Exponential decay 'proof' that $.\overline{9}\neq 1$?

Question

I have doubts about $.\overline{9}$ being equal to 1 due to the following proof:

To get a decimal containing $c$ 9's after the decimal point, the equation f(c) = $1-10^{-c}$ can be used.

For example, f(3) = $1-10^{-3} = 1 - .001 = .999$

With this function, $.\overline{9}$ (or a number containing infinite 9's after the decimal point) can be modeled by $f(\infty)$. This also makes $.\overline{9} = f(\infty)$

This, of course, cannot be evaluated directly. No attempts will be made to perform arithmetic operations using infinity.

To model the difference between 1 and a number with a varying amount of nines, this equation can be used:

$1 - f(c)$

This can be simplified as follows:

$1 - (1 - (10^{-c}))$

$1 - (1 - 10^{-c})$

$1 - 1 + 10^{-c}$

$10^{-c}$

This clearly follows the format of an exponential decay function. This means that the original function of $1 - f(c)$ also follows the rules of exponential decay.

There are two main properties of exponential decay functions relative to this proof:

1. the domain stretches from $-\infty$ to $+\infty$

2. the asymptote is 0 (that is, it approaches but never reaches 0)

Since this function does not reach 0, even at infinite bounds, that would mean that $1 - f(\infty) \neq 0$

From here, $f(\infty)$ can be replaced with $.\overline{9}$ as shown before. The equation becomes $1 - .\overline{9} \neq 0$

A final simplification by adding $.\overline{9}$ to both sides yields:

$1 \neq .\overline{9}$

This seems to refute many $.\overline{9} = 1$ proofs, but not having an extensive math education, I do not know if there is anything wrong with the proof. It seems as if there may be something very wrong with the proof, as basically everyone agrees that $.\overline{9} = 1$. Is there a problem, and if so, what?

Answer 1

The statement beginning "Since this function does not reach 0, even at infinite bounds..." is nonsensical. That's one big flaw in the "proof" since your entire conclusion hinges on it.

In fact, exactly the opposite is true: $$\lim_{n \to\infty}(1-f(n)) =0$$

Answer 2

You either accept that you can treat infinity as a value or you do not. It is true that $10^{-c}>0$ for all finite $c$. That said, we may make $10^{-c}$ as close to $0$ as desired by taking sufficiently large $c$. In particular, if you do choose to give $f(+\infty)$ any relevant meaning, it must be that $f(+\infty)=0$.

Answer 3

If $0.\overline{9}$ equals anything it must equal $1$. To see this we will use the $f$ given in the proof, i.e. $f(n)=1-10^{-n}$. As stated in the proof, for a given $n$, $f(n)=\underbrace{.9999...9}_{n \text{ times}}$

Since $10^{-n}\gt 0$ for all $n$, $f(n)\lt 1$ and so $.\overline{9}\le 1$. But for any real number $r\lt 1$ we can find $m$ large enough so that $f(m)=\underbrace{.9999...9}_{m \text{ times}}\gt r$. Since we can do that for all $r\lt 1$, it follows that $0.\overline{9}\ge 1$

Therefore, $0.\overline{9} = 1$

Answer 4

You have shown that the difference between $1$ and $f(c)$ is $10^{-c}$. This leads to the following "exchange":

I claim that $1 = 0.\overline{9}$. You say no, this is not the case: we have $1 \ne 0.\overline{9}$. In fact, we have $1 > 0.\overline{9}$. We can write this alternatively as: $$ 1 = 0.\overline{9} + \varepsilon $$ for some nonzero $\varepsilon$.

But I can find a $c > 0$ so that $1 - f(c) < \varepsilon$. This is easy, because $$ 1 - f(c) = 10^{-c}, $$ so if I take $c > - \log \varepsilon$, it follows that \begin{align*} -c &< \log \varepsilon \\ 10^{-c} &< \varepsilon. \end{align*} Why is this bad? Notice: \begin{align*} 10^{-c} &< \varepsilon \\ 1 - (1 - 10^{-c}) &< \varepsilon \\ 1 - f(c) &< \varepsilon \\ 1 &< f(c) + \varepsilon. \end{align*} But you claimed that $1 = 0.\overline{9} + \varepsilon$, which clearly contradicts the last line!

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This is the problem. You're right when you say that $10^{-c}$ is positive for all $c$. However, no matter what, $10^{-c}$ can be forced to be smaller than any choice of offset (in the proof above, this was what $\varepsilon$ represents). $10^{-c}$ is eventually smaller than every positive real number. So as $c \to \infty$, $10^{-c} \to 0$, because $0$ is smaller than every positive real number. As a consequence, $f(c) \to 1$.