Repeating decimal

Recurring decimals are a way of representing as decimals certain fractions which are not, in lowest terms, of the form p/(2ⁿ5^m). These decimal representations include an infinitely repeated pattern at the end of the fraction (this repeated pattern may be as short as a single digit).

To indicate the part of the sequence that extends infinitely, dots should be placed above the numerals to be repeated. Where this is 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...
• 2/3 = 0.666666666666...
• 7/12 = 0.58333333333...

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)

9x = 3 (subtracting the 1st line from the second)

x = 1/3 (simplifying)

Another example:

x = 0.18181818...

100x = 18.181818...

99x = 18

x = 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.

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 .99999...

The method of calculating fractions from repeated decimals, especially the case of 1 = .99999..., is sometimes contested by the mathematically naive:

      x = .99999...
    10x = 9.9999...
10x - x = 9.9999... - .99999...
     9x = 9
      x = 1

Some argue that, in the second step of the equation given above, 10x is 9.9999...0 and not 0.999... but this is not the case; the RHS does not terminate (it is recurring) and so there is no end to which a zero can be appended.

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

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

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

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

It follows that

$\lim _{n\to \infty }{10^{n}-1 \over 10^{n}}=.9999...$

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...

See also: Decimal