- Repeating decimal
A

decimal representation of areal number is called a**repeating decimal**(or**recurring decimal**) if at some point it becomes periodic: there is some finite sequence of digits that is repeated indefinitely. For example, the decimal representation of nowrap|1=frac|1|3 = 0.3333333... (spoken as "0.3 repeating") becomes periodic just after the decimal point, repeating the single-digit sequence "3" indefinitely. A somewhat more complicated example is nowrap|1=frac|3227|555 = 5.8144144144..., where the decimal representation becomes periodic at the second digit after the decimal point, repeating the sequence of digits "144" forever.A real number has an ultimately periodic decimal representation if and only if it is a

rational number . Rational numbers are numbers that can be expressed in the form frac|"a"|"b" where "a" and "b" areintegers and "b" is non-zero. This form is known as avulgar fraction . On the one hand, the decimal representation of a rational number is ultimately periodic because it can be determined by along division process, which must ultimately become periodic as there are only finitely many different remainders and so eventually it will find a remainder that has occurred before. On the other hand, each repeating decimal number satisfies alinear equation with integral coefficients, and its unique solution is a rational number. To illustrate the latter point, the number nowrap|1=α = 5.8144144144... above satisfies the equation nowrap|1=10000α − 10α = 58144.144144... − 58.144144... = 58086, whose solution is nowrap|1=α = frac|58086|9990 = frac|3227|555.A decimal representation written with a repeating final 0 is said to "terminate" before the first final 0 (because it is not necessary to explicitly write that there is a repeating 0; instead of "1.585000..." one simply writes "1.585"). Such

**terminating decimals**may be classified as repeating, though they are not always. [*Page 67 of Courant, R. and Robbins, H. "What Is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed." Oxford, England: Oxford University Press, pp. 66-68, 1996.*] These terminating decimals representrational number s whose fractionsin lowest terms are of the form frac|"k"|2^{"n"}5^{"m"}. For example, nowrap|1=1.585 = frac|317|200 = frac|317|2^{3}5^{2}. They can be written as adecimal fraction : nowrap|1=frac|317|200 = frac|1585|1000. However, these numbers still also have a representation as a repeating decimal, obtained by decreasing the final (nonzero) digit by one and appending an indefinitely repeating sequence of digits "9" (e.g. nowrap|1=1 = 0.999999...; 1.585 = 1.584999999...). See The case of 0.99999... below.The remaining type of decimal representations is formed by decimal representations that neither terminate nor repeat. A decimal representation that neither terminates nor repeats represents an

irrational number (which cannot be expressed as the ratio of two integers), such as thesquare root of 2 and the numberπ . Conversely, an irrational number has a non-terminating non-repeating decimal representation. This is true in other bases than 10 as well.**Notation**One convention to indicate a repeating decimal is to put a horizontal line (known as a vinculum) above the repeated numerals ($scriptstyle\; frac\{1\}\{3\}=,\; 0.overline\{3\}$). Another convention is to place dots above the outermost numerals of the repeating digits. Where these methods are impossible, the extension may be represented by an ellipsis (...), although this may introduce uncertainty as to exactly which digits should be repeated. Another notation, used for example in Europe and China, encloses the repeating digits in brackets.

**Fractions with prime denominators****General**A fraction

in lowest terms with a prime denominator other than 2 or 5 (i.e.coprime to 10) always produces a repeating decimal. The period of the repeating decimal, frac|1|"p", where "p" is prime, is either "p" − 1 (the first group) or a divisor of "p" − 1 (the second group).Examples of fractions of the first group are:

* = 0.142857 ; 6 repeating digits

* = 0.0588235294117647 ; 16 repeating digits

* = 0.052631578947368421 ; 18 repeating digits

* = 0.0434782608695652173913 ; 22 repeating digits

* = 0.0344827586206896551724137931 ; 28 repeating digitsThe list can go on to include the fractions frac|1|47, frac|1|59, frac|1|61, frac|1|97, frac|1|109, etc.

The following multiplications exhibit an interesting property:

* = 2 × 0.142857... = 0.285714...

* = 3 × 0.142857... = 0.428571...

* = 4 × 0.142857... = 0.571428...

* = 5 × 0.142857... = 0.714285...

* = 6 × 0.142857... = 0.857142...That is, these multiples can be obtained from rotating the digits of the original decimal of frac|1|7. The reason for the rotating behaviour of the digits is apparent from an arithmetics exercise of finding the decimal of frac|1|7.

Of course 142857 × 7 = 999999, and 142 + 857 = 999.

Decimals of other prime fractions, such as frac|1|17, frac|1|19, frac|1|23, frac|1|29, frac|1|47, frac|1|59, frac|1|61, frac|1|97, and frac|1|109, each exhibit the same property.

Fractions of the second group are:

* = 0.333... which has 1 repeating digit.

* = 0.090909... which has 2 repeating digits.

* = 0.076923... which has 6 repeating digits.Note that the following multiples of frac|1|13 exhibit the discussed property of rotating digits:

* = 0.076923...

* = 0.230769...

* = 0.307692...

* = 0.692307...

* = 0.769230...

* = 0.923076...And similarly these multiples:

* = 0.153846...

* = 0.384615...

* = 0.461538...

* = 0.538461...

* = 0.615384...

* = 0.846153...Again, 076923 × 13 = 999999, and 076 + 923 = 999.

The period of the repeating decimal of frac|1|"p" is equal to the order of 10 modulo "p". The period is equal to "p"-1 if 10 is a primitive root modulo "p".

**Recursion and interesting characteristics explained**Take 1/7 as an example, the arithmetic division yields sequential remainders:1, 3, 2, 6, 4, 5 then back to the same sequence again. These remainders can be rearranged to become 1, 2, 3, 4, 5 and 6, i.e. complete consecutive numbers between 0 and 7.

The arithmetic division of 2/7 yields sequential remainders:2, 6, 4, 5, 1, 3 which form the same cyclic sequence except for a different starting point. This explains the rotating or cyclic behavior of the number 0.142857 in multiplications.

Take 1/17 as another example, the sequential remainders of the arithmetic division are:1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12then back to the same sequence again. These remainders can be rearranged to become complete consecutive numbers between 0 and 17.

Similarly, the arithmetic division of 2/17 yields sequential remainders:2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4, 6, 9, 5, 16, 7 which form the same cyclic sequence except for a different starting point. This explains the rotating or cyclic behavior of the number 0.05888235297117647 in multiplications.

**Connection with Fermat's little theorem**The above deduces that

: 10

^{6}-1 = 999,999 (6 digits of 9) is divisible by 7;: 10

^{12}-1 = 999,999,999,999 (12 digits of 9) is divisible by 13;: 10

^{16}-1 = 9,999,999,999,999,999 (16 digits of 9) is divisible by 17;: 10

^{18}-1 = 999,999,999,999,999,999 (18 digits of 9) is divisible by 19; etcwhich can also be expressed in the form of

modular arithmetic ,: 10

^{(7 - 1)}≡ 1 mod 7;: 10

^{(13 - 1)}≡ 1 mod 13;: 10

^{(17 - 1)}≡ 1 mod 17;: 10

^{(19 - 1)}≡ 1 mod 19; etcThese agree with

Fermat's little theorem , which says that:: a

^{(p -1)}≡ 1 mod p**Alternative proof**To prove that 10

^{(p-1)}-1 ≡ 0 mod p where p is a prime number ≠ 2 or 5Proof:

Use numerical value 7 as an example, i.e. to proof:

: 10

^{(7 - 1)}-1 ≡ 0 mod 7or

: 999,999 (i.e. 6 digits of 9) is divisible by 7;

LHS of the expression is,

: [10

^{(7 - 1)}-1] mod 7 ≡ 9 * (10^{5}+ 10^{4}+ 10^{3}+ 10^{2}+ 10^{1}+ 10^{0}) mod 7Since 9 mod 7 ≠ 0, thus the proof is complete if:

: (10

^{0}+ 10^{1}+ 10^{2}+ 10^{3}+ 10^{4}+ 10^{5}) mod 7 ≡ 0 mod 7The LHS of which is

: (1 + 3 + 3

^{2}+ 3^{3}+ 3^{4}+ 3^{5}) mod 7Which reduces to

: (1 + 3 + 9 + 27 + 81 + 243) mod 7 :: ≡ (1 + 3 + 2 + 6 + 4 + 5) mod 7:: ≡ (1 + 2 + 3 + 4 + 5 + 6) mod 7:: ≡ [6 (6 +1) /2] mod 7:: ≡ 0 mod 7

This can be extended to any prime number p for the following expression

: [10

^{(p -1)}-1] mod p :: ≡ Σ(1, 2, 3 ... p-1) mod p:: ≡ [(p-1)p /2] mod p:: ≡ 0 mod pSee also

proofs of Fermat's little theorem **Fraction from repeating decimal**Given a repeating decimal, it is possible to calculate the fraction that produced it. For example:

: $egin\{alignat\}2\; x\; =\; 0.333333ldots\backslash \; 10x\; =\; 3.333333ldotsquadmbox\{(multiplying\; each\; side\; of\; the\; above\; line\; by\; 10)\}\backslash \; 9x\; =\; 3\; mbox\{(subtracting\; the\; 1st\; line\; from\; the\; 2nd)\}\backslash \; x\; =\; 3/9\; =\; 1/3\; mbox\{(simplifying)\}\backslash end\{alignat\}$

Another example:

: $egin\{align\}\; x\; =\; 0.836363636ldots\backslash \; 100x\; =\; 83.636363636ldots\backslash \; 99x\; =\; 82.8\; \backslash \; x\; =\; frac\{82.8\}\{99\}\; =\; frac\{5\; imes82.8\}\{5\; imes99\}\; =\; frac\{414\}\{495\}\; =\; frac\{9\; imes\; 46\}\{9\; imes\; 55\}\; =\; frac\{46\}\{55\}.end\{align\}$

**A shortcut**The above argument can be applied in particular if the repeating sequence has "n" digits, all of which are 0 except the final one which is 1. For instance for "n" = 7:

: $egin\{align\}\; x\; =\; 0.000000100000010000001ldots\; \backslash \; 10^7x\; =\; 1.000000100000010000001ldots\; \backslash \; (10^7-1)x=9999999x\; =\; 1\; \backslash \; x\; =\; \{1\; over\; 10^7-1\}\; =\; \{1\; over9999999\}end\{align\}$

So this particular repeating decimal corresponds to the fraction 1/(10

^{"n"}− 1), where the denominator is the number written as "n" digits 9. Knowing just that, a general repeating decimal can be expressed as a fraction without having to solve an equation. For example, one could reason:: $egin\{align\}7.48181818ldots\; =\; 7.3\; +\; 0.18181818ldots\; \backslash \; \backslash \; =\; frac\{73\}\{10\}+frac\{18\}\{99\}\; =\; frac\{73\}\{10\}\; +\; frac\{9\; imes2\}\{9\; imes\; 11\}\; =\; frac\{73\}\{10\}\; +\; frac\{2\}\{11\}\; \backslash \; \backslash \; =\; frac\{11\; imes73\; +\; 10\; imes2\}\{10\; imes\; 11\}\; =\; frac\{823\}\{110\}end\{align\}$

More explicitly one gets the following cases.

If the repeating decimal is between 0 and 1, and the repeating block is "n" digits long, first occurring right after the decimal point, then the fraction (not necessarily reduced) will be the integer number represented by the "n"-digit block divided by the one represented by "n" digits 9. For example,

* 0.444444... = frac|4|9 since the repeating block is 4 (a 1-digit block),

* 0.565656... = frac|56|99 since the repeating block is 56 (a 2-digit block),

* 0.012012... = frac|12|999 since the repeating block is 012 (a 3-digit block), and this further reduces to frac|4|333.If the repeating decimal is as above, except that there are "k" (extra) digits 0 between the decimal point and the repeating "n"-digit block, then one can simply add "k" digits 0 after the "n" digits 9 of the denominator (and as before the fraction may subsequently be simplified). For example,

* 0.000444... = frac|4|9000 since the repeating block is 4 and this block is preceded by 3 zeros,

* 0.005656... = frac|56|9900 since the repeating block is 56 and it is preceded by 2 zeros,

* 0.00012012... = frac|12|99900= frac|2|16650 since the repeating block is 012 and it is preceded by 2 (!) zeros.Any repeating decimal not of the form described above can be written as a sum of a terminating decimal and a repeating decimal of one of the two above types (actually the first type suffices, but that could require the terminating decimal to be negative). For example,

* 1.23444... = 1.23 + 0.00444... = frac|123|100 + frac|4|900 = frac|1107|900 + frac|4|900 = frac|1111|900 or alternatively 1.23444... = 0.79 + 0.44444... = frac|79|100 + frac|4|9 = frac|711|900 + frac|400|900 = frac|1111|900

* 0.3789789... = 0.3 + 0.0789789... = frac|3|10 + frac|789|9990 = frac|2997|9990 + frac|789|9990 = frac|3786|9990 = frac|631|1665 or alternatively 0.3789789... = −0.6 + 0.9789789... = frac|−6|10 + frac|978|999 = frac|−5994|9990 + frac|9780|9990 = frac|3786|4995 = frac|631|1665It follows that any repeating decimal with period "n", and "k" digits after the decimal point that do not belong to the repeating part, can be written as a (not necessarily reduced) fraction whose denominator is (10

^{"n"}− 1)10^{"k"}.Conversely the period of the repeating decimal of a fraction frac|"c"|"d" will be (at most) the smallest number "n" such that 10

^{"n"}− 1 is divisible by "d".For example, the fraction frac|2|7 has "d" = 7, and the smallest "k" that makes 10

^{"k"}− 1 divisible by 7 is "k" = 6, because 999999 = 7 × 142857. The period of the fraction frac|2|7 is therefore 6.**Repeating decimals as an infinite series**Repeating decimals can also be expressed as an

infinite series . That is, repeating decimals can be shown to be a sum of asequence of numbers. To take the simplest example, ::$sum\_\{n=1\}^inftyfrac\{1\}\{10^n\}\; =\; \{1\; over\; 10\}\; +\; \{1\; over\; 100\}\; +\; \{1\; over\; 1000\}\; +\; cdots\; =\; 0.overline\{1\}$The above series is a

geometric series with the first term as 1/10 and the common factor 1/10. Because the absolute value of the common factor is less than 1, we can say that the geometric series converges and find the exact value in the form of a fraction by using the following formula where "a" is the first term of the series and "r" is the common factor.::$frac\{a\}\{1-r\}\; =\; frac\{frac\{1\}\{10\{1-frac\{1\}\{10\; =\; frac\{1\}\{9\}\; =\; 0.overline\{1\}$**How a repeating or terminating decimal expansion is found**In order to convert a

rational number represented as a fraction into decimal form, one may uselong division . For example, consider the rational number frac|5|74:__0.0675__74 ) 5.00000__4.44__560__518__420__370__500etc. Observe that at each step we have a remainder; the successive remainders displayed above are 56, 42, 50. When we arrive at 50 as the remainder, and bring down the "0", we find ourselves dividing 500 by 74, which is the same problem we began with. Therefore the decimal repeats: 0.0675675675....

**Why every rational number has a repeating or terminating decimal expansion**Only finitely many different remainders — in the example above, 74 possible remainders: 0, 1, 2, ..., 73 — can occur. If the remainder is 0, then the expansion terminates. If 0 never occurs as a remainder, then only finitely many other possible remainders exist — in the example above they are 1, 2, ,3, ..., 73. Therefore eventually a remainder must occur that has occurred before. The same remainder implies the same new digit in the result and the same new remainder. Therefore the whole sequence repeats itself.

**The case of 0.99999...**A proof that 1 = 0.99999..., using the method of calculating fractions from repeating decimals, follows these steps.

$egin\{alignat\}2\; x\; =\; 0.99999ldots\backslash \; 10x\; =\; 9.99999ldots\backslash \; 10x\; -\; x\; =\; 9.99999ldots\; -\; 0.99999ldots\backslash \; 9x\; =\; 9\backslash \; x\; =\; 1\backslash end\{alignat\}$

The second step is not 10"x" = 9.999...0, because the

right-hand side does not terminate (it is repeating) and so there is no end to which a zero can be appended. [*"*]0.999...#Skepticism in education " documents the reasons students of mathematics often reject the equality of 0.999... and 1.One can also think of this as the sum of a

geometric series .:$S\_a\; =\; sum\_\{n=0\}^\{a\}\; frac\{0.9\}\{10^n\}$

:$S\_a\; =\; 0.9\; sum\_\{n=0\}^\{a\}\; frac\{1\}\{10^n\}$

By a standard result,

:$S\_a\; =\; 0.9\; frac\{10^\{-a-1\}\; -\; 1\}\{10^\{-1\}-1\}.$

From the definition,

:$lim\_\{a\; ightarrow\; infty\}\; S\_a\; =\; 0.99999\; ldots$

So applying this on the sum of the geometric series:

:$lim\_\{a\; ightarrow\; infty\}\; 0.9\; frac\{10^\{-a-1\}\; -\; 1\}\{10^\{-1\}-1\}\; =\; 0.9\; frac\{-1\}\{-0.9\}$

:$0.9\; frac\{-1\}\{-0.9\}\; =\; 1.$

Therefore

:$0.99999\; ldots\; =\; 1.,$

For a less persuasive but more formal-"looking" proof, consider the formula

:$x\; =\; \{10^n-1\; over\; 10^n\},$

:$n\; =\; 1:\; x\; =\; \{9\; over\; 10\}\; =\; 0.9,$

:$n\; =\; 2:\; x\; =\; \{99over\; 100\}\; =\; 0.99.$

It follows that

:$lim\_\{n\; o\; infty\}\{10^n-1\; over\; 10^n\}\; =\; 0.9999dots,.$

On the other hand we can evaluate this limit easily as 1, also, by dividing top and bottom by 10

^{"n"}."The above exposition using formal mathematical notation looks more impressive than the arithmetic proof but it is not persuasive as the crucial step, the division by 10

^{n}, is not actually performed. But even were the proof using limits properly completed the arithmetic proof is adequate and simpler and can be followed by those without the proper understanding of limits."Generalising this, any nonzero number with a finite decimal expression (a

decimal fraction ) can be written in a second way as a repeating decimal.For example, frac|3|4 = 0.75 = 0.750000000... = 0.74999999... .

**ee also***142857

*Decimal sequences for cryptography

*Cyclic number

*Midy's theorem **Notes****External links*** [

*http://mathworld.wolfram.com/RepeatingDecimal.html Mathworld*]

*Wikimedia Foundation.
2010.*

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**Repeating decimal**— Repeating Re*peat ing, a. Doing the same thing over again; accomplishing a given result many times in succession; as, a repeating firearm; a repeating watch. [1913 Webster] {Repeating circle}. See the Note under {Circle}, n., 3. {Repeating… … The Collaborative International Dictionary of English**repeating decimal**— Decimal Dec i*mal, n. A number expressed in the scale of tens; specifically, and almost exclusively, used as synonymous with a decimal fraction. [1913 Webster] {Circulating decimal}, or {Circulatory decimal}, a decimal fraction in which the same… … The Collaborative International Dictionary of English**repeating decimal**— n. a decimal in which beyond a certain point some digit or group of digits is repeated indefinitely (Ex.: 0.47382382382...) … English World dictionary**repeating decimal**— noun a decimal with a sequence of digits that repeats itself indefinitely • Syn: ↑circulating decimal, ↑recurring decimal • Hypernyms: ↑decimal fraction, ↑decimal * * * reˈpeating decimal 7 [repeating decimal] … Useful english dictionary**repeating decimal**— decimal that recurs in the same order infinitely … English contemporary dictionary**repeating decimal**— Math. a decimal numeral that, after a certain point, consists of a group of one or more digits repeated ad infinitum, as 2.33333 ... . or 23.0218181818 ... . Also called circulating decimal, periodic decimal, recurring decimal. [1765 75] * * * … Universalium**repeating decimal**— noun Date: 1773 a decimal in which after a certain point a particular digit or sequence of digits repeats itself indefinitely compare terminating decimal … New Collegiate Dictionary**repeating decimal**— repeat′ing dec′imal n. math. a decimal that, after a certain point, includes a group of one or more digits repeated ad infinitum, as 2.33333 … or 23.02181818 … Also called recurring decimal … From formal English to slang**repeating decimal**— noun A decimal representation of a real number that, at some point, becomes periodic (and repeats the same sequence of digits indefinitely) … Wiktionary**repeating decimal**— noun a recurring decimal … English new terms dictionary