Expert: Josh - 10/16/2008

Question
How do you transform a repeating decimal into a fraction?  I just don't get how to.  Please help.  Thanks.

Answer
I happened to have answered this question in the past, so I thought this might be useful to know.

You can attack this problem using the sum of geometric series.
e.g., Consider the repeating decimal 0.111111... We can write this as 0.1 + 0.01 + 0.001 + .... or 1/10 + 1/100 + 1/1000 +... and so forth.

A BRIEF REVIEW OF GEOMETRIC SERIES:
Recall that an n-term geometric series consists of T(0)=a, T(1)=a*r, T(2)=a*r^2 .... and T(n)=a*r^n. The formula for its sum is given by a*(1-r^n)/(1-r), where "a" is the first term, and "r" is the ratio T(n)/T(n-1) between successive terms.

In the limit as we take an infinite number of terms, this is exactly the case for a recurring decimal, the GS sum formula simplifies to S=a/(1-r).

In the example above, we identify "a" as 1/10, the ratio "r" is given by 1/10 also, thus, the "0.1+0.01+0.001+..." part is simply given by a/(1-r)=(1/10)/(1-1/10)=1/9 as expected.

Let us consider another example, say N=0.32474747...
In this case, we break the representation into two parts. Obviously, we can write 0.32 as 32/100. For the recurring part, we have 0.00474747=0.0047+0.000047+0.00000047... Again, we let a=47/10000, r=1/100 (the geometric ratio between successive terms). The GS sum formula gives (47/10000)/(1-1/100)=47/9900. Finally, adding the two components together, 0.32 + 0.00474747... = (32/100)+(47/9900) = (32*99+47)/9900 = 3215/9900 <<<< This is the exact form.

As a slight variation of this theme, consider N=0.0584215842158421...
The part "58421" is repeated indefinitely.
We have N=58421/10^6 + 58421/10^11 + 58421/10^16 + etc. Using the GS infinite sum formula, where a=58421/10^6 and r=1/10^5, N=a/(1-r)=58421/(10^6-10)=58421/999990.