Midy's theorem

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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 recurring decimal expansion with an even period. If the period of the decimal representation of a/p is 2n, so that

\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 recurring decimal period are the 9s complement of the corresponding digits in its first half. In other words

ai + ai + n = 9
a_1\dots a_n+a_{n+1}\dots a_{2n}=10^n-1.

For example

\frac{1}{17}=0.\overline{0588235294117647}\mbox{ and }05882352+94117647=99999999.

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[edit] 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 10k − 1.

For example

\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

052631 + 578947 + 368421 = 999999
052+631+578+947+368+421=2997=3\times999.

[edit] 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

\frac{1}{19}=0.\overline{032745}_8
0328 + 7458 = 7778
038 + 278 + 458 = 778.

[edit] 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 theorm 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

\frac{a}{p}=[0.\overline{a_1a_2\dots a_l}]_b
\Rightarrow\frac{a}{p}b^l=[a_1a_2\dots a_l.\overline{a_1a_2\dots a_l}]_b
\Rightarrow\frac{a}{p}b^l=N+[0.\overline{a_1a_2\dots a_l}]_b=N+\frac{a}{p}
\Rightarrow\frac{a}{p}=\frac{N}{b^l-1}

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

Note that bl − 1 is a multiple of p because (bl−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 bl−1 is a multiple of bk − 1. Say bl − 1 = m(bk − 1), so

\frac{a}{p}=\frac{N}{m(b^k-1)}.

But bl−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

\frac{am}{p}=\frac{N}{b^k-1}

is an integer. In other words

N\equiv0\pmod{b^k-1}.

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

N_{h-1}=[a_1\dots a_k]_b
N_{h-2}=[a_{k+1}\dots a_2k]_b
.
.
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 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

N=\sum_{i=0}^{h-1}N_ib^{ik}=\sum_{i=0}^{h-1}N_i(b^{k})^i
\Rightarrow N \equiv \sum_{i=0}^{h-1}N_i \pmod{b^k-1}
\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

0 \leq N_i \leq b^k-1.

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

0 < N0 + N1 < 2(bk − 1)

and since N0 + N1 is a multiple of bk − 1, it follows that

N0 + N1 = bk − 1.

[edit] 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

[edit] External links

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