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?)

  • 1
    $\begingroup$ What's your definition of real numbers? And the answer is yes but the representation isn't unique. $\endgroup$
    – xavierm02
    Commented Jun 2, 2013 at 21:45
  • $\begingroup$ Related. $\endgroup$ Commented Jun 3, 2013 at 3:17

5 Answers 5


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).


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 Concise Introduction to Pure Mathematics, Fourth Edition, by Martin Liebeck, pg 22.

  • 1
    $\begingroup$ I like this answer but shouldn't that +1/10 on the RHS of the last inequality be removed? $\endgroup$
    – Elsa
    Commented Mar 3, 2020 at 3:47
  • $\begingroup$ @Elsa: Yes I agree and have edited it accordingly $\endgroup$
    – user46234
    Commented Mar 4, 2020 at 3:38

The answer is yes. The fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ gives us the start line to reach every real number as a sequence of increasing long sequence of digits. ($ 1, \ 1.4, \ 1.41, \ 1.414, \ \dots \rightarrow \sqrt{2} $)

to compute even more digits we have lots of algorithms.

if your question is "I think about $x$ random real number, what's the decimal representation?" I will answer, "is between $a$ and $a+1$ ($a \in \mathbb{N}$) ? If yes, is between $a,1$ and $a,2$ for example (if no I will ask you for $a,2$ and $a,3$ ...) and so on.

At every step I will build a sequence of digits that converge to your number. :)

Obviously there are different sequences that converge to the same number ($1=0,\overline{9}$) and I use a intuitively means of convergence.

  • 1
    $\begingroup$ It's not the fact that $\mathbb Q$ is dense in $\mathbb R$ though, is it? It's the fact that numbers with a terminating decimal expansion are dense in $\mathbb R$. $\endgroup$
    – TonyK
    Commented Jun 2, 2013 at 22:57
  • $\begingroup$ Ok, but numbers with a terminating (=finite?) decimal expansion are rational no? $\endgroup$
    – Riccardo
    Commented Jun 2, 2013 at 23:12
  • $\begingroup$ Yes, but the converse is not true. $\endgroup$
    – TonyK
    Commented Jun 2, 2013 at 23:15
  • $\begingroup$ Yes of course :) $\endgroup$
    – Riccardo
    Commented Jun 2, 2013 at 23:23
  • $\begingroup$ Aren't the rational numbers exactly the set of numbers with a terminating expansion in some base? $\endgroup$
    – Andrew
    Commented Oct 28, 2015 at 19:55

Irrational numbers were known to the ancient Greeks, as I expect you know. But it took humankind another 2000 years to come up with a satisfactory definition of them. This was mainly because nobody realised that a satisfactory definition was lacking.

Once humankind realised this, various suggestions were proposed. One suggestion (Dedekind's) defined a real number as two infinite sets of rational numbers, which 'sandwiched' the real number; another suggestion (Cauchy's) defined a real number as an equivalence class of sequences obeying a certain convergence criterion. The details are available in many places.

But the important point is that all of the reasonable definitions turned out to be equivalent -- the set of real numbers according to Dedekind's definition was 'the same' as the set of real numbers according to Cauchy's definition, although the definitions look completely different.

Now, to your question: another reasonable definition of a real number is a non-terminating decimal expansion (we say non-terminating just to clear up the ambiguity that arises between e.g. 123.4599999... and 123.46 -- only the first is allowed). It turns out that this definition is equivalent to all the others.

So your intuition is correct. But strictly speaking, your question is flawed: instead of asking whether every real number can be represented in this way, you should ask whether this representation of the real numbers is a valid one. And it is.


Yes, every real number has a decimal representation.

Numbers of the form $n/10^k$ where $n$ and $k$ are integers and $n \neq 0$ have two representations each (e.g. 1.000... and 0.999... represent the same real number). Depending on your conventions, zero has either one or two representations: e.g. $0$ and $-0$ in the latter case. Every other real number has only one decimal representation.

Conversely, every decimal number that is all zeroes on the left is the representation of some real number. e.g. the decimal

$$ ...000011.000... $$

is the number eleven. (recall that any digit that isn't written is zero, so the decimal "11" really has zeroes to its left and right, as notated explicitly above)

In standard notation, decimals that are not all zeroes on the left such as

$$ ...11111.000... $$

do not represent real numbers.

There are generalized notations that allow some left-infinite decimals to represent real numbers (the number above would be -1/9). There are also other number systems that can be represented with decimals, such as the 10-adic numbers. It may be confusing to try and learn about such things at this point, though....

  • $\begingroup$ "It may be confusing to try and learn about such things at this point, though...." Well, exactly. I was wondering why you brought it up here. $\endgroup$
    – TonyK
    Commented Jun 2, 2013 at 22:55
  • 2
    $\begingroup$ @TonyK: Because it serves several useful purposes, such as: (1) it points out that curiosity of this sort is useful for mathematicians, (2) it pre-emptively answers any readers who do think about such things, (3) it quashes the idea that left-infinite decimals are an inherently stupid idea that should never have been brought up, (4) gives the interested reader (possibly including the OP in the future) something to look into, (5) exemplifies the point that notation is what we choose it to be, rather than something with inherent meaning, .... $\endgroup$
    – user14972
    Commented Jun 3, 2013 at 10:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .