0.999... is a repeating decimal that represents the number 1. The three dots represent an infinite list of "9" digits. Following the standard rules for representing real numbers in decimal notation, its value is the smallest number greater than or equal to every number in the sequence 0.9, 0.99, 0.999, and so on. It can be proved that this number is 1; that is,
0.999 … = 1. {\displaystyle 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" represent exactly the same number. There are many ways of showing this equality, from intuitive arguments to mathematically rigorous proofs. The intuitive arguments are generally based on properties of finite decimals that are extended without proof to infinite decimals. An elementary but rigorous proof is given below that involves only elementary arithmetic and the Archimedean property: for each real number, there is a natural number that is greater (for example, by rounding up). Other proofs generally involve basic properties of real numbers and methods of calculus, such as series and limits. Why some people reject this equality is a question studied in mathematics education. The equality 0.999... = 1 generalizes to the fact that every non-zero terminating decimal has two equal representations (for example, 8.32000... and 8.31999...). Everything that preceeds refers to the decimal number system, that is to base 10. For other bases, everything remains true with 9 replaced with the largest digit. For example, in the binary number system, one has 0.111... = 1. In some other number systems, 0.999... can have the same meaning, a different definition, or be undefined.
Elementary proof
It is possible to prove the equation 0.999... = 1 using just the mathematical tools of comparison and addition of (finite) decimal numbers, without any reference to more advanced topics. The proof given below is a direct formalization of the intuitive fact that, if one draws 0.9, 0.99, 0.999, etc. on the number 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 so 0.999... = 1.
Intuitive explanation
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. For any number x {\displaystyle 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 {\displaystyle 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)n the number 0.999...9, with n {\displaystyle n} nines after the decimal point. Thus 0.(9)1 = 0.9, 0.(9)2 = 0.99, 0.(9)3 = 0.999, and so on. One has 1 − 0.(9)1 = 0.1 = 1 10 {\displaystyle \textstyle {\frac {1}{10}}} , 1 − 0.(9)2 = 0.01 = 1 10 2 {\displaystyle \textstyle {\frac {1}{10^{2}}}} , and so on; that is, 1 − 0.(9)n = 1 10 n {\textstyle {\frac {1}{10^{n}}}} for every natural number n {\displaystyle n} . Let x {\displaystyle x} be a number not greater than 1 and greater than 0.9, 0.99, 0.999, etc.; that is, 0.(9)n < x {\displaystyle x} ≤ 1, for every n {\displaystyle n} . By subtracting these inequalities from 1, one gets 0 ≤ 1 − x {\displaystyle x} < 1 10 n {\displaystyle \textstyle {\frac {1}{10^{n}}}} . The end of the proof requires that there be no positive number that is less than 1 10 n {\textstyle {\frac {1}{10^{n}}}} for all n {\displaystyle n} . This follows from the Archimedean property, which can be expressed as, "for every real number, there is a natural number that is greater". By computing the reciprocal, this implies that for every positive real number, there are natural numbers whose reciprocals are smaller. Therefore, for any positive real number, there must be some n {\displaystyle n} such that 1 10 n {\textstyle {\frac {1}{10^{n}}}} is smaller. This property implies that if 1 − x {\displaystyle x} < 1 10 n {\displaystyle \textstyle {\frac {1}{10^{n}}}} for all n {\displaystyle n} , then 1 − x {\displaystyle x} can only be equal to 0. So, x {\displaystyle x} = 1 and 1 is the smallest number that is greater than all 0.9, 0.99, 0.999, etc. That is, 1 = 0.999..., as claimed. This proof relies on the Archimedean property of rational 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, 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 a least upper bound of 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 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 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 1 3 {\textstyle {\frac {1}{3}}} = 0.333..., so, ignoring all essential subtleties, "multiplying" this identity by 3 gives 1 = 0.999.... He further says that this argument is unconvincing, because of an unresolved ambiguity over the meaning of the equals 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 [i.e., that 1 3 {\textstyle {\frac {1}{3}}} = 0.333...] without thinking", but also suggests that the argument may lead skeptics to question this assumption. Byers also presents the following argument.
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 and completeness. Baldwin & Norton (2012), citing Katz & 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 by Richman (1999), who notes that skeptics may question whether x {\displaystyle x} is cancellable – that is, whether it makes sense to subtract x {\displaystyle 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 analysis is the study of the logical underpinnings of calculus, including the behavior of sequences and series of real numbers. The proofs in this section establish 0.999... = 1 using techniques familiar from real analysis.
Infinite series and sequences
A common development of decimal expansions is to define them as infinite series. In general
b 0 . b 1 b 2 b 3 b 4 … = b 0 + b 1 ( 1 10 ) + b 2 ( 1 10 ) 2 + b 3 ( 1 10 ) 3 + b 4 ( 1 10 ) 4 + ⋯ . {\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, stating that if | r | {\displaystyle \vert r\vert } < 1, then
a r + a r 2 + a r 3 + ⋯ = a r 1 − r . {\displaystyle ar+ar^{2}+ar^{3}+\cdots ={\frac {ar}{1-r}}.}
Since 0.999... is such a sum with a = 9 {\displaystyle a=9} and common ratio r = 1 10 {\displaystyle \textstyle r={\frac {1}{10}}} , the theorem makes short work of the question:
0.999 … = 0 + 9 ( 1 10 ) + 9 ( 1 10 ) 2 + 9 ( 1 10 ) 3 + ⋯ = 9 ( 1 10 ) 1 − 1 10 = 1. {\displaystyle 0.999\ldots =0+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.
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.... 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 several proof-based introductions to calculus or analysis. A sequence ( x 0 {\displaystyle x_{0}} , x 1 {\displaystyle x_{1}} , x 2 {\displaystyle x_{2}} , ...) has the value x {\displaystyle x} as its limit if the distance | x − x n | {\displaystyle \left\vert x-x_{n}\right\vert } becomes arbitrarily small as n {\displaystyle n} increases. The statement that 0.999... = 1 can itself be interpreted and proven as a limit:
0.999 … = d e f lim n → ∞ 0. 99 … 9 ⏟ n = d e f lim n → ∞ ∑ k = 1 n 9 10 k = lim n → ∞ ( 1 − 1 10 n ) = 1 − lim n → ∞ 1 10 n = 1 − 0 = 1. {\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 10−n approaches 0 as n {\displaystyle n} approaches infinity ( ∞ {\displaystyle \infty } ), 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". Such heuristics are often incorrectly interpreted by students as implying that 0.999... itself is less than 1.
Nested intervals and least upper bounds
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 {\displaystyle x} is known to lie in the closed 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 {\displaystyle 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 b 1 {\displaystyle b_{1}} , b 2 {\displaystyle b_{2}} , b 3 {\displaystyle b_{3}} , ..., and one writes
x = b 0 . b 1 b 2 b 3 … . {\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. 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 b 1 {\displaystyle b_{1}} , b 2 {\displaystyle b_{2}} , b 3 {\displaystyle b_{3}} , ... is defined to be the unique number contained within all the intervals [ b 0 {\displaystyle b_{0}} , b 0 {\displaystyle b_{0}} + 1], [ b 0 . b 1 {\displaystyle b_{0}.b_{1}} , b 0 . b 1 {\displaystyle 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 of least upper bounds or suprema. To directly exploit these objects, one may define b 0 . b 1 b 2 b 3 {\displaystyle b_{0}.b_{1}b_{2}b_{3}} ... to be the least upper bound of the set of approximants b 0 {\displaystyle b_{0}} , b 0 . b 1 {\displaystyle b_{0}.b_{1}} , b 0 . b 1 b 2 {\displaystyle b_{0}.b_{1}b_{2}} , .... 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."
Proofs from the construction of the real numbers
Some approaches explicitly define real numbers to be certain structures built upon the rational numbers, using axiomatic set theory. The natural 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 the integers, and to further extend to ratios, giving the rational numbers. These number systems are accompanied by the arithmetic of addition, subtraction, multiplication, and division. More subtly, they include ordering, 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 cuts and Cauchy sequences. Proofs that 0.999... = 1 that 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 In the Dedekind cut approach, each real number x {\displaystyle x} is defined as the infinite set of all rational numbers less than x {\displaystyle 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 {\displaystyle r} such that r {\displaystyle r} < 0, or r {\displaystyle r} < 0.9, or r {\displaystyle r} < 0.99, or r {\displaystyle r} is less than some other number of the form
1 − 1 10 n = 0. ( 9 ) n = 0. 99 … 9 ⏟ n nines . {\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
a b < 1 , {\displaystyle {\frac {a}{b}}<1,}
with b > 0 {\displaystyle b>0} and b > a {\displaystyle b>a} . This implies
1 − a b = b − a b ≥ 1 b > 1 10 b , {\displaystyle 1-{\frac {a}{b}}={\frac {b-a}{b}}\geq {\frac {1}{b}}>{\frac {1}{10^{b}}},}
and thus
a b < 1 − 1 10 b . {\displaystyle {\frac {a}{b}}<1-{\frac {1}{10^{b}}}.}
Since
1 − 1 10 b = 0. ( 9 ) b < 0.999 … , {\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. 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. 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 {\displaystyle x} | x {\displaystyle x} < 1} to be a cut but not { x {\displaystyle x} | x {\displaystyle 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 1 3 {\textstyle {\frac {1}{3}}} has no representation; see § Alternative number systems below.
Cauchy sequences 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 {\displaystyle x} and y {\displaystyle y} is defined as the absolute value | x − y | {\displaystyle \left\vert x-y\right\vert } , where the absolute value | z | {\displaystyle \left\vert z\right\vert } is defined as the maximum of z {\displaystyle z} and − z {\displaystyle -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 {\displaystyle x_{0}} , x 1 {\displaystyle x_{1}} , x 2 {\displaystyle x_{2}} , ..., a mapping from natural numbers to rationals, for any positive rational δ {\displaystyle \delta } there is an N {\displaystyle N} such that | x m − x n | ≤ δ {\displaystyle \left\vert x_{m}-x_{n}\right\vert \leq \delta } for all m , n > N {\displaystyle m,n>N} ; the distance between terms becomes smaller than any positive rational. If ( x n ) {\displaystyle (x_{n})} and ( y n ) {\displaystyle (y_{n})} are two Cauchy sequences, then they are defined to be equal as real numbers if the sequence ( x n − y n ) {\displaystyle (x_{n}-y_{n})} has the limit 0. Truncations of the decimal number b 0 . b 1 b 2 b 3 {\displaystyle 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
( 1 − 0 , 1 − 9 10 , 1 − 99 100 , … ) = ( 1 , 1 10 , 1 100 , … ) {\displaystyle \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 {\displaystyle n} th term of the sequence, for n ∈ N {\displaystyle n\in \mathbb {N} } , it must therefore be shown that
lim n → ∞ 1 10 n = 0. {\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{10^{n}}}=0.}
This can be proved by the definition of a 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. 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.
Infinite decimal representation 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. In other words, the equality 0.999... = 1 holding true is a necessary condition for strings of digits to behave as real numbers should.
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 0.999... 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 that 0.999... = 1 generalizes 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 are dense. Second, a comparable theorem applies in each radix (base). For example, in base 2 (the binary numeral system) 0.111... equals 1, and in base 3 (the ternary numeral system) 0.222... equals 1. In general, any terminating base b {\displaystyle b} expression has a counterpart with repeated trailing digits equal to b − 1 {\displaystyle 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 the golden ratio base, the two standard representations are 1.000... and 0.101010..., and there are infinitely many more representations that include adjacent 1s. Ge