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Rational Numbers

  1. Jul 29, 2008 #1
    It can easily be shown that the recurring decimal x = 1.123123... is rational, as follows:
    [tex]10^{3}x-x = 1123.123...-1.123123...=1122[/tex] => [tex]x = \frac{1122}{999} \in Q[/tex]

    Show that the recurring decimals 0.3712437127... and 0.9999999...are rational numbers.

    3. The attempt at a solution

    I'm not quite sure what the question is asking as I had never seen a question like this before!
    Does the question mean what devided by what equals 0.3712437127... and 0.9999999...?

    I don't know the method for this & I appreciate some guidance if anyone here knows how to do it.
  2. jcsd
  3. Jul 29, 2008 #2


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    Yes, you have to find out fraction that will give given recurring decimal. You are already given a hint - try to understand example you were given. Use exactly the same approach.

    Note: I am assuming it is just a typo, but your first number (0.37....) is not a recurring decimal.

    Expect surprises with 0.99999... :wink:
  4. Jul 29, 2008 #3


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    I presume you are missing a "3" and that should be 0.3712371237.. so that the "3712" is the recurring block.

    They gave an exaple of the method just before asking the quetion! But here's another example, with more detail:

    0.234323432343... where it is the "2343" that repeats.

    The first thing I do is note that the recurring block has 4 digits- and I know that I can "move the decimal point 4 places" by multiplying by 104= 10000. If I write x= 0.23432343... and multiply by 10000 I get 10000x= 2342.23432343... Notice that the first "2343" block has been moved in front of the decimal point and its place has been taken by the second "2343" block. And, because the repetition never stops, every "2343" block is replaced by another identical block: the decimal part of 10000x is still just .23432343... . If we subtract the equation x= 0.23432343... from 10000x= 2343.23432343..., on the right we get 10000x= 999x and on the right the whole number 2343: 999x= 2343 so
    x= 2343/999. That can be reduced- both numerator and denominator are divisible by 3- but at this point we have shown that the number can be written as a fraction.

    Here's another, slightly harder example: 0.153221622162216... where it is the "2216" that repeats. The reason this is slightly harder is that we must take care of the "153" that we have before the repeating part. That has 3 digits and multiplying by 103= 1000 will move the decimal point 3 places:
    If we write x= 0.15322162216...
    and multiply the equation by 1000 we get
    1000x= 153.221622162216...
    Now we have only the repeating block after the decimal point- and that repeating block has, again, 4 digits. If we multiply again by 104= 10000 we get

    10000000x= 1532216.22162216....
    Again, that each block in that infinitely repeating decimal has been replaced by the next one. The decimal parts of 1532216.22162216... and 153.22162216... are exactly the same.
    Subracting 1000x= 153.22162216... from 10000000x= 1532216.22162216..., the decimal parts cancel and we have
    (10000000- 1000)x= 1532216- 153 or
    9999000x= 1532063.

    Dividing both sides by 9999000 we get x= 1632064/9999000 which perhaps can be reduced but we have succeeded in writing the number as a fraction, proving that it is a rational number.

    Now apply the same ideas to x= 0.3712371237.. where "3712" is repeating and
    x= 0.99999999.... where "9" is repeating.
  5. Jul 29, 2008 #4
    Well, thank you, I try one of them now.

    In 0.3712437124... it is the 37124 which is repeating, it has five digits so we multiply by [tex]10^{5}[/tex] ie. 100000 to get one of those on the other side: 37124.37124...
    Now we subtract: 37124.37124... - 0.3712437124 = 37124

    Now what should we devide it by? 100000?

    In your examples I didn't quite understand how you got the 999 bit!
  6. Jul 29, 2008 #5


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    Subtract one equation from another. If x= .3712437124... then 100000x= 37124.371234.... Subtract one equation from another.

  7. Jul 29, 2008 #6
    That's what I did!
    => [tex]10^{5}x-x = 37124.37124... - 0.3712437124...= 37124[/tex]

    Now I need to devide 37124 by something & that's where I'm stuck. :frown:
  8. Jul 29, 2008 #7


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  9. Jul 29, 2008 #8
    Do you mean;

    [tex]10^5x-x= 37124.37124... - 0.3712437124...[/tex]

    [tex]100000x-x= 37124[/tex]

    [tex]99999x= 37124[/tex]

    [tex]x= \frac{37124}{99999} \in Q[/tex]

    Am I right or what?
  10. Jul 29, 2008 #9


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    Looks OK. Now 0.99999...
  11. Jul 30, 2008 #10

    Yes, now I have to show that 0.9999999 is rational. I want to see the surprises you were talking about!

    i don't see any recurring blocks, so i think i should just take 9 because it is 9 that is repeating.

    [tex]10x-x= 9.9999999... - 0.9999999...[/tex]
    [tex]9x= 9[/tex]
    [tex]x= \frac{9}{9}[/tex]
    = 1 ???

    Does this mean it is not rational?Help
  12. Jul 30, 2008 #11
    Re: 0.99999....

    1 is certainly a rational number, so on what are you basing your conclusion that .999... is irrational?
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