-0.000 000 000 855 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 855(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 855(10) to 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:

|-0.000 000 000 855| = 0.000 000 000 855


2. 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;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 855.

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) 0.000 000 000 855 × 2 = 0 + 0.000 000 001 71;
  • 2) 0.000 000 001 71 × 2 = 0 + 0.000 000 003 42;
  • 3) 0.000 000 003 42 × 2 = 0 + 0.000 000 006 84;
  • 4) 0.000 000 006 84 × 2 = 0 + 0.000 000 013 68;
  • 5) 0.000 000 013 68 × 2 = 0 + 0.000 000 027 36;
  • 6) 0.000 000 027 36 × 2 = 0 + 0.000 000 054 72;
  • 7) 0.000 000 054 72 × 2 = 0 + 0.000 000 109 44;
  • 8) 0.000 000 109 44 × 2 = 0 + 0.000 000 218 88;
  • 9) 0.000 000 218 88 × 2 = 0 + 0.000 000 437 76;
  • 10) 0.000 000 437 76 × 2 = 0 + 0.000 000 875 52;
  • 11) 0.000 000 875 52 × 2 = 0 + 0.000 001 751 04;
  • 12) 0.000 001 751 04 × 2 = 0 + 0.000 003 502 08;
  • 13) 0.000 003 502 08 × 2 = 0 + 0.000 007 004 16;
  • 14) 0.000 007 004 16 × 2 = 0 + 0.000 014 008 32;
  • 15) 0.000 014 008 32 × 2 = 0 + 0.000 028 016 64;
  • 16) 0.000 028 016 64 × 2 = 0 + 0.000 056 033 28;
  • 17) 0.000 056 033 28 × 2 = 0 + 0.000 112 066 56;
  • 18) 0.000 112 066 56 × 2 = 0 + 0.000 224 133 12;
  • 19) 0.000 224 133 12 × 2 = 0 + 0.000 448 266 24;
  • 20) 0.000 448 266 24 × 2 = 0 + 0.000 896 532 48;
  • 21) 0.000 896 532 48 × 2 = 0 + 0.001 793 064 96;
  • 22) 0.001 793 064 96 × 2 = 0 + 0.003 586 129 92;
  • 23) 0.003 586 129 92 × 2 = 0 + 0.007 172 259 84;
  • 24) 0.007 172 259 84 × 2 = 0 + 0.014 344 519 68;
  • 25) 0.014 344 519 68 × 2 = 0 + 0.028 689 039 36;
  • 26) 0.028 689 039 36 × 2 = 0 + 0.057 378 078 72;
  • 27) 0.057 378 078 72 × 2 = 0 + 0.114 756 157 44;
  • 28) 0.114 756 157 44 × 2 = 0 + 0.229 512 314 88;
  • 29) 0.229 512 314 88 × 2 = 0 + 0.459 024 629 76;
  • 30) 0.459 024 629 76 × 2 = 0 + 0.918 049 259 52;
  • 31) 0.918 049 259 52 × 2 = 1 + 0.836 098 519 04;
  • 32) 0.836 098 519 04 × 2 = 1 + 0.672 197 038 08;
  • 33) 0.672 197 038 08 × 2 = 1 + 0.344 394 076 16;
  • 34) 0.344 394 076 16 × 2 = 0 + 0.688 788 152 32;
  • 35) 0.688 788 152 32 × 2 = 1 + 0.377 576 304 64;
  • 36) 0.377 576 304 64 × 2 = 0 + 0.755 152 609 28;
  • 37) 0.755 152 609 28 × 2 = 1 + 0.510 305 218 56;
  • 38) 0.510 305 218 56 × 2 = 1 + 0.020 610 437 12;
  • 39) 0.020 610 437 12 × 2 = 0 + 0.041 220 874 24;
  • 40) 0.041 220 874 24 × 2 = 0 + 0.082 441 748 48;
  • 41) 0.082 441 748 48 × 2 = 0 + 0.164 883 496 96;
  • 42) 0.164 883 496 96 × 2 = 0 + 0.329 766 993 92;
  • 43) 0.329 766 993 92 × 2 = 0 + 0.659 533 987 84;
  • 44) 0.659 533 987 84 × 2 = 1 + 0.319 067 975 68;
  • 45) 0.319 067 975 68 × 2 = 0 + 0.638 135 951 36;
  • 46) 0.638 135 951 36 × 2 = 1 + 0.276 271 902 72;
  • 47) 0.276 271 902 72 × 2 = 0 + 0.552 543 805 44;
  • 48) 0.552 543 805 44 × 2 = 1 + 0.105 087 610 88;
  • 49) 0.105 087 610 88 × 2 = 0 + 0.210 175 221 76;
  • 50) 0.210 175 221 76 × 2 = 0 + 0.420 350 443 52;
  • 51) 0.420 350 443 52 × 2 = 0 + 0.840 700 887 04;
  • 52) 0.840 700 887 04 × 2 = 1 + 0.681 401 774 08;
  • 53) 0.681 401 774 08 × 2 = 1 + 0.362 803 548 16;
  • 54) 0.362 803 548 16 × 2 = 0 + 0.725 607 096 32;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 000 855(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1100 0001 0101 0001 10(2)

6. Positive number before normalization:

0.000 000 000 855(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1100 0001 0101 0001 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 855(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1100 0001 0101 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1100 0001 0101 0001 10(2) × 20 =

1.1101 0110 0000 1010 1000 110(2) × 2-31


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): -31

Mantissa (not normalized):
1.1101 0110 0000 1010 1000 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1011 0000 0101 0100 0110 =

110 1011 0000 0101 0100 0110


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) =
0110 0000

Mantissa (23 bits) =
110 1011 0000 0101 0100 0110


Decimal number -0.000 000 000 855 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 1011 0000 0101 0100 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