base conversion $$\large (0.\overline{1210})_3 = (?)_{10}$$

Question

inkyvoyd · Answer

oops looks like my logic is right by my arithmetic was wrong

inkyvoyd · Answer

http://en.wikipedia.org/wiki/Repeating_decimal#Converting_repeating_decimals_to_fractions

inkyvoyd · Answer

from there you can apply the same logic to base 2 to get
.1210 repeating= 1210_3/2222_3 = 48_10/80_10=3/5=.6

rational · Answer

wow is that a shortcut XD i thought u would expand it and use geometric series formula

inkyvoyd · Answer

@rational I am so glad my algebra teacher taught me this trick a long time ago lol

inkyvoyd · Answer

it is basically geometric series but presented in a way that is more simple than it is normally taught

rational · Answer

That looks neat! so for a repeating pattern of length $r$, do we have
$$\large (0.\overline{abcd\ldots })_b = \dfrac{(abcd\ldots)_b}{(b^r-1)_{10}}$$

inkyvoyd · Answer

yeah, I learned it as a rule a long time ago, and we used algebra to prove it, for the decimal case at least...
I figured that the logic was behind radix representation of numbers, which is why it would just as readily apply to any base as well

inkyvoyd · Answer

the base 10 should actually be b I believe...
so you would see (call c=b-1)
ccccc... for the length of abcd...

rational · Answer

$$(ccc\ldots )_{b} =  (c(111\ldots))_b \~\= (c(b^{r-1} + b^{r-2}+\cdots + b^1+1))_{10}  \~\= \left(c\dfrac{b^r-1}{b-1}\right)_{10} \~\ = (b^r-1)_{10}$$

inkyvoyd · Answer

hahaha oh