How do you add and subtract repeating decimals?

Question

I am currently in $8$th grade and I need to know if it’s like normal subtraction or it’s different because the number is infinite. For example: $$. \overline {5} - . \overline {05}$$ Would that equal $.45$ or $.\overline {45}$? I need to know, because I've searched everywhere for an answer and I cannot find anything on at at all.

Answer 1

If you try it with decimals directly,

$0. \overline{5} - 0.\overline{05} = .55555555.... - .05050505....=.50505050.... \rightarrow 0.\overline {50}$

However, if you recognize that $0. \overline{5}$ is $\dfrac {5}{9}$ and $0. \overline{05} = \dfrac {5}{99}$, you can subtract the fractions to get $$\dfrac {5}{9} - \dfrac {5}{99} \rightarrow \dfrac {55}{99} - \dfrac {5}{99} \rightarrow \dfrac {50}{99} = 0.\overline {50}$$

To wit: anything that repeats in a single digit will have a denominator of $9$; anything that repeats with two digits will end in $99$, and anything repeating with $n$ digits will have a denominator of $10^{n}-1$.

ETA: There is also a trick for determining repeating decimals from Rapid Calculations (from Google Books...)

For the denominator take as many nines as there are recurring figures in the decimal, and as many zeroes as there are non-recurring figures. For the numerator take the entire number and deduct all the non-recurring figures.

So for $0. \overline{5}$, the denominator would be $9$ (only one recurring digit) and the numerator would be $5$, so we get $\dfrac {5}{9}$. For $0. \overline{50}$, the denominator is $99$ (two recurring digits) and the numerator is $50$, giving us $\dfrac {50}{99}$. For $0.4 \overline{25}$, the denominator will be $990$ (two recurring digits, one non-recurring digit) and the numerator will be $421 (425-4)$ to give us $\dfrac {421}{990}$.