Is there a binary fraction with finite decimal expansion that does not end in $5$? I'm trying to come up with the finite decimal fraction not ending with $5$ which can be finitely expressed in binary. At the moment, I don't see how's that possible. Since decimal fractions can only be expressed finitely if the denominator is the exponent of $10$, and binary - if the exponent of $2$, I can't seem to find such number. Whenever I take a decimal fraction with the power of $2$ in denominator and multiply both numerator and denominator by $125$ to get the power of $10$ in denominator, I get the number ending in $5$ in numerator:
$$ \frac{1\cdot125}{8\cdot125} = \frac{125}{1000} = 0.125 $$
$$ \frac{2\cdot125}{8\cdot125} = \frac{250}{1000} = 0.25 $$
 A: The numbers you want are of the form $$\sum_{n=1}^k\frac{\alpha_n} {2^n}, $$ where $\alpha_n\in \{0,1\} $. We can rewrite as $$ \frac {\sum_{n=1}^k{\alpha_n}\,2^{k-n}} {2^n} =\frac {\sum_{n=1}^k{\alpha_n}\,2^{k-n}5^n} {10^n}  . $$ As the numerator is a multiple of five, the decimal expression will always end in five.
A: No (unless you count integers). A nonintegral binary number with terminating representation has the form $$a_m \cdots a_0 . a_{-1} \cdots a_{-n}$$
where $a_m, \cdots, a_{-n+1} \in \{0, 1\}$ and $a_{-n} = 1$, and has value
$$2^m a_m + \cdots + 2^{-n + 1} a_{-n + 1} + 2^{-n} .$$
Now, $2^{-n} = \frac{5^n}{10^n}$, and in particular the numerator of this quantity ends in $5$, so this term contributes a $5$ in the $n$th decimal place after the decimal point. We can also conclude from this that none of the other summands contribute to the $n$th decimal place.
A: There is no such fraction, because if
$$ A = a_na_{n-1}\dots a_1a_0.a_{-1}a_{-2}\dots a_{-m} $$
is your finite binary representation then the number is
$$ A = \sum_{k=-m}^n a_k 2^k $$
and $\frac12=0.5$.
When multiplying numbers you can get the lowest digit by taking the lowest digit of the product of the lowest digits of the factors. E.g. the lowest digit of the product of $1274.135497$ and $351654.69874$ will be $8$ because $4\cdot7=28$.
Since $\frac12=0.5$ it holds that the lowest digit of $(\frac12)^k$ will  (for positive $k$) always be $5$, since $5\cdot5=25$.
This means $A$ will always have $5$ as lower the digit.
