Question

Determine the IEEE single and double floating point representation of the following numbers:

a) (15/2) x 2^50

b) - (15/2) x 2^-50

c) 1/5

Answer 1

a) (15/2) * 2^50

First convert the number to binary. Here is an example how to do the conversion. Rest of the conversions can be done similarly.

(15/2) = 7.5

(7)₁₀ = (111)₂  

(0.5)₁₀ = (0.1)₂

(15/2) * 2^50 = (111.1) * 2⁵⁰ = (1.111) * 2⁵² [Since the base is 2]

Thus exponent , e = 52

In single precision,

exponent, e' = e+127 = 52+127 = 179 = (10110011)₂

Mantissa:- .11100000000000000000000 [Since mantissa is 23 bits in ieee 754 single precision]

Also the number is positive. So sign bit = 0.

Final representation:- 0 10110011 11100000000000000000000 [32 bit]

In double precision,

exponent, e' = e+1023 = 52+1023 = 1075 = (1000110011)₂  [11 bits]

Mantissa:- .1110000000000000000000000000000000000000000000000000 [Since mantissa is 52 bits in ieee 754 double precision]

Also the number is positive. So sign bit = 0.

Final representation:- 0 1000110011 1110000000000000000000000000000000000000000000000000 [64 bit]

b) (-15/2) * (2^-50)

15/2 = 7.5 = (111.1)₂

(-15/2) * (2^-50) = (111.1)*2^-50 = (1.111)*2^-48

In single precision,

exponent, e' = e+127 = -48+127 = 79 = (01001111)₂

Mantissa:- .11100000000000000000000 [Since mantissa is 23 bits in ieee 754 single precision]

Also the number is negative. So sign bit = 1.

Final representation:- 1 01001111 11100000000000000000000 [32 bit]

In double precision,

exponent, e' = e+1023 = -48+1023 = 975 = (01111001111)₂  [11 bits]

Mantissa:- .1110000000000000000000000000000000000000000000000000 [Since mantissa is 52 bits in ieee 754 double precision]

Also the number is negative. So sign bit = 1.

Final representation:- 1 01111001111 1110000000000000000000000000000000000000000000000000 [64 bit]

c) 1/5 = 0.2

= (0.2)₁₀ = (0.0011...)₂

Convert to normal form:- 1.10011001100110011... * 2^-3

In single precision,

exponent, e' = e+127 = -3+127 = 124 = (01111010)₂

Mantissa:- .10011001100110011001100 [Since mantissa is 23 bits in ieee 754 single precision]

Also the number is positive. So sign bit = 0.

Final representation:- 0 01111010 10011001100110011001100 [32 bit]

In double precision,

exponent, e' = e+1023 = -3+1023 = 1020 = (01111111100)₂  [11 bits]

Mantissa:- .1001100110011001100110011001100110011001100110011001 [Since mantissa is 52 bits in ieee 754 double precision]

Also the number is positive. So sign bit = 0.

Final representation:- 0 01111111100 1001100110011001100110011001100110011001100110011001 [64 bit]