Repeating decimal

"Repeating fraction" redirects here. It is not to be confused with Continued fraction.

A repeating or recurring decimal is a way of representing rational numbers in base 10 arithmetic. The decimal representation of a number is said to be repeating if it becomes periodic (repeating its values at regular intervals) and the infinitely-repeated portion is not zero. For example, the decimal representation of ⅓ becomes periodic just after the decimal point, repeating the single digit "3" forever, i.e. 0.333…. A more complicated example is 3227/555, whose decimal becomes periodic after the second digit following the decimal point and then repeats the sequence "144" forever, i.e. 5.8144144144…. At present, there is no single universally accepted notation or phrasing for repeating decimals.

The infinitely-repeated digit sequence is called the repetend or reptend. If the repetend is a zero, this decimal representation is called a terminating decimal rather than a repeating decimal, since the zeros can be omitted and the decimal terminates before these zeros. Every terminating decimal representation can be written as a decimal fraction, a fraction whose divisor is a power of 10 (e.g. 1.585 = 1585/1000); it may also be written as a ratio of the form k/2ⁿ5^m (e.g. 1.585 = 317/2³5²). However, every number with a terminating decimal representation also trivially has a second, alternative representation as a repeating decimal whose repetend is the digit 9. This is obtained by decreasing the final non-zero digit by one and appending a repetend of 9, a fact that some find puzzling. 1.000... = 0.999… and 1.585000... = 1.584999… are two examples of this. (This type of repeating decimal can be obtained by long division if one uses a modified form of the usual division algorithm.)

Any number that cannot be expressed as a ratio of two integers is said to be irrational. Their decimal representation neither terminates nor infinitely repeats but extends forever without regular repetition. Examples of such irrational numbers are the square root of 2 and pi.

Background

Notation

While there are several notational conventions for representing repeating decimals, none of them are accepted universally. In the United States, the convention is generally to indicate a repeating decimal by drawing a horizontal line (a vinculum) above the repetend ($\tfrac{1}{3}=0.\overline{3}$). In the United Kingdom and mainland China, the convention is to place dots above the outermost numerals of the repetend ($\tfrac{1}{3}=0.\dot{3}$). Another notation employed in parts of Europe is to enclose the repetend in parentheses ($\tfrac{1}{3}=0.(3)$). Repeating decimals may also be represented by three periods (an ellipsis, e.g., 0.333…), although this method introduces uncertainty as to which digits should be repeated or even whether repetition is occurring at all, since such ellipses are also employed for irrational decimals such as 3.14159…

Fraction	Ellipsis	Vinculum	Dots	Parentheses
1/9	0.111…	0.1	$0.\dot{1}$	0.(1)
1/3	0.333…	0.3	$0.\dot{3}$	0.(3)
2/3	0.666…	0.6	$0.\dot{6}$	0.(6)
9/11	0.8181…	0.81	$0.\dot{8}\dot{1}$	0.(81)
7/12	0.58333…	0.583	$0.58\dot{3}$	0.58(3)
1/81	0.012345679…	0.012345679	$0.\dot{0}1234567\dot{9}$	0.(012345679)
22/7	3.142857142857…	3.142857	$3.\dot{1}4285\dot{7}$	3.(142857)

In English, there are various ways to read repeating decimals aloud. Some common ones (for ⅓) include "zero point three repeating", "zero point three repeated", "zero point three recurring", and "zero point three into infinity". Mention of the initial zero may also be omitted.

Decimal expansion and recurrence sequence

In order to convert a rational number represented as a fraction into decimal form, one may use long division. For example, consider the rational number 5/74:

           .  .
        0.0675
   74 ) 5.00000
        4.44
          560
          518
           420
           370
            500

etc. 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: 0.0675 675 675 ….

Every rational number is either a terminating or repeating decimal

For any given divisor, only finitely many different remainders can occur. In the example above, the 74 possible remainders are 0, 1, 2, …, 73. If at any point in the division the remainder is 0, the expansion terminates at that point. If 0 never occurs as a remainder, then the division process continues forever, and eventually a remainder must occur that has occurred before. The next step in the division will yield the same new digit in the quotient, and the same new remainder, as the previous time the remainder was the same. Therefore, the following division will repeat the same results.

Every repeating or terminating decimal is a rational number

Each repeating decimal number satisfies a linear equation with integer coefficients, and its unique solution is a rational number. To illustrate the latter point, the number α = 5.8144144144… above satisfies the equation 10000α − 10α = 58144.144144… − 58.144144… = 58086, whose solution is α = 58086/9990 = 3227/555. The process of how to find these integer coefficients is described below.

Table of values

Fraction	Value	Period length	Fraction	Value	Period length	Fraction	Value	Period length
1/2	0.5	0	1/17	0.0588235294117647	16	1/32	0.03125	0
1/3	0.3	1	1/18	0.05	1	1/33	0.03	2
1/4	0.25	0	1/19	0.052631578947368421	18	1/34	0.02941176470588235	16
1/5	0.2	0	1/20	0.05	0	1/35	0.0285714	6
1/6	0.16	1	1/21	0.047619	6	1/36	0.027	1
1/7	0.142857	6	1/22	0.045	2	1/37	0.027	3
1/8	0.125	0	1/23	0.0434782608695652173913	22	1/38	0.0263157894736842105	18
1/9	0.1	1	1/24	0.0416	1	1/39	0.025641	6
1/10	0.1	0	1/25	0.04	0	1/40	0.025	0
1/11	0.09	2	1/26	0.0384615	6	1/41	0.02439	5
1/12	0.083	1	1/27	0.037	3	1/42	0.0238095	6
1/13	0.076923	6	1/28	0.03571428	6	1/43	0.023255813953488372093	21
1/14	0.0714285	6	1/29	0.0344827586206896551724137931	28	1/44	0.0227	2
1/15	0.06	1	1/30	0.03	1	1/45	0.02	1
1/16	0.0625	0	1/31	0.032258064516129	15	1/46	0.02173913043478260869565	22

The period length of 1/n are

0, 0, 1, 0, 0, 1, 6, 0, 1, 0, 2, 1, 6, 6, 1, 0, 16, 1, 18, 0, 6, 2, 22, 1, 0, 6, 3, 6, 28, 1, 15, 0, 2, 16, 6, 1, 3, 18, 6, 0, 5, 6, 21, 2, 1, 22, 46, 1, 42, 0, 16, 6, 13, 3, 2, 6, 18, 28, 58, 1, 60, 15, 6, 0, 6, 2, 33, 16, 22, 6, 35, 1, 8, 3, 1, ... (sequence A051626 in OEIS)

The periodic part of 1/n are

0, 0, 3, 0, 0, 6, 142857, 0, 1, 0, 09, 3, 076923, 714285, 6, 0, 0588235294117647, 5, 052631578947368421, 0, 047619, 45, 0434782608695652173913, 6, 0, 384615, 037, 571428, 0344827586206896551724137931, 3, ... (sequence A036275 in OEIS)

The period length of 1/(nth prime) are

0, 1, 0, 6, 2, 6, 16, 18, 22, 28, 15, 3, 5, 21, 46, 13, 58, 60, 33, 35, 8, 13, 41, 44, 96, 4, 34, 53, 108, 112, 42, 130, 8, 46, 148, 75, 78, 81, 166, 43, 178, 180, 95, 192, 98, 99, 30, 222, 113, 228, 232, 7, 30, 50, 256, 262, 268, 5, 69, 28, ... (sequence A002371 in OEIS)

The least prime p which 1/p with period length n are

3, 11, 37, 101, 41, 7, 239, 73, 333667, 9091, 21649, 9901, 53, 909091, 31, 17, 2071723, 19, 1111111111111111111, 3541, 43, 23, 11111111111111111111111, 99990001, 21401, 859, 757, 29, 3191, 211, ... (sequence A007138 in OEIS)

The least prime p which k/p has n different cycles (1≤k≤p-1) are

7, 3, 103, 53, 11, 79, 211, 41, 73, 281, 353, 37, 2393, 449, 3061, 1889, 137, 2467, 16189, 641, 3109, 4973, 11087, 1321, 101, 7151, 7669, 757, 38629, 1231, ... (sequence A054471 in OEIS)

Fractions with prime denominators

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 length of the repetend (period of the repeating decimal) of 1/p is equal to the order of 10 modulo p. If 10 is a primitive root modulo p, the repetend length is equal to p − 1; if not, the repetend length is a factor of p − 1. This result can be deduced from Fermat's little theorem, which states that 10^p−1 = 1 (mod p).

The base-10 repetend of the reciprocal of any prime number greater than 5 is divisible by 9.

If the repetend length of 1/p for prime p is equal to p − 1 then the repetend, expressed as an integer, is called a cyclic number.

Cyclic numbers

Main article: Cyclic number

Examples of fractions belonging to this group are:

• 1/7 = 0.142857, 6 repeating digits
• 1/17 = 0.05882352 94117647, 16 repeating digits
• 1/19 = 0.052631578 947368421, 18 repeating digits
• 1/23 = 0.04347826086 95652173913, 22 repeating digits
• 1/29 = 0.0344827 5862068 9655172 4137931, 28 repeating digits
• 1/47 = 0.02127659574468085106382 97872340425531914893617, 46 repeating digits
• 1/59 = 0.01694915254237288135593220338 98305084745762711864406779661, 58 repeating digits
• 1/61 = 0.0163934426 2295081967 2131147540 9836065573 7704918032 7868852459, 60 repeating digits
• 1/97 = 0.01030927 83505154 63917525 77319587 62886597 93814432 98969072 16494845 36082474 22680412 37113402 06185567, 96 repeating digits

The list can go on to include the fractions 1/109, 1/113, 1/131, 1/149, 1/167, 1/179, 1/181, 1/193, etc. (sequence A001913 in OEIS)

Every proper multiple of a cyclic number (that is, a multiple having the same number of digits) is a rotation.

• 1/7 = 1 × 0.142857… = 0.142857…
• 2/7 = 2 × 0.142857… = 0.285714…
• 3/7 = 3 × 0.142857… = 0.428571…
• 4/7 = 4 × 0.142857… = 0.571428…
• 5/7 = 5 × 0.142857… = 0.714285…
• 6/7 = 6 × 0.142857… = 0.857142…

The reason for the cyclic behavior is apparent from an arithmetic exercise of long division of ​¹⁄₇: the sequential remainders are the cyclic sequence {1, 3, 2, 6, 4, 5}. See also the article 142857 for more properties of this cyclic number.

A proper prime is a prime p which ends in the digit 1 in base 10 and whose reciprocal in base 10 has a repetend with length p-1. In such primes, each digit 0, 1, ..., 9 appears in the repeating sequence the same number of times as does each other digit (namely, (p-1)/10 times). They are:

61, 131, 181, 461, 491, 541, 571, 701, 811, 821, 941, 971, 1021, 1051, 1091, 1171, 1181, 1291, 1301, 1349, 1381, 1531, 1571, 1621, 1741, 1811, 1829, 1861, ... (sequence A073761 in OEIS)

A prime is a proper prime if and only if it is a full reptend prime and congruent to 1 mod 10.

If a prime p is both full reptend prime and safe prime, then 1/p will produce a stream of p − 1 pseudo-random digits. Those primes are

7, 23, 47, 59, 167, 179, 263, 383, 503, 863, 887, 983, 1019, 1367, 1487, 1619, 1823, ... (sequence A000353 in OEIS)

Other reciprocals of primes

Some reciprocals of primes that do not generate cyclic numbers are:

• 1/3 = 0.3, which has a period of 1.
• 1/11 = 0.09, which has a period of 2.
• 1/13 = 0.076923, which has a period of 6.
• 1/31 = 0.032258064516129, which has a period of 15.
• 1/37 = 0.027, which has a period of 3.
• 1/41 = 0.02439, which has a period of 5.
• 1/43 = 0.023255813953488372093, which has a period of 21.
• 1/53 = 0.0188679245283, which has a period of 13.

(sequence A006559 in OEIS)

The reason is that 3 is a divisor of 9, 11 is a divisor of 99, 41 is a divisor of 99999, etc. To find the period of 1/p, we can check whether the prime p divides some number 99…9 in which the number of digits divides p - 1. Since the period is never greater than p - 1, we can obtain this by calculating $\frac{10^{p-1}-1}{p}.$ For example, for 11 we get

$\frac{10^{11-1}-1}{11}= 909090909$

and then by inspection find the repetend 09 and period of 2.

Those reciprocals of primes can be associated with several sequences of repeating decimals. For example, the multiples of 1/13 can be divided into two sets, with different repetends. The first set is:

• 1/13 = 0.076923…
• 10/13 = 0.769230…
• 9/13 = 0.692307…
• 12/13 = 0.923076…
• 3/13 = 0.230769…
• 4/13 = 0.307692…

where the repetend of each fraction is a cyclic re-arrangement of 076923. The second set is:

• 2/13 = 0.153846…
• 7/13 = 0.538461…
• 5/13 = 0.384615…
• 11/13 = 0.846153…
• 6/13 = 0.461538…
• 8/13 = 0.615384…

where the repetend of each fraction is a cyclic re-arrangement of 153846.

In general, the set of proper multiples of reciprocals of a prime p consists of n subsets, each with repetend length k, where nk = p − 1.

Totient rule

For an arbitrary integer n the length $\lambda(n)$ of the repetend of 1/n divides $\phi(n)$, where $\phi$ is the totient function. The length is equal to $\phi(n)$ if and only if 10 is a primitive root modulo n.

In particular, it follows that $\lambda(p)=p-1$ if and only iff p is a prime and 10 is a primitive root modulo p. Then, the decimal expansions of n/p for n = 1, 2, …, p - 1, all have periods of length p - 1 and differ only by a cyclic permutation. Such numbers p are called full repetend primes.

Reciprocals of composite integers coprime to 10

If p is a prime other than 2 or 5, the decimal representation of the fraction $\tfrac{1}{p^2}$ repeats, e.g.:

1/49 = 0.0204081 6326530 6122448 9795918 3673469 3877551

The period (repetend length) must be a factor of λ(49) = 42, where λ(n) is known as the Carmichael function. This follows from Carmichael's theorem, which states that: if n is a positive integer then λ(n) is the smallest integer m such that

$a^m \equiv 1 \pmod{n}$

for every integer a that is coprime to n.

The period of $\tfrac{1}{p^2}$ is usually pT_p where T_p is the period of $\tfrac{1}{p}$. There are three known primes for which this is not true, and for those the period of $\tfrac{1}{p^2}$ is the same as the period of $\tfrac{1}{p}$, because p² divides 10^p−1−1. These three primes are 3, 487 and 56598313 (sequence A045616 in OEIS).

Similarly, the period of $\tfrac{1}{p^k}$ is usually p^k−1T_p

If p and q are primes other than 2 or 5, the decimal representation of the fraction $\tfrac{1}{p \ q}$ repeats. An example is 1/119:

119 = 7 × 17

λ(7 × 17) = LCM(λ(7), λ(17))

= LCM(6, 16)

= 48

where LCM denotes the least common multiple.

The period T of $\tfrac{1}{p \ q}$ is a factor of λ(pq) and it happens to be 48 in this case:

1/119 = 0.00840336 13445378 15126050 42016806 72268907 56302521

The period T of $\tfrac{1}{p \ q}$ is LCM(T_p, T_q) where T_p is the period of $\tfrac{1}{p}$ and T_q is the period of $\tfrac{1}{q}$.

If p , q, r etc. are primes other than 2 or 5, and k , ℓ, m etc. are positive integers, then $\frac{1}{p^k q^\ell r^m \cdots }$ is a repeating decimal with a period of $\mathrm{LCM}(T_{p^k}, T_{q^\ell}, T_{r^m}, \ldots)$ where $T_{p^k},\ T_{q^\ell},\ T_{r^m}$, etc. are respectively the period of the repeating decimals $\frac{1}{p^k},\ \frac{1}{q^\ell},\ \frac{1}{r^m},\$ etc. as defined above.

Reciprocals of integers not co-prime to 10

An integer that is not co-prime to 10 but has a prime factor other than 2 or 5 has a reciprocal that is eventually periodic, but with a non-repeating sequence of digits that precede the repeating part. The reciprocal can be expressed as:

$\frac{1}{2^a 5^b p^k q^\ell \cdots}\, ,$

where a and b are not both zero.

This fraction can also be expressed as:

$\frac{5^{a-b}}{10^a p^k q^\ell \cdots}\, ,$

if a > b, or as

$\frac{2^{b-a}}{10^b p^k q^\ell \cdots}\, ,$

if b > a, or as

$\frac{1}{10^a p^k q^\ell \cdots}\, ,$

if a = b.

The decimal has:

• An initial transient of max(a, b) digits after the decimal point. Some or all of the digits in the transient can be zeros.
• A subsequent repetend which is the same as that for the fraction $\frac{1}{p^k q^\ell \cdots}$.

For example 1/28 = 0.03571428571428…:

• the initial non-repeating digits are 03; and
• the subsequent repeating digits are 571428.

Converting repeating decimals to fractions

Given a repeating decimal, it is possible to calculate the fraction that produced it. For example:

$\begin{alignat}2 x &= 0.333333\ldots\\ 10x &= 3.333333\ldots&\quad&\text{(multiplying each side of the above line by 10)}\\ 9x &= 3 &&\text{(subtracting the 1st line from the 2nd)}\\ x &= 3/9 = 1/3 &&\text{(reducing to lowest terms)}\\ \end{alignat}$

Another example:

$\begin{align} x &= 0.836363636\ldots\\ 10x &= 8.3636363636\ldots\text{(multiplying by a power of 10 to move decimal to start of repetition)}\\ 1000x &= 836.36363636\ldots\text{(multiplying by a power of 100 to move decimal to end of first repeating decimal)}\\ 990x &= 836.36363636\ldots - 8.36363636\ldots = 828 \text{ (subtracting to clear decimals)}\\ x &= \frac{828}{990} = \frac{18 \times 46}{18 \times 55} = \frac{46}{55}. \end{align}$

A shortcut

The procedure below can be applied in particular if the repetend has n digits, all of which are 0 except the final one which is 1. For instance for n = 7:

$\begin{align} x &= 0.000000100000010000001\ldots \\ 10^7x &= 1.000000100000010000001\ldots \\ (10^7-1)x=9999999x &= 1 \\ x &= {1 \over 10^7-1} = {1 \over9999999} \end{align}$

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

$\begin{align} 7.48181818\ldots & = 7.3 + 0.18181818\ldots \\[8pt] & = \frac{73}{10}+\frac{18}{99} = \frac{73}{10} + \frac{9\times2}{9\times 11} = \frac{73}{10} + \frac{2}{11} \\[12pt] & = \frac{11\times73 + 10\times2}{10\times 11} = \frac{823}{110} \end{align}$

It is possible to get a general formula expressing a repeating decimal with an n digit period (repetend length), beginning right after the decimal point, as a fraction:

x = 0.(A₁A₂…A_n)

10ⁿx = A₁A₂…A_n.(A₁A₂…A_n)

(10ⁿ − 1)x = 99…99x = A₁A₂ … A_n

x = A₁A₂…A_n/(10ⁿ − 1)

= A₁A₂…A_n/99…99

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… = 4/9 since the repeating block is 4 (a 1-digit block),
• 0.565656… = 56/99 since the repeating block is 56 (a 2-digit block),
• 0.012012… = 12/999 since the repeating block is 012 (a 3-digit block), and this further reduces to 4/333.
• 0.9999999… = 9/9 = 1, since the repeating block is 9 (also a 1-digit block)

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… = 4/9000 since the repeating block is 4 and this block is preceded by 3 zeros,
• 0.005656… = 56/9900 since the repeating block is 56 and it is preceded by 2 zeros,
• 0.00012012… = 12/99900 = 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… = 123/100 + 4/900 = 1107/900 + 4/900 = 1111/900 or alternatively 1.23444… = 0.79 + 0.44444… = 79/100 + 4/9 = 711/900 + 400/900 = 1111/900
• 0.3789789… = 0.3 + 0.0789789… = 3/10 + 789/9990 = 2997/9990 + 789/9990 = 3786/9990 = 631/1665 or alternatively 0.3789789… = −0.6 + 0.9789789… = −6/10 + 978/999 = −5994/9990 + 9780/9990 = 3786/9990 = 631/1665

It 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 (10ⁿ − 1)10^k.

Conversely the period of the repeating decimal of a fraction c/d will be (at most) the smallest number n such that 10ⁿ − 1 is divisible by d.

For example, the fraction 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 2/7 is therefore 6.

Repeating decimals as infinite series

A repeating decimal can also be expressed as an infinite series. That is, a repeating decimal can be regarded as the sum of an infinite number of rational numbers. To take the simplest example,

$\sum_{n=1}^\infty \frac{1}{10^n} = {1 \over 10} + {1 \over 100} + {1 \over 1000} + \cdots = 0.\overline{1}$

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.

$\ \frac{a}{1-r} = \frac{\frac{1}{10}}{1-\frac{1}{10}} = \frac{1}{9} = 0.\overline{1}$

Multiplication and cyclic permutation

Main article: Cyclic permutation of integer

The cyclic behavior of repeating decimals in multiplication also leads to the construction of integers which are cyclically permuted when multiplied by certain numbers. For example, 102564 x 4 = 410256. Note that 102564 is the repetend of 4/39 and 410256 the repetend of 16/39.

Other properties of repetend lengths

Various properties of repetend lengths (periods) are given by Mitchell and Dickson.

The period of 1/k for integer k is always ≤ k − 1.

If p is prime, the period of 1/p divides evenly into p − 1.

If k is composite, the period of 1/k is strictly less than k − 1.

The period of c/k, for c coprime to k, equals the period of 1/k.

If $k=2^{a}5^{b}n$ where n > 1 and n is not divisible by 2 or 5, then the length of the transient of 1/k is max(a, b), and the period equals r, where r is the smallest integer such that $10^r \equiv 1 \pmod n$.

If p, p', p", … are distinct primes, then the period of 1/(pp'p"…) equals the lowest common multiple of the periods of 1/p, 1/p' ,1/p" , ….

If k and k' have no common prime factors other than 2 and/or 5, then the period of $\frac{1}{kk'}$ equals the least common multiple of the periods of $\frac{1}{k}$ and $\frac{1}{k'}$.

For prime p, if $\text{period}(\tfrac{1}{p})= \text{period}(\tfrac{1}{p^{2}})= \cdots = \text{period}(\tfrac{1}{p^m})$ but $\text{period}(\tfrac{1}{p^{m}}) \ne \text{period}(\tfrac {1}{p^{m+1}})$, then for $c \ge 0$ we have $\text{period}(\tfrac{1}{p^{m+c}}) = p^{c} \cdot \text{period}(\tfrac{1}{p})$.

If p is a proper prime ending in a 1 – that is, if the repetend of 1/p is a cyclic number of length p − 1 and p = 10h + 1 for some h – then each digit 0, 1, …, 9 appears in the repetend exactly h = (p − 1)/10 times.

For some other properties of repetends, see also.

Extension to other bases

Various features of repeating decimals extend to the representation of numbers in all other integer bases, not just base 10:

• Any number can be represented as an integer component followed by a radix point (the generalization of a decimal point to non-decimal systems) followed by a finite or infinite number of digits.

• A rational number has a terminating sequence after the radix point if all the prime factors of the denominator of the fully reduced fractional form are also factors of the base. This terminating representation is equivalent to a representation with a repeating sequence that can be constructed from the terminating form by decreasing the last digit by 1 and appending an infinite sequence of a digit representing a number that is one less than the base.

• A rational number has an infinitely repeating sequence of finite length less than the value of the fully reduced fraction's denominator if the reduced fraction's denominator contains a prime factor that is not a factor of the base. The repeating sequence is preceded after the radix point by a transient of finite length if the reduced fraction also shares a prime factor with the base.

• An irrational number has a representation of infinite length that never repeats itself.

For example, in duodecimal, 1/2 = 0.6, 1/3 = 0.4, 1/4 = 0.3, 1/6 = 0.2, all have no period; 1/5 = 0.2497, it has period 4, unlike the such value in decimal (it is 0.2, with no period); 1/7 = 0.186ᘔ35, it has period 6, just as it does in decimal.

If b is an integer base and k is an integer,

$\begin{align}1/k\ & = \frac{1}{10} + \frac{(b-k)^1}{10^2} + \frac{(b-k)^2}{10^3} + \frac{(b-k)^3}{10^4} + \frac{(b-k)^4}{10^5} + \cdots + \frac{(b-k)^{N-1}}{10^N} + \cdots \\[6pt] \\\end{align}$

For example 1/7 in base 9,

$\begin{align}1/7\ & = \frac{1}{10} + \frac{2}{10^2} + \frac{4}{10^3} + \frac{8}{10^4} + \frac{17}{10^5} + \frac{35}{10^6} \cdots \\[6pt] \\\end{align}$

Which is 0.125125125... (base 9). Note that 10 (base 9) is 9 (base 10), 10^2 (base 9) is 81 (base 10), 17 (base 9) is 16 (base 10), 35 (base 9) is 32 (base 10)...

Applications to cryptography

Repeating decimals (also called decimal sequences) have found cryptographic and error-correction coding applications. In these applications repeating decimals to base 2 are generally used which gives rise to binary sequences. The maximum length binary sequence for 1/p (when 2 is a primitive root of p) is given by:

$a(i) = 2^{i}~\bmod p ~\bmod 2$

These sequences of period p-1 have an autocorrelation function that has a negative peak of -1 for shift of (p-1)/2. The randomness of these sequences has been examined by diehard tests.

See also

• Decimal representation
• Parasitic number
• Midy's theorem
• Full reptend prime
• Unique prime

External links

• Weisstein, Eric W., "Repeating Decimal", MathWorld.
• Online fractions calculator with detailed solution