Decimal representation

A real number of the form

$r=\sum _{i=0}^{\inf }{\frac {a_{i}}{10^{i}}}$

where $a_{0}$ is a nonnegative integer and $a_{1}$, $a_{2}$,... are integers satisfying $0\leq a_{i}\leq 9$, is usually written more briefly as follows:

$r=a_{0}.a_{1}a_{2}\cdots$

This is said to be a decimal representation of r.

Finite decimal approximations

Real numbers 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 finte decimal $r_{n}=a_{0}.a_{1}a_{2}\cdots a_{n}$ such that

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

Proof:

Let S be the set of all nonnegative integers $\leq x$. Then S is nonempty, since $0\in S$, and S is bounded above by x. Therefore S has a supremum, say $a_{0}=supS$. It is easily verified that $a_{0}\in S$, so $a_{0}$ is a nonnegative integer. We call $a_{0}$ the greatest integer in x, and we write $a_{0}=[x]$. Clearly, we have$a_{0}\leq x<a_{0}+1$.

Now let $a_{1}=[10x-10a_{0}]$, the greatest integer in $10x-10a_{0}$. Since $0\leq 10x-10a_{0}=10(x-a_{0})<10$, we have $0\leq a_{1}\leq 9$ and $a_{1}\leq 10x-10a_{0}<a_{1}+1$. In other words, $a_{1}$ is the largest integer satisfying the inequalities

$a_{0}+{\frac {a_{1}}{10}}\leq x<a_{0}+{\frac {a_{1}+1}{10}}$.

More generally, having chosen $a_{1},...,a_{n-1}$ with $0\leq a_{i}\leq 9$, let $a_{n}$ be the largest integer satisfying the inequalities

$a_{0}+{\frac {a_{1}}{10}}+\cdots +{\frac {a_{n}}{10^{n}}}\leq x<a_{0}+{\frac {a_{1}}{10}}+\cdots +{\frac {a_{n}+1}{10^{n}}}$.

Then $0\leq a_{n}\leq 9$ and we have

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

where $r_{n}=a_{0}.a_{1}a_{2}\cdots a_{n}$.

It is easy to verify that x is actually the supremum of the set of rational numbers $r_{1}$,$r_{2}$,....

Verification of $a_{0}\in S$

By the approximation property of the supremum of a set of real numbers, for every z>0, there exists x in S such that $a_{0}-z<x$. Therefore, $a_{0}\leq x$ and then $a_{0}\in S$ for $x\leq a_{0}$.

Verify that x is the supremum of the set of rational numbers r₁, r₂, ...

For every $i\in {\mathcal {R}}$, $r_{i}\leq x$, or x is the upper bound of the set of rational numbers r₁,r₂,.... Suppose that there is a real number y such that $r_{i}\leq y$ for every $i\in {\mathcal {R}}$ and $y<x$. Thus, $0<x-y<1/10^{n}$ and then $0<x-y\leq 0$.This is a contradiction.Therefore, x is the least upper bound, or the supremum.

Finite decimal representions

The decimal expansion of 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 nonnegative integers.

Proof:

If the decimal expansion of x will end in zeros, or $x=\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}=\sum _{i=0}^{n}10^{n-i}a_{i}/10^{n}$ 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, $x={\frac {p}{2^{n}5^{m}}}={\frac {2^{m}5^{n}p}{2^{n+m}5^{n+m}}}={\frac {2^{m}5^{n}p}{10^{n+m}}}$ for some p. While x is of the form p/10^k, $p=\sum _{i=0}^{n}10^{i}a_{i}$ for some n. By $x=\sum _{i=0}^{n}10^{n-i}a_{i}/10^{n}=\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}$, x will end in zeros.