# Homework Help: Converting recurring decimals to fractions

1. Jan 3, 2010

### Gringo123

I have no difficulty with converting decimals to fractions generally (0.125 = 1/8)
However, I am a bit stuck when it comes to converting a recurring decimals. My Math book tells me that 0.7777777r is not expressed as 77/100 as a fraction, but as 7/9. How do I calcuate that? Can anyone talk me through the process?

2. Jan 3, 2010

### Dick

Call x=0.7777r. What is 10*x-x?

3. Jan 3, 2010

### TauCrouton

You can use the sum of an infinite geometric sequence formula.

$$S = \frac{a}{1-r}$$

your a value is (in this case) 0.7. You want to make a series that will create 0.77777777777... etc.

To make this series, 0.7 is your first term. It is multiplied by the common ration (r) 0.1 to get 0.07 (your second term), and added to 0.7 to get 0.77. 0.07 is then multiplied by 0.1 to get your third term, 0.007 and then added to 0.77 to get 0.777. This happens an infinite amount of times to get 0.777777777..... Thus the need for the sum of an infinite series formula.

Your r (ratio) between each term is 0.1.

You substitute

$$S = \frac{0.7}{1-0.1}$$

and you get

$$S = \frac{0.7}{0.9}$$

set

$$\frac{0.7}{0.9}$$

to zero and get

$$\frac{0.7}{0.9} = 0$$

multiply both sides by ten and you get

$$\frac{7}{9}$$

4. Jan 3, 2010

### Gringo123

Dick - if x=0.7777r. then 10 * x-x = 7.00

Taucrouton - What about if I wanted to convert 0.5454545454.. etc. to a fraction?
Then there are 2 rations. 0.5 x 0.08 = 0.04, then 0.04 x 0.125 = 0.005 so the 2 rations are 0.08 and 0.125.
Am I going in the right direction here?

5. Jan 3, 2010

### icystrike

Hey Gringo! I'll continue from dick's . you've said x=0.77777r and you have acquired
$$10x-x=9x=7$$,
suggesting that x= ?

6. Jan 3, 2010

### Gringo123

Hi Icystrike. Thanks for helping me out!
if 9x = 7, x = 0.7777777r

7. Jan 3, 2010

### icystrike

$$\frac{9x}{9}=\frac{7}{9}$$
x = ?

8. Jan 3, 2010

### willem2

I don't think so. Try the same trick with 100*x - x

9. Jan 3, 2010

### HallsofIvy

In general, suppose x= A.BCCCCC... where "A" is the entire number before the decimal point, "B" is any decimals before repetition, and "C" is the repeating block. Also let "n" be the number of decimal places in B, "m" the number of decimal places in C.
In your example, 0.7777777..., A= 0, B= 0 (and so n= 0) and C= 7 (and so m= 1).

If x= A.BCCCCC... then $10^nx= AB.CCCC...$ and $10^{n+m}= ABC.CCCCC...$. Subtracting, $(10^{n+m}- 10^n})x= ABC- AB$ Where "ABC- AB" is now an integer. (Note that "ABC" does NOT mean "A times B time C" here!).

In your example, n=0 and m= 1 so $10^{n+m}- 10^n= 10^1- 10^0= 10- 1= 9$ and "ABC" is 7 while AB is 0: 9x= 7. Again, what is x?
(Yes, it is NOT 77/100 because that would be 0.7700000....)

Here's another example: write 423.923425252525... where "25" is the repeating section. The "non-repeating" decimal part is "9234" which has 4 decimal places. Multiply both sides of x= 423.923425252525... by $10^4= 10000$ gives 10000x= 4239234.25252525...[/itex]. The repeating part, "25", has two decimal places so multiply that by $10^2= 100$, which would be the same as multiplying the original equation by $10^{4+2}= 10^6= 1000000$, gives $1000000x= 42393425.252525...$

Now subract those two: $1000000x- 10000x= 990000x= 42393425.252525...-4239234.25252525...= 38154191$. Finally, then, x= 38154191/990000. Reduce that fraction if possible.

Note that the "repeating" was essential there. If that "25" did not continue repeating, the two decimal parts would not be the same and we could not cancel them.

Last edited by a moderator: Jan 4, 2010
10. Jan 3, 2010

### TauCrouton

I think you can just use 0.54 as you first term and use a common ratio of 0.01.

Edit: Yes, it works.
I haven't gone through the other methods posted here, but there is definitely more than one way to go about it. The way I posted is the one that makes the most sense to me, so just find one that works and stick with it.

Last edited: Jan 3, 2010
11. Jan 3, 2010

### ideasrule

0.777777..... = 7/9
0.999999999.... = 9/9
0.11111111....=1/9
0.131313131313....=13/99
0.175175175175175...=175/999
etc

12. Jan 3, 2010

### TauCrouton

0.9999999999 technically cant be 9/9, although that is the answer.

Interesting things, infinite sums.

13. Jan 4, 2010

### diazona

It is, actually, why would you say it's not? (Yes this is weird at first sight)

14. Jan 4, 2010

### HallsofIvy

Are you simply saying that it should be written as "1"? "Technically", 0.999999999... (NOT "0,9999999999") is equal to 9/9 which is the same as 1.

15. Jan 4, 2010

### TauCrouton

I'm not saying its not, I'm saying 0.999999999.... isn't technically 1, just infinitesimally close to 1. (see below)

I meant to say the 0.99999... is a repeating 9 not a discreet value of "0.999999999", if that clears things up.

Either way, it is an interesting technicality. The way I was taught to understand infinite sums of sequences is that they come infinitesimally close to a value, but never actually reach the value. Similar in conundrum to the definition of an asymptote, or a limit. They just come so close to the value that we might as well call it that value.

16. Jan 4, 2010

### Dick

Well, no. It technically IS 1. Writing '...' means you don't mean some finite sum. You mean the whole sum. I.e. the limit of the partial sums. That's 1. In the standard definition of the real numbers, there is no such thing as a number 'infinitesimally' close to 1.

17. Jan 4, 2010

### vela

Staff Emeritus
To put it another way, the decimal representation of a number isn't necessarily unique. One can be represented by 0.999... or 1; they're both valid representations of the same number.

18. Jan 5, 2010

### TauCrouton

True, I see your point. Although if we don't limit ourselves to the definition of real numbers, I suppose there is a theoretical number which does come infinitely close to one. Like if you type 0.999999999... you can hold the 9 button for as long as you like, but it will never truly reach 1.

Anyways, I think we answered the original question wayy back. :P