Question

A computer has a word length of 8 bits (including sign). if 2’s complement is used to represent negative numbers, what range of integers can be stored in the computer? if 1’s complement is used? (express your answers in decimal.)

Answer 1

Two's-complement notation can represent all integers from - 2^(N-1) to 2^(N-1) - 1, where N is the word length of the computer. So, in your example, the computer could store all integers from -128 to 127 inclusive. 

One's-complement notation can represent all integers from - 2^(N-1) - 1 to 2^(N-1) - 1, where N is the word length of the computer. So, in your example, the computer could store all integers from -127 to 127 inclusive

Answer 2

The word length of a computer is simply the number of bits, the computer can process

The word length is given as:

$n = 8$

(a) 2's complements

The smallest integer in the 2's complement is::

$N= -(2^{n-1})$

While the largest integer is:

$N= 2^{n-1} -1$

So, we have:

$Smallest = -(2^{8-1})$

$Smallest = -(2^7})$

$Smallest = -128$

And the largest is:

$Largest= 2^{8-1}-1$

$Largest= 2^{7}-1$

$Largest= 128-1$

$Largest= 127$

Hence, the computer can store all integers from within an inclusive range of -128 to 127, using the 2's complement

(b) 1's complements

The smallest integer in the 1's complement is::

$N= -(2^{n-1}) + 1$

While the largest integer is:

$N= 2^{n-1} -1$

So, we have:

$Smallest = -(2^{8-1})+1$

$Smallest = -(2^7})+1$

$Smallest = -128+1$

$Smallest = -127$

And the largest is:

$Largest= 2^{8-1}-1$

$Largest= 2^{7}-1$

$Largest= 128-1$

$Largest= 127$

Hence, the computer can store all integers from within an inclusive range of -127 to 127, using the 1's complement

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