Prime denominators and nines complement

Let p be a prime. If the repeating decimal for the fraction a/p has even period, the the second half of the decimals are the 9’s complement of the first half. This is known as Midy’s theorem.

For a small example, take

1/7 = 0.142857142857…

and notice that 142 + 857 = 999. That is, 8, 5, and 7 are the nine’s complements of 1, 4, and 2 respectively.

For a larger example, we can use Mathematica to look at the decimal expansion of 6/47:

    In:  N[6/47, 60]
    Out: 0.127659574468085106382978723404255319148936170212765957446809

and we can confirm

    12765957446808510638297 + 
    87234042553191489361702 =
    99999999999999999999999

Let’s do another example with 6/43:

    In:  N[6/43, 50]
    Out: 0.13953488372093023255813953488372093023255813953488

Does the theorem hold here? The the hypothesis of the theorem doesn’t hold because the fraction has period 21, which is not even. Can the conclusion of the theorem hold anyway? No, because we can’t split 21 digits in half! But we can split 21 digits into three parts, and if we do we find

    1395348 + 8372093 + 232558 = 9999999

Let’s finish up with an example in another base. We’ll look at the fraction 3/7 in base 17.

    In:  BaseForm[N[3/7, 24], 17]
    Out: 0.74e9c274e9c274e9c2717

If we split the repeating part of the decimal into two parts we get 74e and 9c2, and these add up to the base 17 analog of all 9’s.

    In:  BaseForm[17^^74e + 17^^9c2, 17]
    Out: ggg17

That is, the base 17 numbers 9, c, and 2, are the g’s complements of 7, 4, and e respectively.

[The Mathematica operator ^^ interprets its right argument as a number whose base is given by the left argument. The BaseForm function takes a number as its first argument and returns its representation in the base given as its second argument.]

Related post: Casting out Z’s

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