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0.999...

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From Wikipedia, the free encyclopedia
Alternative decimal expansion of 1

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

Inmathematics,0.999... (also written as0.9,0..9, or0.(9)) is arepeating decimal that is an alternative way of writing the number1. Following the standard rules for representingnumbers in 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 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" representexactly the same number.

An elementary proof is given below that involves onlyelementary arithmetic and the fact that there is nopositive real number less than all110n{\displaystyle {\tfrac {1}{10^{n}}}}wheren is anatural number, a property that results immediately from theArchimedean property of thereal numbers.

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

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

Elementary proof

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TheArchimedean property: any pointx before the finish line lies between two of the pointsPn (inclusive).

It is possible to prove the equation0.999... = 1 using just the mathematical tools of comparison and addition of (finite)decimal numbers, without any reference to more advanced topics such asseries andlimits. The proof givenbelow is 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

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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 numberx{\displaystyle x} that is less than 1, the sequence 0.9, 0.99, 0.999, and so on will eventually reach a number larger thanx{\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.[1][2]

Rigorous proof

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Denote by 0.(9)n the number 0.999...9, withn{\displaystyle 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 =110{\displaystyle \textstyle {\frac {1}{10}}},1 − 0.(9)2 = 0.01 =1102{\displaystyle \textstyle {\frac {1}{10^{2}}}}, and so on; that is,1 − 0.(9)n =110n{\textstyle {\frac {1}{10^{n}}}} for everynatural numbern{\displaystyle n}.

Letx{\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 everyn{\displaystyle n}. By subtracting these inequalities from 1, one gets0 ≤ 1 −x{\displaystyle x} <110n{\displaystyle \textstyle {\frac {1}{10^{n}}}}.

The end of the proof requires that there is no positive number that is less than110n{\textstyle {\frac {1}{10^{n}}}} for alln{\displaystyle n}. This is one version of theArchimedean property, which is true for real numbers.[3][4] This property implies that if1 −x{\displaystyle x} <110n{\displaystyle \textstyle {\frac {1}{10^{n}}}} for alln{\displaystyle n}, then1 −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....

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).[5][6] 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.[a]

Least upper bounds and completeness

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Part of what this argument shows is that there is aleast 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 theaxioms of thereal number system is thecompleteness axiom, which states that every bounded sequence has a least upper bound.[7][8] 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.[9] 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.[10]

Algebraic arguments

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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 that13{\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.[11]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.[12]

Byers also presents the following argument.

x=0.99910x=9.999by multiplying by 1010x=9+0.999by splitting off integer part10x=9+xby definition of x9x=9by subtracting xx=1by dividing by 9{\displaystyle {\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.[13]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.[14]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.[15] The same argument is also given byRichman (1999), who notes that skeptics may question whetherx{\displaystyle x} iscancellable – that is, whether it makes sense to subtractx{\displaystyle x} from both sides.[12]Eisenmann (2008) similarly argues that both the multiplication and subtraction which removes the infinite decimal require further justification.[16]

Analytic proofs

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Real analysis is the study of the logical underpinnings ofcalculus, including the behavior of sequences and series of real numbers.[17] The proofs in this section establish0.999... = 1 using techniques familiar from real analysis.

Infinite series and sequences

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Further information:Decimal representation

A common development of decimal expansions is to define them as sums ofinfinite series. In general:b0.b1b2b3b4=b0+b1(110)+b2(110)2+b3(110)3+b4(110)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 theconvergence theorem concerninggeometric series, stating that if|r|{\displaystyle \vert r\vert } < 1, then:[18]ar+ar2+ar3+=ar1r.{\displaystyle ar+ar^{2}+ar^{3}+\cdots ={\frac {ar}{1-r}}.}

Since 0.999... is such a sum witha=9{\displaystyle a=9} and common ratior=110{\displaystyle \textstyle r={\frac {1}{10}}}, the theorem makes short work of the question:0.999=9(110)+9(110)2+9(110)3+=9(110)1110=1.{\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 inLeonhard Euler'sElements of Algebra.[19]

Limits: The unit interval, including thebase-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 thealgebraic proof given above, and as late as 1811, Bonnycastle's textbookAn Introduction to Algebra uses such an argument for geometric series to justify the same maneuver on 0.999....[20] A 19th-century reaction against such liberal summation methods resulted in the definition that still dominates today: the sum of a series isdefined 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.[21]

Asequence(x0{\displaystyle x_{0}},x1{\displaystyle x_{1}},x2{\displaystyle x_{2}}, ...) has the valuex{\displaystyle x} as itslimit if the distance|xxn|{\displaystyle \left\vert x-x_{n}\right\vert } becomes arbitrarily small asn{\displaystyle n} increases. The statement that0.999... = 1 can itself be interpreted and proven as a limit:[b]0.999 =def limn0.999n =def limnk=1n910k=limn(1110n)=1limn110n=10=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 asn{\displaystyle n} approaches infinity ({\displaystyle \infty }), is often justified by theArchimedean 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 textbookThe University Arithmetic explains, ".999 +, continued to infinity = 1, because every annexation of a 9 brings the value closer to 1"; the 1895Arithmetic for Schools says, "when a large number of 9s is taken, the difference between 1 and .99999... becomes inconceivably small".[22] Suchheuristics are often incorrectly interpreted by students as implying that 0.999... itself is less than 1.[23]

Nested intervals and least upper bounds

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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 numberx{\displaystyle 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 numberx{\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 ofnested intervals, labeled by an infinite sequence of digitsb1{\displaystyle b_{1}},b2{\displaystyle b_{2}},b3{\displaystyle b_{3}}, ..., and one writesx=b0.b1b2b3.{\displaystyle 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.[24]

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. Sob1{\displaystyle b_{1}},b2{\displaystyle b_{2}},b3{\displaystyle b_{3}}, ... is defined to be the unique number contained within all the intervals[b0{\displaystyle b_{0}},b0{\displaystyle b_{0}} + 1],[b0.b1{\displaystyle b_{0}.b_{1}},b0.b1{\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.[25]

The nested intervals theorem is usually founded upon a more fundamental characteristic of the real numbers: the existence ofleast upper bounds orsuprema. To directly exploit these objects, one may defineb0.b1b2b3{\displaystyle b_{0}.b_{1}b_{2}b_{3}}... to be the least upper bound of the set of approximantsb0{\displaystyle b_{0}},b0.b1{\displaystyle b_{0}.b_{1}},b0.b1b2{\displaystyle b_{0}.b_{1}b_{2}}, ....[26] One can then show that this definition (or the nested intervals definition) is consistent with the subdivision procedure, implying0.999... = 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."[27]

Proofs from the construction of the real numbers

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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.[28][29] 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.[30]

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 andCauchy sequences. Proofs that0.999... = 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.[c]

Dedekind cuts

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Further information:Dedekind cut

In theDedekind cut approach, each real numberx{\displaystyle x} is defined as theinfinite set of allrational numbers less thanx{\displaystyle x}.[d] In particular, the real number 1 is the set of all rational numbers that are less than 1.[e] 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 numbersr{\displaystyle r} such thatr{\displaystyle r} < 0, orr{\displaystyle r} < 0.9, orr{\displaystyle r} < 0.99, orr{\displaystyle r} is less than some other number of the form[31]1110n=0.(9)n=0.999n 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 asab<1,{\displaystyle {\frac {a}{b}}<1,}withb>0{\displaystyle b>0} andb>a{\displaystyle b>a}. This implies1ab=bab1b>110b,{\displaystyle 1-{\frac {a}{b}}={\frac {b-a}{b}}\geq {\frac {1}{b}}>{\frac {1}{10^{b}}},}and thusab<1110b.{\displaystyle {\frac {a}{b}}<1-{\frac {1}{10^{b}}}.}

Since1110b=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 byRichard Dedekind in 1872.[32]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.[12] Richman notes that taking Dedekind cuts in anydense subset of 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{\displaystyle x} |x{\displaystyle x} < 1} to be a cut but not{x{\displaystyle x} |x{\displaystyle x} ≤ 1} (or vice versa).[33] 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 fraction13{\textstyle {\frac {1}{3}}} has no representation; see§ Alternative number systems below.

Cauchy sequences

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Further information:Cauchy sequence

Another approach is to define a real number as the limit of aCauchy sequence of rational numbers. This construction of the real numbers uses the ordering of rationals less directly. First, the distance betweenx{\displaystyle x} andy{\displaystyle y} is defined as the absolute value|xy|{\displaystyle \left\vert x-y\right\vert }, where the absolute value|z|{\displaystyle \left\vert z\right\vert } is defined as the maximum ofz{\displaystyle z} andz{\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 sequencex0{\displaystyle x_{0}},x1{\displaystyle x_{1}},x2{\displaystyle x_{2}}, ..., a mapping from natural numbers to rationals, for any positive rationalδ{\displaystyle \delta } there is anN{\displaystyle N} such that|xmxn|δ{\displaystyle \left\vert x_{m}-x_{n}\right\vert \leq \delta } for allm,n>N{\displaystyle m,n>N}; the distance between terms becomes smaller than any positive rational.[34]

If(xn){\displaystyle (x_{n})} and(yn){\displaystyle (y_{n})} are two Cauchy sequences, then they are defined to be equal as real numbers if the sequence(xnyn){\displaystyle (x_{n}-y_{n})} has the limit 0. Truncations of the decimal numberb0.b1b2b3{\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.[35] Thus in this formalism the task is to show that the sequence of rational numbers(10,1910,199100,)=(1,110,1100,){\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 then{\displaystyle n}th term of the sequence, fornN{\displaystyle n\in \mathbb {N} }, it must therefore be shown thatlimn110n=0.{\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{10^{n}}}=0.}This can be proved by thedefinition of a limit. So again,0.999... = 1.[36]

The definition of real numbers as Cauchy sequences was first published separately byEduard Heine andGeorg Cantor, also in 1872.[32] 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.[37]

Infinite decimal representation

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Commonly insecondary schools' mathematics education, the real numbers are constructed by defining a number using an integer followed by aradix point and an infinite sequence written out as a string to represent thefractional 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 thereal axioms after defining anequivalence relation over the set that defines1 =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 equality0.999... = 1 holding true is a necessary condition for strings of digits to behave as real numbers should.[38][39]

Dense order

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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.[40]

Generalizations

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The result that0.999... = 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 aredense.[41][9]

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 baseb{\displaystyle b} expression has a counterpart with repeated trailing digits equal tob{\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.[42]

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 allq{\displaystyle q} between 1 and 2, there are uncountably manybase-q{\displaystyle q} expansions of 1. In contrast, there are still uncountably manyq{\displaystyle q}, including all natural numbers greater than 1, for which there is only onebase-q{\displaystyle 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 Loreti determined the smallest such base, theKomornik–Loreti constantq{\displaystyle q} = 1.787231650.... In this base,1 = 0.11010011001011010010110011010011...; the digits are given by theThue–Morse sequence, which does not repeat.[43]

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:[44]

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 spaces and noticing that their real representations are given bycontinuous functions.[45]

Applications

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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.[46] Examples include:

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 form0.b1b2b3{\displaystyle 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.[47] Investigations in this direction can motivate such concepts asgreatest common divisors,modular arithmetic,Fermat primes,order ofgroup elements, andquadratic reciprocity.[48]

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

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

Then{\displaystyle n}th digit of the representation reflects the position of the point in then{\displaystyle n}th stage of the construction. For example, the point23{\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 point13{\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.[49]

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 argument to decimal expansions, of theuncountability of 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.[f] 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.[50]

Skepticism in education

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Students of mathematics often reject the equality of 0.999... and 1, for reasons ranging from their disparate appearance to deep misgivings over thelimit concept 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.[g]
  • 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".[51]
  • 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.[52]

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 instructivecounterexamples to better understand 0.999...; see§ In alternative number systems below.

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'".[23]

The elementary argument of multiplying0.333... =13{\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.[53] 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... =13{\textstyle {\frac {1}{3}}} using asupremum definition but then insisted that0.999... < 1 based on her earlier understanding oflong division.[54] Others still can prove that13{\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".[55]

As part of theAPOS Theory of 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 viewing13{\textstyle {\frac {1}{3}}} as a number in its own right and to dealing with the set of natural numbers as a whole.[56]

Cultural phenomenon

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With the rise of theInternet, debates about 0.999... have become commonplace onnewsgroups andmessage boards, including many that nominally have little to do with mathematics. In the newsgroupsci.math in the 1990s, arguing over 0.999... became a "popular sport", and was one of the questions answered in itsFAQ.[57][58] The FAQ briefly covers13{\displaystyle \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 Dope discusses 0.999... via13{\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.[59]

ASlate article reports that the concept of 0.999... is "hotly disputed on websites ranging fromWorld of Warcraft message boards toAyn Rand forums".[60] 0.999... features also inmathematical jokes, such as:[61]

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.[62] 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.[63]

In alternative number systems

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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 Gowers argues inMathematics: A Very Short Introduction that the resulting identity0.999... = 1 is 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.[64]

Infinitesimals

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Main article:Infinitesimal

Some proofs that0.999... = 1 rely on theArchimedean property of 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 analysis provides a number system with a full array of infinitesimals (and their inverses).[h]A. H. Lightstone developed a decimal expansion forhyperreal numbers in(0, 1). Lightstone shows how to associate each number with a sequence of digits,0.d1d2d3;d1dd+1,{\displaystyle 0.d_{1}d_{2}d_{3}\ldots ;\ldots d_{\infty -1}d_{\infty }d_{\infty +1}\ldots ,}indexed by thehypernatural numbers. While he does not directly discuss 0.999..., he shows the real number13{\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.[65]

The standard definition of the number 0.999... is thelimit of the sequence 0.9, 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.[66] More generally, the hyperreal numberuH{\displaystyle u_{H}} = 0.999...;...999000..., with last digit 9 at infinitehypernatural rankH{\displaystyle H}, satisfies a strict inequalityuH<1{\displaystyle u_{H}<1}. Accordingly, an alternative interpretation for "zero followed by infinitely many 9s" could be[67]0.999H=1110H.{\displaystyle {\underset {H}{0.\underbrace {999\ldots } }}\;=1\;-\;{\frac {1}{10^{H}}}.}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....[i] Along withKatz & Katz (2010b),Ely (2010) also questions the assumption that students' ideas about0.999... < 1 are erroneous intuitions about the real numbers, interpreting them rather asnonstandard intuitions that could be valuable in the learning of calculus.[68]

Hackenbush

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Combinatorial game theory provides a generalized concept of number that encompasses the real numbers and much more besides.[69] For example, in 1974,Elwyn Berlekamp described a correspondence between strings of red and blue segments inHackenbush and binary expansions of real numbers, motivated by the idea ofdata compression. For example, the value of the Hackenbush string LRRLRLRL... is0.010101...2 =13{\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 number1ω{\textstyle {\frac {1}{\omega }}}, whereω{\displaystyle \omega } is the firstinfinite ordinal; the relevant game is LRRRR... or 0.000...2.[j]

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 to34{\displaystyle \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

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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.[12]

First,Richman (1999) defines a nonnegativedecimal number to be a literal decimal expansion. He defines thelexicographical order and an addition operation, noting that0.999... < 1 simply because0 < 1 in the ones place, but for any nonterminatingx{\displaystyle x}, one has0.999... +x{\displaystyle x} = 1 +x{\displaystyle x}. So one peculiarity of the decimal numbers is that addition cannot always be canceled; another is that no decimal number corresponds to13{\displaystyle \textstyle {\frac {1}{3}}}. After defining multiplication, the decimal numbers form a positive, totally ordered, commutative semiring.[70]

In the process of defining multiplication, Richman also defines another system he calls "cutD{\displaystyle D}", which is the set ofDedekind cuts of decimal fractions. Ordinarily, this definition leads to the real numbers, but for a decimal fractiond{\displaystyle d} he allows both the cut({\displaystyle -\infty },d{\displaystyle d}) and the "principal cut"({\displaystyle -\infty },d{\displaystyle 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 cutD{\displaystyle 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{\displaystyle x} = 1" has no solution.[k]

p-adic numbers

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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....[71] 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.

Thep{\displaystyle p}-adic numbers are an alternative number system of interest innumber theory. Like the real numbers, thep{\displaystyle p}-adic numbers can be built from the rational numbers viaCauchy sequences; the construction uses a different metric in which 0 is closer top{\displaystyle p}, and much closer topn{\displaystyle p^{n}}, than it is to 1.[72] Thep{\displaystyle p}-adic numbers form afield for primep{\displaystyle p} and aring for otherp{\displaystyle p}, including 10. So arithmetic can be performed in thep{\displaystyle 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.[73] 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:[74]999=9+9(10)+9(10)2+9(10)3+=9110=1.{\displaystyle \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... = 1 but was inspired to take the multiply-by-10 proofabove in the opposite direction: ifx{\displaystyle x} = ...999, then10x{\displaystyle x} = ...990, so10x{\displaystyle x} =x{\displaystyle x} − 9, hencex{\displaystyle x} = −1 again.[73]

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"[75] 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 expansion of the10-adic solenoid, with eventually repeating left ends to represent the real numbers and eventually repeating right ends to represent the 10-adic numbers.[76]

See also

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Notes

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  1. ^For example, one can show this as follows: ifx is any number such that0.(9)nx < 1, then0.(9)n−1 ≤ 10x − 9 <x < 1. Thus ifx has this property for alln, the smaller number10x − 9 does, as well.
  2. ^The limit follows, for example, fromRudin (1976), p. 57, Theorem 3.20e. For a more direct approach, see alsoFinney, Weir & Giordano (2001), section 8.1, example 2(a), example 6(b).
  3. ^The historical synthesis is claimed byGriffiths & Hilton (1970), p. xiv and again byPugh (2002), p. 10; both actually prefer Dedekind cuts to axioms. For the use of cuts in textbooks, seePugh (2002), p. 17 orRudin (1976), p. 17. For viewpoints on logic, seePugh (2002), p. 10,Rudin (1976), p.ix, orMunkres (2000), p. 30.
  4. ^Enderton (1977), p. 113 qualifies this description: "The idea behind Dedekind cuts is that a real numberx can be named by giving an infinite set of rationals, namely all the rationals less thanx. We will in effect definex to be the set of rationals smaller thanx. To avoid circularity in the definition, we must be able to characterize the sets of rationals obtainable in this way ..."
  5. ^Rudin (1976), pp. 17–20,Richman (1999), p. 399, orEnderton (1977), p. 119. To be precise, Rudin, Richman, and Enderton call this cut 1∗, 1, and 1R, respectively; all three identify it with the traditional real number 1. Note that what Rudin and Enderton call a Dedekind cut, Richman calls a "non-principal Dedekind cut".
  6. ^Maor (1987), p. 60 andMankiewicz (2000), p. 151 review the former method; Mankiewicz attributes it to Cantor, but the primary source is unclear.Munkres (2000), p. 50 mentions the latter method.
  7. ^Bunch (1982), p. 119;Tall & Schwarzenberger (1978), p. 6. The last suggestion is due toBurrell (1998), p. 28: "Perhaps the most reassuring of all numbers is 1 ... So it is particularly unsettling when someone tries to pass off 0.9~ as 1."
  8. ^For a full treatment of non-standard numbers, seeRobinson (1996).
  9. ^Stewart (2009), p. 175; the full discussion of 0.999... is spread through pp. 172–175.
  10. ^Berlekamp, Conway & Guy (1982), pp. 79–80, 307–311 discuss 1 and1/3 and touch on1/ω. The game for 0.111...2 follows directly from Berlekamp's Rule.
  11. ^Richman (1999), pp. 398–400.Rudin (1976), p. 23 assigns this alternative construction (but over the rationals) as the last exercise of Chapter 1.

References

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  1. ^Cheng (2023), p. 141.
  2. ^Diamond (1955).
  3. ^Baldwin & Norton (2012).
  4. ^Meier & Smith (2017), §8.2.
  5. ^Stewart (2009), p. 175.
  6. ^Propp (2023).
  7. ^Stillwell (1994), p. 42.
  8. ^Earl & Nicholson (2021), "bound".
  9. ^abRosenlicht (1985), p. 27.
  10. ^Bauldry (2009), p. 47.
  11. ^Byers (2007), p. 39.
  12. ^abcdRichman (1999).
  13. ^Peressini & Peressini (2007), p. 186.
  14. ^Baldwin & Norton (2012);Katz & Katz (2010a).
  15. ^Cheng (2023), p. 136.
  16. ^Eisenmann (2008), p. 38.
  17. ^Tao (2003).
  18. ^Rudin (1976), p. 61, Theorem 3.26;Stewart (1999), p. 706.
  19. ^Euler (1822), p. 170.
  20. ^Grattan-Guinness (1970), p. 69;Bonnycastle (1806), p. 177.
  21. ^Stewart (1999), p. 706;Rudin (1976), p. 61;Protter & Morrey (1991), p. 213;Pugh (2002), p. 180;Conway (1978), p. 31.
  22. ^Davies (1846), p. 175;Smith & Harrington (1895), p. 115.
  23. ^abTall (2000), p. 221.
  24. ^Beals (2004), p. 22;Stewart (2009), p. 34.
  25. ^Bartle & Sherbert (1982), pp. 60–62;Pedrick (1994), p. 29;Sohrab (2003), p. 46.
  26. ^Apostol (1974), pp. 9, 11–12;Beals (2004), p. 22;Rosenlicht (1985), p. 27.
  27. ^Apostol (1974), p. 12.
  28. ^Cheng (2023), pp. 153–156.
  29. ^Conway (2001), pp. 25–27.
  30. ^Rudin (1976), pp. 3, 8.
  31. ^Richman (1999), p. 399.
  32. ^abO'Connor & Robertson (2005).
  33. ^Richman (1999), p. 398–399. "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 equation0.9 = 1 is built in at the beginning."
  34. ^Griffiths & Hilton (1970), p. 386, §24.2 "Sequences".
  35. ^Griffiths & Hilton (1970), pp. 388, 393.
  36. ^Griffiths & Hilton (1970), p. 395.
  37. ^Griffiths & Hilton (1970), pp. viii, 395.
  38. ^Gowers (2001).
  39. ^Li (2011).
  40. ^Artigue (2002), p. 212, "... 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).".
  41. ^Petkovšek (1990), p. 408.
  42. ^Protter & Morrey (1991), p. 503;Bartle & Sherbert (1982), p. 61.
  43. ^Komornik & Loreti (1998), p. 636.
  44. ^Kempner (1936), p. 611;Petkovšek (1990), p. 409.
  45. ^Petkovšek (1990), pp. 410–411.
  46. ^Goodwyn (1802);Dickson (1919), pp. 161.
  47. ^Leavitt (1984), p. 301.
  48. ^Ginsberg (2004), pp. 26–30;Lewittes (2006), pp. 1–3;Leavitt (1967), pp. 669, 673;Shrader-Frechette (1978), pp. 96–98.
  49. ^Pugh (2002), p. 97;Alligood, Sauer & Yorke (1996), pp. 150–152;Protter & Morrey (1991), p. 507;Pedrick (1994), p. 29.
  50. ^Rudin (1976), p. 50;Pugh (2002), p. 98.
  51. ^Tall & Schwarzenberger (1978), pp. 6–7;Tall (2000), p. 221.
  52. ^Tall & Schwarzenberger (1978), p. 6;Tall (2000), p. 221.
  53. ^Tall (1976), pp. 10–14.
  54. ^Pinto & Tall (2001), p. 5;Edwards & Ward (2004), pp. 416–417.
  55. ^Mazur (2005), pp. 137–141.
  56. ^Dubinsky et al. (2005), pp. 261–262.
  57. ^Richman (1999), p. 396.
  58. ^de Vreught (1994).
  59. ^Adams (2003).
  60. ^Ellenberg (2014).
  61. ^Renteln & Dundes (2005), p. 27.
  62. ^Richman (1999);Adams (2003);Ellenberg (2014).
  63. ^Wallace (2003), p. 51;Maor (1987), p. 17.
  64. ^Gowers (2002), p. 60.
  65. ^Lightstone (1972), pp. 245–247.
  66. ^Tao (2012), pp. 156–180.
  67. ^Katz & Katz (2010a).
  68. ^Katz & Katz (2010b);Ely (2010).
  69. ^Conway (2001), pp. 3–5, 12–13, 24–27.
  70. ^Richman (1999), pp. 397–399.
  71. ^Gardiner (2003), p. 98;Gowers (2002), p. 60.
  72. ^Mascari & Miola (1988), p. 83–84.
  73. ^abFjelstad (1995), p. 11.
  74. ^Fjelstad (1995), pp. 14–15.
  75. ^DeSua (1960), p. 901.
  76. ^DeSua (1960), p. 902–903.

Sources

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Further reading

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