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Finding the rational expression of a repeating decimal

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data Express the repeating decimal as a series, and find the rational number that it represents
    1) 3.2[itex]\overline{394}[/itex]

    2. Relevant equations Geometric Series (a/1-r)



    3. The attempt at a solution I tried putting the value into a series by having n=1 as it goes to infinity
    Ʃ3.2 + 394(0.0001)n
    but I have a feeling that is wrong. Is there anyway to just isolate the 0.0394 as the repeating value?

    I also tried using (3.2 + 394) as "a", and (0.0001) as "r"

    putting this into the geometric series formula to find the rational number I got

    ((16/5) +394)/(1-0.0001) which came out to be 1986/0.9999

    The Correct answer for the rational expression is 16,181/4,995

    but I don't know how to get there. I have a feeling that I chose my "a" and "r" incorrectly
     
  2. jcsd
  3. Mar 23, 2012 #2
    Do you know the standard trick for converting repeating decimals? For example if I give you .123123123... where "123" is the repeating block, do you know an *EASY* way to immediately represent that repeating decimal as n/m? If yes, this is an application of that idea; if not ... well, there's a really EASY trick for this. Probably in your class notes somewhere.

    (note -- of course if you don't know the trick, my post wasn't helpful ... but I can't think of a hint that only goes part of the way there.)
     
  4. Mar 23, 2012 #3

    tiny-tim

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    Hi OnceKnown! :smile:
    No, just the 394 …

    you can add the 3.2 later! :wink:
     
  5. Mar 23, 2012 #4
    Hi Steve,

    I wouldn't know about this "Easy trick" you're speaking of sorry lol. Thanks for the help though.
     
  6. Mar 23, 2012 #5
    Hi Tim,

    so just the 394 as "a"

    would represent 394/(1-0.0001), but that would bring me to 394/0.9999 still, which I'm still stuck.
     
  7. Mar 23, 2012 #6

    HallsofIvy

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    It's pretty straightforward. The original number is x= 3.2394394394... so 10x= 32.394394..., the multiplication moving the decimal point one place. Multiplying by another [itex]10^3= 1000[/itex] moves the decimal point another three places: 10000x= 32394.394394394...

    Now subtract: 10000x- 10x= 9990x= 32362. The "decimal part" cancels because of that repetition.
     
  8. Mar 23, 2012 #7
    Thank you Halls,

    This is different approach from what we are learning in calc class but it works.
     
  9. Mar 23, 2012 #8
    Once you know that trick you can shortcut it by just putting the repeating block over the same number of 9's. So .123123123... = 123/999, etc.
     
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