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 are 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 1/3 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 at the second digit following the decimal point and then repeats the sequence "144" forever, i.e. 5.8144144144.... Another example of this is 593/53, which becomes periodic after the decimal point, repeating the 13-digit pattern "1886792452830" forever, i.e. 11.18867924528301886792452830....
The finite digit sequence that is repeated infinitely 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 denominator is a power of 10 (e.g. 1.585 = 1585/1000); it may also be written as a ratio of the form k/2n·5m (e.g. 1.585 = 317/23·52). 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 1.000... = 0.999... and 1.585000... = 1.584999.... (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 § Every rational number is either a terminating or repeating decimal). Examples of such irrational numbers are √2 and π.
Background
Notation
Any textual representation is necessarily finite, which is why special non-decimal notation is required to represent repeating decimals. Below are several notational conventions. None of them are accepted universally.
Vinculum: In the United States, Canada, India, France, Germany, Denmark, the Netherlands, Italy, Switzerland, the Czech Republic, Slovakia, Slovenia, Chile, Taiwan, and Turkey, the convention is to draw a horizontal line (a vinculum) above the repetend. Dots: In some Islamic countries, such as Bangladesh, Malaysia, Morocco, Pakistan, Tunisia, Iran, Algeria and Egypt, as well as the United Kingdom, New Zealand, Australia, South Africa, Japan, Thailand, India, South Korea, Singapore, and the People's Republic of China, the convention is to place dots above the outermost numerals of the repetend. Parentheses: In parts of Europe, incl. Austria, Finland, Norway, Poland, Russia and Ukraine, as well as Vietnam and Israel, the convention is to enclose the repetend in parentheses. This can cause confusion with the notation for standard uncertainty or multiplication. Arc: In Spain and some Latin American countries, such as Argentina, Brazil, and Mexico, the arc notation over the repetend is also used as an alternative to the vinculum and the dots notation. Ellipsis: Informally, repeating decimals are often represented by an ellipsis (three periods, 0.333...), especially when the previous notational conventions are first taught in school. This notation introduces uncertainty as to which digits should be repeated and even whether repetition is occurring at all, since such ellipses are also employed for irrational numbers; π, for example, can be represented as 3.14159.... "r" notation: In online settings or informal writing, where notating any of the above may be impossible, a simple way of writing repeated numbers is to place an r at the start of the repetition. For example, 1.234 can be written 1.2r34. In cases where the repetition starts at the decimal, you can get rid of the decimal point, so 11.1886792452830 would be 11r1886792452830. In English, there are various ways to read repeating decimals aloud. For example, 1.234 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, 11.1886792452830 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".
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: 0.0675675675.... For any integer fraction A/B, the remainder at step k, for any positive integer k, is A × 10k (modulo B).
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. 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". In base 10, a fraction has a repeating decimal if and only if in lowest terms, its denominator has at least a prime factor different from 2 and 5 (a prime denominator is considered as a prime factor of itself), or in other words, the denominator cannot be expressed as 2m5n, where m and n are non-negative integers.
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. In the example above, α = 5.8144144144... satisfies the equation
The process of how to find these integer coefficients is described below.
Formal proof Given a repeating decimal x = a . b c ¯ {\displaystyle x=a.b{\overline {c}}} where a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} are groups of digits, let n = ⌈ log 10 b ⌉ {\displaystyle n=\lceil {\log _{10}b}\rceil } , the number of digits of b {\displaystyle b} . Multiplying by 10 n {\displaystyle 10^{n}} separates the repeating and terminating groups:
10 n x = a b . c ¯ . {\displaystyle 10^{n}x=ab.{\bar {c}}.}
If the decimals terminate ( c = 0 {\displaystyle c=0} ), the proof is complete. For c ≠ 0 {\displaystyle c\neq 0} with k ∈ N {\displaystyle k\in \mathbb {N} } digits, let x = y . c ¯ {\displaystyle x=y.{\bar {c}}} where y ∈ Z {\displaystyle y\in \mathbb {Z} } is a terminating group of digits. Then,
c = d 1 d 2 . . . d k {\displaystyle c=d_{1}d_{2}\,...d_{k}}
where d i {\displaystyle d_{i}} denotes the i-th digit, and
x = y + ∑ n = 1 ∞ c ( 10 k ) n = y + ( c ∑ n = 0 ∞ 1 ( 10 k ) n ) − c . {\displaystyle x=y+\sum _{n=1}^{\infty }{\frac {c}{{(10^{k})}^{n}}}=y+\left(c\sum _{n=0}^{\infty }{\frac {1}{{(10^{k})}^{n}}}\right)-c.}
Since ∑ n = 0 ∞ 1 ( 10 k ) n = 1 1 − 10 − k {\displaystyle \textstyle \sum _{n=0}^{\infty }{\frac {1}{{(10^{k})}^{n}}}={\frac {1}{1-10^{-k}}}} ,
x = y − c + 10 k c 10 k − 1 . {\displaystyle x=y-c+{\frac {10^{k}c}{10^{k}-1}}.}
Since x {\displaystyle x} is the sum of an integer ( y − c {\displaystyle y-c} ) and a rational number ( 10 k c 10 k − 1 {\textstyle {\frac {10^{k}c}{10^{k}-1}}} ), x {\displaystyle x} is also rational.
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 segment) of 1/p 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 10p−1 ≡ 1 (mod p). The base-10 digital root of the repetend of the reciprocal of any prime number greater than 5 is 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
Examples of fractions belonging to this group are:
1/7 = 0.142857, 6 repeating digits 1/17 = 0.0588235294117647, 16 repeating digits 1/19 = 0.052631578947368421, 18 repeating digits 1/23 = 0.0434782608695652173913, 22 repeating digits 1/29 = 0.0344827586206896551724137931, 28 repeating digits 1/47 = 0.0212765957446808510638297872340425531914893617, 46 repeating digits 1/59 = 0.0169491525423728813559322033898305084745762711864406779661, 58 repeating digits 1/61 = 0.016393442622950819672131147540983606557377049180327868852459, 60 repeating digits 1/97 = 0.010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567, 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, 1/223, 1/229, etc. (sequence A001913 in the 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 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 1/7 starts '142' and is followed by '857' while 6/7 (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, 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 the 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, 2063... (sequence A000353 in the OEIS).
Other reciprocals of primes Some reciprocals of primes that do not generate cyclic numbers are:
1/3 = 0.3, which has a period (repetend length) of 1. 1/11 = 0.09, which has a period of two. 1/13 = 0.076923, which has a period of six. 1/31 = 0.032258064516129, which has a period of 15. 1/37 = 0.027, which has a period of three. 1/41 = 0.02439, which has a period of five. 1/43 = 0.023255813953488372093, which has a period of 21. 1/53 = 0.0188679245283, which has a period of 13. 1/67 = 0.014925373134328358208955223880597, which has a period of 33. 1/71 = 0.01408450704225352112676058338028169, which has a period of 35. 1/73 = 0.01369863, which has a period of eight. 1/79 = 0.0126582278481, which has a period of 13. 1/83 = 0.01204819277108433734939759036144578313253, which has a period of 41. 1/89 = 0.01123595505617977528089887640449438202247191, which has a period of 44. (sequence A006559 in the 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 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 10p−1 − 1/p. For example, for 11 we get
10 11 − 1 − 1 11 = 909090909 {\displaystyle {\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 L(n) of the decimal repetend of 1/n divides φ(n), where φ is the totient function. The length is equal to φ(n) if and only if 10 is a primitive root modulo n. In particular, it follows that L(p) = p − 1 if and only if 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 period 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 1/p2 repeats:
1/49 = 0.020408163265306122448979591836734693877551. 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
a m ≡ 1 ( mod n ) {\displaystyle a^{m}\equiv 1{\pmod {n}}}
for every integer a that is coprime to n. The period of 1/p2 is usually pTp, where Tp is the period of 1/p. There are three known primes for which this is not true, and for those the period of 1/p2 is the same as the period of 1/p because p2 divides 10p−1−1. These three primes are 3, 487, and 56598313 (sequence A045616 in the OEIS). Similarly, the period of 1/pk is usually pk–1Tp If p and q are primes other than 2 or 5, the decimal representation of the fraction 1/pq 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 1/pq is a factor of λ(pq) and it happens to be 48 in this case:
1/119 = 0.008403361344537815126050420168067226890756302521. The period T of 1/pq is LCM(Tp, Tq), where Tp is the period of 1/p and Tq is the period of 1/q. If p, q, r, etc. are primes other than 2 or 5, and k, ℓ, m, etc. are positive integers, then
1 p k q ℓ r m ⋯ {\displaystyle {\frac {1}{p^{k}q^{\ell }r^{m}\cdots }}}
is a repeating decimal with a period of
LCM ( T p k , T q ℓ , T r m , … ) {\displaystyle \operatorname {LCM} (T_{p^{k}},T_{q^{\ell }},T_{r^{m}},\ldots )}
where Tpk, Tqℓ, Trm,... are respectively the period of the repeating decimals 1/pk, 1/qℓ, 1/rm,... as defined above.
Reciprocals of integers not coprime to 10 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 2 a ⋅ 5 b p k q ℓ ⋯ , {\displaystyle {\frac {1}{2^{a}\cdot 5^{b}p^{k}q^{\ell }\cdots }}\,,}
where a and b are not both zero. This fraction can also be expressed as:
5 a − b 10 a p k q ℓ ⋯ , {\displaystyle {\frac {5^{a-b}}{10^{a}p^{k}q^{\ell }\cdots }}\,,}
if a > b, or as
2 b − a 10 b p k q ℓ ⋯ , {\displaystyle {\frac {2^{b-a}}{10^{b}p^{k}q^{\ell }\cdots }}\,,}
if b > a, or as
1 10 a p k q ℓ ⋯ , {\displaystyle {\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 1/pk qℓ ⋯. For example 1/28 = 0.03571428:
a = 2, b = 0, and the other factors pk qℓ ⋯ = 7 there are 2 initial non-repeating digits, 03; and there are 6 repeating digits, 571428, the same amount as 1/7 has.
Converting repeating decimals to fractions Given a repeating decimal, it is possible to calculate the fraction that produces it. For example:
Another example:
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:
x = 0.000000100000010000001 … 10 7 x = 1.000000100000010000001 … ( 10 7 − 1 ) x = 9999999 x = 1 x = 1 10 7 − 1 = 1 9999999 {\displaystyle {\begin{aligned}x&=0.000000100000010000001\ldots \\10^{7}x&=1.000000100000010000001\ldots \\\left(10^{7}-1\right)x=9999999x&=1\\x&={\frac {1}{10^{7}-1}}={\frac {1}{9999999}}\end{aligned}}}
So this particular repeating decimal corresponds to the fraction 1/10n − 1, 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:
7.48181818 … = 7.3 + 0.18181818 … = 73 10 + 18 99 = 73 10 + 9 ⋅ 2 9 ⋅ 11 = 73 10 + 2 11 = 11 ⋅ 73 + 10 ⋅ 2 10 ⋅ 11 = 823 110 {\displaystyle {\begin{aligned}7.48181818\ldots &=7.3+0.18181818\ldots \\[8pt]&={\frac {73}{10}}+{\frac {18}{99}}={\frac {73}{10}}+{\frac {9\cdot 2}{9\cdot 11}}={\frac {73}{10}}+{\frac {2}{11}}\\[12pt]&={\frac {11\cdot 73+10\cdot 2}{10\cdot 11}}={\frac {823}{110}}\end{aligned}}}
or
11.18867924528301886792452830 … = 11 + 0.18867924528301886792452830 … = 11 + 10 53 = 11 ⋅ 53 + 10 53 = 593 53 {\displaystyle {\begin{aligned}11.18867924528301886792452830\ldots &=11+0.18867924528301886792452830\ldots \\[8pt]&=11+{\frac {10}{53}}={\frac {11\cdot 53+10}{53}}={\frac {593}{53}}\end{aligned}}}
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. a 1 a 2 ⋯ a n ¯ 10 n x = a 1 a 2 ⋯ a n . a 1 a 2 ⋯ a n ¯ ( 10 n − 1 ) x = 99 ⋯ 99 x = a 1 a 2 ⋯ a n x = a 1 a 2 ⋯ a n 10 n − 1 = a 1 a 2 ⋯ a n 99 ⋯ 99 {\displaystyle {\begin{aligned}x&=0.{\overline {a_{1}a_{2}\cdots a_{n}}}\\10^{n}x&=a_{1}a_{2}\cdots a_{n}.{\overline {a_{1}a_{2}\cdots a_{n}}}\\[5pt]\left(10^{n}-1\right)x=99\cdots 99x&=a_{1}a_{2}\cdots a_{n}\\[5pt]x&={\frac {a_{1}a_{2}\cdots a_{n}}{10^{n}-1}}={\frac {a_{1}a_{2}\cdots a_{n}}{99\cdots 99}}\end{aligned}}}
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,
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); this further reduces to 4/333. 0.999999... = 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 = 1/8325 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 An even faster method is to ignore the decimal point completely and go like this
1.23444... = 1234 − 123/900 = 1111/900 (denominator has one 9 and two 0s because one digit repeats and there are two non-repeating digits after the decimal point) 0.3789789... = 3789 − 3/9990 = 3786/9990 (denominator has three 9s and one 0 because three digits repeat and there is one non-repeating d