In mathematics, Midy's theorem, named after French mathematician E. Midy, is a statement about the decimal expansion of fractions a/p where p is a prime and a/p has a repeating decimal expansion with an even period. If the period of the decimal representation of a/p is 2n, so that
then the digits in the second half of the repeating decimal period are the 9s complement of the corresponding digits in its first half. In other words
Midy's theorem in other bases
Midy's theorem and its extension do not depend on special properties of the decimal expansion, but work equally well in any base b, provided we replace 10k − 1 with bk − 1 and carry out addition in base b. For example, in octal
Proof of Midy's theorem
Let p be a prime and a/p be a fraction between 0 and 1. Suppose the expansion of a/p in base b has a period of ℓ, so
where N is the integer whose expansion in base b is the string a1a2...aℓ.
Note that b ℓ − 1 is a multiple of p because (b ℓ − 1)a/p is an integer. Also bn−1 is not a multiple of p for any value of n less than ℓ, because otherwise the repeating period of a/p in base b would be less than ℓ.
Now suppose that ℓ = hk. Then b ℓ − 1 is a multiple of bk − 1. (To see this, substitute x for bk; then bℓ = xh and x − 1 is a factor of xh − 1. ) Say b ℓ − 1 = m(bk − 1), so
But b ℓ − 1 is a multiple of p; bk − 1 is not a multiple of p (because k is less than ℓ ); and p is a prime; so m must be a multiple of p and
is an integer. In other words
Now split the string a1a2...aℓ into h equal parts of length k, and let these represent the integers N0...Nh − 1 in base b, so that
To prove Midy's extended theorem in base b we must show that the sum of the h integers Ni is a multiple of bk − 1.
Since bk is congruent to 1 modulo bk − 1, any power of bk will also be congruent to 1 modulo bk − 1. So
which proves Midy's extended theorem in base b.
To prove the original Midy's theorem, take the special case where h = 2. Note that N0 and N1 are both represented by strings of k digits in base b so both satisfy
N0 and N1 cannot both equal 0 (otherwise a/p = 0) and cannot both equal bk − 1 (otherwise a/p = 1), so
and since N0 + N1 is a multiple of bk − 1, it follows that
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College of Nantes, France: 1836.
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