Converting a decimal number with non-repeating and repeating digits to an improper fraction

Question

What are the steps to convert $3.2\overline{901234567}$ to an improper fraction?

Welcome to Mathematics Stack Exchange. Note $901234567/999999999=73/81=0.\overline{901234567}$ — J. W. Tanner, Sep 26 '19 at 02:02
Thank you for the welcome! 533/162 is the answer I am looking for. I think I can figure out how you got it. Thanks! — David_M, Sep 26 '19 at 02:43

score 1 · Accepted Answer · edited Sep 27 '19 at 02:12

1

$$x = 3.2\overline{901234567}$$

$$10^9 x = 3290123456.7\overline{901234567}$$

$$(10^9 - 1)x = 329123453.5$$

so

$$x = \frac{3290123453.5}{999999999}$$

edited Sep 27 '19 at 02:12

J. W. Tanner

60,406

answered Sep 26 '19 at 02:06

David G. Stork

29,774

I think you left out a "9" digit in the bottom number – J. W. Tanner Sep 26 '19 at 02:25
Thank you, sir! – David_M Sep 26 '19 at 04:04

J. W. Tanner · Answer 2 · 2019-09-26T02:12:19.470

0

$3.2\overline{901234567}=\dfrac{32}{10}+\dfrac{901234567}{9999999990}=\dfrac{32}{10}+\dfrac{73}{810}.$ Can you take it from here?

edited Sep 26 '19 at 02:12

answered Sep 26 '19 at 02:08

J. W. Tanner

60,406

Your second fraction is off by a factor of $10$. The denominator should be $9999999990$. – kccu Sep 26 '19 at 02:11
Thank you, @kccu; I edited in the "$0$" – J. W. Tanner Sep 26 '19 at 02:14

score 0 · Answer 3 · answered Sep 26 '19 at 02:13

You can re-write this repeating decimal using a infinite geometric series: \begin{align*} 3.2\overline{901234567} &= 3.2+901234567(10^{-10}+10^{-19}+10^{-28}+\cdots)\\ &=3.2 + 901234567\cdot 10^{-10}\sum_{k=0}^\infty (10^{-9})^k\\ &=3.2 + 901234567\cdot \frac{1}{10^{10}}\frac{1}{1-10^{-9}}\\ &=\frac{32}{10}+\frac{901234567}{10^{10}-10} \end{align*} and now you can find a common denominator and add these fractions.

Converting a decimal number with non-repeating and repeating digits to an improper fraction

3 Answers3