# $$.9\overline{9} = 1?$$

# <center>Why every proof you have seen is WRONG</center>

Don't worry, **it is true**. Just the "proofs" you've seen are wrong.

# First, why the common "proofs" are wrong
# Then, an actual proof

# Algebraic "proofs" are ALL wrong

$$
\begin{align}
x &:= .9\overline{9} &&&1/3 &= .3\overline{3} \\
10x &= 9.9\overline{9} &\text {or}&& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
9x &= 9 &&& 1 &= .9\overline{9}\\
x &= 1
\end{align}
$$
All algebraic proofs are wrong because there are algebras where $.9\overline{9} \neq 1$.

$$
\begin{align}
x &:= \bf\color{red}{.9\overline{9}} &&&1/3 &= \bf\color{red}{.3\overline{3}} \\
10x &= 9.9\overline{9} &\text {or}&& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
9x &= 9 &&& 1 &= .9\overline{9}\\
x &= 1
\end{align}
$$
You have already assumed part of the conclusion, that $\color{red}{.9\overline{9}}$ or $\color{red}{.3\overline{3}}$ even *exists*.

$$
\begin{align}
x &:= \bf\color{red}{\overline{9}.} \\
10x &= \overline{9}0.  \\
10x+9 &= x\\
9x &= -9\\
x &= -1
\end{align}
$$
So do you believe that $...99999999.0 = -1$?

$$\color{green}{
\begin{align}
\color{black}{x} &\color{black}{:= .9\overline{9}} &&&1/3 &= .3\overline{3} \\
10x &= 9.9\overline{9} &\color{black}{\text {or}}&& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
9x &= 9 &&& 1 &= .9\overline{9}\\
x &= 1
\end{align}}
$$
Okay, suppose you are willing to believe $.9\overline{9}$ is a real number, what else is wrong?

$$
\begin{align}
&&x &:= .9\overline{9} &&&&&1/3 &= .3\overline{3} \\
\implies&& 10x &= 9.9\overline{9} &\text {or}&&&\implies& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
\implies&&9x &= 9 &&&&\implies& 1 &= .9\overline{9}\\
\implies&& x &= 1
\end{align}
$$

$$
\begin{align}
&&x &:= .9\overline{9} &&&&&1/3 &= .3\overline{3} \\
\color{red}{\implies}&& 10x &= 9.9\overline{9} &\text {or}&&&\implies& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
\implies&&9x &= 9 &&&&\bf\color{red}{\implies}& 1 &= .9\overline{9}\\
\implies&& x &= 1
\end{align}
$$

*Why* does $10 \cdot (.9\overline{9}) = 9.9\overline{9}$?
*Why* does $3 \cdot (.3\overline{3}) = .9\overline{9}$?

$$
\begin{align}
x &:= .9\overline{9} &&&1/3 &\color{red}{=} .3\overline{3} \\
10x &= 9.9\overline{9} &\text {or}&& 3\cdot (1/3) &= 3\cdot(.3\overline{3})\\
9x &= 9 &&& 1 &= .9\overline{9}\\
x &= 1
\end{align}
$$
Isn't $1/3 \color{red}{=} .3\overline{3}$ just as mysterious? Why believe it?

# The real source of confusion: definition

What does $.9\overline{9}$ actually mean?

It is **NOT** a mystical infinite sequence of $9$'s after a decimal point.

The definition of $.9\overline{9}$:
$$
\begin{align}
x_1 &:= .9,\\ 
x_2 &:= .99,\\
x_3 &:= .999, \ldots \\
x_n &:= \sum_{i=1}^n 9 \cdot 10^{-i} = \underbrace{.9999}_{n}.
\end{align}
$$

Then $.9\overline{9}$ is **defined** to be the number that the sequence $$
(x_n)_n = (.9, .99, .999, \cdots)
$$ approaches, **if it exists**. That is, it is by definition the limit of the sequence.

For any repeating decimal, we use the same process to define it.

E.g. for $2.64\overline{51}$ you would use the sequence 
$$
(2.645, 2.6451, 2.64515, 2.645151, 2.6451515, \cdots).
$$

# So what is $.9\overline{9} = 1$ even claiming?


It is simply claiming that $$(.9, .99, .999, \cdots)$$ does approach a real number, and that number is $1$.



```python
import numpy as np
import matplotlib.pyplot as plt

N = np.arange(1, 10)
X = np.cumsum(9 * np.power(10.,-N))

plt.axhline(1.0, color="grey", linestyle="dotted")
plt.plot(N, X, marker="x");
```

# Proof (trivial)

Compute
$$
|x_n - 1| = |  \underbrace{.9999}_{n} - 1| = 10^{-n}.
$$
For any tolerance $\epsilon>0$, for any $n > 1/\epsilon$, we have
$$
|x_n - 1| = 10^{-n} < 1/n < \epsilon. 
$$
This formally shows that $x_n$ approaches 1, but you already knew that.

# Honorable mentions



# Thanks!


```python
like()
comment()
subscribe()
while True:
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        tell_friends()
    except KeyError:
        tell_family()
    await asyncio.sleep(60*60*24)
```

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Qwen/Qwen-72B
1. YES
2. YES