<jupyter_start><jupyter_text>$$.9\overline{9} = 1?$$ Why every proof you have seen is WRONG 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: definitionWhat 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 $1$.<jupyter_code>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");<jupyter_output><empty_output><jupyter_text>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!<jupyter_code>like()
comment()
subscribe()
while True:
    try:
        tell_friends()
    except KeyError:
        tell_family()
    await asyncio.sleep(60*60*24)<jupyter_output><empty_output>
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