0.999...

Inmathematics,0.999...(also written as0.9,0..9,or0.(9)) is arepeating decimalthat is an alternative way of writing the number1.Following the standard rules for representingnumbersin decimal notation, its value is the smallest number greater than or equal to every number in the sequence0.9, 0.99, 0.999,....It can be proved that this number is1; that is,

Stylistic impression of the repeating decimal 0.9999..., representing the digit 9 repeating infinitely

$0.999\ldots =1.$

Despite common misconceptions, 0.999... is not "almost exactly 1" or "very, very nearly but not quite 1"; rather, "0.999..." and "1" representexactlythe same number.

An elementary proof is given below that involves onlyelementary arithmeticand the fact that there is nopositive real numberless than all${\tfrac {1}{10^{n}}}$wherenis anatural number,a property that results immediately from theArchimedean propertyof thereal numbers.

There are many other ways of showing this equality, fromintuitivearguments tomathematically rigorousproofs.The intuitive arguments are generally based on properties offinite decimalsthat are extended without proof to infinite decimals. The proofs are generally based on basic properties of real numbers and methods ofcalculus,such asseriesandlimits.A question studied inmathematics educationis why some people reject this equality.

Inother number systems,0.999... can have the same meaning, a different definition, or be undefined. Every nonzeroterminating decimalhas two equal representations (for example, 8.32000... and 8.31999...). Having values with multiple representations is a feature of allpositional numeral systemsthat represent the real numbers.

Elementary proof

TheArchimedean property:any pointxbefore the finish line lies between two of the pointsP_n(inclusive).

It is possible to prove the equation0.999... = 1using just the mathematical tools of comparison and addition of (finite)decimal numbers,without any reference to more advanced topics such asseriesandlimits.The proof givenbelowis a direct formalization of the intuitive fact that, if one draws 0.9, 0.99, 0.999, etc. on thenumber line,there is no room left for placing a number between them and 1. The meaning of the notation 0.999... is the least point on the number line lying to the right of all of the numbers 0.9, 0.99, 0.999, etc. Because there is ultimately no room between 1 and these numbers, the point 1 must be this least point, and so0.999... = 1.

Intuitive explanation

If one places 0.9, 0.99, 0.999, etc. on thenumber line,one sees immediately that all these points are to the left of 1, and that they get closer and closer to 1. For any number$x$that is less than 1, the sequence 0.9, 0.99, 0.999, and so on will eventually reach a number larger than⁠$x$⁠.So, it does not make sense to identify 0.999... with any number smaller than 1. Meanwhile, every number larger than 1 will be larger than any decimal of the form 0.999...9 for any finite number of nines. Therefore, 0.999... cannot be identified with any number larger than 1, either. Because 0.999... cannot be bigger than 1 or smaller than 1, it must equal 1 if it is to be any real number at all.

Rigorous proof

Denote by 0.(9)_nthe number 0.999...9, with$n$nines after the decimal point. Thus0.(9)₁= 0.9,0.(9)₂= 0.99,0.(9)₃= 0.999,and so on. One has1 − 0.(9)₁= 0.1 =⁠$\textstyle {\frac {1}{10}}$⁠,1 − 0.(9)₂= 0.01 =⁠$\textstyle {\frac {1}{10^{2}}}$⁠,and so on; that is,1 − 0.(9)_n=${\textstyle {\frac {1}{10^{n}}}}$for everynatural number⁠$n$⁠.

Let$x$be a number not greater than 1 and greater than 0.9, 0.99, 0.999, etc.; that is,0.(9)_n<$x$≤ 1,for every⁠$n$⁠.By subtracting these inequalities from 1, one gets0 ≤ 1 −$x$<⁠$\textstyle {\frac {1}{10^{n}}}$⁠.

The end of the proof requires that there is no positive number that is less than${\textstyle {\frac {1}{10^{n}}}}$for all⁠$n$⁠.This is one version of theArchimedean property,which is true for real numbers.This property implies that if1 −$x$<⁠$\textstyle {\frac {1}{10^{n}}}$⁠for all⁠$n$⁠,then1 −$x$can only be equal to 0. So,$x$= 1and 1 is the smallest number that is greater than all 0.9, 0.99, 0.999, etc. That is,1 = 0.999....

This proof relies on the Archimedean property of rational and real numbers. Real numbers may be enlarged intonumber systems,such ashyperreal numbers,with infinitely small numbers (infinitesimals) and infinitely large numbers (infinite numbers).When using such systems, the notation 0.999... is generally not used, as there is no smallest number among the numbers larger than all 0.(9)_n.

Least upper bounds and completeness

Part of what this argument shows is that there is aleast upper boundof the sequence 0.9, 0.99, 0.999, etc.: the smallest number that is greater than all of the terms of the sequence. One of theaxiomsof thereal number systemis thecompleteness axiom,which states that every bounded sequence has a least upper bound.This least upper bound is one way to define infinite decimal expansions: the real number represented by an infinite decimal is the least upper bound of its finite truncations.The argument here does not need to assume completeness to be valid, because it shows that this particular sequence of rational numbers has a least upper bound and that this least upper bound is equal to one.

Algebraic arguments

Simple algebraic illustrations of equality are a subject of pedagogical discussion and critique.Byers (2007)discusses the argument that, in elementary school, one is taught that${\textstyle {\frac {1}{3}}}$= 0.333...,so, ignoring all essential subtleties, "multiplying" this identity by 3 gives1 = 0.999....He further says that this argument is unconvincing, because of an unresolved ambiguity over the meaning of theequals sign;a student might think, "It surely does not mean that the number 1 is identical to that which is meant by the notation 0.999...‍."Most undergraduate mathematics majors encountered by Byers feel that while 0.999... is" very close "to 1 on the strength of this argument, with some even saying that it is" infinitely close ", they are not ready to say that it is equal to 1.Richman (1999)discusses how "this argument gets its force from the fact that most people have been indoctrinated to accept the first equation without thinking", but also suggests that the argument may lead skeptics to question this assumption.

Byers also presents the following argument.

${\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots &&{\text{by multiplying by }}10\\10x&=9+0.999\ldots &&{\text{by splitting off integer part}}\\10x&=9+x&&{\text{by definition of }}x\\9x&=9&&{\text{by subtracting }}x\\x&=1&&{\text{by dividing by }}9\end{aligned}}$

Students who did not accept the first argument sometimes accept the second argument, but, in Byers's opinion, still have not resolved the ambiguity, and therefore do not understand the representation of infinite decimals.Peressini & Peressini (2007),presenting the same argument, also state that it does not explain the equality, indicating that such an explanation would likely involve concepts of infinity andcompleteness.Baldwin & Norton (2012),citingKatz & Katz (2010a),also conclude that the treatment of the identity based on such arguments as these, without the formal concept of a limit, is premature.Cheng (2023)concurs, arguing that knowing one can multiply 0.999... by 10 by shifting the decimal point presumes an answer to the deeper question of how one gives a meaning to the expression 0.999... at all.The same argument is also given byRichman (1999),who notes that skeptics may question whether$x$iscancellable– that is, whether it makes sense to subtract$x$from both sides.Eisenmann (2008)similarly argues that both the multiplication and subtraction which removes the infinite decimal require further justification.

Analytic proofs

Real analysisis the study of the logical underpinnings ofcalculus,including the behavior of sequences and series of real numbers.The proofs in this section establish0.999... = 1using techniques familiar from real analysis.

Infinite series and sequences

Further information:Decimal representation

A common development of decimal expansions is to define them as sums ofinfinite series.In general: $$b_{0}.b_{1}b_{2}b_{3}b_{4}\ldots =b_{0}+b_{1}\left({\tfrac {1}{10}}\right)+b_{2}\left({\tfrac {1}{10}}\right)^{2}+b_{3}\left({\tfrac {1}{10}}\right)^{3}+b_{4}\left({\tfrac {1}{10}}\right)^{4}+\cdots.$$

For 0.999... one can apply theconvergencetheorem concerninggeometric series,stating that if⁠$\vert r\vert$⁠< 1,then: $$ar+ar^{2}+ar^{3}+\cdots ={\frac {ar}{1-r}}.$$

Since 0.999... is such a sum with$a=9$and common ratio⁠$\textstyle r={\frac {1}{10}}$⁠,the theorem makes short work of the question: $$0.999\ldots =9\left({\tfrac {1}{10}}\right)+9\left({\tfrac {1}{10}}\right)^{2}+9\left({\tfrac {1}{10}}\right)^{3}+\cdots ={\frac {9\left({\tfrac {1}{10}}\right)}{1-{\tfrac {1}{10}}}}=1.$$ This proof appears as early as 1770 inLeonhard Euler'sElements of Algebra.

Limits: The unit interval, including thebase-4fraction sequence(.3,.33,.333,...)converging to 1.

The sum of a geometric series is itself a result even older than Euler. A typical 18th-century derivation used a term-by-term manipulation similar to thealgebraic proofgiven above, and as late as 1811, Bonnycastle's textbookAn Introduction to Algebrauses such an argument for geometric series to justify the same maneuver on 0.999...‍.A 19th-century reaction against such liberal summation methods resulted in the definition that still dominates today: the sum of a series isdefinedto be the limit of the sequence of its partial sums. A corresponding proof of the theorem explicitly computes that sequence; it can be found in several proof-based introductions to calculus or analysis.

Asequence($x_{0}$,$x_{1}$,$x_{2}$,...)has the value$x$as itslimitif the distance$\left\vert x-x_{n}\right\vert$becomes arbitrarily small as$n$increases. The statement that0.999... = 1can itself be interpreted and proven as a limit: $$0.999\ldots \ {\overset {\underset {\mathrm {def} }{}}{=}}\ \lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ \lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}=\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)=1-\lim _{n\to \infty }{\frac {1}{10^{n}}}=1-0=1.$$ The first two equalities can be interpreted as symbol shorthand definitions. The remaining equalities can be proven. The last step, that 10^-napproaches 0 as$n$approaches infinity (⁠$\infty$⁠), is often justified by theArchimedean propertyof the real numbers. This limit-based attitude towards 0.999... is often put in more evocative but less precise terms. For example, the 1846 textbookThe University Arithmeticexplains, ".999 +, continued to infinity = 1, because every annexation of a 9 brings the value closer to 1"; the 1895Arithmetic for Schoolssays, "when a large number of 9s is taken, the difference between 1 and.99999... becomes inconceivably small".Suchheuristicsare often incorrectly interpreted by students as implying that 0.999... itself is less than 1.

Nested intervals and least upper bounds

Further information:Nested intervals

Nested intervals: in base 3,1 = 1.000... = 0.222....

The series definition above defines the real number named by a decimal expansion. A complementary approach is tailored to the opposite process: for a given real number, define the decimal expansion(s) to name it.

If a real number$x$is known to lie in theclosed interval[0, 10](that is, it is greater than or equal to 0 and less than or equal to 10), one can imagine dividing that interval into ten pieces that overlap only at their endpoints:[0, 1],[1, 2],[2, 3],and so on up to[9, 10].The number$x$must belong to one of these; if it belongs to[2, 3],then one records the digit "2" and subdivides that interval into[2, 2.1],[2.1, 2.2],...,[2.8, 2.9],[2.9, 3].Continuing this process yields an infinite sequence ofnested intervals,labeled by an infinite sequence of digits⁠$b_{1}$⁠,⁠$b_{2}$⁠,⁠$b_{3}$⁠,..., and one writes $$x=b_{0}.b_{1}b_{2}b_{3}\ldots \,.$$

In this formalism, the identities1 = 0.999...and1 = 1.000...reflect, respectively, the fact that 1 lies in both[0, 1].and[1, 2],so one can choose either subinterval when finding its digits. To ensure that this notation does not abuse the "=" sign, one needs a way to reconstruct a unique real number for each decimal. This can be done with limits, but other constructions continue with the ordering theme.

One straightforward choice is thenested intervals theorem,which guarantees that given a sequence of nested, closed intervals whose lengths become arbitrarily small, the intervals contain exactly one real number in theirintersection.So⁠$b_{1}$⁠,⁠$b_{2}$⁠,⁠$b_{3}$⁠,... is defined to be the unique number contained within all the intervals[$b_{0}$,$b_{0}$+ 1],[$b_{0}.b_{1}$,$b_{0}.b_{1}$+ 0.1],and so on. 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.

The nested intervals theorem is usually founded upon a more fundamental characteristic of the real numbers: the existence ofleast upper boundsorsuprema.To directly exploit these objects, one may define⁠$b_{0}.b_{1}b_{2}b_{3}$⁠... to be the least upper bound of the set of approximants⁠$b_{0}$⁠,⁠$b_{0}.b_{1}$⁠,⁠$b_{0}.b_{1}b_{2}$⁠,...‍.One can then show that this definition (or the nested intervals definition) is consistent with the subdivision procedure, implying0.999... = 1again.Tom Apostolconcludes, "the fact that a real number might have two different decimal representations is merely a reflection of the fact that two different sets of real numbers can have the same supremum."

Proofs from the construction of the real numbers

Further information:Construction of the real numbers

Some approaches explicitly define real numbers to be certainstructures built upon the rational numbers,usingaxiomatic set theory.Thenatural numbers{0, 1, 2, 3,...}begin with 0 and continue upwards so that every number has a successor. One can extend the natural numbers with their negatives to give all theintegers,and to further extend to ratios, giving therational numbers.These number systems are accompanied by the arithmetic of addition, subtraction, multiplication, and division.More subtly, they includeordering,so that one number can be compared to another and found to be less than, greater than, or equal to another number.

The step from rationals to reals is a major extension. There are at least two popular ways to achieve this step, both published in 1872:Dedekind cutsandCauchy sequences.Proofs that0.999... = 1that directly uses these constructions are not found in textbooks on real analysis, where the modern trend for the last few decades has been to use an axiomatic analysis. Even when a construction is offered, it is usually applied toward proving the axioms of the real numbers, which then support the above proofs. However, several authors express the idea that starting with a construction is more logically appropriate, and the resulting proofs are more self-contained.

Dedekind cuts

Further information:Dedekind cut

In theDedekind cutapproach, each real number$x$is defined as theinfinite setof allrational numbersless than⁠$x$⁠.In particular, the real number 1 is the set of all rational numbers that are less than 1.Every positive decimal expansion easily determines a Dedekind cut: the set of rational numbers that are less than some stage of the expansion. So the real number 0.999... is the set of rational numbers$r$such that$r$< 0,or$r$< 0.9,or$r$< 0.99,or$r$is less than some other number of the form $$1-{\frac {1}{10^{n}}}=0.(9)_{n}=0.\underbrace {99\ldots 9} _{n{\text{ nines}}}.$$

Every element of 0.999... is less than 1, so it is an element of the real number 1. Conversely, all elements of 1 are rational numbers that can be written as $${\frac {a}{b}}<1,$$ with$b>0$and⁠$b>a$⁠.This implies $$1-{\frac {a}{b}}={\frac {b-a}{b}}\geq {\frac {1}{b}}>{\frac {1}{10^{b}}},$$ and thus $${\frac {a}{b}}<1-{\frac {1}{10^{b}}}.$$

Since $$1-{\frac {1}{10^{b}}}=0.(9)_{b}<0.999\ldots,$$ by the definition above, every element of 1 is also an element of 0.999..., and, combined with the proof above that every element of 0.999... is also an element of 1, the sets 0.999... and 1 contain the same rational numbers, and are therefore the same set, that is,0.999... = 1.

The definition of real numbers as Dedekind cuts was first published byRichard Dedekindin 1872. The above approach to assigning a real number to each decimal expansion is due to an expository paper titled "Is0.999... = 1?"by Fred Richman inMathematics Magazine.Richman notes that taking Dedekind cuts in anydense subsetof the rational numbers yields the same results; in particular, he usesdecimal fractions,for which the proof is more immediate. He also notes that typically the definitions allow{$x$|$x$< 1}to be a cut but not{$x$|$x$≤ 1}(or vice versa).A further modification of the procedure leads to a different structure where the two are not equal. Although it is consistent, many of the common rules of decimal arithmetic no longer hold, for example, the fraction${\textstyle {\frac {1}{3}}}$has no representation; see§ Alternative number systemsbelow.

Cauchy sequences

Further information:Cauchy sequence

Another approach is to define a real number as the limit of aCauchy sequenceof rational numbers. This construction of the real numbers uses the ordering of rationals less directly. First, the distance between$x$and$y$is defined as the absolute value⁠$\left\vert x-y\right\vert$⁠,where the absolute value$\left\vert z\right\vert$is defined as the maximum of$z$and⁠$-z$⁠,thus never negative. Then the reals are defined to be the sequences of rationals that have the Cauchy sequence property using this distance. That is, in the sequence⁠$x_{0}$⁠,⁠$x_{1}$⁠,⁠$x_{2}$⁠,..., a mapping from natural numbers to rationals, for any positive rational$\delta$there is an$N$such that$\left\vert x_{m}-x_{n}\right\vert \leq \delta$for all⁠$m,n>N$⁠;the distance between terms becomes smaller than any positive rational.

If$(x_{n})$and$(y_{n})$are two Cauchy sequences, then they are defined to be equal as real numbers if the sequence$(x_{n}-y_{n})$has the limit 0. Truncations of the decimal number⁠$b_{0}.b_{1}b_{2}b_{3}$⁠... generate a sequence of rationals, which is Cauchy; this is taken to define the real value of the number.Thus in this formalism the task is to show that the sequence of rational numbers $$\left(1-0,1-{9 \over 10},1-{99 \over 100},\ldots \right)=\left(1,{1 \over 10},{1 \over 100},\ldots \right)$$ has a limit 0. Considering the⁠$n$⁠th term of the sequence, for⁠$n\in \mathbb {N}$⁠,it must therefore be shown that $$\lim _{n\rightarrow \infty }{\frac {1}{10^{n}}}=0.$$ This can be proved by thedefinition of a limit.So again,0.999... = 1.

The definition of real numbers as Cauchy sequences was first published separately byEduard HeineandGeorg Cantor,also in 1872.The above approach to decimal expansions, including the proof that0.999... = 1,closely follows Griffiths & Hilton's 1970 workA comprehensive textbook of classical mathematics: A contemporary interpretation.

Infinite decimal representation

Commonly insecondary schools' mathematics education, the real numbers are constructed by defining a number using an integer followed by aradix pointand an infinite sequence written out as a string to represent thefractional partof any given real number. In this construction, the set of any combination of an integer and digits after the decimal point (or radix point in non-base 10 systems) is the set of real numbers. This construction can be rigorously shown to satisfy all of thereal axiomsafter defining anequivalence relationover the set that defines1 =_eq0.999...as well as for any other nonzero decimals with only finitely many nonzero terms in the decimal string with its trailing 9s version. In other words, the equality0.999... = 1holding true is a necessary condition for strings of digits to behave as real numbers should.

Dense order

Further information:Dense order

One of the notions that can resolve the issue is the requirement that real numbers be densely ordered. Dense ordering implies that if there is no new element strictly between two elements of the set, the two elements must be considered equal. Therefore, if 0.99999... were to be different from 1, there would have to be another real number in between them but there is none: a single digit cannot be changed in either of the two to obtain such a number.

Generalizations

The result that0.999... = 1generalizes readily in two ways. First, every nonzero number with a finite decimal notation (equivalently, endless trailing 0s) has a counterpart with trailing 9s. For example, 0.24999... equals 0.25, exactly as in the special case considered. These numbers are exactly the decimal fractions, and they aredense.

Second, a comparable theorem applies in eachradix(base). For example, in base 2 (thebinary numeral system) 0.111... equals 1, and in base 3 (theternary numeral system) 0.222... equals 1. In general, any terminating base$b$expression has a counterpart with repeated trailing digits equal to$b$− 1.Textbooks of real analysis are likely to skip the example of 0.999... and present one or both of these generalizations from the start.

Alternative representations of 1 also occur in non-integer bases. For example, in thegolden ratio base,the two standard representations are 1.000... and 0.101010..., and there are infinitely many more representations that include adjacent 1s. Generally, foralmost all$q$between 1 and 2, there are uncountably manybase-$q$expansions of 1. In contrast, there are still uncountably many⁠$q$⁠,including all natural numbers greater than 1, for which there is only onebase-$q$expansion of 1, other than the trivial 1.000...‍.This result was first obtained byPaul Erdős,Miklos Horváth, and István Joó around 1990. In 1998 Vilmos Komornik andPaola Loretidetermined the smallest such base, theKomornik–Loreti constant$q$= 1.787231650...‍.In this base,1 = 0.11010011001011010010110011010011...;the digits are given by theThue–Morse sequence,which does not repeat.

A more far-reaching generalization addressesthe most general positional numeral systems.They too have multiple representations, and in some sense, the difficulties are even worse. For example:

• In thebalanced ternarysystem,${\textstyle {\frac {1}{2}}}$= 0.111... = 1.111...‍.
• In the reversefactorial number system(using bases 2!, 3!, 4!,... for positionsafterthe decimal point),1 = 1.000... = 0.1234...‍.

Petkovšek (1990)has proven that for any positional system that names all the real numbers, the set of reals with multiple representations is always dense. He calls the proof "an instructive exercise in elementarypoint-set topology";it involves viewing sets of positional values asStone spacesand noticing that their real representations are given bycontinuous functions.

Applications

One application of 0.999... as a representation of 1 occurs in elementarynumber theory.In 1802, H. Goodwyn published an observation on the appearance of 9s in the repeating-decimal representations of fractions whose denominators are certainprime numbers.Examples include:

• ${\textstyle {\frac {1}{7}}}$= 0.142857and142 + 857 = 999.
• ${\textstyle {\frac {1}{73}}}$= 0.01369863and0136 + 9863 = 9999.

E. Midy proved a general result about such fractions, now calledMidy's theorem,in 1836. The publication was obscure, and it is unclear whether his proof directly involved 0.999..., but at least one modern proof by William G. Leavitt does. If it can be proved that if a decimal of the form⁠$0.b_{1}b_{2}b_{3}$⁠... is a positive integer, then it must be 0.999..., which is then the source of the 9s in the theorem.Investigations in this direction can motivate such concepts asgreatest common divisors,modular arithmetic,Fermat primes,orderofgroupelements, andquadratic reciprocity.

Positions of⁠1/4⁠,⁠2/3⁠,and 1 in theCantor set

Returning to real analysis, the base-3 analogue0.222... = 1plays a key role in the characterization of one of the simplestfractals,the middle-thirdsCantor set:a point in theunit intervallies in the Cantor set if and only if it can be represented in ternary using only the digits 0 and 2.

The⁠$n$⁠th digit of the representation reflects the position of the point in the⁠$n$⁠th stage of the construction. For example, the point${\textstyle {\frac {2}{3}}}$is given the usual representation of 0.2 or 0.2000..., since it lies to the right of the first deletion and the left of every deletion thereafter. The point${\textstyle {\frac {1}{3}}}$is represented not as 0.1 but as 0.0222..., since it lies to the left of the first deletion and the right of every deletion thereafter.

Repeating nines also turns up in yet another of Georg Cantor's works. They must be taken into account to construct a valid proof, applyinghis 1891 diagonal argumentto decimal expansions, of theuncountabilityof the unit interval. Such a proof needs to be able to declare certain pairs of real numbers to be different based on their decimal expansions, so one needs to avoid pairs like 0.2 and 0.1999... A simple method represents all numbers with nonterminating expansions; the opposite method rules out repeating nines.A variant that may be closer to Cantor's original argument uses base 2, and by turning base-3 expansions into base-2 expansions, one can prove the uncountability of the Cantor set as well.

Skepticism in education

Students of mathematics often reject the equality of 0.999... and 1, for reasons ranging from their disparate appearance to deep misgivings over thelimitconcept and disagreements over the nature ofinfinitesimals.There are many common contributing factors to the confusion:

• Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal". Seeing two manifestly different decimals representing the same number appears to be aparadox,which is amplified by the appearance of the seemingly well-understood number 1.
• 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".
• Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999..." as meaning the sequence rather than its limit.

These ideas are mistaken in the context of the standard real numbers, although some may be valid in other number systems, either invented for their general mathematical utility or as instructivecounterexamplesto better understand 0.999...; see§ In alternative number systemsbelow.

Many of these explanations were found byDavid Tall,who has studied characteristics of teaching and cognition that lead to some of the misunderstandings he has encountered with his college students. Interviewing his students to determine why the vast majority initially rejected the equality, he found that "students continued to conceive of 0.999... as a sequence of numbers getting closer and closer to 1 and not a fixed value, because 'you haven't specified how many places there are' or 'it is the nearest possible decimal below 1'".

The elementary argument of multiplying0.333... =${\textstyle {\frac {1}{3}}}$by 3 can convince reluctant students that 0.999... = 1. Still, when confronted with the conflict between their belief in the first equation and their disbelief in the second, some students either begin to disbelieve the first equation or simply become frustrated.Nor are more sophisticated methods foolproof: students who are fully capable of applying rigorous definitions may still fall back on intuitive images when they are surprised by a result in advanced mathematics, including 0.999...‍.For example, one real analysis student was able to prove that0.333... =${\textstyle {\frac {1}{3}}}$using asupremumdefinition but then insisted that0.999... < 1based on her earlier understanding oflong division.Others still can prove that${\textstyle {\frac {1}{3}}}$= 0.333...,but, upon being confronted by thefractional proof,insist that "logic" supersedes the mathematical calculations.

Mazur (2005)tells the tale of an otherwise brilliant calculus student of his who "challenged almost everything I said in class but never questioned his calculator", and who had come to believe that nine digits are all one needs to do mathematics, including calculating the square root of 23. The student remained uncomfortable with a limiting argument that9.99... = 10,calling it a "wildly imagined infinite growing process".

As part of theAPOS Theoryof mathematical learning,Dubinsky et al. (2005)propose that students who conceive of 0.999... as a finite, indeterminate string with an infinitely small distance from 1 have "not yet constructed a complete process conception of the infinite decimal". Other students who have a complete process conception of 0.999... may not yet be able to "encapsulate" that process into an "object conception", like the object conception they have of 1, and so they view the process 0.999... and the object 1 as incompatible. They also link this mental ability of encapsulation to viewing${\textstyle {\frac {1}{3}}}$as a number in its own right and to dealing with the set of natural numbers as a whole.

Cultural phenomenon

With the rise of theInternet,debates about 0.999... have become commonplace onnewsgroupsandmessage boards,including many that nominally have little to do with mathematics. In the newsgroupsci.mathin the 1990s, arguing over 0.999... became a "popular sport", and was one of the questions answered in itsFAQ.The FAQ briefly covers⁠$\textstyle {\frac {1}{3}}$⁠,multiplication by 10, and limits, and alludes to Cauchy sequences as well.

A 2003 edition of the general-interest newspaper columnThe Straight Dopediscusses 0.999... via${\textstyle {\frac {1}{3}}}$and limits, saying of misconceptions,

The lower primate in us still resists, saying:.999~ doesn't really represent anumber,then, but aprocess.To find a number we have to halt the process, at which point the.999~ = 1 thing falls apart. Nonsense.

ASlatearticle reports that the concept of 0.999... is "hotly disputed on websites ranging fromWorld of Warcraftmessage boards toAyn Randforums ". 0.999... features also inmathematical jokes,such as:

Q: How many mathematicians does it take toscrew in a lightbulb?
A: 0.999999....

The fact that 0.999... is equal to 1 has been compared toZeno's paradox of the runner.The runner paradox can be mathematically modeled and then, like 0.999..., resolved using a geometric series. However, it is not clear whether this mathematical treatment addresses the underlying metaphysical issues Zeno was exploring.

In alternative number systems

Although the real numbers form an extremely usefulnumber system,the decision to interpret the notation "0.999..." as naming a real number is ultimately a convention, andTimothy Gowersargues inMathematics: A Very Short Introductionthat the resulting identity0.999... = 1is a convention as well:

However, it is by no means an arbitrary convention, because not adopting it forces one either to invent strange new objects or to abandon some of the familiar rules of arithmetic.

Infinitesimals

Main article:Infinitesimal

Some proofs that0.999... = 1rely on theArchimedean propertyof the real numbers: that there are no nonzeroinfinitesimals.Specifically, the difference1 − 0.999...must be smaller than any positive rational number, so it must be an infinitesimal; but since the reals do not contain nonzero infinitesimals, the difference is zero, and therefore the two values are the same.

However, there are mathematically coherent orderedalgebraic structures,including various alternatives to the real numbers, which are non-Archimedean.Non-standard analysisprovides a number system with a full array of infinitesimals (and their inverses).A. H. Lightstonedeveloped a decimal expansion forhyperreal numbersin(0, 1)^∗.Lightstone shows how to associate each number with a sequence of digits, $$0.d_{1}d_{2}d_{3}\ldots;\ldots d_{\infty -1}d_{\infty }d_{\infty +1}\ldots,$$ indexed by thehypernaturalnumbers. While he does not directly discuss 0.999..., he shows the real number${\textstyle {\frac {1}{3}}}$is represented by 0.333...;...333..., which is a consequence of thetransfer principle.As a consequence the number0.999...;...999... = 1.With this type of decimal representation, not every expansion represents a number. In particular "0.333...;...000..." and "0.999...;...000..." do not correspond to any number.

The standard definition of the number 0.999... is thelimit of the sequence0.9, 0.99, 0.999,...‍.A different definition involves anultralimit,i.e., the equivalence class[(0.9, 0.99, 0.999,...)]of this sequence in theultrapower construction,which is a number that falls short of 1 by an infinitesimal amount.More generally, the hyperreal number$u_{H}$= 0.999...;...999000...,with last digit 9 at infinitehypernaturalrank⁠$H$⁠,satisfies a strict inequality⁠$u_{H}<1$⁠.Accordingly, an alternative interpretation for "zero followed by infinitely many 9s" could be $${\underset {H}{0.\underbrace {999\ldots } }}\;=1\;-\;{\frac {1}{10^{H}}}.$$ All such interpretations of "0.999..." are infinitely close to 1.Ian Stewartcharacterizes this interpretation as an "entirely reasonable" way to rigorously justify the intuition that "there's a little bit missing" from 1 in 0.999....Along withKatz & Katz (2010b),Ely (2010)also questions the assumption that students' ideas about0.999... < 1are erroneous intuitions about the real numbers, interpreting them rather asnonstandardintuitions that could be valuable in the learning of calculus.

Hackenbush

Combinatorial game theoryprovides a generalized concept of number that encompasses the real numbers and much more besides.For example, in 1974,Elwyn Berlekampdescribed a correspondence between strings of red and blue segments inHackenbushand binary expansions of real numbers, motivated by the idea ofdata compression.For example, the value of the Hackenbush string LRRLRLRL... is0.010101...₂=${\textstyle {\frac {1}{3}}}$.However, the value of LRLLL... (corresponding to 0.111...₂) is infinitesimally less than 1. The difference between the two is thesurreal number${\textstyle {\frac {1}{\ Omega }}}$,where$\ Omega$is the firstinfinite ordinal;the relevant game is LRRRR... or 0.000...₂.

This is true of the binary expansions of many rational numbers, where the values of the numbers are equal but the corresponding binary tree paths are different. For example,0.10111...₂= 0.11000...₂,which are both equal to⁠$\textstyle {\frac {3}{4}}$⁠,but the first representation corresponds to the binary tree path LRLRLLL..., while the second corresponds to the different path LRLLRRR...‍.

Revisiting subtraction

Another manner in which the proofs might be undermined is if1 − 0.999...simply does not exist because subtraction is not always possible. Mathematical structures with an addition operation but not a subtraction operation includecommutativesemigroups,commutative monoids,andsemirings.Richman (1999)considers two such systems, designed so that0.999... < 1.

First,Richman (1999)defines a nonnegativedecimal numberto be a literal decimal expansion. He defines thelexicographical orderand an addition operation, noting that0.999... < 1simply because0 < 1in the ones place, but for any nonterminating⁠$x$⁠,one has0.999... +$x$= 1 +⁠$x$⁠.So one peculiarity of the decimal numbers is that addition cannot always be canceled; another is that no decimal number corresponds to⁠$\textstyle {\frac {1}{3}}$⁠.After defining multiplication, the decimal numbers form a positive, totally ordered, commutative semiring.

In the process of defining multiplication, Richman also defines another system he calls "cut⁠$D$⁠",which is the set ofDedekind cutsof decimal fractions. Ordinarily, this definition leads to the real numbers, but for a decimal fraction$d$he allows both the cut(⁠$-\infty$⁠,⁠$d$⁠)and the "principal cut"(⁠$-\infty$⁠,⁠$d$⁠].The result is that the real numbers are "living uneasily together with" the decimal fractions. Again0.999... < 1.There are no positive infinitesimals in cut⁠$D$⁠,but there is "a sort of negative infinitesimal", 0⁻,which has no decimal expansion. He concludes that0.999... = 1 + 0⁻,while the equation "0.999... +$x$= 1"has no solution.

p-adic numbers

Main article:p-adic number

When asked about 0.999..., novices often believe there should be a "final 9", believing1 − 0.999...to be a positive number which they write as "0.000...1". Whether or not that makes sense, the intuitive goal is clear: adding a 1 to the final 9 in 0.999... would carry all the 9s into 0s and leave a 1 in the ones place. Among other reasons, this idea fails because there is no "final 9" in 0.999...‍.However, there is a system that contains an infinite string of 9s including a last 9.

The 4-adic integers (black points), including the sequence(3, 33, 333,...)converging to −1. The 10-adic analogue is...999 = −1.

The$p$-adic numbersare an alternative number system of interest innumber theory.Like the real numbers, the$p$-adic numbers can be built from the rational numbers viaCauchy sequences;the construction uses a different metric in which 0 is closer to⁠$p$⁠,and much closer to⁠$p^{n}$⁠,than it is to 1.The$p$-adic numbers form afieldfor prime$p$and aringfor other⁠$p$⁠,including 10. So arithmetic can be performed in the$p$-adics, and there are no infinitesimals.

In the 10-adic numbers, the analogues of decimal expansions run to the left. The 10-adic expansion...999 does have a last 9, and it does not have a first 9. One can add 1 to the ones place, and it leaves behind only 0s after carrying through:1 +...999 =...000 = 0,and so...999 = −1.Another derivation uses a geometric series. The infinite series implied by "...999" does not converge in the real numbers, but it converges in the 10-adics, and so one can re-use the familiar formula: $$\ldots 999=9+9(10)+9(10)^{2}+9(10)^{3}+\cdots ={\frac {9}{1-10}}=-1.$$

Compare with the series in thesection above.A third derivation was invented by a seventh-grader who was doubtful over her teacher's limiting argument that0.999... = 1but was inspired to take the multiply-by-10 proofabovein the opposite direction: if$x$=...999,then10$x$=...990,so10$x$=$x$− 9,hence$x$= −1again.

As a final extension, since0.999... = 1(in the reals) and...999 = −1(in the 10-adics), then by "blind faith and unabashed juggling of symbols"one may add the two equations and arrive at...999.999... = 0.This equation does not make sense either as a 10-adic expansion or an ordinary decimal expansion, but it turns out to be meaningful and true in thedoubly infinitedecimal expansionof the10-adic solenoid,with eventually repeating left ends to represent the real numbers and eventually repeating right ends to represent the 10-adic numbers.

See also

• Finitism
• Informal mathematics

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External links

• .999999... = 1?fromCut-the-Knot
• Why does 0.9999... = 1?
• Proof of the equality based on arithmeticfrom Math Central
• David Tall'sresearch on mathematics cognition
• What is so wrong with thinking of real numbers as infinite decimals?
• Theorem 0.999...onMetamath