0,999...

Trong toán học, số thập phân tuần hoàn '0,999... hay còn được viết $0.{\bar {9}},0,{\dot {9}}$ hoặc $0,(9)\,\!$ là một số thực bằng 1. Nói cách khác: kí hiệu 0,999... và 1 đều thể hiệu cùng một số thực. Điều này đã được nhiều giáo sư toán học trên thế giới công nhận và được giảng dạy trong nhiều sách giáo khoa. Nhiều cách chứng minh đã được trình bày với các mức độ chính xác khác nhau, dựa vào nhiều phép tính toán trên các số thực, các kiến thức được thừa nhận, bối cảnh lịch sử và mục tiêu của người đọc.

Thực tế là số thực có thực có thể được đại diện bởi một dãy số thập phân vô hạn.

The fact that certain real numbers can be represented by more than one digit string is not limited to the decimal system. The same phenomenon occurs in all integer bases, and mathematicians have also quantified the ways of writing 1 in non-integer bases. Nor is this phenomenon unique to 1: every non-zero, terminating decimal has a twin with trailing 9s, such as 28.3287 and 28.3286999…. For simplicity, the terminating decimal is almost always the preferred representation, contributing to a misconception that it is the only representation. Even more generally, any positional numeral system contains infinitely many numbers with multiple representations. These various identities have been applied to better understand patterns in the decimal expansions of fractions and the structure of a simple fractal, the Cantor set. They also occur in a classic investigation of the infinitude of the entire set of real numbers.

In the last few decades, researchers of mathematics education have studied the reception of this equality among students, many of whom initially question or reject this equality. Many are persuaded by textbooks, teachers and arithmetic reasoning as below to accept that the two are equal. However, they are often uneasy enough that they offer further justification. The students' reasoning for denying or affirming the equality is typically based on one of a few common erroneous intuitions about the real numbers; for example that each real number has a unique decimal expansion, that nonzero infinitesimal real numbers should exist, or that the expansion of 0.999… eventually terminates. Number systems that bear out some of these intuitions can be constructed, but only outside the standard real number system used in elementary, and most higher, mathematics. Indeed, some settings contain numbers that are "just shy" of 1; these are generally unrelated to 0.999…, but they are of considerable interest in mathematical analysis.

Chứng minh

Số học

Phân số và phép chia

One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like ¹⁄₃ becomes a recurring decimal, 0.333…, in which the digits repeat without end. This decimal yields a quick proof for 0.999… = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 × 0.333… equals 0.999…. And 3 × ¹⁄₃ equals 1, so 0.999… = 1.

Another form of this proof multiplies ¹/₉ = 0.111… by 9.

${\begin{aligned}0.333\dots &{}={\frac {1}{3}}\\3\times 0.333\dots &{}=3\times {\frac {1}{3}}={\frac {3\times 1}{3}}\\0.999\dots &{}=1\end{aligned}}$

${\begin{aligned}0.111\dots &{}={\frac {1}{9}}\\9\times 0.111\dots &{}=9\times {\frac {1}{9}}={\frac {9\times 1}{9}}\\0.999\dots &{}=1\end{aligned}}$

A more compact version of the same proof is given by the following equations:

$1={\frac {9}{9}}=9\times {\frac {1}{9}}=9\times 0.111\dots =0.999\dots$

Since both equations are valid, 0.999… must equal 1 (by the transitive property). Similarly, ³/_{3 = 1, and ³/3 = 0.999…. So, 0.999… must equal 1.}

Biến đổi số học

Another kind of proof more easily adapts to other repeating decimals. When a number in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.999… equals 9.999…, which is 9 greater than the original number.

To see this, consider that subtracting 0.999… from 9.999… can proceed digit by digit; in each of the digits after the decimal separator the result is 9 − 9, which is 0. But trailing zeros do not change a number, so the difference is exactly 9. The final step uses algebra. Let the decimal number in question, 0.999…, be called x. Then 10x − x = 9. This is the same as 9x = 9. Dividing both sides by 9 completes the proof: x = 1. Written as a sequence of equations,

${\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots \\10x-x&=9.999\ldots -0.999\ldots \\9x&=9\\x&=1\\0.999\ldots &=1\end{aligned}}$

The validity of the digit manipulations in the above two proofs does not have to be taken on faith or as an axiom; it follows from the fundamental relationship between decimals and the numbers they represent. This relationship, which can be developed in several equivalent manners, already establishes that the decimals 0.999… and 1.000... both represent the same number.

Giải tích

Since the question of 0.999… does not affect the formal development of mathematics, it can be postponed until one proves the standard theorems of real analysis. One requirement is to characterize real numbers that can be written in decimal notation, consisting of an optional sign, a finite sequence of any number of digits forming an integer part, a decimal separator, and a sequence of digits forming a fractional part. For the purpose of discussing 0.999…, the integer part can be summarized as b₀ and one can neglect negatives, so a decimal expansion has the form

$b_{0}.b_{1}b_{2}b_{3}b_{4}b_{5}\dots$

It is vital that the fraction part, unlike the integer part, is not limited to a finite number of digits. This is a positional notation, so for example the 5 in 500 contributes ten times as much as the 5 in 50, and the 5 in 0.05 contributes one tenth as much as the 5 in 0.5.

Chuỗi vô hạn

Xem thêm thông tin: [[:Decimal representation]]

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

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

For 0.999… one can apply the convergence theorem concerning geometric series:

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

Since 0.999… is such a sum with a common ratio $r=\textstyle {\frac {1}{10}}$, the theorem makes short work of the question:

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

This proof (actually, that 10 equals 9.999…) appears as early as 1770 in Leonhard Euler's Elements of Algebra.

Limits: The unit interval, including the base-4 decimal 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 algebra 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 any proof-based introduction to calculus or analysis.

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:

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

The last step — that $\lim _{n\to \infty }{\frac {1}{10^{n}}}=0$ — is often justified by the axiom that the real numbers have the Archimedean property. 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 interpreted by students as implying that 0.999… itself is less than 1.

Chia khoảng và tính bị chặn

Xem thêm thông tin: [[:Nested intervals]]

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

x = b₀.b₁b₂b₃…

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 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.

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, …}. 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.

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