Question

Do you guys agree with this valid math statement:
0.2 = 0.19999999...

Answer 1

\[0.1999...=0.1+\frac{9}{100}(1+\frac{1}{10}+\frac{1}{10^{2}}+...)=0.1+\frac{9}{100}\sum_{i=0}^{\infty}(\frac{1}{10})^{i}\]
and
\[\sum_{n=1}^{\infty}(r)^{n}=\frac{1}{1-r}\]
then
\[0.1999...=0.1+\frac{9}{100}\frac{1}{1-\frac{1}{10}}=0.1+0.1=0.2\]
lol

Answer 2

Easier to just do this:
\( \begin{array}{l}
10x\ =1.9999...\\
100x\ =\ 19.9999...\\
\\
100x\ -\ 10x\ =\ 18\\
90x\ =\ 18\\
x=\dfrac{18}{90}\\
=\dfrac{1}{5}\\
=0.2
\end{array}\)

Answer 3

Wow that's so much simpler

Answer 4

indeed

Answer 5

Imagine how annoying that'd be if instead of writing
3 + 2 = 5
I'd start writing
2.99999... + 1.9999... = 4.99999.. and give my answer to teachers ahaha