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Proof of n/7, n=1,2,3,4,5,6

  1. Jun 4, 2004 #1
    Hi there,

    There was an interesting problem proposed to me by some office mates a couple of days ago: "Prove that n/7, where n= 1, 2, 3, 4, 5, 6, is a repeating decimal where the digits repeat in a cyclical manner." I presented a more general proof of any fraction where the base is not 2^x*5^y will be repeating and non-terminating. However, the proof presented did not satisfy their cyclical requirement. Any suggestions on explaining to them that my proof is satisfactory or what should I add to a proof that would show the cyclical nature of all repeating decimals?

    Thanks

    Ken
     
  2. jcsd
  3. Jun 4, 2004 #2

    arildno

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    It is not true that all repeating decimals are cyclic in the way that the n/7-set is cyclic

    For example,
    1/9=0.11...
    2/9=0.22..

    There is no "cyclic" connection between 1/9 and 2/9!

    What you should do, is simply to perform the divisions 1/7,2/7,3/7,4/7,5/7,6/7
    and identify the cyclic connection between these decimal expansions.
     
  4. Jun 4, 2004 #3

    Gokul43201

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    If the decimal representation of p/q has repeating blocks of 'n' digits, then
    (p/q)*(10^n -1) =k, an integer.

    In the case of n/7, the repeating block has 6 digits. So all you need is for 999999 to be divisible by 7....and it is !

    From this you can see that if q is even you won't have repeating blocks. Also, no 999...9 will be divisible by 5. So if q=5 there will not be repeating blocks. And since 9 is divisible by both 3 and 9, the repeating block for those denominators will only be 1 digit long. This, I imagine does part of the job of your general proof for q=2^x*5^y.

    i.e: For q<10, repeating blocks will seen only if q=3,7,9, and the size of the repeating blocks are 1,6,1 respectively.
     
    Last edited: Jun 4, 2004
  5. Jun 4, 2004 #4

    Gokul43201

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    Oops...didn't realize that you were supposed to show that the all numbers of the form n/7 had cyclical repetitions of the SAME block of digits.

    Hmmm... it doesn't really say this anywhere in your post...
     
  6. Jun 4, 2004 #5

    arildno

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    Well that's how I interpreted "in a cyclical manner" (since I happen to know about this curious property of the n/7-fractions)

    However, it seems you have a lot more general knowledge than me about number theory;
    so I'd like to ask you:
    What conditions must be present in order for the fractions m/n (1<=m<n) represent a "cyclic" permutation of the same block of digits?

    I have a sneaking suspicion that n=7 is the only number where we have this property, but I'm not sure..
     
  7. Jun 4, 2004 #6

    arildno

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    Just for the record, we have:
    1/7=0.142857..
    2/7=0.285714..
    3/7=0.428571..
    4/7=0.571428..
    5/7=0.714285..
    6/7=0.857142..
     
  8. Jun 4, 2004 #7

    Gokul43201

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    I have a vanishingly small knowledge of number theory, and even that little bit is self taught...but I shall see where this goes...

    It's not hard to show that n/q and m/q can be written with repeating blocks (of size k) of the same numbers if there exists some r<k satisfying
    m*10^r == n (mod q) for all 0<n<m<q. {where a==b (mod c) means a-b is divisible by c}

    If your suspicion is right, r's can be found only for q=7.

    (I actually wrote the proof down till this point...and then I accidentally hit some button and lost everything...will do it again if required.)

    If we restrict q to [1..10], then it's easy to eliminate q=2,3,4,5,6,8,9,10.

    For example, take q=3 : 10==1 (mod 3) => 10^r == 1 (mod 3) for all r => m*10^r==m (mod 3).
    So if we pick some (m,n) with m<>n, then there will be no r that satisfies the required condition.

    Finding all q's in Z for which r's can be found seems harder. It is easy to show that q must be odd. Further, q will need to be prime for n/q to have a decimal representation with repeating blocks.

    But for now, this is as far as I will try to go.
     
    Last edited: Jun 4, 2004
  9. Jun 4, 2004 #8

    Gokul43201

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    Also, the above condition can be shown to be satisfied by q=7. There are only 36 pairs of (m,n) to check and clearly the condition for (m,m) is satisfied by r=6, so there's ONLY 30 pairs to go !

    So that answers imathgeek's original question. Perhaps his general proof is the way to go. That would be nice to see.

    I'm sure there's a nicer way to prove this.
     
    Last edited: Jun 4, 2004
  10. Jun 4, 2004 #9

    Gokul43201

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    Uh Oh...check out 17,47.
     
  11. Jun 4, 2004 #10

    arildno

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    Cool, I'll ponder on this for a while..:smile:
     
  12. Jun 4, 2004 #11
    oops.

    yeah, you're right. I wasn't thinking too well before that first cup of coffee. Now, i am back to the proof.

    Pure math is fun, but I still enjoy applied math. :)

    Have fun,

    Ken
     
    Last edited: Jun 4, 2004
  13. Jun 4, 2004 #12
    I have done that. Thanks for the tip, though.
     
  14. Jun 4, 2004 #13

    CTS

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  15. Jun 4, 2004 #14

    Gokul43201

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    "No general method is known for finding full reptend primes. "

    Hmmm...a few wasted minutes, but I did enjoy them.
     
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