Question

A computer word has n=36 bits and 9 bits are used for the exponent of real numbers. (a) What is the range of real numbers for this computer? (b) What is the precision of real numbers? (c) What is the smallest positive real number for this 36-bit computer? (d) What is the range of unsigned integers for this computer? (e) What is the range of signed integers for this computer?

Answer 1

in digitial system bit are represented by 1 or 0 it means a bit can be 1 or 0 at a one time.

and also 1 byte can store 8 bits

according to question we have 36bits where 9 bits are used for exponent of real numbers.

question a:

answer: the range of real numbers for this computer is 0 to (2^n)

= 0 to (2^36)

=0 to 68719476736

so finaly range wil be 0 to 68719476736

question b:

answer: in a digital number system real number after decimal is stored in precision format.

so a n bit computer consists of 3 things to store binary number

1. sign

2. exponents

3. mantisa

in a 32 bit of computer

1 bit is for sign bit

8 bit for exponents

23 bit for mantisa

it is also called single precision

but in double precision contain

1 bit for sign

11 bit for exponents

52 bit for mantisa

and this computer also called 64 bit.

question no c:

answer:  

0 will be smallest positive number in 36 bit computer because it's range will be 0 to 68719476736 so in this range 0 will be the smalles positive number.

question d:

a 36 bit computer can hold 0 to 2^n -1 unsigned integer

and it equals to 68719476735

sp its range is 0 to 68719476735.

question e:

a computer with n bit can hold signed integer range from -2^(n-1) to 2^(n-1)-1

so a 36 bit computer can hold signed integer range from -2^(36-1) to 2^(36-1)-1

= -34359738368 to 34359738367