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Homework Help: Convert repeating decimal to improper fraction

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Convert 23.588 (the 88 is repeating) to an improper fraction.


    2. Relevant equations
    I don't have any.


    3. The attempt at a solution

    I'm not sure of the best way to go about this so I've taken a method that I've seen on some problems that were more simple as they didn't have the 23.5 in front of the repeating decimal.

    I see that this can be broken down into:

    23 + .5 + 8/10 + 8/10(1/10) + 8/10(1/10)^2...etc.

    Although, I'm really not sure how to collapse the infinite progression of 8/10(1/10)^n into an actual number.

    I've tried plugging 8/10 in for 'a' and 1/10 in for 'r' in the infinite progression equation:

    a/(1-r)

    This results in 8/9 then adding 1/2 to that we get 25/18 and then adding 23 to that I get 439/18

    However, this result is not correct.

    Any help/suggestions are welcome. Thank you for your time!
     
  2. jcsd
  3. Apr 28, 2010 #2
    you have:
    23 + .5 + 8/10 + 8/10(1/10) + 8/10(1/10)^2...etc.

    needs to be:
    23 + .5 + 8/100 + 8/100(1/10) + 8/100(1/10)^2...etc.

    This corresponds to a geometric series with a=8/100 and r=1/10

    so the fraction would be the sum of the series (using a/(1-r)) plus 23.5

    23.5+ (8/100)/(1-1/10)=23.5+8/90=47/2+8/90=2123/90
     
  4. Apr 28, 2010 #3
    Ohh, of course I'd make a silly mistake like that...

    Thanks!
     
  5. Apr 28, 2010 #4
    There's an easier way, as a matter of fact.

    Let a=23.5888... (I'm using ellipses to represent repeating decimals.)
    Then 10a=235.888...
    and 100a=2358.888...
    Subtraction will get rid of our repeating decimal.
    Therefore, 90a=2123
    a=2123/90.

    23.5888...=2123/90

    I got this from Numbers, Rational and Irration by Ivan Niven, which I recommend. It doesn't use any advanced mathematics.
     
  6. Apr 28, 2010 #5
    How did you jump from 100a=2358.888 to 90a=2123?
     
  7. Apr 28, 2010 #6
    100a-10a


    good call on the easier way.
     
  8. Apr 28, 2010 #7
    a=8/10

    10*a=8

    You'd obviously get 90a on the left but how do I know to use 10 and multiply by 100 and how does 2358.888888888888 - 8 = 2123?
     
  9. Apr 28, 2010 #8

    Mentallic

    User Avatar
    Homework Helper

    a=23.5888...
    not 8/10.

    I have no idea where you got that from.

    Anyway, if

    a = 23.588...
    10a = 235.88...
    100a = 2358.88...

    Notice now that if we do 100a-10a, the 8's repeating will cancel each other out.

    2358.888...
    -235.888...
    =2123.00...
    =2123

    But 100a-10a=90a, so if 90a=2123 then a=2123/90.

    Try it in the calculator :smile:
     
  10. Apr 29, 2010 #9

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you have a single digit repeating infinitely it can be reduced to finding the representation of .111...

    You should be able to find that with a bit of playing.
     
  11. Apr 29, 2010 #10
    Awesome :D Thank you jtyler, brainy, mentallic, and integral!

    These techniques will all help me significantly on my test/exam coming up.
     
  12. Apr 29, 2010 #11
    No problem. Glad to help.
     
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