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Write # as a ratio of two integers

  1. Apr 18, 2004 #1
    Problem: Write the number 3.1415999999999... as a ratio of two integers.

    In my book, they have a similar example, but using 2.3171717... And this is how they solved that problem.

    2.3171717... = 2.3 + (17/10^3) + (17/10^5) + (17/10^7) + ...

    After the first term we have a geometric series with a = (17/10^3) and r = (1/10^2). Therefore:

    2.3171717... = 2.3 + [(17/10^3) / (1 - (1/10^2))] = 2.3 + [(17/1000)/(99/100)] = (23/10) + (17/990) = 1147/495 == 2.3171717...

    Thinking I could follow the similar steps with a different number, I thought it would work, but it really isn't.

    This is what I did:

    3.1415999999999... = 3.1415 + (99/10^6) + (99/10^8) + (99/10^10)

    a = (99/10^6) and r = (1/10^2)

    3.1415 + [(99/10^6) / (1 - (1/10^2))] = 3.1415 + [(99/1000000)/(99/100) = (31415/10000) + (1/10000) = (31416/10000) = 3.1416 which isn't 3.1415999999999...

    What am I doing wrong?

    Thanks
     
  2. jcsd
  3. Apr 18, 2004 #2

    Hurkyl

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    Actually, it is.


    P.S. any particular reason you were grouping the nines in pairs?
     
  4. Apr 18, 2004 #3
    technically, it is, but is that correct though? and no, there was no reason i paired them up.
     
  5. Apr 18, 2004 #4
    3.1416=3.141599999999... is very true. So any fractional representation of one is a representation of the other. In fact, that's how I would have solved this problem; I wouldn't have bothered with an infinite geometric series in this case.
     
  6. Apr 20, 2004 #5

    HallsofIvy

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    Techically it's true but is it correct??? Is that what you are asking?

    "True" is "true"- there is no "technically"! And if it's true, then it's correct.
     
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