Recurring decimal

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A recurring or repeating decimal is a number which when expressed as a decimal has a set of "final" digits which repeat an infinite number of times. For instance 1/3 = 0.3333333... (spoken as "0.3 recurring").

More fully: a recurring decimal is a rational number whose expression in the (decimal numeral system) has some point after which the same sequence of digits repeats infinitely-many times. The repetition may begin before, at, or after the decimal point and the repeating sequence may consist of just one digit or of any finite number of digits.

A decimal number written with a repeating final 0 is NOT classed as a recurring decimal, and the decimal is said to terminate before the first final 0 (because it is not necessary to explicitly write that there is a repeating "0".

There is one special case where it is not necessary to express the recurrence, but sometimes useful to do so. That is where the repeating part is a single digit 9. In these cases, the final 9s may be omitted, and the previous digit increased by one. (Eg 0.999999... = 1 ; 1.77999999... = 1.78). In general, the recurring form is only used to show how the number was derived, or to express an interesting relationship (1 = 3/3 = 3 × 1/3 = 3 × 0.333333... = 0.99999...). See 0.999....

The other types of decimals are terminating decimals and decimals which neither terminate nor repeat.

Terminating decimals represent rational numbers whose fractions in lowest terms are of the form k/(2ⁿ5^m). A rational number is one that can be expressed in the form a/b where a and b are integers (b must of course be non-zero, as division by zero is undefined), that is, a vulgar fraction.
A non-terminating non-recurring decimal represents an irrational number (which cannot be expressed as the ratio of two integers).

1 Notation
2 Fractions with prime denominators
3 Why all repeating or terminating decimals must be rational numbers
- 3.1 A shortcut
4 Repeating decimals as an infinite series
5 Why all rational numbers must have repeating or terminating decimal expansions
6 The case of 0.99999...
7 See also
8 External links

[edit] Notation

One convention to indicate a recurring decimal is to put a horizontal line above the repeated numerals ( $1/3 = 0.\overline{3}$ ). Another convention is to place dots above the outermost numerals of the recurring 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:

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

Another notation, used for example in Europe, encloses the recurring digits in brackets:

2/3 = 0.(6)
1/7 = 0.(142857)
7/12 = 0.58(3)

[edit] 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 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, 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, 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...

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

[edit] Why all repeating or terminating decimals must be rational numbers

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 10^k − 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.

[edit] A shortcut

If the repeating decimal is between 0.1 and 1, and the repeating block is n digits long occurring right at the decimal point, then the fraction (not necessarily reduced) will be the n-digit block over n digits of 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),
0.789789... = 789/999 since the repeating block is 789 (a 3-digit block), etc.

If the repeating decimal is between 0 and 0.1, and the repeating n-digit block is preceded only by k digits of 0 (all of which are to the right of the decimal point), then the fraction (not necessarily reduced) will be the n-digit block over the integer consists of n digits of 9 followed by k digits of 0. 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,
0.0789789... = 789/9990 since the repeating block is 789 and it is preceded by 1 zero.

For any repeating decimal not prescribed above, it can be written as a sum of a terminating decimal and a repeating decimal of one of the two above types. For example,

1.23444... = 1.23 + 0.00444... = 123/100 + 4/900 = 1107/900 + 4/900 = 1111/900
0.3789789 ... = 0.3 + 0.0789789... = 3/10 + 789/9990 = 2997/9990 + 789/9990 = 3786/9990 = 631/1665

[edit] 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.

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

[edit] Why all 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 ) 5.00000
        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.... 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.

[edit] The case of 0.99999...

Main article: 0.999...

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:

$S_a = \sum_{n=0}^{a} \frac{0.9}{10^n}$

$S_a = 0.9 \sum_{n=0}^{a} \frac{1}{10^n}$

By standard result:

$S_a = 0.9 \frac{10^{-a-1} - 1}{10^{-1}-1}$

From definition:

$\lim_{a \rightarrow \infty} S_a = 0.99999 \ldots$

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:

$.99999 \ldots = 1$

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

$\lim_{n \to \infty}{10^n-1 \over 10^n} = 0.9999\dots$

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 recurring decimal.

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