If a digit is written as $3.\dot{3}$, what level of infinity do the dots continue on for? Can this be proven to be true, or is it just a quirk of the notation?

More specifically, $1/3$ can obviously be broken up like


However, I'm wondering what $\infty$ is actually meaning here. Any help is appreciated.

  • $\begingroup$ The symbol $\infty$ in this notation reminds of the extra point in the one-point compactification of $\Bbb N$ $\endgroup$ Jun 22 '19 at 17:02

This means that there are countably infinite $3$'s after the one's digit. We can see this with the base-$10$ expansion of $3.\dot{3}.$ We have

$$3.\dot{3} = 3.333\dotsc = 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dotsb = \sum_{n=0}^{\infty} \frac{3}{10^{n}}.$$

  • 1
    $\begingroup$ So it could also be that there is $3.333\dots 3$? I mean, a sequence of length $\Bbb N\cup\{\Bbb N\}$ is countably infinite... $\endgroup$
    – Asaf Karagila
    Jun 22 '19 at 17:04
  • $\begingroup$ @AsafKaragila wouldn't that mean that it would be possible to rewrite $3.\dot{3}$ as $$\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}$$ in which case $$3\times\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}$$ would be a countable infinite sequence, so $$10 = 2\times\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}+\sum^{|\mathbb{N}|+1}_{n=0}{\frac{3}{10^n}}$$, yet since Euler's proof states that $9.\dot{9}=10$, $1=0$ at the top of the sum function in the last expression. This is the main reason I was confused about this in the first place, because this disproof is obviously wrong, yet so simply to follow. $\endgroup$ Jun 22 '19 at 18:29
  • $\begingroup$ @Geza: My point is that sequences are indexed by ordinals and not cardinals. The question is not "how many digits are there" but "how long is the sequence of digits". $\endgroup$
    – Asaf Karagila
    Jun 22 '19 at 18:30

There is nothing to prove. That notations is just a convention for $$ 3.333\ldots $$ which is in turn just $$ \sum_{i=0}^\infty 3 \times 10^{-i} = 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots $$

The "$\infty$" is not a real number, nor even a cardinal number. It's part of the convention we use to write this sum of countably many real numbers, one for each natural number. You could equally well write $$ \sum_{i \in \mathbb{N}} $$ (since the order of summation does not matter for a sum of positive terms).

  • $\begingroup$ I've edited the question to make it a bit clearer. $\endgroup$ Jun 22 '19 at 16:55
  • $\begingroup$ @GezaKerecsenyi See my edit. $\endgroup$ Jun 22 '19 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.