Decimal representation

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This article gives a mathematical definition. For a more accessible article see Decimal.

A decimal representation of a non-negative real number r is an expression of the form

$r=\sum_{i=0}^\infty \frac{a_i}{10^i}$

where $a 0$ is a nonnegative integer, and $a_1, a_2, \dots$ are integers satisfying $0\leq a_i\leq 9$ ; this is usually written more briefly as

$r=a_0.a_1 a_2 a_3\dots.$

That is to say, $a 0$ is the integer part of $r$ , not necessarily between 0 and 9, and $a_1, a_2, a_3,\dots$ are the digits forming the fractional part of $r .$

1 Finite decimal approximations
2 Multiple decimal representations
3 Finite decimal representations
4 Recurring decimal representations
5 See also
6 External links

[edit] Finite decimal approximations

Any real number can be approximated to any desired degree of accuracy by rational numbers with finite decimal representations.

Assume $x\geq 0$ . Then for every integer $n\geq 1$ there is a finite decimal $r_n=a_0.a_1a_2\cdots a_n$ such that

$r_n\leq x < r_n+\frac{1}{10^n}.\,$

Proof:

Let $r_n = \textstyle\frac{p}{10^n}$ , where $p = \lfloor 10^nx\rfloor$ . Then $p \leq 10^nx < p+1$ , and the result follows from dividing all sides by $10 n$ . (The fact that $r n$ has a finite decimal representation is easily established.)

[edit] Multiple decimal representations

Main article: Proof that 0.999... equals 1

Some real numbers have two infinite decimal representations. For example, the number 1 may be equally represented by 1.00000... as by 0.99999... (where for the sake of brevity the infinite sequences of digits 0 and 9, respectively, have been replaced by "..."). Conventionally, the version with zero digits is preferred; by omitting the infinite sequence of zero digits, removing any final zero digits and a possible final decimal point, a normalized finite decimal representation is obtained.

[edit] Finite decimal representations

The decimal expansion of non-negative real number x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form 2ⁿ5^m, where m and n are non-negative integers.

Proof:

If the decimal expansion of x will end in zeros, or $x=\sum_{i=0}^n\frac{a_i}{10^i}=\sum_{i=0}^n10^{n-i}a_i/10^n$ for some n, then the denominator of x is of the form 10ⁿ = 2ⁿ5ⁿ.

Conversely, if the denominator of x is of the form 2ⁿ5^m, $x=\frac{p}{2^n5^m}=\frac{2^m5^np}{2^{n+m}5^{n+m}}= \frac{2^m5^np}{10^{n+m}}$ for some p. While x is of the form $\textstyle\frac{p}{10^k}$ , $p=\sum_{i=0}^{n}10^ia_i$ for some n. By $x=\sum_{i=0}^n10^{n-i}a_i/10^n=\sum_{i=0}^n\frac{a_i}{10^i}$ , x will end in zeros.