Midy's theorem

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

${\frac {a}{p}}=0.{\overline {a_{1}a_{2}a_{3}\dots a_{n}a_{n+1}\dots a_{2n}}}$

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

$a_{i}+a_{i+n}=9$

$a_{1}\dots a_{n}+a_{n+1}\dots a_{2n}=10^{n}-1.$

For example

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

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 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 $10^{k}-1$.

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

$052631+578947+368421=999999$

$052+631+578+947+368+421=2997=3\times 999.$

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 $10^{k}-1$ with $b^{k}-1$ and carry out addition in base b. For example, in octal

${\frac {1}{19}}=0.{\overline {032745}}_{8}$

$032_{8}+745_{8}=777_{8}$

$03_{8}+27_{8}+45_{8}=77_{8}.$

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 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

${\frac {a}{p}}=[0.{\overline {a_{1}a_{2}\dots a_{l}}}]_{b}$

$\Rightarrow {\frac {a}{p}}b^{l}=[a_{1}a_{2}\dots a_{l}.{\overline {a_{1}a_{2}\dots a_{l}}}]_{b}$

$\Rightarrow {\frac {a}{p}}b^{l}=N+[0.{\overline {a_{1}a_{2}\dots a_{l}}}]_{b}=N+{\frac {a}{p}}$

$\Rightarrow {\frac {a}{p}}={\frac {N}{b^{l}-1}}$

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

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

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

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

is an integer. In other words

$N\equiv 0{\pmod {b^{k}-1}}.$

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

$N_{h-1}=[a_{1}\dots a_{k}]_{b}$

$N_{h-2}=[a_{k+1}\dots a_{2}k]_{b}$

$.$

$.$

$N_{0}=[a_{l-k+1}\dots a_{l}]_{b}$

To prove Midy's extended theorem in base b we must show that the sum of the h integers 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

$N=\sum _{i=0}^{h-1}N_{i}b^{ik}=\sum _{i=0}^{h-1}N_{i}(b^{k})^{i}$

$\Rightarrow N\equiv \sum _{i=0}^{h-1}N_{i}{\pmod {b^{k}-1}}$

$\Rightarrow \sum _{i=0}^{h-1}N_{i}\equiv 0{\pmod {b^{k}-1}}$

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 b^k − 1 (otherwise a/p = 1), so

$0<N_{0}+N_{1}<2(b^{k}-1)$

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

$N_{0}+N_{1}=b^{k}-1.$

External links

• Weisstein, Eric W. "Midy's Theorem". MathWorld.