Recurring decimal

Recurring decimals are a way of representing as decimals certain fractions which are not of the form p/(2ⁿ5^m) in lowest terms. These decimal representations include an infinitely repeated pattern at the end of the fraction (this 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...\n* 1/7 = 0.142857142857...\n* 1/3 = 0.333333333333...\n* 2/3 = 0.666666666666...\n* 7/12 = 0.58333333333...

Table of contents

1 Calculating the fraction 2 Why rational numbers must have repeating or terminating decimal expansions 3 The case of .99999...

Calculating the fraction

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

x = 0.333333...\n:10x = 3.33333...\n:9x = 3 so that x = 1/3

x = 0.18181818...\n:100x = 18.181818...\n:99x = 18 so that 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\n    74 ) 5.000000\n         4 44\n           560\n           518\n            420\n            370\n             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.... But why must we eventually see a remainder that we have seen earlier? The answer is that only finitely many different remainders -- in this case 74 possible remainders: 0, 1, 2, ..., 73 -- can occur. We cannot go on using different remainders forever. What has not been shown here is why the decimals need recur in the same pattern again and again. They must repeat, but this is not shown here. For example, the non-repeating decimal expansions 0.101001000100001000001... and 0.787788777888 are neither recurring decimals and they does not terminate but we do not know from this explanation that they are not rational numbers. The explanation is necessary which is why I haven't deleted this section.

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:\n x = .99999...\n 10x = 9.9999...\n 10x - x = 9.9999... - .99999...\n 9x = 9\n x = 1\nSome 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 perhaps more impressive but no more formal proof, consider the formula:\n:\n:\n: \nIt follows that\n: 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 understading of limits. Generalising this, any number with a finite decimal expression (a decimal fraction) also has an expression as a recurring decimal. For example 3/4 = 0.75 = 0.74999999... See also: Decimal