32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 67 764 239 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 67 764 239(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. 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;
  • 67 764 239 ÷ 2 = 33 882 119 + 1;
  • 33 882 119 ÷ 2 = 16 941 059 + 1;
  • 16 941 059 ÷ 2 = 8 470 529 + 1;
  • 8 470 529 ÷ 2 = 4 235 264 + 1;
  • 4 235 264 ÷ 2 = 2 117 632 + 0;
  • 2 117 632 ÷ 2 = 1 058 816 + 0;
  • 1 058 816 ÷ 2 = 529 408 + 0;
  • 529 408 ÷ 2 = 264 704 + 0;
  • 264 704 ÷ 2 = 132 352 + 0;
  • 132 352 ÷ 2 = 66 176 + 0;
  • 66 176 ÷ 2 = 33 088 + 0;
  • 33 088 ÷ 2 = 16 544 + 0;
  • 16 544 ÷ 2 = 8 272 + 0;
  • 8 272 ÷ 2 = 4 136 + 0;
  • 4 136 ÷ 2 = 2 068 + 0;
  • 2 068 ÷ 2 = 1 034 + 0;
  • 1 034 ÷ 2 = 517 + 0;
  • 517 ÷ 2 = 258 + 1;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


67 764 239(10) =


100 0000 1010 0000 0000 0000 1111(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the left, so that only one non zero digit remains to the left of it:


67 764 239(10) =


100 0000 1010 0000 0000 0000 1111(2) =


100 0000 1010 0000 0000 0000 1111(2) × 20 =


1.0000 0010 1000 0000 0000 0011 11(2) × 226


4. 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 0 (a positive number)


Exponent (unadjusted): 26


Mantissa (not normalized):
1.0000 0010 1000 0000 0000 0011 11


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


26 + 2(8-1) - 1 =


(26 + 127)(10) =


153(10)


6. 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;
  • 153 ÷ 2 = 76 + 1;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


153(10) =


1001 1001(2)


8. 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. 000 0001 0100 0000 0000 0001 111 =


000 0001 0100 0000 0000 0001


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

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
1001 1001


Mantissa (23 bits) =
000 0001 0100 0000 0000 0001


The base ten decimal number 67 764 239 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1001 - 000 0001 0100 0000 0000 0001

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