Why do we divide or multiply by 2 when converting binary? Trying to understand the fundamentals of binary rather than just following steps, I wanted to know why do we multiply by 2 to convert a decimal (0.5, 0.25) to a binary and why do we divide by 2 when we want to convert a whole number (200) by 2? Obviously, it works but how ? 
Take the following example: 
Convert $200_{10}$ to binary: 
Solution: 
     D > B    | Remainder    
-------------------------------------
200 / 2 = 100 | 0    

100 / 2 = 50  | 0             

50/2 = 25     | 0  

25/2 = 12     | 1

12/2 = 6      | 0 

6/2 = 3       | 0 

3/2 = 1       | 1 

1/2 = 0       | 1  

By taking the remainder from bottom-to-top
$200_{10} = 11001000_2$ 

Why this method works ? In other words, what's the secret behind the division by $2$ ?     

Now converting decimals (e.g. 0.5, 0.25) to binary: 
Now Suppose we have a decimal like $0.25$ and we want to convert it to binary, one of the method which I know goes like this: 
Multiplying the decimal by 2 repeatedly: 
0.25 * 2 = {0}.50 | {0}
0.50 * 2 = {1}.00 | {1} 
0.00 
--------------------------
                   .01
                   0.01  

For more details about the above method: 
Decimal to binary conversion with fraction
You can see the the two operations are reversed, to convert a whole number to a binary we divide by $2$ and to convert a fraction (decimal) we used multiplication. Add to that the order in which we take the result from bottom-to-top and from top-to-bottom. How that works? 

(Why division used to convert whole numbers to binary and why multiplication used to convert decimals (e.g. 0.25) to binary?)  

 A: This all boils down to the concept of positional notation. For example, consider the number $19$ (in base $10$), which in base $2$ becomes $10011$. To understand why, you need to understand what this notation mean:
$$
19_{10} = \color{lime}{1}\color{green}{0}\color{olive}{0}\color{grey}{1}1_2 := 1 \cdot 2^0 + \color{grey}{1} \cdot 2^1 + \color{olive}{0} \cdot 2^2 + \color{green}{0} \cdot 2^3 + \color{lime}{1} \cdot 2^4
$$
So, how can we perform this conversion? Observe that repeatedly applying the distributive property of the product over the sum we have
$$
\begin{align}
19_{10} &= 1 + 2 \cdot \color{grey}{1} + 2^2 \cdot \color{olive}{0} + 2^3 \cdot \color{green}{0} + 2^4 \cdot \color{lime}{1} \\
&= 1 + 2 \cdot (\color{grey}{1} + 2 \cdot \color{olive}{0} + 2^2 \cdot \color{green}{0} + 2^3 \cdot \color{lime}{1}) \\
&= 1 + 2 \cdot \big(\color{grey}{1} + 2 \cdot (\color{olive}{0} + 2 \cdot \color{green}{0} + 2^2 \cdot \color{lime}{1})\big) \\
&= 1 + 2 \cdot \Big(\color{grey}{1} + 2 \cdot \big(\color{olive}{0} + 2 \cdot (\color{green}{0} + 2 \cdot \color{lime}{1})\big)\Big)
\end{align}
$$
then you can see that the remainder $r_0$ of $19$ when divided by $2$ is its first binary digit. Then we can perform division by $2$ with remainder on $(19 - r_0)/2$ to find the next digit, $r_1$, and so on.
The case of fractional numbers is similar, after you observe that
$$
0.abc\dotsc_2 := a \cdot 2^{-1} + b \cdot 2^{-2} + c \cdot 2^{-3} + \dotsb
$$

Note that this is not the only possible way to perform this conversion, but it is certainly quite an efficient way. Further, by analogy you can tweak this algorithm to write a number in any integer base $b > 1$, with the integers between $0$ and $b-1$ (included) as digits!

Update: Since this was asked in the comments, here's a quick way of finding the binary digits of a given number's fractional part.
First note that if $f$ is the fractional part, then $0 \leq f < 1$. Furthermore, $0 \leq 2f < 2$, and if you look at the above equation for $0.abc\dotsc_2$ you'll see that the integer part of $2f$ is none other than $a$, the first digit of the binary expansion! If you now take out $a$, you can repeat the process, because the same equation tells us that the integer part of $2(2f-a)$ is $b$. In other words, here's the algorithm:


*

*Let $f$ be given such that $0 \leq f < 1$.

*The next digit is $\lfloor 2f \rfloor$, where $\lfloor \cdot \rfloor$ denotes taking the integer part.

*Let $f' = 2f - \lfloor 2f \rfloor$ be the fractional part of $2f$. If $f'=0$ stop, otherwise continue from 1 with $f = f'$.


For example, consider the number $\frac{5}{8}=0.625_{10}$. Then:


*

*$2*\frac{5}{8} = \frac{5}{4} = 1 + \frac{1}{4}$, so the first digit is $1$.

*$2*\frac{1}{4} = \frac{1}{2}$, so the second digit is $0$.

*$2*\frac{1}{2} = 1$, so the third and final digit is $1$.


In the end, $0.625_{10} = 0.101_{2}$.
For a number with infinite binary expansion, consider $1/3 = 0.333\dotsc_{10}$. Then:


*

*$2*\frac{1}{3} = \frac{2}{3}$, so the first digit is $0$.

*$2*\frac{2}{3} = \frac{4}{3} = 1 + \frac{1}{3}$, so the next digit is $1$.


Since we are left with $\frac{1}{3}$, which is what we started with, we can conclude that the expansion is periodic, thus $0.333\dotsc_{10} = 0.0101\dotsc_{2}$.
A: To convert a decimal number into a binary number, decimal number is repeatedly divided by 2 to count the division steps (k) required to reach terminal stage when further division of previously obtained quotient by 2 is not possible. K parameter defines everything about the conversion. For example, k =  the number of binary positions for binary digits (0, 1) to fill up to represent the decimal number in binary form accurately. The k value also defines the power of 2 to be assigned to binary digit as per its position value in the stretch of binary digits such that ith binary digit (range of i = 1 to k) is given by 2^ (k-i). If 6 divisions were required for a particular conversion then k = 6; there will be 6 binary positions. The range of power of 2 will extend from 0 to k-1. Highest power of 2 possible for any integer is defined by 2^(k-1). Starting from highest power (left extreme) to right extreme (unit position), powers of 2 will run as 2^(k-1), 2^(k-2),…,2^1,2^0.  To fill 6 binary digit positions, powers of 2 to be added run from 2^5 to 2^0 (k= 6).
For instance, converting decimal 35 to binary form, we need 6 repeated divisions of 35 by 2 to reach to stage when quotient obtained resists any further division by 2: 
Step1 {35: quotient = 35//2 =17, residue =35 % 2 =1}
Step2 {17: quotient = 17//2 = 8, residue = 17 % 2 =1}
Step3 {8:  quotient = 8//2 = 4, residue = 8% 2 =0}
Step4 {4:  quotient = 4//2 =2, residue = 4 % 2 =0}
Step5 {2:  quotient = 2//2 = 1, residue = 2 % 2 =0}
Ste6 {1:  quotient = 1//2 = > 0, residue = 1%2 =1}
K= 6.  There are 6 binary digits for decimal number 35.
Therefore, maximum power of 2 in 35 is 2^(k-1) = 2^5 = 32
The six binary positions represented by string of residues from bottom to top are  100011. There are only 3 positions representing power of 2. These are defined by k values: k=6, k=2, k=1 (from left) representing sum of 2^ (6-1) +2^ (2-1) +2^ (1-1) =32+2+1 =35.
To convert any binary number to a decimal number, write down the k values of unity positions in binary digits (counting 1 from right extreme to left), for each position take multiples of 2 for (k-1) times. Applying the rule to cited example: 100011. K=6, 2, 1.  It represents sum of:  2^5 + 2^1 + 2^0 =35. The rule extends to any number base system where residues with be in range 0 through (b-1) where b is number base under consideration (3,4,5,6,8,12,16,30,60,...). There too 0 positions are ignored and positions showing residue other than 0 is considered for multiplication by b^(k-1). This is generalized form of representation in terms of repeated divisions or multiplications employing powers of base, and making multiplications or divisions by b. In present case b=2, and the residue to be considered for multiplication is 1.
