32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -3.477 962 220 57 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -3.477 962 220 57(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:

|-3.477 962 220 57| = 3.477 962 220 57

2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.


3(10) =


11(2)


4. Convert to binary (base 2) the fractional part: 0.477 962 220 57.

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) 0.477 962 220 57 × 2 = 0 + 0.955 924 441 14;
  • 2) 0.955 924 441 14 × 2 = 1 + 0.911 848 882 28;
  • 3) 0.911 848 882 28 × 2 = 1 + 0.823 697 764 56;
  • 4) 0.823 697 764 56 × 2 = 1 + 0.647 395 529 12;
  • 5) 0.647 395 529 12 × 2 = 1 + 0.294 791 058 24;
  • 6) 0.294 791 058 24 × 2 = 0 + 0.589 582 116 48;
  • 7) 0.589 582 116 48 × 2 = 1 + 0.179 164 232 96;
  • 8) 0.179 164 232 96 × 2 = 0 + 0.358 328 465 92;
  • 9) 0.358 328 465 92 × 2 = 0 + 0.716 656 931 84;
  • 10) 0.716 656 931 84 × 2 = 1 + 0.433 313 863 68;
  • 11) 0.433 313 863 68 × 2 = 0 + 0.866 627 727 36;
  • 12) 0.866 627 727 36 × 2 = 1 + 0.733 255 454 72;
  • 13) 0.733 255 454 72 × 2 = 1 + 0.466 510 909 44;
  • 14) 0.466 510 909 44 × 2 = 0 + 0.933 021 818 88;
  • 15) 0.933 021 818 88 × 2 = 1 + 0.866 043 637 76;
  • 16) 0.866 043 637 76 × 2 = 1 + 0.732 087 275 52;
  • 17) 0.732 087 275 52 × 2 = 1 + 0.464 174 551 04;
  • 18) 0.464 174 551 04 × 2 = 0 + 0.928 349 102 08;
  • 19) 0.928 349 102 08 × 2 = 1 + 0.856 698 204 16;
  • 20) 0.856 698 204 16 × 2 = 1 + 0.713 396 408 32;
  • 21) 0.713 396 408 32 × 2 = 1 + 0.426 792 816 64;
  • 22) 0.426 792 816 64 × 2 = 0 + 0.853 585 633 28;
  • 23) 0.853 585 633 28 × 2 = 1 + 0.707 171 266 56;
  • 24) 0.707 171 266 56 × 2 = 1 + 0.414 342 533 12;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


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:


0.477 962 220 57(10) =


0.0111 1010 0101 1011 1011 1011(2)


6. Positive number before normalization:

3.477 962 220 57(10) =


11.0111 1010 0101 1011 1011 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.477 962 220 57(10) =


11.0111 1010 0101 1011 1011 1011(2) =


11.0111 1010 0101 1011 1011 1011(2) × 20 =


1.1011 1101 0010 1101 1101 1101 1(2) × 21


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): 1


Mantissa (not normalized):
1.1011 1101 0010 1101 1101 1101 1


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


1 + 2(8-1) - 1 =


(1 + 127)(10) =


128(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;
  • 128 ÷ 2 = 64 + 0;
  • 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;

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) =


128(10) =


1000 0000(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 removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 101 1110 1001 0110 1110 1110 11 =


101 1110 1001 0110 1110 1110


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 0000


Mantissa (23 bits) =
101 1110 1001 0110 1110 1110


The base ten decimal number -3.477 962 220 57 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 0000 - 101 1110 1001 0110 1110 1110

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