1. ## periodic decimal expansion

How would I express the number with periodic decimal expansion 0.321212121... in the form $\displaystyle \frac{m}{n}$. $\displaystyle m,n \in \mathbb{Z}$ ?

I don't know how this is done.

2. Originally Posted by Jason Bourne
How would I express the number with periodic decimal expansion 0.321212121... in the form $\displaystyle \frac{m}{n}$. $\displaystyle m,n \in \mathbb{Z}$ ?

I don't know how this is done.
This is fast and loose and not too rigorous (but it gives the general idea and can be made more rigorous):

S = 0.321212121

10S = 3.212121... (1)

1000S = 321.212121 .... (2)

(2) - (1) and solve for S.

3. Originally Posted by mr fantastic
This is fast and loose and not too rigorous (but it gives the geneal idea and can be made more rigorous):

S = 0.321212121

10S = 3.212121... (1)

100S = 321.212121 .... (2)

(2) - (1) and solve for S.
thanks. Would this be how to do this generally?

4. Originally Posted by Jason Bourne
How would I express the number with periodic decimal expansion 0.321212121... in the form $\displaystyle \frac{m}{n}$. $\displaystyle m,n \in \mathbb{Z}$ ?
I like to do this with a geometric series.
$\displaystyle 0.321212121... = \frac{3} {{10}} + \frac{{21}} {{10^3 }} + \frac{{21}} {{10^5 }} + \cdots = \frac{3} {{10}} + \sum\limits_{k = 1}^\infty {\frac{{21}} {{10^{2k + 1} }}}$

5. Originally Posted by mr fantastic
This is fast and loose and not too rigorous (but it gives the geneal idea and can be made more rigorous):

S = 0.321212121

10S = 3.212121... (1)

100S = 321.212121 .... (2) <-- I think he meant to write 1000S = ...

(2) - (1) and solve for S.
Originally Posted by Jason Bourne
thanks. Would this be how to do this generally?
Pretty much. The idea is to get rid of the "tail" depend on how many digits are repeating. e.g. In this case there are 2 digit repeating, so the factor will be 1000 and 10.

So say if the question is asking for a expression for 0.3214214214214...
Then, what do you think you should do?