Midy's theorem

In mathematics, Midy's theorem, named after French mathematician E. Midy^[1], 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

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{a}{p}=0.\overline{a_1a_2a_3\dots a_na_{n+1}\dots a_{2n}}

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

Failed to parse (Missing texvc executable; please see math/README to configure.): a_i+a_{i+n}=9

Failed to parse (Missing texvc executable; please see math/README to configure.): a_1\dots a_n+a_{n+1}\dots a_{2n}=10^n-1.

For example

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{17}=0.\overline{0588235294117647}\mbox{ and }05882352+94117647=99999999.

Extended Midy's theorem

If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime) then Midy's theorem can be generalised as follows. The extended Midy's theorem^[2] states that if the repeating portion of the decimal expansion of a/p is divided into blocks of length k then the sum of these blocks will be a multiple of Failed to parse (Missing texvc executable; please see math/README to configure.): 10^k-1 .

For example

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{19}=0.\overline{052631578947368421}

has a period of 18. Dividing the repeating portion into blocks of length 6 or 3 and summing gives

Failed to parse (Missing texvc executable; please see math/README to configure.): 052631+578947+368421=999999

Failed to parse (Missing texvc executable; please see math/README to configure.): 052+631+578+947+368+421=2997=3\times999.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): 10^k-1

with Failed to parse (Missing texvc executable; please see math/README to configure.): b^k-1
and carry out addition in base b. For example, in octal

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{19}=0.\overline{032745}_8

Failed to parse (Missing texvc executable; please see math/README to configure.): 032_8+745_8=777_8

Failed to parse (Missing texvc executable; please see math/README to configure.): 03_8+27_8+45_8=77_8.

Proof of Midy's theorem

Short proofs of Midy's theorem can be given using results from group theory. However, it is also possible to prove Midy's theorem using elementary algebra and modular arithmetic:

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{a}{p}=[0.\overline{a_1a_2\dots a_l}]_b

Failed to parse (Missing texvc executable; please see math/README to configure.): \Rightarrow\frac{a}{p}b^l=[a_1a_2\dots a_l.\overline{a_1a_2\dots a_l}]_b

Failed to parse (Missing texvc executable; please see math/README to configure.): \Rightarrow\frac{a}{p}b^l=N+[0.\overline{a_1a_2\dots a_l}]_b=N+\frac{a}{p}

Failed to parse (Missing texvc executable; please see math/README to configure.): \Rightarrow\frac{a}{p}=\frac{N}{b^l-1}

where N is the integer whose expansion in base b is the string a₁a₂...a_l.

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 b^k − 1. Say b^l − 1 = m(b^k − 1), so

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{a}{p}=\frac{N}{m(b^k-1)}.

But b^l−1 is a multiple of p; b^k−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

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{am}{p}=\frac{N}{b^k-1}

is an integer. In other words

Failed to parse (Missing texvc executable; please see math/README to configure.): N\equiv0\pmod{b^k-1}.

Now split the string a₁a₂...a_l into h equal parts of length k, and let these represent the integers N₀...N_h − 1 in base b, so that

Failed to parse (Missing texvc executable; please see math/README to configure.): N_{h-1}=[a_1\dots a_k]_b

Failed to parse (Missing texvc executable; please see math/README to configure.): N_{h-2}=[a_{k+1}\dots a_{2k}]_b

Failed to parse (Missing texvc executable; please see math/README to configure.): .

Failed to parse (Missing texvc executable; please see math/README to configure.): .

Failed to parse (Missing texvc executable; please see math/README to configure.): N_0=[a_{l-k+1}\dots a_l]_b

To prove Midy's extended theorem in base b we must show that the sum of the h integers N_i is a multiple of b^k − 1.

Since b^k is congruent to 1 modulo b^k−1, any power of b^k will also be congruent to 1 modulo b^k − 1. So

Failed to parse (Missing texvc executable; please see math/README to configure.): N=\sum_{i=0}^{h-1}N_ib^{ik}=\sum_{i=0}^{h-1}N_i(b^{k})^i

Failed to parse (Missing texvc executable; please see math/README to configure.): \Rightarrow N \equiv \sum_{i=0}^{h-1}N_i \pmod{b^k-1}

Failed to parse (Missing texvc executable; please see math/README to configure.): \Rightarrow \sum_{i=0}^{h-1}N_i \equiv 0 \pmod{b^k-1}

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

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 \leq N_i \leq b^k-1.

N₀ and N₁ cannot both equal 0 (otherwise a/p = 0) and cannot both equal b^k − 1 (otherwise a/p = 1), so

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 < N_0+N_1 < 2(b^k-1)

and since N₀ + N₁ is a multiple of b^k − 1, it follows that

Failed to parse (Missing texvc executable; please see math/README to configure.): N_0+N_1 = b^k-1.

References

1. ^ A Theorem on Repeating Decimals; W. G. Leavitt; American Mathematical Monthly, Vol. 74, No. 6 (Jun. - Jul., 1967) , pp. 669-673
2. ^ Extended Midy's Theorem, Bassam Abdul-Baki, 2005

External links

• Eric W. Weisstein, Midy's Theorem at MathWorld.fr:Théorème de Midy