Thursday, 21 November 2013

number systems - Proof that every repeating decimal is rational



Wikipedia claims that every repeating decimal represents a rational number.



According to the following definition, how can we prove that fact?



Definition: A number is rational if it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.


Answer




Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then



$$10^mx=10^ma+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and



$$10^{m+p}x=10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}.\overline{d_{m+1}\dots d_{m+p}}\tag{2}\;.$$



Subtract $(1)$ from $(2)$:



$$10^{m+p}x-10^mx=(10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots d_{m+p})-(10^ma+d_1d_2\dots d_m)\;.\tag{3}$$




The righthand side of $(3)$ is the difference of two integers, so it’s an integer; call it $N$. The lefthand side is $\left(10^{m+p}-10^m\right)x$, so



$$x=\frac{N}{10^{m+p}-10^m}=\frac{N}{10^m(10^p-1)}\;,$$



a quotient of two integers.



Example: $x=2.34\overline{567}$. Then $100x=234.\overline{567}$ and $100000x=234567.\overline{567}$, so



$$99900x=100000x-100x=234567-234=234333\;,$$ and




$$x=\frac{234333}{99900}=\frac{26037}{11100}\;.$$


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