I'm a PreCalculus student trying to find a rigorous proof that $\displaystyle\frac{1}{3} = 0.333\ldots$, but I couldn't find it. I think (just think) that this proof would start by proving that

$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = \frac{1}{3}$. My guesses (assuming that proving that $\displaystyle\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i$ converges is trivial):

$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = 3\cdot\sum_{i = 1}^{\infty}10^{-i} = 3\cdot\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i = 3\cdot\left(\frac{1}{1 - \frac{1}{10}}-1\right) = 3\cdot\left(\frac{10}{9}-1\right) = \frac{1}{3}$.

Questions: is this completely rigorous? Which flaws could be found in this proof? How can I improve it?

PS. I'm not sure how to tag this. Feel free to edit, if necessary.


3 Answers 3


Since $\sum_{i=1}^\infty 3\cdot 10^{-i}$ is what the notation "$0.333...$" means, your argument is perfectly good. It's not just the "start of a proof", it is all there is to it.

Okay, perhaps it is not really trivial to prove that the geometric series converges, but straightforward it is. Just plug in the definition of the sum of a series and crank the handle, using standard tricks to rewrite each of the partial sums in turn.

  • $\begingroup$ Sorry, but I didn't understand what is wrong in using the fact of $10x - x = 3,333\ldots - 0,333\ldots = 3$. It seems like abuse of notation, but I'm not sure. Could you clarify, please? $\endgroup$
    – Ian Mateus
    Jun 13, 2012 at 22:24
  • $\begingroup$ @Ian: The problem with it is that if you want to go that way, you first need to prove that arithmetic on infinite decimals actually works the way you're implicitly using it here. In order to prove that you first need a definition of what those things you're adding and multiplying are in the first place, and there the infinite series (or minor variations on that theme) is essentially the only idea that works. It's doable, but doing it rigorously involves much more work than just showing $0.333...=\frac13$ directly. $\endgroup$ Jun 13, 2012 at 22:31
  • 1
    $\begingroup$ $3.333\ldots = 3\cdot\sum_{i=0}^{\infty}10^{-i} = 3 + 3\cdot\sum_{i=1}^{\infty}10^{-i}$ and $0.333\ldots = 3\cdot\sum_{i=1}^{\infty}10^{-i}$. Subtracting gives us the term $\left(3\cdot\sum_{i=1}^{\infty}10^{-i} - 3\cdot\sum_{i=1}^{\infty}10^{-i}\right)$ which would be $\infty - \infty$ (undefined) if $\sum_{i=1}^{\infty}10^{-i}$ diverges. Is it a "reason", nonsense or just irrelevant? $\endgroup$
    – Ian Mateus
    Jun 13, 2012 at 22:44
  • $\begingroup$ @Ian: Hmmm... you make a good case. Do you have a result available that allows you to subtract series term by term here? $\endgroup$ Jun 13, 2012 at 22:50
  • $\begingroup$ If I got it, defining $3.333\ldots = 3 + 3\cdot\sum_{i=1}^{\infty}10^{-i}$, $0.333\ldots = 3\cdot\sum_{i=1}^{\infty}10^{-i}$, and proving convergence of $3\cdot\sum_{i=1}^{\infty}10^{-i}$ makes it rigorous, right? $\endgroup$
    – Ian Mateus
    Jun 13, 2012 at 22:56

Your proof is correct. Essentially you proved it was the convergence of the geometric series.

Alternatively if you want to avoid using the geometric series formula and just use a very simple limit argument, note that , $(3)(.333...) = .999... = 1$ $\text{ }$ To see this, note that $.999... = \sum_{i = 1}^\infty \frac{9}{10^i}$ converges to $1$. Take a look at partial sums. Thus $.333... = \frac{1}{3}$.


You make some assumptions.

$\sum\limits_{i=0}^{\infty} 3*10^{-i}$. You first assumption is that you assume this string of symbols is meaningful and has a consistent finite real value.

For a pre-calculus class I'll give that to you. We may assume that every finite real number can be written as a decimal and every decimal, even infinitely written ones (infinite in the directions of the negative powers; NOT infinite in the direction of the positive powers) represent a finite real numbers. This is actually not trivial but requires more sophistication than a student will have now[1].

$\sum\limits_{i=0}^{\infty} 3*10^{-i}= 3\sum\limits_{i=0}^{\infty} 10^{-i}$. It is an assumption that you can factor out from an infinite sum. I'll give you that. It's reasonable and it'd be absolutely impossible to prove[2].

$\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i = \left(\frac{1}{1 - \frac{1}{10}}-1\right)$ That's a big assumption. I'm not sure when and how that was proven to you. I don't think I'm willing to give you that one[3].

From there on out the proof is fine.

I'd rather see:

Let $k = \sum\limits_{i=1}^{\infty} 10^{-i}$. Then $10k = 10\sum\limits_{i=1}^{\infty} 10^{-i} = \sum\limits_{i=1}^{\infty}(10*10^{-i})=\sum\limits_{i=1}^{\infty}10^{-i + 1}= \sum\limits_{k=0}^{\infty}10^{-k} = 10^{-0} + \sum\limits_{k=1}^{\infty}10^{-k} = 1 + \sum\limits_{i=1}^{\infty}10^{-i} = 1 + k$.

So $10k = 1 + k$ so $k = \frac 19$. $0.33333 = 3k = 3*\frac 19=\frac 13$.


[1]. The definition of the real numbers is that the real numbers are a number system where if you have any infinite set of real numbers, but there is some value that is larger than all the numbers in the set. Then there is some specific value, called the "least upper bound" which is the smallest real number that is equal or larger than all of the numbers int the set.

So for instance $\{1-\frac 12, 1-\frac 12 + \frac 14 - \frac 18, 1-\frac 12 +\frac 14 -\frac 18 + \frac 1{16} - \frac 1{32},... \}$ are all less that $1$. So by the definition of reals there must be some number $k$ so that $k$ is equal or bigger than all of them, but $k$ is the smallest number that is. As it turns out, $k = \frac 23$. $\frac 23$ is bigger than all of them, but any number smaller than $\frac 23$ will have some terms larger than it.

So consider $a_i*10^{-i}$ were $a_i \in\{0,1,2,3,4,5,6,7,8,9\}$. Then $0\le a_i*10^{-i}< 10^{-i+1}$. So if we have an infinite list of $a_i$ terms and we consider the infinite set of $\{\sum\limits_{i=1}^m a_i*10^{-i}\}$. Every one of those terms is less than $1$ so there is some value that is the least term that is equal or larger than all of them. We indicate that number as $\sum\limits_{i=1}^\infty a_i*10^{-i}$ or $.a_1a_2a_3......$.

[2]. If all $\sum\limits_{i=1}^m a_i*10^{-i}\le k$ then then all $\sum\limits_{i=1}^m j*a_i*10^{-i}\le j*k$. And is $k$ is the least upper bound of $\{\sum\limits_{i=1}^m a_i*10^{-i}\}$ than $jk$ must be the least upper bound of $\{\sum\limits_{i=1}^m j*a_i*10^{-i}\}$. (Anything smaller then $j*k$ won't be an upper bound.) So $j*k = j(\sum\limits_{i=1}^\infty a_i*10^{-i}) = \sum\limits_{i=1}^\infty j*a_i*10^{-i}$.

[3] This involves infinite borrowing... I'm not willing to accept that.


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