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 $$a=\frac{9}{10}$$, and ratio $$r=\frac{1}{10}$$.

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


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

Proof: Using Fractions
$$ $$ $$ $$

Proof: Using Multiplication by 10
$$ $$ $$ $$ $$

Therefore we have that $$0.999\ldots=1$$.

Proof: Using Long Division
We begin with the knowledge that

$$\frac{9}{9} = \frac{1}{1} = 1 $$

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:

0.9999...   ---   9|9.0000...     8.1     ---       90       81       --        90        81        --         9...

Thus, we are compelled to believe that

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