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A decimal expression is said to be repeating if it ends in a repeating pattern of digits. For example, the following are repeating decimal expressions:

$$.333..., .1231333..., 123121312131213...$$

Show that a real number in $(0,1]$ is rational if and only if it has a repeating decimal representation.

Find all decimal representations for the rational numbers $1/5$ and $10/13$ Do I need to prove this. Can I just say that $1/5 = 0.2$ and $10/13$ is $0.7692307692307...$

Greg Martin
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ayv2
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Any periodic decimal can be written as a geometric series, where the sum formula is then a rational expression.

The other way around, any rational number $\frac mn$ can be rewritten as $10^{-k}\cdot \frac pq$ with $gcd(p,q)=1$ and $q$ containing no factors $2$ or $5$. Set $d=\phi(q)$ the value of Eulers totient function, then per Fermat's little theorem, $10^{d}\equiv 1\pmod{q}$, that is, there is some number $q'$ with $qq'=10^{d}-1$.

This now allows to write the fraction $\frac mn=10^{-k}\cdot\frac {pq'}{10^{d}-1}$ as a periodic decimal with period $d$.

Lutz Lehmann
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    A shorter way of the latter part is recalling the long division algorithm. It goes on with remainders, which are always less than denominator – so there's only $m$ possible remainders, which implies the sequence of calculations must eventually loop with a period less than $m$, consequently giving a repeating representation (be it decimal, hexadecimal, binary or any other positional). – CiaPan May 24 '14 at 20:36
  • @CiaPan I believe that's only shorter if you fill your 'proof' with $\cdots$'s, as proving properties of recursive algorithms oftens takes a bit more of effort. – YoTengoUnLCD Jun 22 '16 at 23:20