the real numbers, lacking properties such as the ability to take limits (which defines the real numbers), to divide (which defines the rational numbers, and thus applies to real numbers), or to add and subtract (which defines the integers, and thus applies to real numbers). Furthermore, we must define what we mean by "an infinitesimal amount." There is no nonzero constant infinitesimal in the real numbers; quantities generally thought of informally as "infinitesimal" include ε, which is not a fixed constant; differentials, which are not numbers at all; differential forms, which are not real numbers and have anticommutativity; 0+, which is not a number, but rather part of the expression $\lim_{x \rightarrow 0^+} f(x)$, the right limit of x (which can also be expressed without the "+" as $\lim_{x \downarrow 0} f(x)$); and values in number systems such as dual numbers and hyperreals. In these systems, 0.999... = 1 still holds due to real numbers being a subfield. As detailed in the main article, there are systems for which 0.999... and 1 are distinct, systems that have both alternative means of notation and alternative properties, and systems for which subtraction no longer holds. These, however, are rarely used and possess little to no practical application. Q: Are you sure 0.999... equals 1 in hyperreals? A: If notation '0.999...' means anything useful in hyperreals, it still means number 1. There are several ways to define hyperreal numbers, but if we use the construction given here, the problem is that almost same sequences give different hyperreal numbers, $0.(9) < 0.9(9) < 0.99(9) < 0.(99) < 0.9(99) < 0.(999) < 1\;$, and even the '()' notation doesn't represent all hyperreals. The correct notation is (0.9; 0.99; 0,999; ...). Q: If it is possible to construct number systems in which 0.999... is less than 1, shouldn't we be talking about those instead of focusing so much on the real numbers? Aren't people justified in believing that 0.999... is less than one when other number systems can show this explicitly? A: At the expense of abandoning many familiar features of mathematics, it is possible to construct a system of notation in which the string of symbols "0.999..." is different than the number 1. This object would represent a different number than the topic of this article, and this notation has no use in applied mathematics. Moreover, it does not change the fact that 0.999... = 1 in the real number system. The fact that 0.999... = 1 is not a "problem" with the real number system and is not something that other number systems "fix". Absent a WP:POV desire to cling to intuitive misconceptions about real numbers, there is little incentive to use a different system. Q: The initial proofs don't seem formal and the later proofs don't seem understandable. Are you sure you proved this? I'm an intelligent person, but this doesn't seem right. A: Yes. The initial proofs are necessarily somewhat informal so as to be understandable by novices. The later proofs are formal, but more difficult to understand. If you haven't completed a course on real analysis, it shouldn't be surprising that you find difficulty understanding some of the proofs, and, indeed, might have some skepticism that 0.999... = 1; this isn't a sign of inferior intelligence. Hopefully the informal arguments can give you a flavor of why 0.999... = 1. If you want to formally understand 0.999..., however, you'd be best to study real analysis. If you're getting a college degree in engineering, mathematics, statistics, computer science, or a natural science, it would probably help you in the future anyway. Q: But I still think I'm right! Shouldn't both sides of the debate be discussed in the article? A: The criteria for inclusion in Wikipedia is for information to be attributable to a reliable published source, not an editor's opinion. Regardless of how confident you may be, at least one published, reliable source is needed to warrant space in the article. Until such a document is provided, including such material would violate Wikipedia policy. Arguments posted on the Talk:0.999.../Arguments page are disqualified, as their inclusion would violate Wikipedia policy on original research.
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## 0.999... = 1 is just a trivial truth about the convergence of a sequence

What do you think of this:

0.999... = 1 is just a trivial truth about the convergence of a sequence.

For the decimal expansion 0.999... is defined as the infinite series 0.9+0.09+0.009+...

And the sum of the series 0.9+0.09+0.009+... is defined as the limit of the associated sequence of partial sums (0.9, 0.99, 0.999, ...)

So 0.999... = 1 just means that the sequence (0.9, 0.99, 0.999, ...) converges to 1. And whereas this is undoubtedly true, there is nothing counterintuitive about it. For even though this sequence converges to 1, it, of course, never reaches 1.

(Note that this "proof" of 0.999... = 1 also holds if we only consider the rational numbers. So it also holds for the rational numbers that 0.999... = 1).

Pulvertaft (talk) 22:08, 7 November 2013 (UTC)

This page is for discussions of proposed suggestions for possible improvements of the page. Do you have a proposed suggestion? Tkuvho (talk) 09:11, 8 November 2013 (UTC)

Yes: https://en.wikipedia.org/w/index.php?title=0.999...&diff=580649687&oldid=580647913 Pulvertaft (talk) 15:29, 8 November 2013 (UTC)

That is an undesirable change. Common problems are thinking that 0.999... is a sequence. It isn't. That it converges. It doesn't. That it is close to but somehow not equal to 1. It isn't. That it has something to do with sequences. It doesn't. 0.999... is just another way of writing 1 and has nothing to do with the convergence of a sequence. Hawkeye7 (talk) 17:33, 8 November 2013 (UTC)
Agreed, it's unnecessary. If we're trying to show that the infinite sum 0.9+0.09+0.009+··· is the limit of the partial sums, or that it is the sum of an infinite geometric series, that's already covered in the "Infinite series and sequences" section. — Loadmaster (talk) 19:27, 8 November 2013 (UTC)

It's all a question about definition. If 0.999... is defined as just another way of writing 1, then no wonder that 0.999=1 ;-). Similarly, if 0.999... is defined as a series, and the sum of a series is defined as the limit of its partial sums, then again it is no wonder that 0.999...=1. I was just trying to spell out that the mystery about how 0.999... can be equal to 1 disappears in this "proof" - which is given in technical terms just above my proposed suggestion: "According to this proof 0.999... = 1 just means that the associated sequence (0.9, 0.99, 0.999, ...) converges to 1. And whereas this is undoubtedly true, there is nothing counterintuitive about it. For even though this sequence converges to 1, it, of course, never reaches 1". As so many people are mystified about how 0.999... can be equal to 1, it seems important to get rid of the mystery where it's possible. — Preceding unsigned comment added by Pulvertaft (talkcontribs) 21:33, 8 November 2013 (UTC)

From my point of view, the "mystery" happens to disappear with every proof which is already in the article and I don't see why what you're describing should be held in distinction. You simply strike me as somebody who is incredibly good at visualising mathematical series :) JaeDyWolf ~ Baka-San (talk) 23:27, 8 November 2013 (UTC)
The last time this was discussed as I recall someone pointed out that the fact that "0.999... = 1 just means that the associated sequence (0.9, 0.99, 0.999, ...) converges to 1" is already mentioned in the page. If it isn't we should certainly add it. This fact can easily be sourced. Tkuvho (talk) 20:21, 9 November 2013 (UTC)
It is mentioned on the page, under "Cauchy sequences". Hawkeye7 (talk) 21:19, 9 November 2013 (UTC)

It is indeed correct that the number 0.999... can be thought as a limit of the sequence (0.9, 0.99, 0.999, ...), just like the number 1.000... (which, by convention, is usually written simply as "1") can be thought as a limit of the (constant) sequence (1.0, 1.00, 1.000, ...). But the key point is that the two sequences have the same limit; in other words, "0.999..." and "1.000..." represent the same real number. - Mike Rosoft (talk) 21:10, 8 January 2014 (UTC)

No. The inverse sequence is 1.000...1 (i.e. an infinite sequence of zeros followed by a one). Because 1 - 0.1 = 0.9, 1 - 0.01 = 0.99 (and so on ad infinitum), so the inverse is 1 + 0.1 = 1.1, 1 + 0.01 = 1.01 (and so on ad infinitum). And 1.0000... != 1.0000....1. Alexander Gras (talk) 19:23, 27 February 2014 (UTC)
1 - 0.1 = 0.9
1 - 0.01 = 0.99
1 - 0.001 = 0.999
1 - 0.000...1 = 0.999...
1 + 0.1 = 1.1
1 + 0.01 = 1.01
1 + 0.001 = 1.001
1 + 0.000...1 = 1.000...1
1 + 0.0 = 1.0
1 + 0.00 = 1.00
1 + 0.000 = 1.000
1 + 0.000... = 1
1 - 0.0 = 1
1 - 0.00 = 1
1 - 0.000 = 1
1 - 0.000... = 1
0 != 1
0.0 != 0.1
0.00 != 0.01
0.000... != 0.000...1
therefore
0.999... != 1.000...
0.000...1 is not well-defined; such a string does not represent a real number. See Talk:0.999.../Arguments for some recent discussions about other number systems and their shortcomings. In particular, what is 1/1.000...1? Anyway, this is getting rather off-topic; if you want to discuss other number systems that may have a number 1.000...1, please do so on the arguments page. Huon (talk) 19:27, 27 February 2014 (UTC)

## Repeating decimal

Given the fact that any number with a repeating decimal representation is always a rational (e.g., 0.142857142857... = 17), would it be useful to mention that since 0.999... is a repeating decimal it therefore must be a rational, and that rational is 1? Or to phrase it another way, 0.999... must be some rational, and there are no other rationals that it could possibly be other than 1. Repeating decimals are mentioned in the "Applications" section, but I don't think this rather obvious fact about 0.999... is mentioned explicitly there. — Loadmaster (talk) 19:26, 4 December 2013 (UTC)

I don't see it illuminates things. All whole numbers are rational numbers but you don't generally make this point when referring to whole number properties, for example in number theory. Yes, it's "some" rational, it's also "some" real number. That number is 1 so it can't be any other rational or real number. But that's just a statement, and a fairly obvious one if you accept 0.999... equals one.--JohnBlackburnewordsdeeds 19:45, 4 December 2013 (UTC)
My point was more specific, that since 0.999... is a repeating decimal, it must therefore be a rational number and not something else. This is meant to be another one of the justifications to present to those who have problems accepting that 0.999... is 1, and a rather obvious justification at that. For those people, they may have in mind that 0.999... is somehow a different kind of number than whole numbers, rationals, etc. Which of course it isn't, it's just an ordinary rational. — Loadmaster (talk) 17:13, 6 December 2013 (UTC)

## Is 0.999... a rational number?

1 certainly seems to be, but how can they be the same if one is irrational and the other is not? 85.252.131.79 (talk) 13:48, 6 January 2014 (UTC)

Not here please—see wp:talk page guidelines. You can try at the wp:Reference desk/Mathematics. Good luck. - DVdm (talk) 13:52, 6 January 2014 (UTC)
You should also take a look at the Repeating decimal article before going any further here. — Loadmaster (talk) 22:01, 6 January 2014 (UTC)

0.999... is indeed a rational number; it is equal to 9/9, which happens to be equal to 1. (Just like 0.888... equals 8/9.) - Mike Rosoft (talk) 18:26, 9 January 2014 (UTC)

## How can 0.999 be 1?

(I hope I got the decimal notation for infinitely small but not zero correct... a guess after not being able to find it. 0.000...1 (infinitely zeroes before the 1)

true or false? 0.999... = 1

true or false? 0.000...1 = 0

true or false? 1 - 0.000...1 = 0.999...

true or false? 0.999... - 0.000...1 = 0.99...8

true or false? 0.999... = 0.99...8 = 0.99...7 = 0.99...6

true or false? the difference between every neighbouring floating point number would be 0.000...1?

what's the total sum of all differences between every neighbouring floating point number? infinity or zero? - ZhuLien

(edited as I had actually meant floating point not real numbers. if I have the term incorrect, I meant a number with a decimal point - there is no preferred or intended 'numbering system' - just the existence of the numbers - I've have moved to the suggested Arguments Talk thread below. ZhuLien 27.32.141.11 (talk) 20:20, 7 March 2014 (UTC))

— Preceding unsigned comment added by 202.168.6.200 (talkcontribs) 04:49, 17 January 2014‎ (UTC)

Please sign your talk page messages with four tildes (~~~~), not with a type name as you did here above. Thanks.
Not here please—see wp:talk page guidelines. You can try at the wp:Reference desk/Mathematics. Good luck. - DVdm (talk) 07:50, 17 January 2014 (UTC)
I'll reply at Talk:0.999.../Arguments. Huon (talk) 21:51, 17 January 2014 (UTC)
The first statement is correct, the rest are meaningless (as 0.000...1, 0.99...8, etc. do not exist in real numbers). - Mike Rosoft (talk) 19:43, 7 February 2014 (UTC)
Well, they do if you regard then as limits. So
$0.000\dots1 = \lim_{n \to \infty} 10^{-n} = 0$
$0.999\dots8 = \lim_{n \to \infty}(1-2\times 10^{-n}) = 1$
So the first five statements are all true, but the sixth is meaningless without a definition of "neighbouring real number". Gandalf61 (talk) 15:42, 5 March 2014 (UTC)