Chapter 2.5, Problem 37ES

To determine

(a)

To show:

When you apply the 8-bit two’s complement procedure to the 8-bit two’s complement for -128, you get the 8-bit two’s complement for -128.

To determine

(b)

To show:

If $a,b$ and $a+b$ are integers in the range 1 through 128, then $(2^8-a)+(2^8-b)=(2^8-(a+b))+2^8\ge 2^8+2^7.$ and to explain the reason for that if integers $a,b$ and $a+b$ are all in the range 1 through 128, then the 8-bit two’s complement of $(-a)+(-b)$ is a negative number.