Rational numbers with repeating decimals in binary

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.

• Rational numbers are rational in any basis, so no. Unless you are trying to distinguish between "repeating" and "terminating". – lulu Apr 2 at 14:35
• 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 $10$. – Brian M. Scott Apr 2 at 14:37
• @lulu just to make sure, a non-terminating decimal is not necessarily repeating? – Kookie Apr 2 at 14:38
• @BrianM.Scott Yes that what I was trying to say. – Kookie Apr 2 at 14:38
• @Kookie Of course. But "repeating" decimals, including the terming ones are precisely the rationals, so the basis doesn't matter. – lulu Apr 2 at 14:39

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.