32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -729.781 25 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -729.781 25(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-729.781 25| = 729.781 25

2. First, convert to binary (in base 2) the integer part: 729.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 729 ÷ 2 = 364 + 1;
  • 364 ÷ 2 = 182 + 0;
  • 182 ÷ 2 = 91 + 0;
  • 91 ÷ 2 = 45 + 1;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


729(10) =


10 1101 1001(2)


4. Convert to binary (base 2) the fractional part: 0.781 25.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.781 25 × 2 = 1 + 0.562 5;
  • 2) 0.562 5 × 2 = 1 + 0.125;
  • 3) 0.125 × 2 = 0 + 0.25;
  • 4) 0.25 × 2 = 0 + 0.5;
  • 5) 0.5 × 2 = 1 + 0;

5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.781 25(10) =


0.1100 1(2)


6. Positive number before normalization:

729.781 25(10) =


10 1101 1001.1100 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


729.781 25(10) =


10 1101 1001.1100 1(2) =


10 1101 1001.1100 1(2) × 20 =


1.0110 1100 1110 01(2) × 29


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)


Exponent (unadjusted): 9


Mantissa (not normalized):
1.0110 1100 1110 01


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


9 + 2(8-1) - 1 =


(9 + 127)(10) =


136(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 136 ÷ 2 = 68 + 0;
  • 68 ÷ 2 = 34 + 0;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


136(10) =


1000 1000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, by adding the necessary number of zeros to the right.


Mantissa (normalized) =


1. 01 1011 0011 1001 0 0000 0000 =


011 0110 0111 0010 0000 0000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (8 bits) =
1000 1000


Mantissa (23 bits) =
011 0110 0111 0010 0000 0000


The base ten decimal number -729.781 25 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 1000 - 011 0110 0111 0010 0000 0000

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