32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 511 850 587 654 33 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 511 850 587 654 33(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. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

2. 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.


0(10) =


0(2)


3. Convert to binary (base 2) the fractional part: 0.000 511 850 587 654 33.

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) 0.000 511 850 587 654 33 × 2 = 0 + 0.001 023 701 175 308 66;
  • 2) 0.001 023 701 175 308 66 × 2 = 0 + 0.002 047 402 350 617 32;
  • 3) 0.002 047 402 350 617 32 × 2 = 0 + 0.004 094 804 701 234 64;
  • 4) 0.004 094 804 701 234 64 × 2 = 0 + 0.008 189 609 402 469 28;
  • 5) 0.008 189 609 402 469 28 × 2 = 0 + 0.016 379 218 804 938 56;
  • 6) 0.016 379 218 804 938 56 × 2 = 0 + 0.032 758 437 609 877 12;
  • 7) 0.032 758 437 609 877 12 × 2 = 0 + 0.065 516 875 219 754 24;
  • 8) 0.065 516 875 219 754 24 × 2 = 0 + 0.131 033 750 439 508 48;
  • 9) 0.131 033 750 439 508 48 × 2 = 0 + 0.262 067 500 879 016 96;
  • 10) 0.262 067 500 879 016 96 × 2 = 0 + 0.524 135 001 758 033 92;
  • 11) 0.524 135 001 758 033 92 × 2 = 1 + 0.048 270 003 516 067 84;
  • 12) 0.048 270 003 516 067 84 × 2 = 0 + 0.096 540 007 032 135 68;
  • 13) 0.096 540 007 032 135 68 × 2 = 0 + 0.193 080 014 064 271 36;
  • 14) 0.193 080 014 064 271 36 × 2 = 0 + 0.386 160 028 128 542 72;
  • 15) 0.386 160 028 128 542 72 × 2 = 0 + 0.772 320 056 257 085 44;
  • 16) 0.772 320 056 257 085 44 × 2 = 1 + 0.544 640 112 514 170 88;
  • 17) 0.544 640 112 514 170 88 × 2 = 1 + 0.089 280 225 028 341 76;
  • 18) 0.089 280 225 028 341 76 × 2 = 0 + 0.178 560 450 056 683 52;
  • 19) 0.178 560 450 056 683 52 × 2 = 0 + 0.357 120 900 113 367 04;
  • 20) 0.357 120 900 113 367 04 × 2 = 0 + 0.714 241 800 226 734 08;
  • 21) 0.714 241 800 226 734 08 × 2 = 1 + 0.428 483 600 453 468 16;
  • 22) 0.428 483 600 453 468 16 × 2 = 0 + 0.856 967 200 906 936 32;
  • 23) 0.856 967 200 906 936 32 × 2 = 1 + 0.713 934 401 813 872 64;
  • 24) 0.713 934 401 813 872 64 × 2 = 1 + 0.427 868 803 627 745 28;
  • 25) 0.427 868 803 627 745 28 × 2 = 0 + 0.855 737 607 255 490 56;
  • 26) 0.855 737 607 255 490 56 × 2 = 1 + 0.711 475 214 510 981 12;
  • 27) 0.711 475 214 510 981 12 × 2 = 1 + 0.422 950 429 021 962 24;
  • 28) 0.422 950 429 021 962 24 × 2 = 0 + 0.845 900 858 043 924 48;
  • 29) 0.845 900 858 043 924 48 × 2 = 1 + 0.691 801 716 087 848 96;
  • 30) 0.691 801 716 087 848 96 × 2 = 1 + 0.383 603 432 175 697 92;
  • 31) 0.383 603 432 175 697 92 × 2 = 0 + 0.767 206 864 351 395 84;
  • 32) 0.767 206 864 351 395 84 × 2 = 1 + 0.534 413 728 702 791 68;
  • 33) 0.534 413 728 702 791 68 × 2 = 1 + 0.068 827 457 405 583 36;
  • 34) 0.068 827 457 405 583 36 × 2 = 0 + 0.137 654 914 811 166 72;

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...)


4. 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:


0.000 511 850 587 654 33(10) =


0.0000 0000 0010 0001 1000 1011 0110 1101 10(2)


5. Positive number before normalization:

0.000 511 850 587 654 33(10) =


0.0000 0000 0010 0001 1000 1011 0110 1101 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 511 850 587 654 33(10) =


0.0000 0000 0010 0001 1000 1011 0110 1101 10(2) =


0.0000 0000 0010 0001 1000 1011 0110 1101 10(2) × 20 =


1.0000 1100 0101 1011 0110 110(2) × 2-11


7. 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): -11


Mantissa (not normalized):
1.0000 1100 0101 1011 0110 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


-11 + 2(8-1) - 1 =


(-11 + 127)(10) =


116(10)


9. 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;
  • 116 ÷ 2 = 58 + 0;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


116(10) =


0111 0100(2)


11. 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, only if necessary (not the case here).


Mantissa (normalized) =


1. 000 0110 0010 1101 1011 0110 =


000 0110 0010 1101 1011 0110


12. 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) =
0111 0100


Mantissa (23 bits) =
000 0110 0010 1101 1011 0110


The base ten decimal number 0.000 511 850 587 654 33 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0100 - 000 0110 0010 1101 1011 0110

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

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

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111