# Recurring decimal

A recurring decimal is an expression representing a real number in the decimal numeral system, in which after some point the same sequence of digits repeats infinitely many times. The repetition may begin before, at, or after the decimal point. The repeating sequence may consist of just one digit or of any finite number of digits. If the repeating sequence is merely a repeating "0", then the decimal is said to terminate because it is not necessary to explicitly write that there is a repeating "0". Such terminating decimals represent rational numbers whose fractions in lowest terms are of the form k/(2n5m).

One convention to indicate a recurring decimal is to put a horizontal line above the repeated numerals. Another convention is to place dots above the numerals. 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:

• 1/9 = 0.111111111111...
• 1/7 = 0.142857142857...
• 1/3 = 0.333333333333...
• 1/81 = 0.0123456790...
• 2/3 = 0.666666666666...
• 7/12 = 0.58333333333...
 Contents

## 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 recurring decimal. The period of the recurring 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.

This is to say:

• 106 − 1 = 999,999 (6 digits of 9) is divisible by 7;
• 1016 − 1 = 9,999,999,999,999,999 (16 digits of 9) is divisible by 17;
• 1018 − 1 = 999,999,999,999,999,999 (18 digits of 9) is divisible by 19; etc.

These can be deduced from Fermat's little theorem.

It can also be generalised to say that p − 1 digits of 1 (or 2, 3, 4, 5, 6, 7, 8, 9) is divisible by p, which is a prime number other than than 2 or 5.

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

Decimals of other prime fractions such as 1/17, 1/19, 1/23, 1/29, 1/47, 1/59, 1/61, 1/97, 1/109 all 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...

## Calculating the fraction

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

  x = 0.333333...
10x = 3.33333...  (multiplying each side of the above line by 10)
9x = 3           (subtracting the 1st line from the 2nd)
x = 3/9 = 1/3   (simplifying)


Another example:

   x = 0.18181818...
100x = 18.181818...
99x = 18
x = 18/99 = 2/11


From this kind of argument, we can see that the period of the repeating decimal of a fraction n/d will be (at most) the smallest number k such that 10k − 1 is divisible by d.

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

## Why rational numbers must have repeating or terminating decimal expansions

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 ) 5.000000
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.... We eventually see a remainder that we have seen earlier because only finitely many different remainders -- in this case 74 possible remainders: 0, 1, 2, ..., 73 -- can occur. As soon as we only bring down zeros, the same remainder implies the same new digit in the result and the same new remainder. Therefore the whole sequence repeats itself, again and again.

## The case of 0.99999...

The method of calculating fractions from repeated decimals, especially the case of 1 = 0.99999..., is sometimes contested by the mathematically naive:

      x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1


Some argue that, in the second step of the equation given above, 10x is 9.999...0 and not 9.999... but this is not the case: the right-hand side does not terminate (it is recurring) 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 progression. Where:

[itex]S_a = \sum_{n=0}^{a} \frac{0.9}{10^n}[itex]
[itex]S_a = 0.9 \sum_{n=0}^{a} \frac{1}{10^n}[itex]

By standard result:

[itex]S_a = 0.9 \frac{10^{-n} - 1}{10^{-1}-1}[itex]

From definition:

[itex]\lim_{a \rightarrow \infty} S_a = 0.99999 \ldots[itex]

So applying this on the sum of the geometric series:

[itex]\lim_{a \rightarrow \infty} 0.9 \frac{10^{-n} - 1}{10^{-1}-1} = 0.9 \frac{-1}{-0.9}[itex]
[itex]0.9 \frac{-1}{-0.9} = 1[itex]

Therefore:

[itex].99999 \ldots = 1[itex]

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

[itex]x = {10^n-1 \over 10^n}[itex]
[itex]n = 1: x = {9 \over 10} = 0.9[itex]
[itex]n = 2: x = {99\over 100} = 0.99[itex]

It follows that

[itex]\lim_{n \to \infty}{10^n-1 \over 10^n} = 0.9999...[itex]

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 number with a finite decimal expression (a decimal fraction) can be written in a second way as a recurring decimal.

For example 3/4 = 0.75 = 0.750000000... = 0.74999999 ...

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