Question

Represent the following decimal numbers using IEEE-754 floating point representation.

A. -0.375

B. -Infinity

C. 17

D. 5.25

Answer 1

a) -0.375
-0.375
Converting 0.375 to binary
   Convert decimal part first, then the fractional part
   > First convert 0 to binary
   Divide 0 successively by 2 until the quotient is 0
   Read remainders from the bottom to top as
   So, 0 of decimal is  in binary
   > Now, Convert 0.37500000 to binary
      > Multiply 0.37500000 with 2.  Since 0.75000000 is < 1. then add 0 to result
      > Multiply 0.75000000 with 2.  Since 1.50000000 is >= 1. then add 1 to result
      > Multiply 0.50000000 with 2.  Since 1.00000000 is >= 1. then add 1 to result
      > This is equal to 1, so, stop calculating
   0.375 of decimal is .011 in binary
   so, 0.375 in binary is 00000000.011
-0.375 in simple binary => .011
so, -0.375 in normal binary is .011 => 1.1 * 2^-2

single precision:
--------------------
sign bit is 1(-ve)
exponent bits are (127-2=125) => 01111101
   Divide 125 successively by 2 until the quotient is 0
      > 125/2 = 62, remainder is 1
      > 62/2 = 31, remainder is 0
      > 31/2 = 15, remainder is 1
      > 15/2 = 7, remainder is 1
      > 7/2 = 3, remainder is 1
      > 3/2 = 1, remainder is 1
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 1111101
   So, 125 of decimal is 1111101 in binary
frac/significant bits are 10000000000000000000000

so, -0.375 in single-precision format is 1 01111101 10000000000000000000000
in hexadecimal it is 0xBEC00000
Answer: 1 01111101 10000000000000000000000

b) -Infinity
Sign bit = 1
exponent bits = 11111111
mantissa bits = 00000000000000000000000
Answer: 1 11111111 00000000000000000000000

c) 17
17
Converting 17.0 to binary
   Convert decimal part first, then the fractional part
   > First convert 17 to binary
   Divide 17 successively by 2 until the quotient is 0
      > 17/2 = 8, remainder is 1
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10001
   So, 17 of decimal is 10001 in binary
   > Now, Convert 0.00000000 to binary
      > Multiply 0.00000000 with 2.  Since 0.00000000 is < 1. then add 0 to result
      > This is equal to 1, so, stop calculating
   0.0 of decimal is .0 in binary
   so, 17.0 in binary is 00010001.0
17.0 in simple binary => 10001.0
so, 17.0 in normal binary is 10001.0 => 1.0001 * 2^4

single precision:
--------------------
sign bit is 0(+ve)
exponent bits are (127+4=131) => 10000011
   Divide 131 successively by 2 until the quotient is 0
      > 131/2 = 65, remainder is 1
      > 65/2 = 32, remainder is 1
      > 32/2 = 16, remainder is 0
      > 16/2 = 8, remainder is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10000011
   So, 131 of decimal is 10000011 in binary
frac/significant bits are 00010000000000000000000

so, 17.0 in single-precision format is 0 10000011 00010000000000000000000
in hexadecimal it is 0x41880000
Answer: 0 10000011 00010000000000000000000

d) 5.25
5.25
Converting 5.25 to binary
   Convert decimal part first, then the fractional part
   > First convert 5 to binary
   Divide 5 successively by 2 until the quotient is 0
      > 5/2 = 2, remainder is 1
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 101
   So, 5 of decimal is 101 in binary
   > Now, Convert 0.25000000 to binary
      > Multiply 0.25000000 with 2.  Since 0.50000000 is < 1. then add 0 to result
      > Multiply 0.50000000 with 2.  Since 1.00000000 is >= 1. then add 1 to result
      > This is equal to 1, so, stop calculating
   0.25 of decimal is .01 in binary
   so, 5.25 in binary is 00000101.01
5.25 in simple binary => 101.01
so, 5.25 in normal binary is 101.01 => 1.0101 * 2^2

single precision:
--------------------
sign bit is 0(+ve)
exponent bits are (127+2=129) => 10000001
   Divide 129 successively by 2 until the quotient is 0
      > 129/2 = 64, remainder is 1
      > 64/2 = 32, remainder is 0
      > 32/2 = 16, remainder is 0
      > 16/2 = 8, remainder is 0
      > 8/2 = 4, remainder is 0
      > 4/2 = 2, remainder is 0
      > 2/2 = 1, remainder is 0
      > 1/2 = 0, remainder is 1
   Read remainders from the bottom to top as 10000001
   So, 129 of decimal is 10000001 in binary
frac/significant bits are 01010000000000000000000

so, 5.25 in single-precision format is 0 10000001 01010000000000000000000
in hexadecimal it is 0x40A80000
Answer: 0 10000001 01010000000000000000000