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There's a question that I've been thinking about for quite some time now. We all know that numbers with infinite decimal expansion such as $0.\overline{3}$ or $0.\overline{1}$ are not necessarily irrational. $$0.\overline{3}=0.333333\dots=\frac{1}{3}$$ $$0.\overline{1}=0.111111\ldots=\frac{1}{9}$$ Therefore we can say that for instance $0.\overline{4}$ is rational because we have: $$0.\overline{4}=4\cdot 0.\overline{1}=\frac{4}{9}$$ My question is: can we generalize these results to any number that has a repeating pattern in its decimal expansion? Can we claim for instance that $0.\overline{123}=0.123123123\ldots$ is rational? This question boils down to asking whether or not expressions like $0.\overline{01}$ and $0.\overline{001}$ etcetera are rational. If so, in the case of my example we could say: $$0.\overline{123}=123\cdot 0.\overline{001}$$ A naive thought of mine was that $0.\overline{01}$ is simply $\frac{1}{10}\cdot 0.\overline{1}$ but this is of course not true. Then I thought that maybe we can represent $0.\overline{01}$ as $\sum\limits_{i=1}^{\infty}\frac{1}{10^{2i}}$ which converges by the ratio test. However this does not tell us if it is rational or not.

This is not for homework or anything, just something I've been thinking about and I was wondering if any of you have some knowledge about this issue. Thanks.

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Yes, all repeating decimal numbers are rational. I'm sure someone here will give a good explanation but here's one link that'll help:… – Alraxite Feb 8 '13 at 11:22
Note that $\sum\limits_{i=1}^{\infty}\dfrac{1}{10^{2i}}=\frac{1}{100}\frac{1}{1-\frac{1}{1‌​00}}=\dfrac{1}{99}$, which is rational. In general $\sum\limits_{i=1}^{\infty}\dfrac{1}{10^{si}}=\frac{1}{10^s}\frac{1}{1-\frac{1}{‌​10^s}}=\dfrac{1}{99\ldots9}$. See this – P.. Feb 8 '13 at 11:32
With a little more work, you can show that eventually repeating decimals (e.g., $.7346262626262\dots$) are rational, and, conversely, every rational has a decimal that is (eventually) repeating (or terminating). – Gerry Myerson Feb 8 '13 at 11:34
Because we could just multiply it by 10 to some power and then be left with a number plus some repeating decimal right? In this case $.7346262626262= \frac{734}{1000} +\frac{1}{1000}\cdot 0.62626262$ which is the sum of two rational numbers and thus rational. – Slugger Feb 8 '13 at 11:37
up vote 6 down vote accepted

As you mentioned, any infinite repeating decimal with $k$ digits is form: $$q = r + 10^{-k}r+10^{-2k}r+..$$ This series always converges to: $$q = \frac{r 10^k}{10^k-1}$$ Which is of course rational. For example: $$0.123123123.. = \frac{0.123\cdot 10^3}{10^3-1}=\frac{123}{999}$$ Or a longer one: $$0.571428571428... = \frac{0.571428\cdot 10^{6}}{10^{6}-1}=\frac{4}{7}$$

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Ah great thanks! – Slugger Feb 8 '13 at 11:33

Let's take some number with repeating decimal expansion. I'll choose $x = 0.\overline{142857}$ for this. You might recognize this a $\frac{1}{7}$, but the way we're going to find out goes for any repeating decimal expansion.

First, take the equation above and multiply by $10^6$, since there are $6$ repeating decimals (this $6$ is the only thing that depends on the specific number you're interested in; everything else is the same for every number): $$ 1,\!000,\!000x = 142,\!857.\overline{142857} $$ Then we subtract $x = 0.\overline{142857}$ from both sides, like so: $$ 999,\!999x = 142,\!857 $$ and we get $$ x = \frac{ 142,\!857}{999,\!999} $$ which is the ratio of two integers, and thus a rational number.

This way you can express any number with repeating decimals as a fraction where the denominator is a string of nines. Of course, most likely you can shorten the fraction to get a nicer number, but that is not neccessary if you just want to prove rationality.

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Ah cool thanks, I forgot about that trick. It's the same trick people often use to show that $0.999999\ldots =1$. – Slugger Feb 8 '13 at 11:35
Random fact: $142857$ is a very fun number. Multiply it by $2$, $3$, $4$, $5$ and $6$ to see why it's so fun. – Lazar Ljubenović Feb 8 '13 at 11:38
@TeunVerstraaten And strictly speaking, it's the exact same trick as the other answer, only in the other answer it's hidden away in the general formula for the sum of a convergent, infinite geometric series. – Arthur Feb 8 '13 at 13:37
@LazarLjubenović $0,\!588,\!235,\!294,\!117,\!647$ is the same, but for multiplication by any integer up to 16. The leading zero is, unfortunately, necessary. – Arthur Feb 8 '13 at 13:42

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