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.