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?
On
$$.3333... = \dfrac{1}{3}$$
So
$$10x-x = \left(53+\frac{1}{3}\right)-\left(5+\frac{1}{3}\right) = 53-5$$
On
Let x=5.333333.... Multiplying x by 10: $$10x = 53.333333...=53+0.333333....\\$$ And $$10x-53 = 0.333333...\\$$ Multiplying x by 100: $$100x=533.333333... \\$$ $$100x= 533+0.333333....\\$$ $$100x=533+10x-53\\$$ $$\implies100x-10x = 533-53 =480\\$$ $$\implies90x=480\implies x=\frac{480}{90}=\frac{16}{3}\\$$
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 $$ \sum_{k\ge1}\frac{23}{10^{2k}}=\frac{23}{100}\frac{1}{1-\dfrac{1}{100}} =\frac{23}{99} $$