Repeating decimal

Citations

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 10^p−1 ≡ 1 (mod p).

The base-10 digital root of the repetend of the reciprocal of any prime number greater than 5 is 9.[1]

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.

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

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

for every integer a that is coprime to n.

The period of 1/p² is usually pT_p, where T_p 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 10^p−1−1. These three primes are 3, 487, and 56598313 (sequence A045616 in the OEIS).[4]

Similarly, the period of 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 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(T_p, T_q), where T_p is the period of 1/p and T_q 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

${\displaystyle {\frac {1}{p^{k}q^{\ell }r^{m}\cdots }}}$

is a repeating decimal with a period of

${\displaystyle \operatorname {LCM} (T_{p^{k}},T_{q^{\ell }},T_{r^{m}},\ldots )}$

where T_p^k, T_q^ℓ, T_r^m,... are respectively the period of the repeating decimals 1/p^k, 1/q^ℓ, 1/r^m,... 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:

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

where a and b are not both zero.

This fraction can also be expressed as:

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

if a > b, or as

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

if b > a, or as

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

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/p^k q^ℓ ⋯.

For example 1/28 = 0.03571428:

• a = 2, b = 0, and the other factors p^k q^ℓ ⋯ = 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:

${\displaystyle x}$	${\displaystyle =0.333333\ldots }$
${\displaystyle 10x}$	${\displaystyle =3.333333\ldots }$	(multiply each side of the above line by 10)
${\displaystyle 9x}$	${\displaystyle =3}$	(subtract the 1st line from the 2nd)
${\displaystyle x}$	${\displaystyle ={\frac {3}{9}}={\frac {1}{3}}}$	(reduce to lowest terms)

Another example:

${\displaystyle x}$	${\displaystyle =\ \ \ \ 0.836363636\ldots }$
${\displaystyle 10x}$	${\displaystyle =\ \ \ \ 8.36363636\ldots }$	(move decimal to start of repetition = move by 1 place = multiply by 10)
${\displaystyle 1000x}$	${\displaystyle =836.36363636\ldots }$	(collate 2nd repetition here with 1st above = move by 2 places = multiply by 100)
${\displaystyle 990x}$	${\displaystyle =828}$	(subtract to clear decimals)
${\displaystyle x}$	${\displaystyle ={\frac {828}{990}}={\frac {18\cdot 46}{18\cdot 55}}={\frac {46}{55}}}$	(reduce to lowest terms)

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:

${\displaystyle {\begin{aligned}x&=0.000000100000010000001\ldots \\10^{7}x&=1.000000100000010000001\ldots \\\left(10^{7}-1\right)x=9999999x&=1\\x&={\frac {1}{10^{7}-1}}={\frac {1}{9999999}}\end{aligned}}}$

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:

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

or

${\displaystyle {\begin{aligned}11.18867924528301886792452830\ldots &=11+0.18867924528301886792452830\ldots \\[8pt]&=11+{\frac {10}{53}}={\frac {11\cdot 53+10}{53}}={\frac {593}{53}}\end{aligned}}}$

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:

${\displaystyle {\begin{aligned}x&=0.{\overline {a_{1}a_{2}\cdots a_{n}}}\\10^{n}x&=a_{1}a_{2}\cdots a_{n}.{\overline {a_{1}a_{2}\cdots a_{n}}}\\[5pt]\left(10^{n}-1\right)x=99\cdots 99x&=a_{1}a_{2}\cdots a_{n}\\[5pt]x&={\frac {a_{1}a_{2}\cdots a_{n}}{10^{n}-1}}={\frac {a_{1}a_{2}\cdots a_{n}}{99\cdots 99}}\end{aligned}}}$

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 digit after the decimal point)

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.

In compressed form

The following picture suggests kind of compression of the above shortcut. Thereby ${\displaystyle \mathbf {I} }$ represents the digits of the integer part of the decimal number (to the left of the decimal point), ${\displaystyle \mathbf {A} }$ makes up the string of digits of the preperiod and ${\displaystyle \#\mathbf {A} }$ its length, and ${\displaystyle \mathbf {P} }$ being the string of repeated digits (the period) with length ${\displaystyle \#\mathbf {P} }$ which is nonzero.

Formation Rule

In the generated fraction, the digit ${\displaystyle 9}$ will be repeated ${\displaystyle \#\mathbf {P} }$ times, and the digit ${\displaystyle 0}$ will be repeated ${\displaystyle \#\mathbf {A} }$ times.

Note that in the absence of an integer part in the decimal, ${\displaystyle \mathbf {I} }$ will be represented by zero, which being to the left of the other digits, will not affect the final result, and may be omitted in the calculation of the generating function.

Examples:

$${\displaystyle {\begin{array}{lllll}3.254444\ldots &=3.25{\overline {4}}&={\begin{Bmatrix}\mathbf {I} =3&\mathbf {A} =25&\mathbf {P} =4\\&\#\mathbf {A} =2&\#\mathbf {P} =1\end{Bmatrix}}&={\dfrac {3254-325}{900}}&={\dfrac {2929}{900}}\\\\0.512512\ldots &=0.{\overline {512}}&={\begin{Bmatrix}\mathbf {I} =0&\mathbf {A} =\emptyset &\mathbf {P} =512\\&\#\mathbf {A} =0&\#\mathbf {P} =3\end{Bmatrix}}&={\dfrac {512-0}{999}}&={\dfrac {512}{999}}\\\\1.09191\ldots &=1.0{\overline {91}}&={\begin{Bmatrix}\mathbf {I} =1&\mathbf {A} =0&\mathbf {P} =91\\&\#\mathbf {A} =1&\#\mathbf {P} =2\end{Bmatrix}}&={\dfrac {1091-10}{990}}&={\dfrac {1081}{990}}\\\\1.333\ldots &=1.{\overline {3}}&={\begin{Bmatrix}\mathbf {I} =1&\mathbf {A} =\emptyset &\mathbf {P} =3\\&\#\mathbf {A} =0&\#\mathbf {P} =1\end{Bmatrix}}&={\dfrac {13-1}{9}}&={\dfrac {12}{9}}&={\dfrac {4}{3}}\\\\0.3789789\ldots &=0.3{\overline {789}}&={\begin{Bmatrix}\mathbf {I} =0&\mathbf {A} =3&\mathbf {P} =789\\&\#\mathbf {A} =1&\#\mathbf {P} =3\end{Bmatrix}}&={\dfrac {3789-3}{9990}}&={\dfrac {3786}{9990}}&={\dfrac {631}{1665}}\end{array}}}$$

The symbol ${\displaystyle \emptyset }$ in the examples above denotes the absence of digits of part ${\displaystyle \mathbf {A} }$ in the decimal, and therefore ${\displaystyle \#\mathbf {A} =0}$ and a corresponding absence in the generated fraction.

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,

${\displaystyle 0.{\overline {1}}={\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\cdots =\sum _{n=1}^{\infty }{\frac {1}{10^{n}}}}$

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.

${\displaystyle {\frac {a}{1-r}}={\frac {\frac {1}{10}}{1-{\frac {1}{10}}}}={\frac {1}{10-1}}={\frac {1}{9}}}$

Similarly,

${\displaystyle {\begin{aligned}0.{\overline {142857}}&={\frac {142857}{10^{6}}}+{\frac {142857}{10^{12}}}+{\frac {142857}{10^{18}}}+\cdots =\sum _{n=1}^{\infty }{\frac {142857}{10^{6n}}}\\[6px]\implies &\quad {\frac {a}{1-r}}={\frac {\frac {142857}{10^{6}}}{1-{\frac {1}{10^{6}}}}}={\frac {142857}{10^{6}-1}}={\frac {142857}{999999}}={\frac {1}{7}}\end{aligned}}}$

Multiplication and cyclic permutation

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 × 4 = 410256. 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[5] and Dickson.[6]

• 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^bn 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 multiplicative order of 10 mod n, that is the smallest integer such that 10^r ≡ 1 (mod n).
• If p, p′, p″,... are distinct primes, then the period of 1/p p′ 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 or 5, then the period of 1/k k′ equals the least common multiple of the periods of 1/k and 1/k′.
• For prime p, if

${\displaystyle {\text{period}}\left({\frac {1}{p}}\right)={\text{period}}\left({\frac {1}{p^{2}}}\right)=\cdots ={\text{period}}\left({\frac {1}{p^{m}}}\right)}$

for some m, but

${\displaystyle {\text{period}}\left({\frac {1}{p^{m}}}\right)\neq {\text{period}}\left({\frac {1}{p^{m+1}}}\right),}$

then for c ≥ 0 we have

${\displaystyle {\text{period}}\left({\frac {1}{p^{m+c}}}\right)=p^{c}\cdot {\text{period}}\left({\frac {1}{p}}\right).}$

• 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.[7]

Extension to other bases

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

• Every real number can be represented as an integer part followed by a radix point (the generalization of a decimal point to non-decimal systems) followed by a finite or infinite number of digits.
• If the base is an integer, a terminating sequence obviously represents a rational number.
• A rational number has a terminating sequence if all the prime factors of the denominator of the fully reduced fractional form are also factors of the base. These numbers make up a dense set in Q and R.
• If the positional numeral system is a standard one, that is it has base

${\displaystyle b\in \mathbb {Z} \smallsetminus \{-1,0,1\}}$

combined with a consecutive set of digits

${\displaystyle D:=\{d_{1},d_{1}+1,\dots ,d_{r}\}}$

with r := |b|, d_r := d₁ + r − 1 and 0 ∈ D, 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 D into some closed interval of the reals, which maps the strings 0.A₁A₂...A_nd_b and 0.A₁A₂...(A_n+1)d₁ with A_i ∈ D and A_n ≠ d_b to the same real number – and there are no other duplicate images. In the decimal system, for example, there is 0.9 = 1.0 = 1; in the balanced ternary system there is 0.1 = 1.T = 1/2.

• A rational number has an indefinitely repeating sequence of finite length l, if the reduced fraction's denominator contains a prime factor that is not a factor of the base. If q is the maximal factor of the reduced denominator which is coprime to the base, l is the smallest exponent such that q divides b^ℓ − 1. It is the multiplicative order ord_q(b) of the residue class b mod q which is a divisor of the Carmichael function λ(q) which in turn is smaller than q. The repeating sequence is preceded by a transient of finite length if the reduced fraction also shares a prime factor with the base. A repeating sequence

${\displaystyle \left(0.{\overline {A_{1}A_{2}\ldots A_{\ell }}}\right)_{b}}$

represents the fraction

${\displaystyle {\frac {(A_{1}A_{2}\ldots A_{\ell })_{b}}{b^{\ell }-1}}.}$

• An irrational number has a representation of infinite length that is not, from any point, an indefinitely repeating sequence of finite length.

For example, in duodecimal, 1/2 = 0.6, 1/3 = 0.4, 1/4 = 0.3 and 1/6 = 0.2 all terminate; 1/5 = 0.2497 repeats with period length 4, in contrast with the equivalent decimal expansion of 0.2; 1/7 = 0.186A35 has period 6 in duodecimal, just as it does in decimal.

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

${\displaystyle {\frac {1}{k}}={\frac {1}{b}}+{\frac {(b-k)^{1}}{b^{2}}}+{\frac {(b-k)^{2}}{b^{3}}}+{\frac {(b-k)^{3}}{b^{4}}}+\cdots +{\frac {(b-k)^{N-1}}{b^{N}}}+\cdots ={\frac {1}{b}}{\frac {1}{1-{\frac {b-k}{b}}}}.}$

For example 1/7 in duodecimal:

$${\displaystyle {\frac {1}{7}}=\left({\frac {1}{10^{\phantom {1}}}}+{\frac {5}{10^{2}}}+{\frac {21}{10^{3}}}+{\frac {A5}{10^{4}}}+{\frac {441}{10^{5}}}+{\frac {1985}{10^{6}}}+\cdots \right)_{\text{base 12}}}$$

which is 0.186A35_base12. 10_base12 is 12_base10, 10²_base12 is 144_base10, 21_base12 is 25_base10, A5_base12 is 125_base10.

function b_adic(b,p,q) // b ≥ 2; 0 < p < q
  static digits = "0123..."; // up to the digit with value b–1
begin
  s = "";  // the string of digits
  pos = 0; // all places are right to the radix point
  while not defined(occurs[p]) do
    occurs[p] = pos; // the position of the place with remainder p
    bp = b*p;
    z = floor(bp/q); // index z of digit within: 0 ≤ z ≤ b-1
    p = b*p − z*q;   // 0 ≤ p < q
    if p = 0 then L = 0;
      if not z = 0 then
        s = s . substring(digits, z, 1) 
      end if
      return (s);
    end if
    s = s . substring(digits, z, 1); // append the character of the digit
    pos += 1;
  end while
  L = pos - occurs[p]; // the length of the repetend (being < q)
  // mark the digits of the repetend by a vinculum:
  for i from occurs[p] to pos-1 do
    substring(s, i, 1) = overline(substring(s, i, 1));
  end for
  return (s);
end function

Applications to cryptography

Repeating decimals (also called decimal sequences) have found cryptographic and error-correction coding applications.[8] 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:[9]

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

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.[10]

See also

• Decimal representation
• Full reptend prime
• Midy's theorem
• Parasitic number
• Trailing zero
• Unique prime
• 0.999..., a repeating decimal equal to one
• Pigeonhole principle

References and remarks

External links

• Weisstein, Eric W. "Repeating Decimal". MathWorld.