0.999...

If one places 0.9, 0.99, 0.999, etc. on the number line, one sees immediately that all these points are to the left of 1, and that they get closer and closer to 1.

More precisely, the distance from 0.9 to 1 is 0.1 = 1/10, the distance from 0.99 to 1 is 0.01 = 1/10², and so on. The distance to 1 from the nth point (the one with n 9s after the decimal point) is 1/10ⁿ.

Therefore, if 1 were not the smallest number greater than 0.9, 0.99, 0.999, etc., then there would be a point on the number line that lies between 1 and all these points. This point would be at a positive distance from 1 that is less than 1/10ⁿ for every integer n. In the standard number systems (the rational numbers and the real numbers), there is no positive number that is less than 1/10ⁿ for all n. This is (one version of) the Archimedean property, which can be proven to hold in the system of rational numbers. Therefore, 1 is the smallest number that is greater than all 0.9, 0.99, 0.999, etc., and so 1 = 0.999....

Part of what this argument shows is that there is a least upper bound of the sequence 0.9, 0.99, 0.999, etc.: a smallest number that is greater than all of the terms of the sequence. One of the axioms of the real number system is the completeness 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 in fact has a least upper bound, and that this least upper bound is equal to one.

The previous explanation is not a proof, as one cannot define properly the relationship between a number and its representation as a point on the number line. For the accuracy of the proof, the number 0.999...9, with n nines after the decimal point, is denoted 0.(9)_n. Thus 0.(9)₁ = 0.9, 0.(9)₂ = 0.99, 0.(9)₃ = 0.999, and so on. As 1/10ⁿ = 0.0...01, with n digits after the decimal point, the addition rule for decimal numbers implies

${\displaystyle 0.(9)_{n}+1/10^{n}=1,}$

and

${\displaystyle 0.(9)_{n}<1,}$

for every positive integer n.

One has to show that 1 is the smallest number that is no less than all 0.(9)_n. For this, it suffices to prove that, if a number x is not larger than 1 and no less than all 0.(9)_n, then x = 1. So let x such that

${\displaystyle 0.(9)_{n}\leq x\leq 1,}$

for every positive integer n. Therefore,

${\displaystyle 1-1\leq 1-x\leq 1-0.(9)_{n}.}$

which, using basic arithmetic and the first equality established above, simplifies to

${\displaystyle 0\leq 1-x\leq 1/10^{n}.}$

This implies that the difference between 1 and x is less than the inverse of any positive integer. Thus this difference must be zero, and, thus x = 1; that is

${\displaystyle 0.999\ldots =1.}$

This proof relies on the fact that zero is the only nonnegative number that is less than all inverses of integers, or equivalently that there is no number that is larger than every integer. This is the Archimedean property, that is verified for rational numbers and real numbers. Real numbers may be enlarged into number systems, such as hyperreal numbers, with infinitely small numbers (infinitesimals) and infinitely large numbers (infinite numbers). When using such systems, notation 0.999... is generally not used, as there is no smallest number that is no less than all 0.(9)_n. (This is implied by the fact that 0.(9)_n ≤ x < 1 implies 0.(9)_n–1 ≤ 2x – 1 < x < 1).

A common development of decimal expansions is to define them as sums of infinite series. In general:

${\displaystyle b_{0}.b_{1}b_{2}b_{3}b_{4}\ldots =b_{0}+b_{1}\left({\tfrac {1}{10}}\right)+b_{2}\left({\tfrac {1}{10}}\right)^{2}+b_{3}\left({\tfrac {1}{10}}\right)^{3}+b_{4}\left({\tfrac {1}{10}}\right)^{4}+\cdots .}$

For 0.999... one can apply the convergence theorem concerning geometric series:[3]

If ${\displaystyle |r|<1}$ then ${\displaystyle ar+ar^{2}+ar^{3}+\cdots ={\frac {ar}{1-r}}.}$

Since 0.999... is such a sum with a = 9 and common ratio r = 1⁄10, the theorem makes short work of the question:

${\displaystyle 0.999\ldots =9\left({\tfrac {1}{10}}\right)+9\left({\tfrac {1}{10}}\right)^{2}+9\left({\tfrac {1}{10}}\right)^{3}+\cdots ={\frac {9\left({\tfrac {1}{10}}\right)}{1-{\tfrac {1}{10}}}}=1.}$

This proof appears as early as 1770 in Leonhard Euler's Elements of Algebra.[4]

Limits: The unit interval, including the base-4 fraction 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 the algebraic proof given above, and as late as 1811, Bonnycastle's textbook An Introduction to Algebra uses such an argument for geometric series to justify the same maneuver on 0.999...[5] A 19th-century reaction against such liberal summation methods resulted in the definition that still dominates today: the sum of a series is defined to 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 any proof-based introduction to calculus or analysis.[6]

A sequence (x₀, x₁, x₂, ...) has a limit x if the distance |x − x_n| becomes arbitrarily small as n increases. The statement that 0.999... = 1 can itself be interpreted and proven as a limit:[7]

${\displaystyle 0.999\ldots \ {\overset {\underset {\mathrm {def} }{}}{=}}\ \lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ \lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}\ =\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)=1-\lim _{n\to \infty }{\frac {1}{10^{n}}}=1\,-\,0=1.}$

The first two equalities can be interpreted as symbol shorthand definitions. The remaining equalities can be proven. The last step, that 1⁄10ⁿ → 0 as n → ∞, is often justified by the Archimedean property of the real numbers. This limit-based attitude towards 0.999... is often put in more evocative but less precise terms. For example, the 1846 textbook The University Arithmetic explains, ".999 +, continued to infinity = 1, because every annexation of a 9 brings the value closer to 1"; the 1895 Arithmetic for Schools says, "when a large number of 9s is taken, the difference between 1 and .99999... becomes inconceivably small".[8] Such heuristics are often incorrectly interpreted by students as implying that 0.999... itself is less than 1.

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

The series definition above is a simple way to define 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 the closed interval [0, 10] (i.e., 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 of nested intervals, labeled by an infinite sequence of digits b₀, b₁, b₂, b₃, ..., and one writes

${\displaystyle x=b_{0}.b_{1}b_{2}b_{3}\ldots }$

In this formalism, the identities 1 = 0.999... and 1 = 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.[9]

One straightforward choice is the nested 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 their intersection. So b₀.b₁b₂b₃... is defined to be the unique number contained within all the intervals [b₀, b₀ + 1], [b₀.b₁, b₀.b₁ + 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.[10]

The Nested Intervals Theorem is usually founded upon a more fundamental characteristic of the real numbers: the existence of least upper bounds or suprema. To directly exploit these objects, one may define b₀.b₁b₂b₃... to be the least upper bound of the set of approximants {b₀, b₀.b₁, b₀.b₁b₂, ...}.[11] One can then show that this definition (or the nested intervals definition) is consistent with the subdivision procedure, implying 0.999... = 1 again. Tom Apostol concludes,

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.[12]

In the Dedekind cut approach, each real number x is defined as the infinite set of all rational numbers less than x.[14] In particular, the real number 1 is the set of all rational numbers that are less than 1.[15] Every positive decimal expansion easily determines a Dedekind cut: the set of rational numbers which 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[16]

${\displaystyle 1-{\frac {1}{10^{n}}}=0.(9)_{n}=0.\underbrace {99\ldots 9} _{n{\text{ nines}}}.}$

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

${\displaystyle {\frac {a}{b}}<1,}$

with b > 0 and b > a. This implies

${\displaystyle 1-{\frac {a}{b}}={\frac {b-a}{b}}\geq {\frac {1}{b}}>{\frac {1}{10^{b}}},}$

and thus

${\displaystyle {\frac {a}{b}}<1-{\frac {1}{10^{b}}}.}$

and since

${\displaystyle 1-{\frac {1}{10^{b}}}=0.(9)_{b}<0.999\ldots }$

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 by Richard Dedekind in 1872.[17] The above approach to assigning a real number to each decimal expansion is due to an expository paper titled "Is 0.999 ... = 1?" by Fred Richman in Mathematics Magazine,[18] which is targeted at teachers of collegiate mathematics, especially at the junior/senior level, and their students.[19] Richman notes that taking Dedekind cuts in any dense subset of the rational numbers yields the same results; in particular, he uses decimal 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) "Why do that? Precisely to rule out the existence of distinct numbers 0.9* and 1. [...] So we see that in the traditional definition of the real numbers, the equation 0.9* = 1 is built in at the beginning."[20] 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 1⁄3 has no representation; see "Alternative number systems" below.

Another approach is to define a real number as the limit of a Cauchy sequence of 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 |x − y|, where the absolute value |z| 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 (x₀, x₁, x₂, ...), a mapping from natural numbers to rationals, for any positive rational δ there is an N such that |x_m − x_n| ≤ δ for all m, n > N. (The distance between terms becomes smaller than any positive rational.)[21]

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 b₀.b₁b₂b₃... generate a sequence of rationals which is Cauchy; this is taken to define the real value of the number.[22] Thus in this formalism the task is to show that the sequence of rational numbers

${\displaystyle \left(1-0,1-{9 \over 10},1-{99 \over 100},\dots \right)=\left(1,{1 \over 10},{1 \over 100},\dots \right)}$

has the limit 0. Considering the nth term of the sequence, for n ∈ ${\displaystyle \mathbb {N} }$, it must therefore be shown that

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

This limit is plain[23] if one understands the definition of limit. So again 0.999... = 1.

The definition of real numbers as Cauchy sequences was first published separately by Eduard Heine and Georg Cantor, also in 1872.[17] The above approach to decimal expansions, including the proof that 0.999... = 1, closely follows Griffiths & Hilton's 1970 work A comprehensive textbook of classical mathematics: A contemporary interpretation. The book is written specifically to offer a second look at familiar concepts in a contemporary light.[24]

Commonly in secondary schools' mathematics education, the real numbers are constructed by defining a number using an integer followed by a radix point and an infinite sequence written out as a string to represent the fractional part of 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 the real axioms after defining an equivalence relation over the set that defines 1 =_eq 0.999... as well as for any other nonzero decimals with only finitely many nonzero terms in the decimal string with its trailing 9s version.[25] With this construction of the reals, all proofs of the statement "1 = 0.999..." can be viewed as implicitly assuming the equality when any operations are performed on the real numbers.

One of the notions that can resolve the issue is the requirement that real numbers are densely ordered. Students are taking for granted that ${\displaystyle 0.99999...}$ is before ${\displaystyle 1}$ while this kind of intuitive ordering is better defined as purely lexicographical.

"... the ordering of the real numbers is recognized as a dense order. However, depending on the context, students can reconcile this property with the existence of numbers just before or after a given number (0.999... is thus often seen as the predecessor of 1)."[26]

Dense order requires that there is a third real value strictly between ${\displaystyle 0.99999...}$ and ${\displaystyle 1}$, but there is none: we cannot change a single digit in either of the two to obtain such a number. If ${\displaystyle 0.99999...}$ and ${\displaystyle 1}$ are to represent real numbers they have to be equal. Dense ordering implies that if there is no new element strictly between two elements of the set, the two elements must be considered equal.

That all these different number systems suffer from multiple representations for some real numbers can be attributed to a fundamental difference between the real numbers as an ordered set and collections of infinite strings of symbols, ordered lexicographically. Indeed, the following two properties account for the difficulty:

• If an interval of the real numbers is partitioned into two non-empty parts L, R, such that every element of L is (strictly) less than every element of R, then either L contains a largest element or R contains a smallest element, but not both.
• The collection of infinite strings of symbols taken from any finite "alphabet", lexicographically ordered, can be partitioned into two non-empty parts L, R, such that every element of L is less than every element of R, while L contains a largest element and R contains a smallest element. Indeed, it suffices to take two finite prefixes (initial substrings) p₁, p₂ of elements from the collection such that they differ only in their final symbol, for which symbol they have successive values, and take for L the set of all strings in the collection whose corresponding prefix is at most p₁, and for R the remainder, the strings in the collection whose corresponding prefix is at least p₂. Then L has a largest element, starting with p₁ and choosing the largest available symbol in all following positions, while R has a smallest element obtained by following p₂ by the smallest symbol in all positions.

The first point follows from basic properties of the real numbers: L has a supremum and R has an infimum, which are easily seen to be equal; being a real number it either lies in R or in L, but not both since L and R are supposed to be disjoint. The second point generalizes the 0.999.../1.000... pair obtained for p₁ = "0", p₂ = "1". In fact one need not use the same alphabet for all positions (so that for instance mixed radix systems can be included) or consider the full collection of possible strings; the only important points are that at each position a finite set of symbols (which may even depend on the previous symbols) can be chosen from (this is needed to ensure maximal and minimal choices), and that making a valid choice for any position should result in a valid infinite string (so one should not allow "9" in each position while forbidding an infinite succession of "9"s). Under these assumptions, the above argument shows that an order preserving map from the collection of strings to an interval of the real numbers cannot be a bijection: either some numbers do not correspond to any string, or some of them correspond to more than one string.

Marko Petkovšek 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 elementary point-set topology"; it involves viewing sets of positional values as Stone spaces and noticing that their real representations are given by continuous functions.[31]

Some proofs that 0.999... = 1 rely on the Archimedean property of the real numbers: that there are no nonzero infinitesimals. Specifically, the difference 1 − 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 therefore zero, and therefore the two values are the same.

However, there are mathematically coherent ordered algebraic structures, including various alternatives to the real numbers, which are non-Archimedean. Non-standard analysis provides a number system with a full array of infinitesimals (and their inverses).[51] A. H. Lightstone developed a decimal expansion for hyperreal numbers in (0, 1)^∗.[52] Lightstone shows how to associate to each number a sequence of digits,

${\displaystyle 0.d_{1}d_{2}d_{3}\dots$ ;\dots d_{\infty -1}d_{\infty }d_{\infty +1}\dots ,}

indexed by the hypernatural numbers. While he does not directly discuss 0.999..., he shows the real number 1⁄3 is represented by 0.333...;...333... which is a consequence of the transfer principle. As a consequence the number 0.999...;...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 the limit of the sequence 0.9, 0.99, 0.999, ... A different definition involves what Terry Tao refers to as ultralimit, i.e., the equivalence class [(0.9, 0.99, 0.999, ...)] of this sequence in the ultrapower 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 infinite hypernatural rank H, satisfies a strict inequality u_H < 1. Accordingly, an alternative interpretation for "zero followed by infinitely many 9s" could be

${\displaystyle {\underset {H}{0.\underbrace {999\ldots } }}\;=1\;-\;{\frac {1}{10^{H}}}.}$[53]

All such interpretations of "0.999..." are infinitely close to 1. Ian Stewart characterizes this interpretation as an "entirely reasonable" way to rigorously justify the intuition that "there's a little bit missing" from 1 in 0.999....[54] Along with Katz & Katz, Robert Ely also questions the assumption that students' ideas about 0.999... < 1 are erroneous intuitions about the real numbers, interpreting them rather as nonstandard intuitions that could be valuable in the learning of calculus.[55][56]

Combinatorial game theory provides alternative reals as well, with infinite Blue-Red Hackenbush as one particularly relevant example. In 1974, Elwyn Berlekamp described a correspondence between Hackenbush strings and binary expansions of real numbers, motivated by the idea of data compression. For example, the value of the Hackenbush string LRRLRLRL... is 0.010101₂... = 1⁄3. However, the value of LRLLL... (corresponding to 0.111...₂) is infinitesimally less than 1. The difference between the two is the surreal number 1⁄ω, where ω is the first infinite ordinal; the relevant game is LRRRR... or 0.000...₂.[57]

This is in fact 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 3/4, but the first representation corresponds to the binary tree path LRLRLLL... while the second corresponds to the different path LRLLRRR....

Another manner in which the proofs might be undermined is if 1 − 0.999... simply does not exist, because subtraction is not always possible. Mathematical structures with an addition operation but not a subtraction operation include commutative semigroups, commutative monoids and semirings. Richman considers two such systems, designed so that 0.999... < 1.

First, Richman defines a nonnegative decimal number to be a literal decimal expansion. He defines the lexicographical order and an addition operation, noting that 0.999... < 1 simply because 0 < 1 in the ones place, but for any nonterminating x, one has 0.999... + x = 1 + x. So one peculiarity of the decimal numbers is that addition cannot always be cancelled; another is that no decimal number corresponds to 1⁄3. After defining multiplication, the decimal numbers form a positive, totally ordered, commutative semiring.[58]

In the process of defining multiplication, Richman also defines another system he calls "cut D", which is the set of Dedekind cuts of decimal fractions. Ordinarily this definition leads to the real numbers, but for a decimal fraction d he allows both the cut (−∞, d) and the "principal cut" (−∞, d]. The result is that the real numbers are "living uneasily together with" the decimal fractions. Again 0.999... < 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 that 0.999... = 1 + 0⁻, while the equation "0.999... + x = 1" has no solution.[59]

When asked about 0.999..., novices often believe there should be a "final 9", believing 1 − 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....[60] 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 numbers are an alternative number system of interest in number theory. Like the real numbers, the p-adic numbers can be built from the rational numbers via Cauchy sequences; the construction uses a different metric in which 0 is closer to p, and much closer to pⁿ, than it is to 1. The p-adic numbers form a field for prime p and a ring for 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.[61] 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:

${\displaystyle \ldots 999=9+9(10)+9(10)^{2}+9(10)^{3}+\cdots ={\frac {9}{1-10}}=-1.}$[62]

(Compare with the series above.) A third derivation was invented by a seventh-grader who was doubtful over her teacher's limiting argument that 0.999... = 1 but was inspired to take the multiply-by-10 proof above in the opposite direction: if x = ...999 then 10x =  ...990, so 10x = x − 9, hence x = −1 again.[61]

As a final extension, since 0.999... = 1 (in the reals) and ...999 = −1 (in the 10-adics), then by "blind faith and unabashed juggling of symbols"[63] 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 the doubly infinite decimal expansion of the 10-adic solenoid, with eventually repeating left ends to represent the real numbers[64] and eventually repeating right ends to represent the 10-adic numbers.