Converting a repeating decimal to ratio of integers

[paulmdrdo]

0.17777777777 convert into a ratio.

 

[M R]

Re: converting a repeating decimal to ratio of integers

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

Or [tex] \text{Let } x=0.0777..[/tex] so that [tex]10x=0.777..[/tex].

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

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

 

[paulmdrdo]

Re: converting a repeating decimal to ratio of integers

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

Or [tex] \text{Let } x=0.0777..[/tex] so that [tex]10x=0.777..[/tex].

Subtracting gives [tex]9x=0.7[/tex] and so [tex]x=\frac{7}{90}[/tex]. Now just add [tex]\frac{1}{10}+\frac{7}{90}[/tex] 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"?

 

[M R]

Re: converting a repeating decimal to ratio of integers

what do you mean by "GP"?

Sorry, I have to stop using abbreviations.

A GP is a geometric progression: [tex]a, ar, ar^2, ar^3...[/tex].

If you haven't met this then the second method I posted is fine.

 

[soroban]

Re: converting a repeating decimal to ratio of integers

Hello, paulmdrdo!

[tex]\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}[/tex]

[tex]\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}[/tex]

Therefore: .[tex]x \;=\;\frac{16}{90} \;=\;\frac{8}{45}[/tex]

 

[paulmdrdo]

how would I decide what appropriate power of ten should i use?

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

 

[M R]

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.

 

[paulmdrdo]

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.

 

[M R]

"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

 

[Prove It]

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?

 

[MarkFL]

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.) \(\displaystyle x=0.1\overline{7}\)

\(\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}\)

\(\displaystyle x=\frac{16}{90}=\frac{8}{45}\)

2.) \(\displaystyle x=3.5\overline{47}\)

\(\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}\)

\(\displaystyle x=\frac{3512}{990}=\frac{1756}{495}\)