2
$\begingroup$

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

$$\sum^{\infty}_{n=0}{\frac{3}{10^n}}$$

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

$\endgroup$
1
  • $\begingroup$ The symbol $\infty$ in this notation reminds of the extra point in the one-point compactification of $\Bbb N$ $\endgroup$
    – Hagen von Eitzen
    Jun 22, 2019 at 17:02

2 Answers 2

Reset to default
6
$\begingroup$

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

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

$\endgroup$
3
  • 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, 2019 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. $\endgroup$
    – Geza Kerecsenyi
    Jun 22, 2019 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, 2019 at 18:30
4
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ I've edited the question to make it a bit clearer. $\endgroup$
    – Geza Kerecsenyi
    Jun 22, 2019 at 16:55
  • $\begingroup$ @GezaKerecsenyi See my edit. $\endgroup$
    – Ethan Bolker
    Jun 22, 2019 at 17:13

You must log in to answer this question.

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