Question

Show that the integer m with two's complement representation $\left(a_{n-1}{a}_{n-2},\dots \dots ,a_1{a}_0\right)$can be found using the equation

$m=-a_{n-1}{2}^{n-1}+-a_{n-2}^{\mathrm{\prime }}{2}^{n-2}+\dots \dots +a_1^{\mathrm{\prime }}2+a_0$

Short answer

In symbols, this says that$2^{n-1}-(-m)=m=a_{i=0}^{n-2}{a}_i{2}^i$. Solving for m gives us the equation we are trying to provelocalid="1668502689818" $\text{(since}\left.a_{n-1}=1\right)$

Step-by-step solution

Step 1

If m is positive (or 0 ), then the leading bit $\left(a_{n-1}\right)$is 0 , so the formula reads simply $m=a^{n-2}{a}_{i=0}{2}^i$which is clearly correct, since this is the binary expansion of${\mathrm{f}}^{n-1}-(-m)$. Now suppose that is negative.

The two's complement expansion for m has its leading bit equal to 1 .

By the definition of two's complement, the remaining n - 1 bits are binary expansion of$2^{n-1}-(-m)$. In symbols, this says that$2^{n-1}-(-m)=m=a_{i=0}^{n-2}{a}_i{2}^i$. Solving for m gives us the equation we are trying to prove (since data-custom-editor="chemistry" $a_{n-1}=1$).