Can every real number be represented by a (possibly infinite) decimal?

Does every real number have a representation within our decimal system? The reason I ask is because, from beginning a mathematics undergraduate degree a lot of 'mathematical facts' I had previously assumed have been consistently modified, or altogether stripped away. I'm wondering if my subconscious assumption that every real number can be represented in such a way is in fact incorrect?

If so, is there a proof? If not, why not?

(Also I'm not quite sure how to tag this question?)

Solution 1:

Suppose $x$ is a positive real number. It can be shown that $n\mapsto 10^n$ is a map that is unbounded above in the integers, so by Archimedean property, there is some integer $n$ with $10^{n+1}\ge x$. Take the least such $n$ (why must one exist?), and let $a_{-n}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $a\cdot 10^n<x$ (why must one exist?). Now, given $a_{-n},a_{-n+1},...,a_{m}$ for some $m\in\Bbb Z$ with $m\ge-n,$ we let $a_{m+1}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $$a\cdot 10^{-(m+1)}<x-\sum_{j=-m}^na_{-j}\cdot10^j$$ (why must one exist?) Recursively, this determines a sequence $a_{-n},a_{-n+1},...$ of elements of $\{0,1,2,...,8,9\}$ such that for all integers $m\ge-n$ we have $$S_m:=\sum_{j=-m}^na_{-j}\cdot 10^j<x.$$ In fact, $S_{-n},S_{-n+1},...,S_m,...$ is a non-decreasing sequence of positive numbers, so since bounded above by $x,$ this sequence converges to some number no greater than $x$ by the Monotone Convergence Theorem. We can even do better than that, and show that the sequence of partial sums $S_n$ converges to $x$ (why?). The series thus determined is the infinite decimal expansion of $x$.

If $x$ had been negative, we could acquire a decimal expansion for $-x$ in this way, and then the opposite of that would be a decimal expansion for $x$. $0$ has a decimal expansion, too, so all real numbers have a decimal expansion, and moreover, all of them (except arguably $0$) have an infinite decimal expansion (though some also have a finite decimal expansion).

Solution 2:

Every real number $x$ has a decimal expression

$$x = a_0.a_1a_2a_3\ldots. $$

Proof: Picture $x$ on the real line. Certainly $x$ lines between two consecutive integers; let $a_0$ be the lower of these, so that

$$ a_0 \le x < a_0+1.$$

Now divide the line between $a_0$ and $a_0+1$ into ten equal sections. Certainly $x$ lies in one of these sections, so we can find $a_1$ between $0$ and $9$ such that

$$ a_0 + \frac{a_1}{10} \le x < a_0 + \frac{a_1+1}{10} .$$

Similarly, we can find $a_2$ such that

$$ a_0 + \frac{a_1}{10} + \frac{a_2}{10^2} \le x < a_0 + \frac{a_1}{10} + \frac{a_2+1}{10^2} ,$$

and so on. If we do this enough times, the sum $a_0 + \frac{a_1}{10} + \frac{a_2}{10^2} + \ldots$ gets as close as we like to $x$.

From A Conscise Introduction to Pure Mathematics, Fourth Edition, by Martin Liebeck, pg 22.