Example: Converting to IEEE 754 Form

Put 0.085 in single-precision format

1. The first step is to look at the sign of the number.
   Because 0.085 is positive, the sign bit =0.
   
   (-1)⁰ = 1.




2. Write 0.085 in base-2 scientific notation.
   This means that we must factor it into a number in the range [1 <= n < 2] and a power of 2.
   
   0.085 = (-1)⁰  *  (1+fraction)   * 2 ^power,    or:
   0.085 / 2^power = (1+fraction).
   
   So we can divide 0.085 by a power of 2 to get the (1 + fraction).
   
   0.085 / 2^-1 = 0.17
   0.085 / 2^-2 = 0.34
   0.085 / 2^-3 = 0.68
   0.085 / 2^-4 = 1.36
   
   Therefore, 0.085 = 1.36 * 2^-4




3. Find the exponent.
   The power of 2 is -4, and the bias for the single-precision format is 127. This means that the exponent = 123_ten, or 01111011_bin




4. Write the fraction in binary form
   The fraction = 0.36 . Unfortunately, this is not a "pretty" number, like those shown in the book. The best we can do is to approximate the value. Single-precision format allows 23 bits for the fraction.
   
   Binary fractions look like this:
   
   0.1 = (1/2) = 2^-1
   0.01 = (1/4) = 2^-2
   0.001 = (1/8) = 2^-3
   
   To approximate 0.36, we can say:
   
   0.36 = (0/2) + (1/4) + (0/8) + (1/16) + (1/32) +...
   0.36 = 2^-2 + 2^-4 + 2^-5+...
   
   0.36_ten ~ 0.01011100001010001111011_bin .
   
   The binary string we need is: 01011100001010001111011.
   
   It's important to notice that you will not get 0.36 exactly. This is why floating-point numbers have error when you put them in IEEE 754 format.




5. Now put the binary strings in the correct order -
   1 bit for the sign, followed by 8 for the exponent, and 23 for the fraction. The answer is:
   
   

   	Sign	Exponent	Fraction
   Decimal	0	123	0.36
   Binary	0	01111011	01011100001010001111011

Example: IEEE 754 to Float ->