Theorem of Midy

In Mathematical, the Theorem of Midy , thus called in homage to the Mathematician French E. Midy, is a statement concerning the decimal development of the Fraction S has / p with p a Prime number and has / p is the development in recurring Décimale with a even Period. If the period of the decimal representation of has / p is 2 N , then

$\ frac {has} {p} =0. \ overline {a_1a_2a_3 \ dowries a_na_ {n+1} \ dowries a_ {2n}}$

and the figures in second half of the decimal periodic development period are the complement compared to 9 of the corresponding figures in first half. In other words:

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

a_1 \ dowries a_n+a_ {n+1} \ dowries a_ {2n} =10^n-1.

For example

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

Theorem of wide Midy

If K is a unspecified Diviseur of the period of the decimal expansion of has / p (with p still first) then the Theorem of Midy can be generalized in the following way. The Théorème of wide Midy states that if one period of the decimal representation of has / p is divided into blocks of size K then the sum of these blocks is a multiple of

10^k-1

. Who more is, if K is worth 2 or 3, the sum of the blocks is worth

10^k-1

exactly.

For example

$\ frac {1} {19} =0. \overline{052631578947368421}$

one period 18 has. By dividing one period into blocks of size 6 or 3 and into summoning, one finds:

$052631+578947+368421=999999$

052+631+578+947+368+421=2997=3\times999.

Theorem of Midy in other bases

The theorem of Midy and its extensions do not depend on particular properties of the decimal expansion, because it still goes in any bases B , has condition of replacing

10^k-1

b^k-1

and of carrying out the operations of addition in the 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 the Theorem of Midy

Short evidence can be given by using results of the Theorie of the groups. However, one can also show this theorem by using the Elementary algebra and the modular Arithmétique:

Either p a prime number and has / p a fraction ranging between 0 and 1. Let us suppose that the expansion of has / p bases B of it is of period L , then

$\ frac {has} {p} = a_l} _b$

\ Rightarrow \ frac {has} {p} b^l= a_l. \ overline {a_1a_2 \ dowries a_l} _b

\ Rightarrow \ frac {has} {p} b^l=N+ a_l} _b=N+ \ frac {has} {p}

\ Rightarrow \ frac {has} {p} = \ frac {NR} {b^l-1}

or NR is the entirety whose writing bases B of it is thereafter defined has ₁ has ₂… has _L.

Let us note that B ^L − 1 is a Multiple of p because ( B ^L− 1) has / p is an entirety. Moreover, B ^N− 1 is not not a multiple of p for all the values of N smaller than L , because if not the period of the expansion in base B of has / p would be smaller than L .

Now let us suppose that L = HK . Then B ^L− 1 is a multiple of B ^K − 1. Let us pose B ^L − 1 = m ( B ^K − 1), then

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

But B ^L− 1 is a multiple of p ; B ^K− 1 is not not a multiple of p (because K is smaller than L ); and p is first; sonc m must be a multiple of p and

$\ frac {amndt} {p} = \ frac {NR} {b^k-1}$

is an entirety. In other words:

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

Now, let us cut out has ₁ has ₂… has _l in H shares of equal size has K , and pose that these shares are the writing bases B of it entireties NR ₀… NR _{H − 1}, then

$N_ {h-1} = a_k_b$

N_ {H2} = a_2k_b

.

.

N_0= a_l_b

To show the Theorem of Midy extended in base B we must show that the sum of the whole H NR _i is a multiple of B ^K − 1.

As B ^K is adequate has 1 modulo B ^K− 1, any power of B ^K will be also adequate has 1 modulo B ^K − 1. thus

$N= \ sum_ {i=0} ^ {h-1} N_ib^ {ik} = \ sum_ {i=0} ^ {h-1} N_i (b^ {K}) ^i$

\ Rightarrow NR \ 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}

what proves Théorème of Midy extended in base B .

To prove the theorem of original Midy, it is enough to take the particular case where H = 2. NR ₀ and NR ₁ is all the 2 represented by a sequence of K figures in base B , therefore they satisfy all 2

$0 \ Leq N_i \ Leq b^k-1.$

NR ₀ and NR ₁ cannot all 2 be null (if not has / p = 0) and cannot all 2 be equal has B ^K − 1 (if not has / p = 1), therefore

$0 < N_0+N_1 < 2 (b^k-1)$

and as NR ₀ + NR ₁ is a multiple of B ^K − 1, it comes that

$N_0+N_1 = b^k-1.$

References

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