Converting recurring decimals to fractionsby Gringo123 Tags: converting, decimals, fractions, recurring 

#1
Jan310, 12:36 AM

P: 141

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
Jan310, 12:45 AM

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




#3
Jan310, 01:01 AM

P: 9

You can use the sum of an infinite geometric sequence formula.
[tex] S = \frac{a}{1r} [/tex] 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 [tex] S = \frac{0.7}{10.1} [/tex] and you get [tex] S = \frac{0.7}{0.9} [/tex] set [tex] \frac{0.7}{0.9} [/tex] to zero and get [tex] \frac{0.7}{0.9} = 0 [/tex] multiply both sides by ten and you get [tex] \frac{7}{9} [/tex] 



#4
Jan310, 01:23 AM

P: 141

Converting recurring decimals to fractions
Dick  if x=0.7777r. then 10 * xx = 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
Jan310, 02:11 AM

P: 439

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



#6
Jan310, 02:26 AM

P: 141

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



#7
Jan310, 02:35 AM

P: 439

[tex]\frac{9x}{9}=\frac{7}{9}[/tex]
x = ? 



#8
Jan310, 06:18 AM

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#9
Jan310, 06:19 AM

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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 [itex]10^nx= AB.CCCC...[/itex] and [itex]10^{n+m}= ABC.CCCCC...[/itex]. Subtracting, [itex](10^{n+m} 10^n})x= ABC AB[/itex] 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 [itex]10^{n+m} 10^n= 10^1 10^0= 10 1= 9[/itex] 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 "nonrepeating" decimal part is "9234" which has 4 decimal places. Multiply both sides of x= 423.923425252525... by [itex]10^4= 10000[/itex] gives 10000x= 4239234.25252525...[/itex]. The repeating part, "25", has two decimal places so multiply that by [itex]10^2= 100[/itex], which would be the same as multiplying the original equation by [itex]10^{4+2}= 10^6= 1000000[/itex], gives [itex]1000000x= 42393425.252525...[/itex] Now subract those two: [itex]1000000x 10000x= 990000x= 42393425.252525...4239234.25252525...= 38154191[/itex]. 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. 



#10
Jan310, 04:14 PM

P: 9

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. 



#11
Jan310, 11:15 PM

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



#12
Jan310, 11:51 PM

P: 9

0.9999999999 technically cant be 9/9, although that is the answer. Interesting things, infinite sums. 



#13
Jan410, 12:06 AM

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#14
Jan410, 04:52 AM

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#15
Jan410, 09:50 PM

P: 9

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
Jan410, 10:23 PM

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#17
Jan410, 11:49 PM

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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
Jan510, 01:28 AM

P: 9

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


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