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Proof: Numbers with repeating blocks of digits are rational

  1. Jun 4, 2004 #1
    Hi everyone,

    I need to prove that any number with a repeating block of digits is a rational number. Someone told me I should first find a method of constructing a rational number in the form a/b from a number with repeating blocks of digits (and to do it with very 'easy' numbers first). I'm still stumped though.

    For example, given 0.33333..., how do I show that it equals 1/3?
     
  2. jcsd
  3. Jun 4, 2004 #2

    jcsd

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    You can express 0.333.. as a geometric series:

    [tex]\sum^{\infty}_{n=1} \frac{3}{10^n} = 3\sum^{\infty}_{n=1} \left(\frac{1}{10}\right)^n[/tex]

    use:

    [tex]\sum^{\infty}_{n=1}r^n = \frac{r}{1-r}[/tex]

    [tex]3\left(\frac{\frac{1}{10}}{1 - \frac{1}{10}}\right) = \frac{1}{3}[/tex]
     
  4. Jun 4, 2004 #3
    Nice. I had totally forgotten about the geometric series. My head has overloaded with maths. Thanks again.
     
  5. Jun 4, 2004 #4

    Gokul43201

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    n=0.3333...
    10n=3.3333...
    =>10n-n=9n=3.0
    =>n=3/9=1/3
     
  6. Jun 4, 2004 #5

    Gokul43201

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    Let n == 0.abc...kabc...kabc... repeating blocks of (abc...k), each block having r digits
    Then n*10^r == abc...k(point)abc...kabc...kabc... i.e. move the decimal point r places to the right.
    Now subtract, n*(10^r - 1)==abc...k digits after the decimal point vanish
    So n== abc...k/(10^r - 1)= p/q, a rational number

    QED

    Plz excuse the freedom I've exercised with notation.
     
  7. Jun 4, 2004 #6

    Gokul43201

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    Alternatively, jcsd's geometric sum method can be generalized for repeating blocks of any size.

    PS: Also look at recent post on n/7, n=1,2,...6
     
  8. Jun 5, 2004 #7

    CTS

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    .ABC...Z (with repeating length L)=
    [tex]A*\sum_{k=1}^\infty \frac{1}{10^k^L}+B*\sum_{k=1}^\infty \frac{1}{10^k^L*10}+C*\sum_{k=1}^\infty \frac{1}{10^k^L*10^2}+...+[/tex][tex]Z*\sum_{k=1}^\infty \frac{1}{10^k^L*10^L/10}[/tex]

    Because A, B,..., Z, are rational and [tex]\sum_{k=1}^\infty \frac{1}{10^k^X}[/tex] is rational for any X, the above sum is also rational.
     
  9. Jun 9, 2004 #8
    have you tried using different bases other than 10?
     
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