Round the world!

Let $x=1.999\cdots$

$\implies 10x=19.999\cdots$

Subtracting second equation from first,

$10x-x=19.999\cdots-1.999\cdots=18$

$\implies 9x=18$

$\implies x=2$

$x=1.999\cdots$

$x=2$

$\implies 1.999\cdots=2$

The best solution :) $1.8+1.8\cdot 10^{-1}+1.8\cdot 10^{-2}+1.8\cdot 10^{-3}+1.8\cdot 10^{-4}+\cdots$ $1.8(10^{0}+10^{-1}+10^{-2}+10^{-3}+10^{-4}+\cdots)$ $\text{From the sum of infinite geometric series:}$ $1.8(\frac{10}{10-1})$ $\frac{1.8\cdot 10}{9}$ $\frac{18}{9}$ $\boxed{2}$

1.99999.....=1+0.99999999...=1+0.11111...... 9=1+(1/9) 9=1+1=2

$2 = 2$

$1 + 1 = 2$

$1 + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 2$

$\boxed{\frac{1}{3} = 0.333...}$

$1 + 0.333... + 0.333... + 0.333... = 2$

$\text{So, therefore...}$

$\boxed{1.999... = 2}$

i used the wrong formula but somehow got the correct answer i misread the title as round the word so assumed the question was about rounding so i chose 2

I can't write LATEX, so I'll just use normal numbers and words.

1.999... = 2

Subtract one from each side.

0.999... = 1

Next, assume x = 0.999...

10x = 9.999...

10x - x = 9.999... - 0.999...

9x = 9

Divide both sides by 9

x = 1

x = 1 and 0.999..., so 1 + x = 2