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Decimal representation

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(Redirected fromFinite decimal)
Expression of numbers as sequences of digits
This article is about decimal expansion of real numbers. For finite decimal representation, seeDecimal.
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Adecimal representation of anon-negativereal numberr is its expression as asequence of symbols consisting ofdecimal digits traditionally written with a single separator:r=bkbk1b0.a1a2{\displaystyle r=b_{k}b_{k-1}\cdots b_{0}.a_{1}a_{2}\cdots }Here. is thedecimal separator,k is anonnegative integer, andb0,,bk,a1,a2,{\displaystyle b_{0},\cdots ,b_{k},a_{1},a_{2},\cdots } aredigits, which are symbols representing integers in the range 0, ..., 9.

Commonly,bk0{\displaystyle b_{k}\neq 0} ifk1.{\displaystyle k\geq 1.} The sequence of theai{\displaystyle a_{i}}—the digits after the dot—is generallyinfinite. If it is finite, the lacking digits are assumed to be 0. If allai{\displaystyle a_{i}} are0, the separator is also omitted, resulting in a finite sequence of digits, which represents anatural number.

The decimal representation represents theinfinite sum:r=i=0kbi10i+i=1ai10i.{\displaystyle r=\sum _{i=0}^{k}b_{i}10^{i}+\sum _{i=1}^{\infty }{\frac {a_{i}}{10^{i}}}.}

Every nonnegative real number has at least one such representation; it has two such representations (withbk0{\displaystyle b_{k}\neq 0} ifk>0{\displaystyle k>0})if and only if one has a trailing infinite sequence of0, and the other has a trailing infinite sequence of9. For having a one-to-one correspondence between nonnegative real numbers and decimal representations, decimal representations with a trailing infinite sequence of9 are sometimes excluded.[1]

Integer and fractional parts

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The natural numberi=0kbi10i{\textstyle \sum _{i=0}^{k}b_{i}10^{i}}, is called theinteger part ofr, and is denoted bya0 in the remainder of this article. The sequence of theai{\displaystyle a_{i}} represents the number0.a1a2=i=1ai10i,{\displaystyle 0.a_{1}a_{2}\ldots =\sum _{i=1}^{\infty }{\frac {a_{i}}{10^{i}}},}which belongs to theinterval[0,1),{\displaystyle [0,1),} and is called thefractional part ofr (except when allai{\displaystyle a_{i}} are equal to9).

Finite decimal approximations

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Any real number can be approximated to any desired degree of accuracy byrational numbers with finite decimal representations.

Assumex0{\displaystyle x\geq 0}. Then for every integern1{\displaystyle n\geq 1} there is a finite decimalrn=a0.a1a2an{\displaystyle r_{n}=a_{0}.a_{1}a_{2}\cdots a_{n}} such that:

rnx<rn+110n.{\displaystyle r_{n}\leq x<r_{n}+{\frac {1}{10^{n}}}.}

Proof:Letrn=p10n{\displaystyle r_{n}=\textstyle {\frac {p}{10^{n}}}}, wherep=10nx{\displaystyle p=\lfloor 10^{n}x\rfloor }.Thenp10nx<p+1{\displaystyle p\leq 10^{n}x<p+1}, and the result follows from dividing all sides by10n{\displaystyle 10^{n}}.(The fact thatrn{\displaystyle r_{n}} has a finite decimal representation is easily established.)

Non-uniqueness of decimal representation and notational conventions

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Main article:0.999...

Some real numbersx{\displaystyle x} have two infinite decimal representations. For example, the number 1 may be equally represented by 1.000... as by0.999... (where the infinite sequences of trailing 0's or 9's, respectively, are represented by "..."). Conventionally, the decimal representation without trailing 9's is preferred. Moreover, in thestandard decimal representation ofx{\displaystyle x}, an infinite sequence of trailing 0's appearing after thedecimal point is omitted, along with the decimal point itself ifx{\displaystyle x} is an integer.

Certain procedures for constructing the decimal expansion ofx{\displaystyle x} will avoid the problem of trailing 9's. For instance, the following algorithmic procedure will give the standard decimal representation: Givenx0{\displaystyle x\geq 0}, we first definea0{\displaystyle a_{0}} (theinteger part ofx{\displaystyle x}) to be the largest integer such thata0x{\displaystyle a_{0}\leq x} (i.e.,a0=x{\displaystyle a_{0}=\lfloor x\rfloor }). Ifx=a0{\displaystyle x=a_{0}} the procedure terminates. Otherwise, for(ai)i=0k1{\textstyle (a_{i})_{i=0}^{k-1}} already found, we defineak{\displaystyle a_{k}} inductively to be the largest integer such that:

a0+a110+a2102++ak10kx.{\displaystyle a_{0}+{\frac {a_{1}}{10}}+{\frac {a_{2}}{10^{2}}}+\cdots +{\frac {a_{k}}{10^{k}}}\leq x.}*

The procedure terminates wheneverak{\displaystyle a_{k}} is found such that equality holds in (*); otherwise, it continues indefinitely to give an infinite sequence of decimal digits. It can be shown thatx=supk{i=0kai10i}{\textstyle x=\sup _{k}\left\{\sum _{i=0}^{k}{\frac {a_{i}}{10^{i}}}\right\}}[2] (conventionally written asx=a0.a1a2a3{\displaystyle x=a_{0}.a_{1}a_{2}a_{3}\cdots }), wherea1,a2,a3{0,1,2,,9},{\displaystyle a_{1},a_{2},a_{3}\ldots \in \{0,1,2,\ldots ,9\},} and the nonnegative integera0{\displaystyle a_{0}} is represented indecimal notation. This construction is extended tox<0{\displaystyle x<0} by applying the above procedure tox>0{\displaystyle -x>0} and denoting the resultant decimal expansion bya0.a1a2a3{\displaystyle -a_{0}.a_{1}a_{2}a_{3}\cdots }.

Types

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Finite

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The decimal expansion of non-negative real numberx will end in zeros (or in nines) if, and only if,x is a rational number whose denominator is of the form 2n5m, wherem andn are non-negative integers.

Proof:

If the decimal expansion ofx will end in zeros, orx=i=0nai10i=i=0n10niai/10n{\textstyle x=\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}=\sum _{i=0}^{n}10^{n-i}a_{i}/10^{n}}for somen, then the denominator ofx is of the form 10n = 2n5n.

Conversely, if the denominator ofx is of the form 2n5m,x=p2n5m=2m5np2n+m5n+m=2m5np10n+m{\displaystyle x={\frac {p}{2^{n}5^{m}}}={\frac {2^{m}5^{n}p}{2^{n+m}5^{n+m}}}={\frac {2^{m}5^{n}p}{10^{n+m}}}}for somep.Whilex is of the formp10k{\displaystyle \textstyle {\frac {p}{10^{k}}}},p=i=0n10iai{\displaystyle p=\sum _{i=0}^{n}10^{i}a_{i}} for somen.Byx=i=0n10niai/10n=i=0nai10i{\displaystyle x=\sum _{i=0}^{n}10^{n-i}a_{i}/10^{n}=\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}},x will end in zeros.

Infinite

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Repeating decimal representations

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Main article:Repeating decimal

Some real numbers have decimal expansions that eventually get into loops, endlessly repeating a sequence of one or more digits:

13 = 0.33333...
17 = 0.142857142857...
1318185 = 7.1243243243...

Every time this happens the number is still arational number (i.e. can alternatively be represented as a ratio of an integer and a positive integer).Also the converse is true: The decimal expansion of a rational number is either finite, or endlessly repeating.

Finite decimal representations can also be seen as a special case of infinite repeating decimal representations. For example,3625 = 1.44 = 1.4400000...; the endlessly repeated sequence is the one-digit sequence "0".

Non-repeating decimal representations

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Other real numbers have decimal expansions that never repeat. These are precisely theirrational numbers, numbers that cannot be represented as a ratio of integers. Some well-known examples are:

2 = 1.41421356237309504880...
  e  = 2.71828182845904523536...
  π  = 3.14159265358979323846...

Conversion to fraction

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Further information:Fraction § Arithmetic with fractions

Every decimal representation of a rational number can be converted to a fraction by converting it into a sum of the integer, non-repeating, and repeating parts and then converting that sum to a single fraction with a common denominator.

For example, to convert±8.1234567¯{\textstyle \pm 8.123{\overline {4567}}} to a fraction one notes the lemma:0.0004567¯=4567×0.0000001¯=4567×0.0001¯×1103=4567×19999×1103=45679999×1103=4567(1041)×103The exponents are the number of non-repeating digits after the decimal point (3) and the number of repeating digits (4).{\displaystyle {\begin{aligned}0.000{\overline {4567}}&=4567\times 0.000{\overline {0001}}\\&=4567\times 0.{\overline {0001}}\times {\frac {1}{10^{3}}}\\&=4567\times {\frac {1}{9999}}\times {\frac {1}{10^{3}}}\\&={\frac {4567}{9999}}\times {\frac {1}{10^{3}}}\\&={\frac {4567}{(10^{4}-1)\times 10^{3}}}&{\text{The exponents are the number of non-repeating digits after the decimal point (3) and the number of repeating digits (4).}}\end{aligned}}}

Thus one converts as follows:±8.1234567¯=±(8+123103+4567(1041)×103)from above=±8×(1041)×103+123×(1041)+4567(1041)×103common denominator=±812264449999000multiplying, and summing the numerator=±203066112499750reducing{\displaystyle {\begin{aligned}\pm 8.123{\overline {4567}}&=\pm \left(8+{\frac {123}{10^{3}}}+{\frac {4567}{(10^{4}-1)\times 10^{3}}}\right)&{\text{from above}}\\&=\pm {\frac {8\times (10^{4}-1)\times 10^{3}+123\times (10^{4}-1)+4567}{(10^{4}-1)\times 10^{3}}}&{\text{common denominator}}\\&=\pm {\frac {81226444}{9999000}}&{\text{multiplying, and summing the numerator}}\\&=\pm {\frac {20306611}{2499750}}&{\text{reducing}}\\\end{aligned}}}

If there are no repeating digits one assumes that there is a forever repeating 0, e.g.1.9=1.90¯{\displaystyle 1.9=1.9{\overline {0}}}, although since that makes the repeating term zero the sum simplifies to two terms and a simpler conversion.

For example:±8.1234=±(8+1234104)=±8×104+1234104common denominator=±8123410000multiplying, and summing the numerator=±406175000reducing{\displaystyle {\begin{aligned}\pm 8.1234&=\pm \left(8+{\frac {1234}{10^{4}}}\right)&\\&=\pm {\frac {8\times 10^{4}+1234}{10^{4}}}&{\text{common denominator}}\\&=\pm {\frac {81234}{10000}}&{\text{multiplying, and summing the numerator}}\\&=\pm {\frac {40617}{5000}}&{\text{reducing}}\\\end{aligned}}}

See also

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References

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  1. ^Knuth, Donald Ervin (1973).The Art of Computer Programming. Vol. 1: Fundamental Algorithms.Addison-Wesley. p. 21.
  2. ^Rudin, Walter (1976).Principles of Mathematical Analysis. New York:McGraw-Hill. p. 11.ISBN 0-07-054235-X.

Further reading

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