MHB Converting a repeating decimal to ratio of integers

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To convert the repeating decimal 0.177777... into a ratio of integers, it can be expressed as the sum of 0.1 and 0.0777..., leading to the fraction 16/90, which simplifies to 8/45. The method involves setting x equal to the decimal, multiplying by powers of ten to align the repeating parts, and then subtracting to eliminate the decimals. For decimals with two repeating digits, such as 3.5474747474..., the same principle applies, where the appropriate powers of ten are chosen to cancel the repeating digits during subtraction. This approach effectively demonstrates how to convert repeating decimals into simplified fractions.
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0.17777777777 convert into a ratio.
 
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Re: converting a repeating decimal to ratio of integers

Hi,
This is 0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+... where you have a GP to sum.

Or \text{Let } x=0.0777.. so that 10x=0.777...

Subtracting gives 9x=0.7 and so x=\frac{7}{90}. Now just add \frac{1}{10}+\frac{7}{90} and simplify.

I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
 
Re: converting a repeating decimal to ratio of integers

M R said:
Hi,
This is 0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+... where you have a GP to sum.

Or \text{Let } x=0.0777.. so that 10x=0.777...

Subtracting gives 9x=0.7 and so x=\frac{7}{90}. Now just add \frac{1}{10}+\frac{7}{90} and simplify.

I should also say that we can write a decimal as a fraction but we can't write it as a ratio.

what do you mean by "GP"?
 
Re: converting a repeating decimal to ratio of integers

paulmdrdo said:
what do you mean by "GP"?

Sorry, I have to stop using abbreviations. :)

A GP is a geometric progression: a, ar, ar^2, ar^3....

If you haven't met this then the second method I posted is fine.
 
Re: converting a repeating decimal to ratio of integers

Hello, paulmdrdo!

\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}
\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}

Therefore: .x \;=\;\frac{16}{90} \;=\;\frac{8}{45}
 
how would I decide what appropriate power of ten should i use?

for example i have 3.5474747474... how would you convert this one?
 
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.

If you use 1000 and 10 you will get

1000x=3547.474747...

10x=35.474747...

So 990x=3512 and x=3512/990=1756/495.

I'm adopting Soroban's approach as I prefer it to what I did earlier.
 
M R said:
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.

If you use 1000 and 10 you will get

1000x=3547.474747...

10x=35.474747...

So 990x=3512 and x=3512/990=1756/495.

I'm adopting Soroban's approach as I prefer it to what I did earlier.

"a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.
 
Last edited:
paulmdrdo said:
"a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me.

No problem.

We have 10^3 and 10^1.

The difference between 3 and 1 is 3-1=2
 
  • #10
paulmdrdo said:
how would I decide what appropriate power of ten should i use?

for example i have 3.5474747474... how would you convert this one?

You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.

So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..."

What powers of 10 will enable this?
 
  • #11
A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.

1.) $$x=0.1\overline{7}$$

$$10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$

$$x=\frac{16}{90}=\frac{8}{45}$$

2.) $$x=3.5\overline{47}$$

$$10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$

$$x=\frac{3512}{990}=\frac{1756}{495}$$
 
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