Subtraction of two repeating decimals When I was looking at the proof that every repeating decimal is rational, I came across this example:
$x=5.33333333\ldots$ ($3$ repeat indefinitely)
$10x=53.3333333\ldots$ ($3$ repeat indefinitely)
My question is why $10x-x = 53-5$?
Why is it possible to subtract the infinite part of $x$ and $10x$?
Why is $0.33333\ldots - 0.3333\ldots = 0.0000\ldots$ is this really a zero? 
 A: This argument is actually not a proof, unless you have already proved that repeating decimal alignments really define a rational number and that you can do multiplication by $10$ just by shifting the decimal point.
This is an indication that if $5.(3)$ (this is how a repeating decimal is often represented) defines a rational number $x$ and if multiplication by $10$ on these objects obeys the same rules as for finite decimal numbers, then
$$
10x-x=53-5=48
$$
so the only possible choice of $x$ is $x=48/9=16/3$.
The repeating decimals are instead associated to a number with the concept of a series:
$$
5.(3)=5+\sum_{k\ge1}\frac{3}{10^k}
$$
just like the finite decimal $5.333$ is
$$
5+\frac{3}{10}\frac{3}{100}+\frac{3}{1000}
$$
The series above is a geometric series and its sum is known to be
$$
\frac{3}{10}\frac{1}{1-\dfrac{1}{10}}=\frac{3}{10}\frac{1}{\dfrac{9}{10}}=
\frac{3}{10}\frac{10}{9}=\frac{1}{3}
$$
so that
$$
5.(3)=5+\frac{1}{3}=\frac{16}{3}
$$
by definition.
If $0<r<1$, the sum of the series
$$
\sum_{k\ge1}ar^k=ar\frac{1}{1-r}
$$
For $0.(23)$, the series would be
$$
\sum_{k\ge1}\frac{23}{10^{2k}}=\frac{23}{100}\frac{1}{1-\dfrac{1}{100}}
=\frac{23}{99}
$$
A: $$.3333... = \dfrac{1}{3}$$
So
$$10x-x = \left(53+\frac{1}{3}\right)-\left(5+\frac{1}{3}\right) = 53-5$$
A: Let x=5.333333....
Multiplying x by 10:
$$10x = 53.333333...=53+0.333333....\\$$
And
$$10x-53 = 0.333333...\\$$
Multiplying x by 100:
$$100x=533.333333... \\$$
$$100x= 533+0.333333....\\$$
$$100x=533+10x-53\\$$
$$\implies100x-10x = 533-53 =480\\$$
$$\implies90x=480\implies x=\frac{480}{90}=\frac{16}{3}\\$$
