0.999...=1

Theorem: 0.9999...=1

Proof 1
$$.33...=(1/3)$$

$$(3)(.33...)=(1/3)(3)$$

$$(.99...)= (3/3)=1$$

Proof 2
$$(.11...)=(1/9)$$

$$(9).11...=(1/9)(9)$$

$$(.99..)= (9/9)=1$$

Proof 3
c=.99... 10c=9.9.. 10c-c=(9.9..)-(.99...) 9c=9 c=1 .99...=1

Proof 4
Two numbers are only distinct numbers if a third number can be place between them. Since no number can be fit between .99... and 1 they must be equal.

Using Geometric Series
We can represent $$0.9999...$$ as a geometric series with first term $$a=\frac{9}{10}$$, and ratio $$r=\frac{1}{10}$$. Since our ratio is less then 1, then we know that $$\sum_{n=0}^{\infty}\frac{9}{10}(\frac{1}{10})^n$$ must converge to: $$\frac{a}{1-r}=\frac{\frac{1}{9}}{1-\frac{1}{10}}=\frac{\frac{1}{9}}{\frac{1}{9}}=1$$ QED