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:

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

$\blacksquare$

Proof: Using Fractions

	\(\displaystyle \)	\(\displaystyle 0.333\ldots\)	\(=\)	\(\displaystyle 1/3\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle 3(0.333\ldots)\)	\(=\)	\(\displaystyle 3(1/3)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle 0.999\ldots\)	\(=\)	\(\displaystyle 3/3\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)

$\blacksquare$

Proof: Using Multiplication by 10

	\(\displaystyle \)	\(\displaystyle \text{Let } c\)	\(=\)	\(\displaystyle 0.999\ldots\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle 10c\)	\(=\)	\(\displaystyle (9.999\ldots)\)	\(\displaystyle \)	multiplying $c$ by $10$
	\(\displaystyle \)	\(\displaystyle 10c-c\)	\(=\)	\(\displaystyle (9.999\ldots)-(0.999\ldots)\)	\(\displaystyle \)	subtracting $c$ from each side
	\(\displaystyle \)	\(\displaystyle 9c\)	\(=\)	\(\displaystyle 9\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle c\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)

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

$\blacksquare$

Proof: Using Long Division

We begin with the knowledge that:

\(\displaystyle \)

\(\displaystyle \frac 9 9\)

\(=\)

\(\displaystyle \frac 1 1 = 1\)

\(\displaystyle \)

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

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

$\blacksquare$