# 0.999...=1/Proof 4

## Theorem

$0.999 \ldots = 1$

## Proof

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:

$0.999\ldots = \dfrac 9 9 = 1$

$\blacksquare$