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 $\frac{1}{3}$

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 $N\to\infty;(\frac{1}{10})^{N+1}\to0$
$$=3\cdot \lim_{N\to\infty}\frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$
$$=3\cdot \frac{10}{9}$$
$$=\frac{10}{3}$$
$$\Rightarrow 3+0.33333...=\frac{10}{3}$$
$$\Rightarrow 0.33333...=\frac{10}{3}-3$$
$$\Rightarrow 0.33333...=\frac{10-9}{3}$$
$$\Rightarrow 0.33333...=\frac{1}{3}$$
