Math Help - rational number representation

1. rational number representation

I have a question that I can't recall how to do. Here it is.

Determine and state the rational number representation for the following decimal. Put the rational number in lowest terms.

( line is above the numbers
1485) ____

n= 0. 0 1485

n=

I would greatly appreciate your assistance.

Thank you.

Melissa

2. Originally Posted by mdoodles
I have a question that I can't recall how to do. Here it is.

Determine and state the rational number representation for the following decimal. Put the rational number in lowest terms.

( line is above the numbers
1485) ____

n= 0. 0 1485
Here is one way to do it:

$n=0.0\overline{1485}$

So,

$
\begin{tabular}{rcr@{.}l}
10000n & = & 148&514851485...\\
- n & = & 0&014851485...\\
\hline
9999n & = & 148&5\\
\end{tabular}
$

Now,

$9999n = 148.5$

$\Rightarrow n = \frac{148.5}{9999} = \frac{1485}{99990}$

Then reduce. Alternatively, you can represent the number as an infinite series.

3. Thank you so much. I remember we did this in class but I couldn't find it in my notebook. I just found it in my other notebook. It was driving me crazy. Thank you so much for your assistance. I appreciate it.

Melissa

4. Originally Posted by mdoodles
Thank you so much. I remember we did this in class but I couldn't find it in my notebook. I just found it in my other notebook. It was driving me crazy. Thank you so much for your assistance. I appreciate it.

Melissa
No problem. In case you are still interested, here is how to do it with infinite series:

$n=0.0\overline{1485}=0.01485+\frac{0.01485}{10000} +\frac{0.01485}{10000^2}+\cdots=\sum_{n=0}^\infty \frac{0.01485}{10000^n}$

$=\frac{1485}{10^5}\sum_{n=0}^\infty\left(\frac{1}{ 10^{4n}}\right)=\frac{1485}{10^5}\sum_{n=0}^\infty \left(\frac{1}{10^4}\right)^n$

$=\frac{1485}{10^5}\left(\frac1{1-\frac1{10^4}}\right)=\frac1{\left(\frac{9999}{10^4 }\right)}$

$=\frac{1485}{10^5}\left(\frac{10^4}{9999}\right)=\ frac{1485}{99990}=\frac{3^3\cdot5\cdot11}{2\cdot3^ 2\cdot5\cdot11\cdot101}=\frac3{202}$