32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -805 288 653 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -805 288 653(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:

|-805 288 653| = 805 288 653

2. 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;
  • 805 288 653 ÷ 2 = 402 644 326 + 1;
  • 402 644 326 ÷ 2 = 201 322 163 + 0;
  • 201 322 163 ÷ 2 = 100 661 081 + 1;
  • 100 661 081 ÷ 2 = 50 330 540 + 1;
  • 50 330 540 ÷ 2 = 25 165 270 + 0;
  • 25 165 270 ÷ 2 = 12 582 635 + 0;
  • 12 582 635 ÷ 2 = 6 291 317 + 1;
  • 6 291 317 ÷ 2 = 3 145 658 + 1;
  • 3 145 658 ÷ 2 = 1 572 829 + 0;
  • 1 572 829 ÷ 2 = 786 414 + 1;
  • 786 414 ÷ 2 = 393 207 + 0;
  • 393 207 ÷ 2 = 196 603 + 1;
  • 196 603 ÷ 2 = 98 301 + 1;
  • 98 301 ÷ 2 = 49 150 + 1;
  • 49 150 ÷ 2 = 24 575 + 0;
  • 24 575 ÷ 2 = 12 287 + 1;
  • 12 287 ÷ 2 = 6 143 + 1;
  • 6 143 ÷ 2 = 3 071 + 1;
  • 3 071 ÷ 2 = 1 535 + 1;
  • 1 535 ÷ 2 = 767 + 1;
  • 767 ÷ 2 = 383 + 1;
  • 383 ÷ 2 = 191 + 1;
  • 191 ÷ 2 = 95 + 1;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the positive number.

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


805 288 653(10) =


10 1111 1111 1111 1011 1010 1100 1101(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the left, so that only one non zero digit remains to the left of it:


805 288 653(10) =


10 1111 1111 1111 1011 1010 1100 1101(2) =


10 1111 1111 1111 1011 1010 1100 1101(2) × 20 =


1.0111 1111 1111 1101 1101 0110 0110 1(2) × 229


5. 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): 29


Mantissa (not normalized):
1.0111 1111 1111 1101 1101 0110 0110 1


6. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


29 + 2(8-1) - 1 =


(29 + 127)(10) =


156(10)


7. 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;
  • 156 ÷ 2 = 78 + 0;
  • 78 ÷ 2 = 39 + 0;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


156(10) =


1001 1100(2)


9. 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 removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 011 1111 1111 1110 1110 1011 00 1101 =


011 1111 1111 1110 1110 1011


10. 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) =
1001 1100


Mantissa (23 bits) =
011 1111 1111 1110 1110 1011


The base ten decimal number -805 288 653 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1001 1100 - 011 1111 1111 1110 1110 1011

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