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Is there a proof that any number of the form $a_1.a_2\overline{a_3}$ where we are contatenating them and they are all finite sequences of digits is rational?

Regards.

Asinomás
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2 Answers2

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Any periodic (as you call it) number can be written in the form $ r + \sum_{n=0}^\infty b 10^{k-nl}$ where $r$ is some rational number, $b,k$ are integers and $l$ is a positive integer.

Example 1.23454545... has $r=1.23, b=45, k=-4, l=2$.

But $\sum_{n=0}^\infty b 10^{k-nl}$ is sum of geometric series with a common ratio of $10^{-l}$.

Since $l\geq1$, we get $10^{l}\geq10 $ or $0<\frac1{10^{l}}\leq\frac1{10}<1$ and thus we have the convergent series

$$\sum_{n=0}^\infty b 10^{k-nl} = b 10^k \sum_{n=0}^\infty \left(10^{-l}\right)^n = b 10^k \frac1{1-10^{-l}} $$

That is certainly a rational number.

Andrew
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tom
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If $a_3$ consists of $k$ digits, multiply the number with $10^k$ and subtract. The difference is a termininating decimal, i.e. a rational number with at most a power of $10$ as denominator.

Hagen von Eitzen
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