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:

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

Thus, we are compelled to believe that:

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

$\blacksquare$