# n repeating digit of decimal number

• Dec 22nd 2011, 04:43 AM
Bokas
n repeating digit of decimal number
I have a problem how to convert number 0.3333....3 where 3 repeats N times to fraction.
for N=1 0.3
N=2 0.33 and so on.

I know how to convert number where digits repeats forever, but how can I convert this to fraction to get correct result for any N.

• Dec 22nd 2011, 04:47 AM
Prove It
Re: n repeating digit of decimal number
Quote:

Originally Posted by Bokas
I have a problem how to convert number 0.3333....3 where 3 repeats N times to fraction.
for N=1 0.3
N=2 0.33 and so on.

I know how to convert number where digits repeats forever, but how can I convert this to fraction to get correct result for any N.

\displaystyle \displaystyle \begin{align*} x &= 0.\dot{3} \\ 10x &= 3.\dot{3} \\ 10x - x &= 3.\dot{3} - 0.\dot{3} \\ 9x &= 3 \\ x &= \frac{3}{9} \\ x &= \frac{1}{3} \\ 0.\dot{3} &= \frac{1}{3} \end{align*}
• Dec 22nd 2011, 06:24 AM
HallsofIvy
Re: n repeating digit of decimal number
Quote:

Originally Posted by Bokas
I have a problem how to convert number 0.3333....3 where 3 repeats N times to fraction.
for N=1 0.3
N=2 0.33 and so on.

I know how to convert number where digits repeats forever, but how can I convert this to fraction to get correct result for any N.

A terminating decimal is easier than a repeating decimal (which is what ProveIt did). If x= 0.33333..3 where the "3" repeats N times, then $\displaystyle 10^Nx= 3333...3$ so that $\displaystyle x= \frac{3333...3}{10^N}$. Because of that "$\displaystyle 10^N$" in the denominator, a terminating decimal, converted to a fraction, has only powers of 2 and 5 in its denominator.