A repeating decimal or recurring decimal is a decimal representation of a number whose digits are eventually periodic (that is, after some place, the same sequence of digits is repeated forever); if this sequence consists only of zeros (that is if there is only a finite number of nonzero digits), the decimal is said to be terminating, and is not considered as repeating. It can be shown that a number is rational if and only if its decimal representation is repeating or terminating. For example, the decimal representation of becomes periodic just after the decimal point, repeating the single digit "3" forever, i.e. 0.333.... A more complicated example is, whose decimal becomes periodic at the second digit following the decimal point and then repeats the sequence "144" forever, i.e. 5.8144144144.... Another example of this is, which becomes periodic after the decimal point, repeating the 13-digit pattern "1886792452830" forever, i.e. 11.18867924528301886792452830....
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.[1] Every terminating decimal representation can be written as a decimal fraction, a fraction whose denominator is a power of 10 (e.g.); it may also be written as a ratio of the form (e.g.). 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 (rightmost) non-zero digit by one and appending a repetend of 9. Two examples of this are and . (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 repetition (see). Examples of such irrational numbers are and .[2]
There are several notational conventions for representing repeating decimals. None of them are accepted universally.
| 0. | 0. | 0.(1) | 0. | 0.111... | |||
| = | 0. | 0. | 0.(3) | 0. | 0.333... | ||
| = | 0. | 0. | 0.(6) | 0. | 0.666... | ||
| = | 0. | 0. | 0.(81) | 0. | 0.8181... | ||
| = | 0.58 | 0.58 | 0.58(3) | 0.58 | ... | ||
| = | 0. | 0.4285 | 0.(142857) | 0. | ... | ||
| = | 0. | 0.1234567 | 0.(012345679) | 0. | ... | ||
| = | 3. | 3.4285 | 3.(142857) | 3. | ... | ||
| = | 11. | 11.88679245283 | 11.(1886792452830) | 11. | ... |
In English, there are various ways to read repeating decimals aloud. For example, 1.2 may be read "one point two repeating three four", "one point two repeated three four", "one point two recurring three four", "one point two repetend three four" or "one point two into infinity three four". Likewise, 11. may be read "eleven point repeating one double eight six seven nine two four five two eight three zero", "eleven point repeated one double eight six seven nine two four five two eight three zero", "eleven point recurring one double eight six seven nine two four five two eight three zero" "eleven point repetend one double eight six seven nine two four five two eight three zero" or "eleven point into infinity one double eight six seven nine two four five two eight three zero".
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 : 0.0 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: ....
For any integer fraction, the remainder at step k, for any positive integer k, is A × 10k (modulo B).
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. Then the length of the repetend, also called "period", is defined to be 0.
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. The repeating sequence of digits is called "repetend" which has a certain length greater than 0, also called "period".[3]
In base 10, a fraction has a repeating decimal if and only if in lowest terms, its denominator has any prime factors besides 2 or 5, or in other words, cannot be expressed as 2m 5n, where m and n are non-negative integers.
Each repeating decimal number satisfies a linear equation with integer coefficients, and its unique solution is a rational number. In the example above, satisfies the equation
| 10000α − 10α | = 58144.144144... − 58.144144... | |
| 9990α | = 58086 | |
| Therefore, α | = = |
Given a repeating decimal
x=a.b\overline{c}
a
b
c
n=\lceil{log10b}\rceil
b
10n
10nx=ab.\bar{c}.
If the decimals terminate (
c=0
c ≠ 0
k\inN
x=y.\bar{c}
y\inZ
c=d1d2...dk
where
di
| infty | |
| x=y+\sum | |
| n=1 |
| c | |
| {(10k) |
n}=y
| infty | |
| +\left(c\sum | |
| n=0 |
| 1 | |
| {(10k) |
n}\right)-c.
Since
| infty | |
| style\sum | |
| n=0 |
| 1 | |
| {(10k) |
n}=
| 1 | |
| 1-10-k |
x=y-c+
| 10kc | |
| 10k-1 |
.
Since
x
y-c
x
| decimal expansion | binary expansion | ||||
|---|---|---|---|---|---|
| 0.5 | 0 | 0.1 | 0 | ||
| 0. | 1 | 0. | 2 | ||
| 0.25 | 0 | 0.01 | 0 | ||
| 0.2 | 0 | 0. | 4 | ||
| 0.1 | 1 | 0.0 | 2 | ||
| 0. | 6 | 0. | 3 | ||
| 0.125 | 0 | 0.001 | 0 | ||
| 0. | 1 | 0. | 6 | ||
| 0.1 | 0 | 0.0 | 4 | ||
| 0. | 2 | 0. | 10 | ||
| 0.08 | 1 | 0.00 | 2 | ||
| 0. | 6 | 0. | 12 | ||
| 0.0 | 6 | 0.0 | 3 | ||
| 0.0 | 1 | 0. | 4 | ||
| 0.0625 | 0 | 0.0001 | 0 |
| decimal expansion | |||
|---|---|---|---|
| 0. | 16 | ||
| 0.0 | 1 | ||
| 0. | 18 | ||
| 0.05 | 0 | ||
| 0. | 6 | ||
| 0.0 | 2 | ||
| 0. | 22 | ||
| 0.041 | 1 | ||
| 0.04 | 0 | ||
| 0.0 | 6 | ||
| 0. | 3 | ||
| 0.03 | 6 | ||
| 0. | 28 | ||
| 0.0 | 1 | ||
| 0. | 15 |
| decimal expansion | |||
|---|---|---|---|
| 0.03125 | 0 | ||
| 0. | 2 | ||
| 0.0 | 16 | ||
| 0.0 | 6 | ||
| 0.02 | 1 | ||
| 0. | 3 | ||
| 0.0 | 18 | ||
| 0. | 6 | ||
| 0.025 | 0 | ||
| 0. | 5 | ||
| 0.0 | 6 | ||
| 0. | 21 | ||
| 0.02 | 2 | ||
| 0.0 | 1 | ||
| 0.0 | 22 | ||
| 0. | 46 |
Thereby fraction is the unit fraction and ℓ10 is the length of the (decimal) repetend.
The lengths ℓ10(n) of the decimal repetends of, n = 1, 2, 3, ..., 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, 18, 6, 6, 13, 0, 9, 5, 41, 6, 16, 21, 28, 2, 44, 1, 6, 22, 15, 46, 18, 1, 96, 42, 2, 0... .
For comparison, the lengths ℓ2(n) of the binary repetends of the fractions, n = 1, 2, 3, ..., are:
0, 0, 2, 0, 4, 2, 3, 0, 6, 4, 10, 2, 12, 3, 4, 0, 8, 6, 18, 4, 6, 10, 11, 2, 20, 12, 18, 3, 28, 4, 5, 0, 10, 8, 12, 6, 36, 18, 12, 4, 20, 6, 14, 10, 12, 11, ... (=[''n''], if n not a power of 2 else =0).
The decimal repetends of, n = 1, 2, 3, ..., 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, 032258064516129, 0, 03, 2941176470588235, 285714... .
The decimal repetend lengths of, p = 2, 3, 5, ... (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, 141, 146, 153, 155, 312, 79... .
The least primes p for which has decimal repetend length n, n = 1, 2, 3, ..., 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, 2791, 353, 67, 103, 71, 999999000001, 2028119, 909090909090909091, 900900900900990990990991, 1676321, 83, 127, 173... .
The least primes p for which has n different cycles, n = 1, 2, 3, ..., 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, 49663, 12289, 859, 239, 27581, 9613, 18131, 13757, 33931... .
See main article: Reciprocals of primes. 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 segment) of is equal to the order of 10 modulo p. If 10 is a primitive root modulo p, then the repetend length is equal to p - 1; if not, then the repetend length is a factor of p - 1. This result can be deduced from Fermat's little theorem, which states that .
The base-10 digital root of the repetend of the reciprocal of any prime number greater than 5 is 9.[7]
If the repetend length of for prime p is equal to p - 1 then the repetend, expressed as an integer, is called a cyclic number.
See main article: Cyclic number.
Examples of fractions belonging to this group are:
The list can go on to include the fractions,,,,,,,,,, etc. .
Every proper multiple of a cyclic number (that is, a multiple having the same number of digits) is a rotation:
The reason for the cyclic behavior is apparent from an arithmetic exercise of long division of : the sequential remainders are the cyclic sequence . See also the article 142,857 for more properties of this cyclic number.
A fraction which is cyclic thus has a recurring decimal of even length that divides into two sequences in nines' complement form. For example starts '142' and is followed by '857' while (by rotation) starts '857' followed by its nines' complement '142'.
The rotation of the repetend of a cyclic number always happens in such a way that each successive repetend is a bigger number than the previous one. In the succession above, for instance, we see that 0.142857... < 0.285714... < 0.428571... < 0.571428... < 0.714285... < 0.857142.... This, for cyclic fractions with long repetends, allows us to easily predict what the result of multiplying the fraction by any natural number n will be, as long as the repetend is known.
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, times). They are:[8]
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,... .
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 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, 2063... .
Some reciprocals of primes that do not generate cyclic numbers are:
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, we can check whether the prime p divides some number 999...999 in which the number of digits divides p − 1. Since the period is never greater than p − 1, we can obtain this by calculating . For example, for 11 we get
| 1011-1-1 | |
| 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 can be divided into two sets, with different repetends. The first set is:
where the repetend of each fraction is a cyclic re-arrangement of 076923. The second set is:
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.
For an arbitrary integer n, the length L(n) of the decimal repetend of divides φ(n), where φ is the totient function. The length is equal to if and only if 10 is a primitive root modulo n.[9]
In particular, it follows that if and only if p is a prime and 10 is a primitive root modulo p. Then, the decimal expansions of for n = 1, 2, ..., p − 1, all have period p − 1 and differ only by a cyclic permutation. Such numbers p are called full repetend primes.
If p is a prime other than 2 or 5, the decimal representation of the fraction repeats:
= 0..
The period (repetend length) L(49) 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
am\equiv1\pmodn
for every integer a that is coprime to n.
The period of is usually pTp, where Tp is the period of . There are three known primes for which this is not true, and for those the period of is the same as the period of because p2 divides 10p-1-1. These three primes are 3, 487, and 56598313 .[10]
Similarly, the period of is usually pk–1Tp
If p and q are primes other than 2 or 5, the decimal representation of the fraction repeats. An example is :
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 is a factor of λ(pq) and it happens to be 48 in this case:
= 0..
The period T of is LCM(Tp, Tq), where Tp is the period of and Tq is the period of .
If p, q, r, etc. are primes other than 2 or 5, and k, ℓ, m, etc. are positive integers, then
| 1 | |
| pkq\ellrm … |
| \operatorname{LCM}(T | |
| pk |
,
| T | |
| q\ell |
,
| T | |
| rm |
,\ldots)
An integer that is not coprime 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:
| 1 | |
| 2a ⋅ 5bpkq\ell … |
,
where a and b are not both zero.
This fraction can also be expressed as:
| 5a-b | |
| 10apkq\ell … |
,
if a > b, or as
| 2b-a | |
| 10bpkq\ell … |
,
if b > a, or as
| 1 | |
| 10apkq\ell … |
,
if a = b.
The decimal has:
For example = 0.03:
Given a repeating decimal, it is possible to calculate the fraction that produces it. For example:
x | =0.333333\ldots | ||||||||
10x | =3.333333\ldots | (multiply each side of the above line by 10) | |||||||
9x | =3 | (subtract the 1st line from the 2nd) | |||||||
x | =
| (reduce to lowest terms) |
Another example:
x | = 0.836363636\ldots | |||||||||||
10x | = 8.36363636\ldots | (move decimal to start of repetition = move by 1 place = multiply by 10) | ||||||||||
1000x | =836.36363636\ldots | (collate 2nd repetition here with 1st above = move by 2 places = multiply by 100) | ||||||||||
990x | =828 | (subtract to clear decimals) | ||||||||||
x | =
=
=
| (reduce to lowest terms) |
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\\ 107x&=1.000000100000010000001\ldots\\ \left(107-1\right)x=9999999x&=1\\ x&=
| 1 | |
| 107-1 |
=
| 1 | |
| 9999999 |
\end{align}
So this particular repeating decimal corresponds to the fraction, where the denominator is the number written as n 9s. 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] &=
| 73 | + | |
| 10 |
| 18 | |
| 99 |
=
| 73 | |
| 10 |
+
| 9 ⋅ 2 | |
| 9 ⋅ 11 |
=
| 73 | |
| 10 |
+
| 2 | |
| 11 |
\\[12pt] &=
| 11 ⋅ 73+10 ⋅ 2 | |
| 10 ⋅ 11 |
=
| 823 | |
| 110 |
\end{align}
\begin{align} 11.18867924528301886792452830\ldots&=11+0.18867924528301886792452830\ldots\\[8pt] &=11+
| 10 | |
| 53 |
=
| 11 ⋅ 53+10 | |
| 53 |
=
| 593 | |
| 53 |
\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:
\begin{align} x&=0.\overline{a1a2 … an}\\ 10nx&=a1a2 … an.\overline{a1a2 … an}\\[5pt] \left(10n-1\right)x=99 … 99x&=a1a2 … an\\[5pt] x&=
| a1a2 … an | |
| 10n-1 |
=
| a1a2 … an | |
| 99 … 99 |
\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 9s. For example,
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,
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,
An even faster method is to ignore the decimal point completely and go like this
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)10k.
Conversely the period of the repeating decimal of a fraction will be (at most) the smallest number n such that 10n − 1 is divisible by d.
For example, the fraction has d = 7, and the smallest k that makes 10k − 1 divisible by 7 is k = 6, because 999999 = 7 × 142857. The period of the fraction is therefore 6.
The following picture suggests kind of compression of the above shortcut.Thereby
I
A
\#A
P
\#P
In the generated fraction, the digit
9
\#P
0
\#A
Note that in the absence of an integer part in the decimal,
I
Examples:
The symbol
\emptyset
A
\#A=0
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,
0.\overline{1}=
| 1 | |
| 10 |
+
| 1 | |
| 100 |
+
| 1 | |
| 1000 |
+ … =
| infty | |
| \sum | |
| n=1 |
| 1 | |
| 10n |
The above series is a geometric series with the first term as and the common factor . 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.
| a | |
| 1-r |
=
| ||||
|
=
| 1 | |
| 10-1 |
=
| 1 | |
| 9 |
Similarly,
\begin{align} 0.\overline{142857}&=
| 142857 | |
| 106 |
+
| 142857 | |
| 1012 |
+
| 142857 | |
| 1018 |
+ … =
| infty | |
| \sum | |
| n=1 |
| 142857 | |
| 106n |
\\[6px] \implies&
| a | |
| 1-r |
=
| ||||
|
=
| 142857 | |
| 106-1 |
=
| 142857 | |
| 999999 |
=
| 17 \end{align} | |
See main article: Transposable 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 is the repetend of and 410256 the repetend of .
Various properties of repetend lengths (periods) are given by Mitchell[11] and Dickson.[12]
| period\left( | 1 |
| p |
\right)=period\left(
| 1 | |
| p2 |
\right)= … =period\left(
| 1 | |
| pm |
\right)
for some m, but
| period\left( | 1 |
| pm |
\right)\neperiod\left(
| 1 | |
| pm+1 |
\right),
then for c ≥ 0 we have
| period\left( | 1 |
| pm+c |
\right)=pc ⋅ period\left(
| 1 | |
| p |
\right).
For some other properties of repetends, see also.[13]
Various features of repeating decimals extend to the representation of numbers in all other integer bases, not just base 10:
b\in\Z\smallsetminus\{-1,0,1\}
combined with a consecutive set of digits
D:=\{d1,d1+1,...,dr\}
with, and, then a terminating sequence is obviously equivalent to the same sequence with non-terminating repeating part consisting of the digit 0. If the base is positive, then there exists an order homomorphism from the lexicographical order of the right-sided infinite strings over the alphabet into some closed interval of the reals, which maps the strings and with and to the same real number – and there are no other duplicate images. In the decimal system, for example, there is 0. = 1. = 1; in the balanced ternary system there is 0. = 1. = .
\left(0.\overline{A1A2\ldotsA\ell}\right)b
represents the fraction
| (A1A2\ldotsA\ell)b | |
| b\ell-1 |
.
For example, in duodecimal, = 0.6, = 0.4, = 0.3 and = 0.2 all terminate; = 0. repeats with period length 4, in contrast with the equivalent decimal expansion of 0.2; = 0. has period 6 in duodecimal, just as it does in decimal.
If is an integer base and is an integer, then
| 1 | |
| k |
=
| 1 | |
| b |
+
| (b-k)1 | |
| b2 |
+
| (b-k)2 | |
| b3 |
+
| (b-k)3 | |
| b4 |
+ … +
| (b-k)N-1 | |
| bN |
+ … =
| 1b | ||||||
|
For example 1/7 in duodecimal:
which is 0.base12. 10base12 is 12base10, 102base12 is 144base10, 21base12 is 25base10, A5base12 is 125base10.
For a rational (and base) there is the following algorithm producing the repetend together with its length:
The subsequent line calculates the new remainder of the division modulo the denominator . As a consequence of the floor function floor we have
| bp | |
| q |
-1 < z=\left\lfloor
| bp | |
| q |
\right\rfloor \le
| bp | |
| q |
,
bp-q<zq \implies p':=bp-zq<q
zq\lebp \implies 0\lebp-zq=:p'.
while loop. Such a recurrence is detected by the associative array occurs. The new digit is formed in the yellow line, where is the only non-constant. The length of the repetend equals the number of the remainders (see also section Every rational number is either a terminating or repeating decimal).Repeating decimals (also called decimal sequences) have found cryptographic and error-correction coding applications.[14] In these applications repeating decimals to base 2 are generally used which gives rise to binary sequences. The maximum length binary sequence for (when 2 is a primitive root of p) is given by:[15]
a(i)=2i\bmodp\bmod2
\operatorname{ord}n(b):=min\{L\in\N\midbL\equiv1\bmodn\}
λ(n):=max\{\operatorname{ord}n(b)\mid\gcd(b,n)=1\}