1

I am trying to prove that $.\dot 9 = 1$. I've come up with something that seems intuitively like a proof, but I believe it is structured incorrectly. Furthermore, it contains a term that is undefined. I think this "proof" may only intuitively prove the above statement, but it may be mathematically nonsense.

Here it is:

$$1-0.9 = \frac{1}{10^1}$$ $$1-0.99 = \frac{1}{10^2}$$ $$1-0.999 = \frac{1}{10^3}$$ $$...$$ $$1-0.\overline{99} = \frac{1}{\infty}$$

Or, in other terms:

$$1-\sum_{k=1}^n \frac{9}{10^k} =\frac{1}{10^n}$$

The final $1-0.\overline{99} = \frac{1}{\infty}$ is supposed to show that the difference is nothing, and therefore that $.\dot9 = 1$, but $\frac{1}{\infty}$ is not $0$, but rather undefined. However, even though that last term is undefined, maybe it's a way of showing that $\frac{1}{10^k} \rightarrow 0$ as $k \rightarrow \infty$?

Now, I believe there are other problems with this proof beyond the use of an undefined term. Problems pertaining to its structure. So, to anyone "critiquing" this, could you look beyond the undefined term and show what more is wrong with the proof. Let's say something divided by infinity isn't undefined; how is this proof still invalid?

3 Answers3

3

Yes, the way your proof would be formalized is by saying $1 - \sum_{k = 1}^{n}9 \cdot 10^{-k} = 10^{-n}$, and then showing $\lim_{n \to \infty}10^{-n} = 0$. The "$\infty$" can be formalized by the fact that $\frac{1}{10^n} \to 0$ if and only if $10^n \to \infty$.

Mason
  • 10,415
  • I see, not very familiar with the limit function, but I see now that this is how one formalizes it. Is this proof the most common one? I feel like it was a pretty obvious way to approach the problem, but maybe there is a different, perhaps simpler way? –  Sep 10 '21 at 02:24
  • @ChemLad The most common way to evaluate a sum of the form $\sum_{n = 0}^{\infty}r^n$ is to just use the geometric series theorem which states that if $r \in \mathbb{C}$, $\sum_{n = 0}^{\infty}r^n$ converges if and only if $|r| < 1$, and the limit is $\frac{1}{1 - r}$. The proof is fairly elementary (compute the partial sums then take the limit) and can be found in most calculus or analysis books. – Mason Sep 10 '21 at 03:45
0

We have that the difference between $1$ and $0.999\ldots 9$ is $\frac{1}{10^n}$.

Since as $n$ gets larger, we have $\lim_{n\to\infty} \frac{1}{10^n}=0$, we have that the sequence converges to $1$.

Alan Abraham
  • 5,142
  • 6
  • 20
-1

There are number systems where this type of argument can be made rigorously, but in these number systems, it is no longer true that $0.\bar{9} = 1$. For instance, in the hyperreal numbers $^*\Bbb{R}$, the expressions $$0.99999... \text{ and } 1.00000...$$ represent legitimately different numbers, their difference really is an infinitesimal, and this infinitesimal really is the reciprocal of an infinite number. However, after taking standard parts, we learn that $0.\bar{9}$ and $1$ "round off" to the same standard real number, namely $1$, and so this still amounts to a proof that $0.\bar{9} = 1$ considered as elements of $\Bbb{R}$.

Rivers McForge
  • 5,787
  • It's more subtle than this - see https://math.stackexchange.com/questions/3686843/hyperreals-other-models-and-1-0-999 In the hyperreals, 0.9 recurring is either ill-defined (if you consider it as a sum over standard naturals n of $9/10^n$) or 1 (if you sum over $\mathbb{N}^*$. – Matthew Towers May 29 '23 at 08:21