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

\(\displaystyle 0.333 \ldots\) \(=\) \(\displaystyle 1 / 3\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 3 \left({0.333 \ldots}\right)\) \(=\) \(\displaystyle 3 \left({1 / 3}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 0.999 \ldots\) \(=\) \(\displaystyle 3 / 3\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ $\quad$

$\blacksquare$


Proof using Multiplication by 10

Let $c = 0.999\ldots$

Then:

\(\displaystyle c\) \(=\) \(\displaystyle 0.999 \ldots\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 10 c\) \(=\) \(\displaystyle \left({9.999 \ldots}\right)\) $\quad$ multiplying $c$ by $10$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 10 c - c\) \(=\) \(\displaystyle \left({9.999 \ldots}\right) - \left({0.999 \ldots}\right)\) $\quad$ subtracting $c$ from each side $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 9 c\) \(=\) \(\displaystyle 9\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle c\) \(=\) \(\displaystyle 1\) $\quad$ $\quad$

It follows that:

$0.999\ldots = 1$

$\blacksquare$


Proof using Long Division

We begin with the knowledge that:

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

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

\((1):\quad\) \(\displaystyle 0 . \underset n {\underbrace {999 \cdots 9} }\) \(=\) \(\displaystyle 1 - 0.1^n\) $\quad$ $\quad$
\(\displaystyle 0.999 \cdots\) \(=\) \(\displaystyle \left[\!\left[{\left\langle{0.9, \, 0.99, \, 0.999, \, \cdots}\right\rangle}\right]\!\right]\) $\quad$ Definition of Real Numbers $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left\langle{1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots}\right\rangle}\right]\!\right]\) $\quad$ from $\left({1}\right)$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left\langle{1, \, 1, \, 1, \, \cdots}\right\rangle}\right]\!\right] - \left[\!\left[{\left\langle{0.1^1, \ , 0.1^2, \, 0.1^3, \, \cdots}\right\rangle}\right]\!\right]\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 - 0\) $\quad$ Power of Number less than One, Definition of Real Numbers $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ $\quad$

$\blacksquare$


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