Does this trick work for fixed point numbers as well? That is$,$ can I take the complement of every digit of a given fixed point number and just add one to the least significant bit $($lsb$)$ to obtain the two's complement fixed point representation? Go back to your answer for the $1\text{'}$s complement representation of $-12\mathrm{.}625$ in $8$ bits with n$=5$ and m$=3$ and add $1$ to the lsb$.$ Is the result the correct $2\text{'}$s complement representation of $-12\mathrm{.}625?$

Group of answer choices

a$.$ we obtain $10011\mathrm{.}011$ which is indeed the two's complement representation of $-12\mathrm{.}625$

b$.$ we obtain $10011\mathrm{.}011$ which is NOT the two's complement representation of $-12\mathrm{.}625$

c$.$ we obtain $10011\mathrm{.}010$ which is indeed the two's complement representation of $-12\mathrm{.}625$

d$.$ None of the above