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Is it possible to prove that there exists a rational number with repeating decimal digits in base-10 representation that isn't repeating in binary?

For example, $0.\overline{0011}_2$ is a binary representation of $0.2_{10}$ which contains repeating digits to the right of the decimal point.

I'm wondering if there exists some $A \in \mathbb{Q}$ in which $A_{10}$ contains repeating decimal digits, but $A_2$ doesn't.

I'm asking this out of curiosity, this is not homework.

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    $\begingroup$ Rational numbers are rational in any basis, so no. Unless you are trying to distinguish between "repeating" and "terminating". $\endgroup$ – lulu Apr 2 at 14:35
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    $\begingroup$ Do you mean an $A$ whose decimal expansion is non-terminating, but whose binary expansion terminates? If so, the answer is no. If the binary expansion terminates, $A$ can be written as a fraction whose denominator is a power of $2$, and in that case its decimal expansion will also terminate, since $2$ is a factor of $10$. $\endgroup$ – Brian M. Scott Apr 2 at 14:37
  • $\begingroup$ @lulu just to make sure, a non-terminating decimal is not necessarily repeating? $\endgroup$ – Kookie Apr 2 at 14:38
  • $\begingroup$ @BrianM.Scott Yes that what I was trying to say. $\endgroup$ – Kookie Apr 2 at 14:38
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    $\begingroup$ @Kookie Of course. But "repeating" decimals, including the terming ones are precisely the rationals, so the basis doesn't matter. $\endgroup$ – lulu Apr 2 at 14:39
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If "repeating" refers to "non-terminating", such numbers do not exist. Any non-repeating or terminating number in binary has the form $a/2^n$ for some integral $a,n$. Multiplying top and bottom by $5^n$ yields $5^na/10^n$, showing that the number is terminating in base 10 as well. Hence, by contraposition, any number repeating in decimal is also repeating in binary.

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