0.999...=1

Theorem

 * $0.999 \ldots = 1$

Proof: Using Geometric Series
We can represent $0.999 \ldots$ as the sum of an infinite geometric progression with first term $\displaystyle a = \frac 9 {10}$, and ratio $\displaystyle r = \frac 1 {10}$.

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


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

Proof: Using Multiplication by 10
Therefore we have that $0.999\ldots=1$.

Proof: Using Long Division
We begin with the knowledge that:

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: 0.9999...   ---   9|9.0000...     8.1     ---       90       81       --        90        81        --         9... Thus, we are compelled to believe that

$.999... = \frac 9 9 = 1$