0.999...=1
Theorem
- $0.999 \ldots = 1$
Proof using Geometric Series
By Sum of Infinite Geometric Sequence:
- $0.999 \ldots = \dfrac a {1 - r}$
where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.
Since our ratio is less than $1$, then we know that $\ds \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ 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
| \(\ds 0.333 \ldots\) | \(=\) | \(\ds 1 / 3\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3 \paren {0.333 \ldots}\) | \(=\) | \(\ds 3 \paren {1 / 3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0.999 \ldots\) | \(=\) | \(\ds 3 / 3\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Proof using Multiplication by 10
Let $c = 0.999 \ldots$
Then:
| \(\ds c\) | \(=\) | \(\ds 0.999 \ldots\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 10 c\) | \(=\) | \(\ds \paren {9.999 \ldots}\) | multiplying $c$ by $10$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 10 c - c\) | \(=\) | \(\ds \paren {9.999 \ldots} - \paren {0.999 \ldots}\) | subtracting $c$ from each side | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 9 c\) | \(=\) | \(\ds 9\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c\) | \(=\) | \(\ds 1\) |
It follows that:
- $0.999 \ldots = 1$
$\blacksquare$
Proof using Long Division
We begin with the knowledge that:
| \(\ds \frac 9 9\) | \(=\) | \(\ds \frac 1 1 = 1\) |
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
| \(\text {(1)}: \quad\) | \(\ds 0 . \underset n {\underbrace {999 \cdots 9} }\) | \(=\) | \(\ds 1 - 0.1^n\) | |||||||||||
| \(\ds 0.999 \cdots\) | \(=\) | \(\ds \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}\) | Definition of Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots} } {}\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - 0\) | Sequence of Powers of Number less than One, Definition of Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Also see
- For a video presentation of the contents of this page, visit the Khan Academy.