If you are doing $R$'s complement, to get the negative of a number, change each digit $d$ to $R-1-d$, then add $1$ at the end, you do fine if $R$ is even. If $R$ is odd the divisions by $2$ don't come out even and you can't look at the leading digit to find the sign. For $R=4$, for example, the greatest three digit positive number is $133_4=31_{10}=\frac {4^3}2-1$ and the smallest negative number is $200_4=-32_{10}=-\frac {4^3}2$. For $R=3$ and three digit numbers, you would like the largest positive to be $111_3=13_{10}$ and the smallest negative to be $112_3=-13_{10}$, which fits your equation to within $1$, but you have to look at the whole number to know the sign.
I try to offer my suggestion and it should not be taken as solution because I am not sure it is correct or not.
Notation: $[n]$ is called the greatest integer function. It is used to extract the integral part of the number $n$. Example: $[4.78] = 4$.
Step-1 … Convert $0.375$ to binary first.
--------- Multiply $0.375$ by $2 = 0.75$----record the result of $[0.75]$ as $A$, which is $0$ in this case.
--------- Let $X = 0.75 – [0.75] = 0.75 – 0 = 0.75$.
--------- Multiply $X$ by $2 = 1.5$----record the result of $[1.5]$ as $B$, which is $1$ in this case.
--------- Let $Y = 1.5 – [1.5] = 0.5$
--------- Multiply $Y$ by $2 = 1.0$----record the result of $[1.0]$ as $C$, which is $1$.
--------- Let $Z = 1.0 – [1.0] = 0$ and the process can be stopped as $0$ has been reached.
--------- $0.375_{(10)} = 0.ABC_{(2)} = 0.011_{(2)}$; where $A, B, C$ are digits of a number.
--------- [Check: $0.011_{(2)} = 0*2^{-1} + 1*2^{-2} + 1*2^{-3} = … = 0.375$]
Step-2 … Attach thirteen $0$’s to your result such that $0.375_{(10)} = 0.0110 0000 0000 0000_{(2)}$
Step-3 … Do the negation part (ie. The $2$’s complement part)
--------- Perform an $1$’s complement. Result $= 1.1001 1111 1111 1111_{(2)}$
--------- Add $1$ to the result and get $1. 1010 0000 0000 0000_{(2)}$.
Best Answer
Here's a handy way to take the two's complement of a binary number. Start from the right and copy all digits up to and including the rightmost $1$. Then flip all other bits. It's not difficult to see that this is equivalent to flip and add $1$, but this method is faster and less error-prone.
Applied to your examples, this method tells us that 1101 is the two's complement of 0011. We can also flip and add 1. We get 0010+1= 0011. Same result, as expected.