Quaternary numeral system: fractions I have a question related to the expression of a real number in base 4. Consider the table here: it is clear to me how all columns of the table are obtained except the fourth one: how do they get the positional representation in quaternary base?
 A: Try long division. For example, to calculate $1 \div 7$, do the following. (Note that $7_{10} = 13_4$.)
$$
\require{enclose}
\begin{array}{r}
                    0.02102\ldots  \\[-3pt]
13 \enclose{longdiv}{1.00000\ldots} \\[-3pt]
         \underline{32}\phantom{000000} \\[-3pt]
                    20\phantom{00000}  \\[-3pt]
         \underline{13}\phantom{00000} \\[-3pt]
                    10\phantom{0000}  \\[-3pt]
         \underline{0}\phantom{0000} \\[-3pt]
                    100\phantom{00}  \\[-3pt]
         \underline{32}\phantom{00} \\[-3pt]
\end{array}
$$
So in base $4$, we have $\frac1{13_4} = 0.021021\ldots_4$.
A: The non-repeating ones like $\frac{1}{2},\frac{1}{4}$ are obvious. For the others we repeatedly use $\frac{1}{1-x}=1+x+x^2+x^3+\dots$.
We have $\frac{1}{3}=\frac{1}{4}\frac{1}{1-\frac{1}{4}}=\frac{1}{4}(1+\frac{1}{4}+\left(\frac{1}{4}\right)^2+\dots=\frac{1}{4}+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^2+\dots=0.111\dots$.
Hence $\frac{1}{12}=0.011111\dots$ and $\frac{1}{6}=0.02222\dots$
We have $\frac{1}{15}=\frac{1}{16}\frac{1}{1-\frac{1}{16}}=\frac{1}{16}+\left(\frac{1}{16}\right)^2+\left(\frac{1}{16}\right)^3+\dots=0.010101\dots$ Hence $\frac{1}{5}=0.030303\dots$.
We have $\frac{1}{7}=\frac{9}{63}=\frac{9}{64}\frac{1}{1-\frac{1}{64}}=\frac{9}{64}\left(1+\frac{1}{64}+\left(\frac{1}{64}\right)^2+\dots\right)=0.021021021\dots$. Hence $\frac{2}{7}=0.102102102\dots$ and so $\frac{1}{14}=0.0102102102\dots$.
and so on.
