# Midy's theorem

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 (sequence A028416 in the OEIS). If the period of the decimal representation of a/p is 2n, de modo a {estilo de exibição {fratura {uma}{p}}=0.{overline {uma_{1}uma_{2}uma_{3}dots a_{n}uma_{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. Em outras palavras, {estilo de exibição a_{eu}+uma_{i+n}=9} {estilo de exibição a_{1}dots a_{n}+uma_{n+1}dots a_{2n}=10^{n}-1.} Por exemplo, {estilo de exibição {fratura {1}{13}}=0.{overline {076923}}{texto{ e }}076+923=999.} {estilo de exibição {fratura {1}{17}}=0.{overline {0588235294117647}}{texto{ e }}05882352+94117647=99999999.} Conteúdo 1 Extended Midy's theorem 2 Midy's theorem in other bases 3 Proof of Midy's theorem 4 Corolário 5 Notas 6 Referências 7 External links 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 k-digit numbers, then their sum is a multiple of 10k − 1.

Por exemplo, {estilo de exibição {fratura {1}{19}}=0.{overline {052631578947368421}}} has a period of 18. Dividing the repeating portion into 6-digit numbers and summing them gives {displaystyle 052631+578947+368421=999999.} De forma similar, dividing the repeating portion into 3-digit numbers and summing them gives {displaystyle 052+631+578+947+368+421=2997=3times 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 10k − 1 with bk − 1 and carry out addition in base b.

Por exemplo, in octal {estilo de exibição {começar{alinhado}&{fratura {1}{19}}=0.{overline {032745}}_{8}\[8pt]&032_{8}+745_{8}=777_{8}\[8pt]&03_{8}+27_{8}+45_{8}=77_{8}.fim{alinhado}}} In duodecimal (using inverted two and three for ten and eleven, respectivamente) {estilo de exibição {começar{alinhado}&{fratura {1}{19}}=0.{overline {076{matemática {E}}45}}_{12}\[8pt]&076_{12}+{matemática {E}}45_{12}={matemática {EEE}}_{12}\[8pt]&07_{12}+6{matemática {E}}_{12}+45_{12}={matemática {EE}}_{12}fim{alinhado}}} Proof of Midy's theorem Short proofs of Midy's theorem can be given using results from group theory. No entanto, 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 e 1. Suppose the expansion of a/p in base b has a period of ℓ, assim {estilo de exibição {começar{alinhado}&{fratura {uma}{p}}=[0.{overline {uma_{1}uma_{2}dots a_{bem }}}]_{b}\[6pt]&Rightarrow {fratura {uma}{p}}b^{bem }=[uma_{1}uma_{2}dots a_{bem }.{overline {uma_{1}uma_{2}dots a_{bem }}}]_{b}\[6pt]&Rightarrow {fratura {uma}{p}}b^{bem }=N+[0.{overline {uma_{1}uma_{2}dots a_{bem }}}]_{b}=N+{fratura {uma}{p}}\[6pt]&Rightarrow {fratura {uma}{p}}={fratura {N}{b^{bem }-1}}fim{alinhado}}} 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. (Para ver isso, substitute x for bk; then bℓ = xh and x − 1 is a factor of xh − 1. ) Say b ℓ − 1 = m(bk − 1), assim {estilo de exibição {fratura {uma}{p}}={fratura {N}{m(b^{k}-1)}}.} 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 {estilo de exibição {fratura {am}{p}}={fratura {N}{b^{k}-1}}} is an integer. Em outras palavras, {displaystyle Nequiv 0{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, de modo a {estilo de exibição {começar{alinhado}N_{h-1}&=[uma_{1}dots a_{k}]_{b}\N_{h-2}&=[uma_{k+1}dots a_{2k}]_{b}\&{} vdots \N_{0}&=[uma_{l-k+1}dots a_{eu}]_{b}fim{alinhado}}} 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. Então {displaystyle N=sum _{i=0}^{h-1}N_{eu}b^{ik}=soma _{i=0}^{h-1}N_{eu}(b^{k})^{eu}} {displaystyle Rightarrow Nequiv sum _{i=0}^{h-1}N_{eu}{pmod {b^{k}-1}}} {displaystyle Rightarrow sum _{i=0}^{h-1}N_{eu}equivalente 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 {displaystyle 0leq N_{eu}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), assim {estilo de exibição 0

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