Is 1 divided by 3 equal to 0.333...? I have been taught that $\frac{1}{3}$ is 0.333.... However, I believe that this is not true, as 1/3 cannot actually be represented in base ten; even if you had infinite threes, as 0.333... is supposed to represent, it would not be exactly equal to 1/3, as 10 is not divisible by 3.
0.333... = 3/10 + 3/100 + 3/1000...

This occured to me while I  discussion on one of Zeno's Paradoxes. We were talking about one potential solution to the race between Achilles and the Tortoise, one of Zeno's Paradoxes. The solution stated that it would take Achilles $11\frac{1}{3}$ seconds to pass the tortoise, as 0.111... = 1/9. However, this isn't that case, as, no matter how many ones you add, 0.111... will never equal precisely $\frac{1}{9}$.
Could you tell me if this is valid, and if not, why not? Thanks!
I'm not arguing that $0.333...$ isn't the closest that we can get in base 10; rather, I am arguing that, in base 10, we cannot accurately represent $\frac{1}{3}$
 A: Here is a simple reasoning that $1/3=0.3333...$.
Lets denote $0.333333......$ by $x$. Then
$$x=0.33333.....$$
$$10x=3.33333...$$
Subtracting we get $9x=3$. Thus $x=\frac{3}{9}=\frac{1}{3}$. 
Since $x$ was chosen as $0.3333....$ it means that $0.3333...=\frac{1}{3}$. 
A: You didn't follow the thread here Is it true that $0.999999999\ldots = 1$?.
Well, $\frac{1}{3}=0.33333\ldots$ 
You can use $1$.Geometric Progression. Or $2$. The one N.S suggested. 
A: This question is similar to show that $0.999\ldots=1$.
I give here a proof and tou can see  Does .99999... = 1?
for another proofs. We have
$$0.999\ldots\leq 1\leq 0.999\ldots+\frac{1}{10^n},\, \forall n\in\mathbb{N},$$
so by passing at limit ($n\to\infty$) we find
$$0.999\ldots\leq1\leq 0.999\ldots$$
which allows us to conclude.
A: You can find the sum of $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$ using the formula of sum of infinite geometric progression.
$$a_1 = \frac{3}{10}$$
$$r=\frac{1}{10}$$
$$\sum =\frac{a_1}{1-r}=\frac{3}{10}\times\frac{10}{9} =\frac{1}{3}$$
A: The problematic part of the question is "no matter how many ones you add, 0.111... will never equal precisely 1/9."
In this (imprecise) context $0.111\ldots$ is an infinite sequence of ones; the sequence of ones does not terminate, so there is no place at which to add another one; each one is already followed by another one. Thus, $10\times0.111\ldots=1.111\ldots$ is precise. Therefore, $9\times0.111\ldots=1.000\ldots=1$ is precise, and $0.111\ldots=1/9$.
I say "imprecise" because we also say $\pi=3.14159\dots$ where ... there means an unspecified sequence of digits following. A more precise way of writing what, in the context of this question, we mean by $0.111\dots$ is $0.\overline{1}$ where the group of digits under the bar is to be repeated without end.
In this question, $0.333\ldots=0.\overline{3}$, and just as above, $10\times0.\overline{3}=3.\overline{3}$, and therefore, $9\times0.\overline{3}=3.\overline{0}=3$, which means $0.\overline{3}=3/9=1/3$.
A: The following uses the limiting concept to prove that $0.33333=\frac{1}{3}$
$$3+0.33333...=\sum_{i=0}^N\frac{3}{10^n}\quad\text{where N}\to\infty$$
$$=3\sum_{n=0}^N\frac{1}{10^n}$$
$$=3\sum_{n=0}^N(\frac{1}{10})^n$$
Known
$$1+a^2+a^3....a^n=\frac{a^{n+1}-1}{a-1}$$
So
$$=3\cdot \frac{(\frac{1}{10})^{N+1}-1}{\frac{1}{10}-1}$$
$$=3\cdot \frac{(\frac{1}{10})^{N+1}-1}{\frac{1-10}{10}}$$
$$=3\cdot \frac{((\frac{1}{10})^{N+1}-1)*10}{1-10}$$
$$=3\cdot \frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$
Now apply the limiting concept, if $N\to\infty;(\frac{1}{10})^{N+1}\to0$
$$=3\cdot \lim_{N\to\infty}\frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$
$$=3\cdot \frac{10}{9}$$
$$=\frac{10}{3}$$
$$\Rightarrow 3+0.33333...=\frac{10}{3}$$
$$\Rightarrow 0.33333...=\frac{10}{3}-3$$
$$\Rightarrow 0.33333...=\frac{10-9}{3}$$
$$\Rightarrow 0.33333...=\frac{1}{3}$$
