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Version 19 |
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Every rational number $\frac{m}{n}$, where $m$ and $n$ are positive integers, has an endless \PMlinkescapetext{periodic} {\em decimal expansion} (or {\em decadic expansion} -- according to Greek).\, The decimal expansion of $\frac{m}{n}$ means the series \PMlinkescapetext{presentation}
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Every rational number $\frac{m}{n}$, where $m$ and $n$ are positive integers, has an endless \PMlinkescapetext{periodic} {\em decimal expansion} (or {\em decadic expansion} -- according to Greek). \,The decimal expansion of $\frac{m}{n}$ means the series \PMlinkescapetext{presentation}
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| \begin{align} |
\begin{align} |
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\nu.\nu_1\nu_2\nu_3\ldots = \nu+10^{-1}\nu_1+10^{-2}\nu_2+10^{-3}\nu_3+\cdots
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\nu.\nu_1\nu_2\nu_3... = \nu+10^{-1}\nu_1+10^{-2}\nu_2+10^{-3}\nu_3+...
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| \end{align} |
\end{align} |
| where \,$\nu = \lfloor\frac{m}{n}\rfloor$\, is the \PMlinkname{integer part}{Floor} of $\frac{m}{n}$ and the integers $\nu_j$ satisfy \,$0 \leqq \nu_j < 10$. |
where \,$\nu = \lfloor\frac{m}{n}\rfloor$\, is the \PMlinkname{integer part}{Floor} of $\frac{m}{n}$ and the integers $\nu_j$ satisfy \,$0 \leqq \nu_j < 10$. |
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We may suppose that $m$ and $n$ are coprime (if necessary, reduce the fraction).\, Then the \PMlinkescapetext{length $l$ of the period} depends only on the denominator $n$.\, In the case that\, $\gcd(n,\,10) = 1$,\, the \PMlinkescapetext{period length} is the least positive integer $l$ such that $10^l\equiv 1 \pmod{n}$ (the \PMlinkescapetext{period length} does not change if we multiply the fraction by a suitable power of 10 and then reduce all prime factors of 10 from the denominator).\, In every case, the \PMlinkescapetext{period length} is a factor of the number $\varphi(n)$, where $\varphi$ is the Euler's totient function.
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We may suppose that $m$ and $n$ are coprime (if necessary, reduce the fraction). \,Then the \PMlinkescapetext{length $l$ of the period} depends only on the denominator $n$. \,In the case that \,$\gcd(n,\,10) = 1$, \,the \PMlinkescapetext{period length} is the least positive integer $l$ such that $10^l\equiv 1 \pmod{n}$ (the \PMlinkescapetext{period length} does not change if we multiply the fraction by a suitable power of 10 and then reduce all prime factors of 10 from the denominator). \,In every case, the \PMlinkescapetext{period length} is a factor of the number $\varphi(n)$, where $\varphi$ is the Euler's totient function.
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| \textbf{Examples} |
\textbf{Examples} |
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| $\frac{1}{8} = 0.125000... = 0.124999...$ (one-digit \PMlinkescapetext{periods}; N.B. two possibilities), |
$\frac{1}{8} = 0.125000... = 0.124999...$ (one-digit \PMlinkescapetext{periods}; N.B. two possibilities), |
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| $\frac{1}{12} = 0.08333...$ (one-digit per.), |
$\frac{1}{12} = 0.08333...$ (one-digit per.), |
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| $\frac{1}{37} = 0.'027'027'027'...$ (three-digit per.), |
$\frac{1}{37} = 0.'027'027'027'...$ (three-digit per.), |
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| $\frac{1}{82} = 0.0'12195'12195'12195'...$ (five-digit per.), |
$\frac{1}{82} = 0.0'12195'12195'12195'...$ (five-digit per.), |
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| $\frac{1}{25351} = 0.000039446...$ (hundred-digit per.) |
$\frac{1}{25351} = 0.000039446...$ (hundred-digit per.) |
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The tail of infinitely many 0's (as in 0.125000...) is of course usually not written out.\, Such a tail is possible only when $n$ has no other prime factors except prime factors of the base of the digit system in question.
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The tail of infinitely many 0's (as in 0.125000...) is of course usually not written out. \,Such a tail is possible only when $n$ has no other prime factors except prime factors of the base of the digit system in question.
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| If the tails of 0's are not accepted, then the digital expansion of every positive rational is unique (then e.g. 0.124999... is the only \PMlinkescapetext{expansion} for $\frac{1}{8}$ in the decimal system). |
If the tails of 0's are not accepted, then the digital expansion of every positive rational is unique (then e.g. 0.124999... is the only \PMlinkescapetext{expansion} for $\frac{1}{8}$ in the decimal system). |
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Completely similar results concern the digital expansions in any other positional digit system.\, Let the fraction $\frac{1}{31}$ be an example ($\varphi(31) = 30$); its \PMlinkescapetext{presentation} is
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Completely similar results concern the digital expansions in any other positional digit system. \,Let the fraction $\frac{1}{31}$ be an example ($\varphi(31) = 30$); its \PMlinkescapetext{presentation} is
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in the decadic (decimal) digit system\, $\frac{1}{31} = 0.'032258064516129'..._{\mathrm{ten}}$ \quad(15-digit per.),
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in the decadic (decimal) digit system \,$\frac{1}{31} = 0.'032258064516129'..._{\mathrm{ten}}$ \quad(15-digit per.),
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in the hexadic (senary) digit system\, $\frac{1}{51} = 0.'010545'010545'010545'..._{\mathrm{six}}$ \quad(6-digit per.),
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in the hexadic (senary) digit system \,$\frac{1}{51} = 0.'010545'010545'010545'..._{\mathrm{six}}$ \quad(6-digit per.),
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in the dyadic (\PMlinkescapetext{binary}) digit system\, $\frac{1}{11111} = 0.000010000100001..._{\mathrm{two}}$ \quad(5-digit per.).
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in the dyadic (\PMlinkescapetext{binary}) digit system \,$\frac{1}{11111} = 0.000010000100001..._{\mathrm{two}}$ \quad(5-digit per.).
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| \textbf{Note.} \,Also any irrational number has a unique decimal expansion, but it is non-periodic; for example the \PMlinkname{Liouville's number}{ExampleOfTranscendentalNumber} |
\textbf{Note.} \,Also any irrational number has a unique decimal expansion, but it is non-periodic; for example the \PMlinkname{Liouville's number}{ExampleOfTranscendentalNumber} |
| $$0.110001\,000000\,000000\,000001\,000000\,...$$ |
$$0.110001\,000000\,000000\,000001\,000000\,...$$ |
| which is transcendental over $\mathbb{Q}$. |
which is transcendental over $\mathbb{Q}$. |
