Repeating decimal

A decimal representation of a real 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 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 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 a⁄b where a and b are integers and b is non-zero. This form is known as a vulgar fraction. On the one hand, the decimal representation of a rational number is ultimately periodic because it can be determined by a long 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 a linear equation with integral 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.

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. These terminating decimals represent rational numbers whose fractions in lowest terms are of the form k⁄2ⁿ5^m. For example, 1.585 = 317⁄200 = 317⁄2³5². They can be written as a decimal fraction: 317⁄200 = 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. 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 the square root of 2 and the number π. 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.

Fraction	Decimal	Overline	Dots	Brackets
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)
1⁄7	0.142857142857...	0.142857	$0.{\dot {1}}4285{\dot {7}}$	0.(142857)
1⁄81	0.012345679...	0.012345679	$0.{\dot {0}}1234567{\dot {9}}$	0.(012345679)
7⁄12	0.58333...	0.583	$0.58{\dot {3}}$	0.58(3)

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, 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:

• 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

The list can go on to include the fractions 1⁄47, 1⁄59, 1⁄61, 1⁄97, 1⁄109, etc.

The following multiplications exhibit an interesting property:

• 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...

That is, these multiples can be obtained from rotating the digits of the original decimal of 1⁄7. The reason for the rotating behaviour of the digits is apparent from an arithmetics exercise of finding the decimal of 1⁄7.

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

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

Fractions of the second group are:

• 1⁄3 = 0.333... which has 1 repeating digit.
• 1⁄11 = 0.090909... which has 2 repeating digits.
• 1⁄13 = 0.076923... which has 6 repeating digits.

Note that the following multiples of 1⁄13 exhibit the discussed property of rotating digits:

• 1⁄13 = 0.076923...
• 3⁄13 = 0.230769...
• 4⁄13 = 0.307692...
• 9⁄13 = 0.692307...
• 10⁄13 = 0.769230...
• 12⁄13 = 0.923076...

And similarly these multiples:

• 2⁄13 = 0.153846...
• 5⁄13 = 0.384615...
• 6⁄13 = 0.461538...
• 7⁄13 = 0.538461...
• 8⁄13 = 0.615384...
• 11⁄13 = 0.846153...

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

The period of the repeating decimal of 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 inclusively between 1 and 6.

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, 12 then back to the same sequence again. These remainders can be rearranged to become complete consecutive numbers inclusively between 1 and 16.

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

10⁶ − 1 = 999,999 (i.e. 6 digits of 9) is divisible by 7;

10¹² − 1 = 999,999,999,999 (i.e. 12 digits of 9) is divisible by 13;

10¹⁶ − 1 = 9,999,999,999,999,999 (i.e. 16 digits of 9) is divisible by 17;

10¹⁸ − 1 = 999,999,999,999,999,999 (i.e. 18 digits of 9) is divisible by 19; etc

which 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; etc

These 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 5

Proof:

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

10^{(7 − 1)} − 1 ≡ 0 mod 7

or

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

LHS of the expression is,

[10^{(7 − 1)} − 1] mod 7 ≡ 9 * (10⁵ + 10⁴ + 10³ + 10² + 10¹ + 10⁰) mod 7

Since 9 mod 7 ≠ 0, thus the proof is complete if:

(10⁰ + 10¹ + 10² + 10³ + 10⁴ + 10⁵) mod 7 ≡ 0 mod 7

The LHS of which is

(1 + 3 + 3² + 3³ + 3⁴ + 3⁵) mod 7

Which 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 the prime numbers for the first group of repeating frictions listed above. If p is the prime number, the expression simplify to:

[10^{(p − 1)} − 1] mod p

≡ Σ(1, 2, 3, ..., p − 1) mod p

≡ [(p − 1)p /2] mod p

≡ 0 mod p

See 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:

Failed to parse (unknown function "\begin{alignat}"): {\displaystyle \begin{alignat}2 x &= 0.333333\ldots\\ 10x &= 3.333333\ldots&\quad&\mbox{(multiplying each side of the above line by 10)}\\ 9x &= 3 &&\mbox{(subtracting the 1st line from the 2nd)}\\ x &= 3/9 = 1/3 &&\mbox{(simplifying)}\\ \end{alignat}}

Another example:

${\begin{aligned}x&=0.836363636\ldots \\100x&=83.636363636\ldots \\99x&=82.8\\x&={\frac {82.8}{99}}={\frac {5\times 82.8}{5\times 99}}={\frac {414}{495}}={\frac {9\times 46}{9\times 55}}={\frac {46}{55}}.\end{aligned}}$

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:

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

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:

${\begin{aligned}7.48181818\ldots &=7.3+0.18181818\ldots \\\\&={\frac {73}{10}}+{\frac {18}{99}}={\frac {73}{10}}+{\frac {9\times 2}{9\times 11}}={\frac {73}{10}}+{\frac {2}{11}}\\\\&={\frac {11\times 73+10\times 2}{10\times 11}}={\frac {823}{110}}\end{aligned}}$

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 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.

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 = 3786⁄4995 = 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 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 a sequence of 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.

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

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 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....

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 0.999...

Main article: 0.999...

A proof that 1 = 0.999..., using the method of calculating fractions from repeating decimals, follows these steps.

Failed to parse (unknown function "\begin{alignat}"): {\displaystyle \begin{alignat}2 x &= 0.999\ldots\\ 10x &= 9.999\ldots\\ 10x - x &= 9.999\ldots - 0.999\ldots\\ 9x &= 9\\ x &= 1\\ \end{alignat}}

The second step is not 10x = 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.

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,

$\lim _{a\rightarrow \infty }S_{a}=0.999\ldots$

So applying this on the sum of the geometric series:

$\lim _{a\rightarrow \infty }0.9{\frac {10^{-a-1}-1}{10^{-1}-1}}=0.9{\frac {-1}{-0.9}}$

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

Therefore

$0.999\ldots =1.\,$

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

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

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

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

It follows that

$\lim _{n\to \infty }{10^{n}-1 \over 10^{n}}=0.999\dots \,.$

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

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 10ⁿ, 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, 3⁄4 = 0.75 = 0.75000... = 0.74999... .

See also

• 142857
• Decimal sequences for cryptography
• Cyclic number
• Midy's theorem

External links

• Mathworld