To represent a rational number $a$ as binary means to write it in the form $$d_n2^n + d_{n-1}2^{n-1} + \ldots + d_02^0 + d_{-1}2^{-1} + \ldots$$ where each of the $d_i$ is either $0$ or $1$. For example, the number $5\frac34$ can be represented as $$\color{darkblue}{1}\cdot 2^2 + \color{darkblue}{0}\cdot 2^1 + \color{darkblue}{1}\cdot 2^0 + \color{darkred}{1}\cdot 2^{-1}+ \color{darkred}{1}\cdot 2^{-2}$$ and we usually abbreviate this to just $$\color{darkblue}{101}.\color{darkred}{11}$$
Note that this does not say anything about how to actually find the digits $d_i$; that is a separate matter. It also does not say anything about how the original number $q$ is represented itself; it could be in decimal or it could be in Roman numerals. it doesn't matter; as long as you can do arithmetic with it and somehow compute the correct set of $d_i$, you have converted the number to binary.
Integers
Suppose we have some integer $q$ and we would like to find the correct set of $d_i$ to represent the number in binary. One algorithm we might use is as follows: We want to write
$$q = d_n2^n + d_{n-1}2^{n-1} + \ldots + d_12^1 + d_02^0 \tag{1}$$
and it should be clear that every one of the terms on the right-hand side is even, except possibly the $d_02^0$ term. So if $q$ is even, we must have $d_0=0$, and if $q$ is odd, we must have $d_0=1$. Then $q-d_0$ is even, and $\frac12(q-d_0)$ is a smaller integer. We can write
$$q-d_0 = d_n2^n + d_{n-1}2^{n-1} + \ldots + d_12^1$$
and dividing by 2 we get:
$$q' = \frac12(q-d_0) = d_n2^{n-1} + d_{n-1}2^{n-2} + \ldots + d_12^0.\tag{2}$$
The right-hand expression is in the same form as $(1)$, so is the exact binary expansion of the smaller integer $\frac12(q-d_0)$, which we can call $q'$. Comparing $(1)$ and $(2)$, we see that the binary digits of $q'$ are the same as those of $q$, but with the rightmost one, $d_0$, dropped from the end. So if we could find the binary digits of $q'$, they would be $d_nd_n-1\ldots d_1$ and we could simply append $d_0$ to the end to get the digits of $q$.
Example: $11$
To compute the binary expansion of $q=11$ we first check that $11$ is odd, so $d_0 = 1$, and then calculate $q' = \frac12(11-1) = 5$. The binary expansion of $11$ will be the same as the expansion of $5$, with an extra $1$ on the end.
Repeating the process, we now want to calculate the binary expansion of $q'=5$. This is odd, so $d_1 = 1$ and we take $q'' = \frac12(5-1) = 2$. Then $q''$ is even, so $d_2$ is $0$, and $q''' = \frac12(2-0) = 1$, which is odd so $d_3 = 1$ and $q'''' = \frac12(1-1) = 0$ and we stop. The binary representation of $11$ is $d_3d_2d_1d_0 = 1011$.
Integers with fractions
Suppose we have $$q = d_n2^n + d_{n-1}2^{n-1} + \ldots + d_12^1 + d_02^0 + d_{-1}2^{-1} + \ldots
\tag{1}$$
We will work from the left instead of from the right this time.
The sum of the terms except for the first is $d_{n-1}2^{n-1} + \ldots + d_12^1 + d_02^0$ and can be shown to be strictly less than $2^n$. (There is actually an interesting exception to this that we will disregard to keep the discussion simple.) So $d_n=1$ if and only if $q \ge 2^n$, and conversely $d_n=0$ if and only if $q < 2^n$. So we can determine $d_n$ easily just by comparing $q$ with $2^n$.
Having done so, we can calculate $$q' = q - d_n2^n = d_{n-1}2^{n-1} + \ldots + d_12^1 + d_02^0 + d_{-1}2^{-1} + \ldots$$ and repeat the process for $q'$.
Example: $\frac52$
Let's consider the number $q=\frac52$. We guess that $n=3$ and find that $q<2^3=8$, so $d_3 = 0$. We then subtract $d_32^3 = 0$ from $q$ to obtain $q' = \frac52$ and continue. $q'\lt2^2$ so $d_2 = 0$ and $q'' = q' - 0\cdot2^2 = \frac52$ again. But on the next step something more interesting happens: $q'' \ge 2^1$ so $d_1 = 1$ and $q''' = q'' - 1\cdot 2^1 = \frac12$.
Then $q''' < 2^0$ so $d_0 = 0$ and $q^{iv} = q''' - 0\cdot2^0 = \frac12$ again. And $q^{iv} \ge 2^{-1}$ so $d_{-1} = 1$ and $q^v = q^{iv} - 1\cdot2^{-1} = 0$. $q^v=0$, so we can stop. We have calculated that $(d_3, d_2, d_1, d_0, d_{-1}) = (0,0,1,0,1)$, so the binary expansion of $\frac52$ is $0010.1$; we usually drop the leading zeroes and just write $10.1$.
Instead of stopping, we could continue for as many steps as we wanted, calculating more $d_i$, but it is clear that since $q^v=0$, all the following $d_i$ would also be $0$. So we can also say that $10.1000000000000000$ is a binary representation of $\frac52$. This is also correct; binary expansions, like decimal expansions, are not quite unique.
Note that there was no need to have the $\frac52$ written as an ordinary decimal; the only important thing was that we could compare it with the various $2^i$ and that we could do arithmetic on it. if you need it to be decimal to do those things, then yes, you need it to be decimal. But that it a contingent fact about your knowledge of how to calculate, not anything that is required by the algorithm itself. The algorithm finds a binary representation of a number, not of a decimal numeral.
Numbers less than 1
Suppose our number $q$ is between 0 and 1, so that the $d_i$ are all zero when $i\ge 0$. That is, $$q = d_{-1}2^{-1} + d_{-2}2^{-2} + \ldots.$$
Here we could proceed exactly as in the previous section, checking to see if $q\ge 2^{-1}$, then if the remainder $q'\ge 2^{-2}$, and so on.
But we can simplify the arithmetic a little bit. Instead of testing $q\ge\frac12$, we do the equivalent check to see if $2q\ge 1$. If it is, then $d_0 = 1$, and if not then $d_0 = 0$. Algebraically, we would write:
$$2q = d_{-1}2^{0} + d_{-2}2^{-1} + \ldots$$
and then $d_{-1} = 1$ if and only if $2q\ge 1$. Having found $d_{-1}$, we can as usual subtract it (from $2q$ this time) and continue checking the remainder $2q-d_{-1}$, using the same method.
Example: $\frac13$
For example, let us find the binary expansion of $q=\frac13$. We see that $2q = \frac23 < 1$, so $d_{-1} = 0$, and $q' = 2q - d_{-1} = \frac23$. Then repeating the process we find $2q' = \frac43 \ge 1,$ so $d_{-2} = 1$ and we calculate $q''= 2q'- d_{-2} = \frac13$. At this point we recognize that we are in a loop and the same calculations will keep producing $0$ and then $1$ over and over. Or put another way, the binary expansion of $q$ starts with a $0$, followed by the binary expansion of $q'$, which starts with a 1 and is followed by the binary expansion of $q$ again. So the complete expansion for $\frac13$ is $.0101010101\ldots$.
Example: $\frac\pi4$
We can apply the same technique for an irrational number, say $q=\frac\pi4$:
$$
\def\no{\text{no}}
\def\yes{\text{yes}}
\begin{array}{ccc}
q & q \ge 1? & d_i \\\hline
\frac\pi4 & \no & 0 \\
\frac\pi2 & \yes & 1 \\
\pi-2 & \yes & 1 \\
2\pi-6 & \no & 0 \\
4\pi-12 & \no & 0 \\
8\pi-24 & \yes & 1 \\
16\pi-50 & \no & 0 \\
\end{array}
$$
So the binary expansion of $\frac\pi4$ begins $0.110010$ and continues from there. Notice that again we did not have to represent any of the input numbers in decimal, unless perhaps we needed to do so in order to calculate whether each one was bigger or smaller than 1. (Maybe you use decimal, but I ask the computer, which uses base 256 arithmetic; there is no decimal number anywhere to be seen.)
The only difference between handling a rational and an irrational number is that the procedure will eventually repeat, when applied to a rational number, and if we detect this we can cut if off early. When applied to an irrational number, the procedure will never repeat.
Summary
Each one of these methods starts from the same place: the formula $(1)$ which defines what it means for a sequence of binary digits to be a binary representation for some number $q$. Each one then manipulates the formulas to show how to calculate the particular set of binary digits for a given $q$.
