0.999...=1

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Theorem

$0.999 \ldots = 1$


Proof using Geometric Series

By Sum of Infinite Geometric Progression:

$0.999 \ldots = \dfrac a {1 - r}$

where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.

Since our ratio is less than $1$, then we know that $\displaystyle \sum_{n \mathop = 0}^\infty \frac 9 {10}\left({\frac 1 {10}}\right)^n$ must converge to:

$\dfrac a {1-r} = \dfrac{\frac 9 {10}} {1 - \frac 1 {10}} = \dfrac {\frac 9 {10}} {\frac 9 {10}} = 1$

$\blacksquare$


Proof using Fractions

\(\displaystyle 0.333 \ldots\) \(=\) \(\displaystyle 1 / 3\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 3 \paren {0.333 \ldots}\) \(=\) \(\displaystyle 3 \paren {1 / 3}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0.999 \ldots\) \(=\) \(\displaystyle 3 / 3\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


Proof using Multiplication by 10

Let $c = 0.999 \ldots$

Then:

\(\displaystyle c\) \(=\) \(\displaystyle 0.999 \ldots\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 10 c\) \(=\) \(\displaystyle \paren {9.999 \ldots}\) multiplying $c$ by $10$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 10 c - c\) \(=\) \(\displaystyle \paren {9.999 \ldots} - \paren {0.999 \ldots}\) subtracting $c$ from each side
\(\displaystyle \leadsto \ \ \) \(\displaystyle 9 c\) \(=\) \(\displaystyle 9\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle c\) \(=\) \(\displaystyle 1\)

It follows that:

$0.999 \ldots = 1$

$\blacksquare$


Proof using Long Division

We begin with the knowledge that:

\(\displaystyle \frac 9 9\) \(=\) \(\displaystyle \frac 1 1 = 1\)

Now we divide $9$ by $9$ using the standard process of long division, only instead of stating that $90$ divided by $9$ is $10$, we say that it is "$9$ remainder $9$," yielding the following result:

$ \require{enclose} \phantom{0..}0.9999\ldots\\ 9\enclose{longdiv}{9.0000\ldots}\\ \phantom{0}\underline{\phantom{0}8.1}\\ \phantom{000.}90\\ \phantom{0.}\underline{\phantom{00}81}\\ \phantom{0000.}90\\ \phantom{0.}\underline{\phantom{000}81}\\ \phantom{00000.}9\ldots\\ $

Thus, we are compelled to believe that:

$0.999\ldots = \dfrac 9 9 = 1$

$\blacksquare$


Proof using Sequences

\((1):\quad\) \(\displaystyle 0 . \underset n {\underbrace {999 \cdots 9} }\) \(=\) \(\displaystyle 1 - 0.1^n\)
\(\displaystyle 0.999 \cdots\) \(=\) \(\displaystyle \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}\) Definition of Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots} } {}\) from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - 0\) Sequence of Powers of Number less than One, Definition of Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


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