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| In [[mathematics]], '''Midy's theorem''', named after [[France|French]] [[mathematician]] E. Midy,<ref>{{cite journal|last=Leavitt|first=William G.|title=A Theorem on Repeating Decimals|journal=The American Mathematical Monthly|date=June 1967|volume=74|issue=6|pages=669–673|url=http://digitalcommons.unl.edu/mathfacpub/48/|publisher=Mathematical Association of America|doi=10.2307/2314251}}</ref><ref>{{cite web|last=Kemeny|first=John|title=The Secret Theorem of M. E. Midy = Casting In Nines|url=http://johnkemeny.com/blog/?p=393|accessdate=27 November 2011}}</ref> is a statement about the [[decimal]] expansion of [[Fraction (mathematics)|fraction]]s ''a''/''p'' where ''p'' is a [[prime number|prime]] and ''a''/''p'' has a [[repeating decimal]] expansion with an even period. If the period of the decimal representation of ''a''/''p'' is 2''n'', so that
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| :<math>\frac{a}{p}=0.\overline{a_1a_2a_3\dots a_na_{n+1}\dots a_{2n}}</math>
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| then the digits in the second half of the repeating decimal period are the [[method of complements#Numeric complements|9s complement]] of the corresponding digits in its first half. In other words
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| :<math>a_i+a_{i+n}=9 \, </math>
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| :<math>a_1\dots a_n+a_{n+1}\dots a_{2n}=10^n-1. \, </math>
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| For example
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| :<math>\frac{1}{17}=0.\overline{0588235294117647}\text{ and }05882352+94117647=99999999. \, </math>
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| ==Midy's theorem in other bases==
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| Midy's theorem and its extension do not depend on special properties of the decimal expansion, but work equally well in any [[base (exponentiation)|base]] ''b'', provided we replace 10<sup>''k''</sup> − 1 with ''b''<sup>''k''</sup> − 1 and carry out addition in base ''b''. For example, in [[octal]]
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| :<math>\frac{1}{19}=0.\overline{032745}_8</math>
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| :<math>032_8+745_8=777_8 \, </math>
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| :<math>03_8+27_8+45_8=77_8. \, </math>
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| ==Proof of Midy's theorem==
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| 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]]:
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| 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
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| :<math>
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| \begin{align}
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| & \frac{a}{p} = [0.\overline{a_1a_2\dots a_\ell}]_b \\[6pt]
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| & \Rightarrow\frac{a}{p}b^\ell = [a_1a_2\dots a_\ell.\overline{a_1a_2\dots a_\ell}]_b \\[6pt]
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| & \Rightarrow\frac{a}{p}b^\ell = N+[0.\overline{a_1a_2\dots a_\ell}]_b=N+\frac{a}{p} \\[6pt]
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| & \Rightarrow\frac{a}{p} = \frac{N}{b^\ell-1}
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| \end{align}
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| </math>
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| where ''N'' is the integer whose expansion in base ''b'' is the string ''a''<sub>1</sub>''a''<sub>2</sub>...''a''<sub>''ℓ''</sub>. | |
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| Note that ''b''<sup> ''ℓ''</sup> − 1 is a multiple of ''p'' because (''b''<sup> ''ℓ''</sup> − 1)''a''/''p'' is an integer. Also ''b''<sup>''n''</sup>−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''.
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| Now suppose that ''ℓ'' = ''hk''. Then ''b''<sup> ''ℓ''</sup> − 1 is a multiple of ''b''<sup>''k''</sup> − 1. (To see this, substitute ''x'' for ''b''<sup>''k''</sup>; then ''b''<sup>''ℓ''</sup> = ''x''<sup>''h''</sup> and ''x'' − 1 is a factor of ''x''<sup>''h''</sup> − 1. ) Say ''b''<sup> ''ℓ''</sup> − 1 = ''m''(''b''<sup>''k''</sup> − 1), so
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| :<math>\frac{a}{p}=\frac{N}{m(b^k-1)}.</math>
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| But ''b''<sup> ''ℓ''</sup> − 1 is a multiple of ''p''; ''b''<sup>''k''</sup> − 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
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| :<math>\frac{am}{p}=\frac{N}{b^k-1}</math>
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| is an integer. In other words
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| :<math>N\equiv0\pmod{b^k-1}. \, </math>
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| Now split the string ''a''<sub>1</sub>''a''<sub>2</sub>...''a''<sub>''ℓ''</sub> into ''h'' equal parts of length ''k'', and let these represent the integers ''N''<sub>0</sub>...''N''<sub>''h'' − 1</sub> in base ''b'', so that
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| :<math>
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| \begin{align}
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| N_{h-1} & = [a_1\dots a_k]_b \\
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| N_{h-2} & = [a_{k+1}\dots a_{2k}]_b \\
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| & {}\ \ \vdots \\
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| N_0 & = [a_{l-k+1}\dots a_l]_b
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| \end{align}
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| </math>
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| To prove Midy's extended theorem in base ''b'' we must show that the sum of the ''h'' integers ''N''<sub>''i''</sub> is a multiple of ''b''<sup>''k''</sup> − 1.
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| Since ''b''<sup>''k''</sup> is congruent to 1 modulo ''b''<sup>''k''</sup> − 1, any power of ''b''<sup>''k''</sup> will also be congruent to 1 modulo ''b''<sup>''k''</sup> − 1. So
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| :<math>N=\sum_{i=0}^{h-1}N_ib^{ik}=\sum_{i=0}^{h-1}N_i(b^{k})^i</math> | |
| :<math>\Rightarrow N \equiv \sum_{i=0}^{h-1}N_i \pmod{b^k-1}</math>
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| :<math>\Rightarrow \sum_{i=0}^{h-1}N_i \equiv 0 \pmod{b^k-1}</math>
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| which proves Midy's extended theorem in base ''b''.
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| To prove the original Midy's theorem, take the special case where ''h'' = 2. Note that ''N''<sub>0</sub> and ''N''<sub>1</sub> are both represented by strings of ''k'' digits in base ''b'' so both satisfy
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| :<math>0 \leq N_i \leq b^k-1. \, </math>
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| ''N''<sub>0</sub> and ''N''<sub>1</sub> cannot both equal 0 (otherwise ''a''/''p'' = 0) and cannot both equal ''b''<sup>''k''</sup> − 1 (otherwise ''a''/''p'' = 1), so
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| :<math>0 < N_0+N_1 < 2(b^k-1) \, </math>
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| and since ''N''<sub>0</sub> + ''N''<sub>1</sub> is a multiple of ''b''<sup>''k''</sup> − 1, it follows that
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| :<math>N_0+N_1 = b^k-1. \, </math>
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| ==Notes==
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| <references/>
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| ==References==
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| *Rademacher, H. and Toeplitz, O. The Enjoyment of Mathematics: Selections from Mathematics for the Amateur. Princeton, NJ: Princeton University Press, pp. 158-160, 1957.
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| *E. Midy, De Quelques Propriétés des Nombres et des Fractions Décimales Périodiques.
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| College of Nantes, France: 1836.
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| *[[Kenneth A. Ross|Ross, Kenneth A.]] Repeating decimals: a period piece. Math. Mag. 83 (2010), no. 1, 33–45.
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| ==External links==
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| * {{MathWorld|urlname=MidysTheorem|title=Midy's Theorem}}
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| [[Category:Theorems in number theory]]
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| [[Category:Fractions]]
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| [[Category:Numeral systems]]
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Greetings. Allow me begin by telling you the author's name - Phebe. South Dakota is exactly where I've usually been residing. For years I've been working as a payroll clerk. To gather coins is 1 of over the counter std test - source website, things I love most.