-0.000 000 000 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 81(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 81(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 81| = 0.000 000 000 81


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 81.

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 81 × 2 = 0 + 0.000 000 001 62;
  • 2) 0.000 000 001 62 × 2 = 0 + 0.000 000 003 24;
  • 3) 0.000 000 003 24 × 2 = 0 + 0.000 000 006 48;
  • 4) 0.000 000 006 48 × 2 = 0 + 0.000 000 012 96;
  • 5) 0.000 000 012 96 × 2 = 0 + 0.000 000 025 92;
  • 6) 0.000 000 025 92 × 2 = 0 + 0.000 000 051 84;
  • 7) 0.000 000 051 84 × 2 = 0 + 0.000 000 103 68;
  • 8) 0.000 000 103 68 × 2 = 0 + 0.000 000 207 36;
  • 9) 0.000 000 207 36 × 2 = 0 + 0.000 000 414 72;
  • 10) 0.000 000 414 72 × 2 = 0 + 0.000 000 829 44;
  • 11) 0.000 000 829 44 × 2 = 0 + 0.000 001 658 88;
  • 12) 0.000 001 658 88 × 2 = 0 + 0.000 003 317 76;
  • 13) 0.000 003 317 76 × 2 = 0 + 0.000 006 635 52;
  • 14) 0.000 006 635 52 × 2 = 0 + 0.000 013 271 04;
  • 15) 0.000 013 271 04 × 2 = 0 + 0.000 026 542 08;
  • 16) 0.000 026 542 08 × 2 = 0 + 0.000 053 084 16;
  • 17) 0.000 053 084 16 × 2 = 0 + 0.000 106 168 32;
  • 18) 0.000 106 168 32 × 2 = 0 + 0.000 212 336 64;
  • 19) 0.000 212 336 64 × 2 = 0 + 0.000 424 673 28;
  • 20) 0.000 424 673 28 × 2 = 0 + 0.000 849 346 56;
  • 21) 0.000 849 346 56 × 2 = 0 + 0.001 698 693 12;
  • 22) 0.001 698 693 12 × 2 = 0 + 0.003 397 386 24;
  • 23) 0.003 397 386 24 × 2 = 0 + 0.006 794 772 48;
  • 24) 0.006 794 772 48 × 2 = 0 + 0.013 589 544 96;
  • 25) 0.013 589 544 96 × 2 = 0 + 0.027 179 089 92;
  • 26) 0.027 179 089 92 × 2 = 0 + 0.054 358 179 84;
  • 27) 0.054 358 179 84 × 2 = 0 + 0.108 716 359 68;
  • 28) 0.108 716 359 68 × 2 = 0 + 0.217 432 719 36;
  • 29) 0.217 432 719 36 × 2 = 0 + 0.434 865 438 72;
  • 30) 0.434 865 438 72 × 2 = 0 + 0.869 730 877 44;
  • 31) 0.869 730 877 44 × 2 = 1 + 0.739 461 754 88;
  • 32) 0.739 461 754 88 × 2 = 1 + 0.478 923 509 76;
  • 33) 0.478 923 509 76 × 2 = 0 + 0.957 847 019 52;
  • 34) 0.957 847 019 52 × 2 = 1 + 0.915 694 039 04;
  • 35) 0.915 694 039 04 × 2 = 1 + 0.831 388 078 08;
  • 36) 0.831 388 078 08 × 2 = 1 + 0.662 776 156 16;
  • 37) 0.662 776 156 16 × 2 = 1 + 0.325 552 312 32;
  • 38) 0.325 552 312 32 × 2 = 0 + 0.651 104 624 64;
  • 39) 0.651 104 624 64 × 2 = 1 + 0.302 209 249 28;
  • 40) 0.302 209 249 28 × 2 = 0 + 0.604 418 498 56;
  • 41) 0.604 418 498 56 × 2 = 1 + 0.208 836 997 12;
  • 42) 0.208 836 997 12 × 2 = 0 + 0.417 673 994 24;
  • 43) 0.417 673 994 24 × 2 = 0 + 0.835 347 988 48;
  • 44) 0.835 347 988 48 × 2 = 1 + 0.670 695 976 96;
  • 45) 0.670 695 976 96 × 2 = 1 + 0.341 391 953 92;
  • 46) 0.341 391 953 92 × 2 = 0 + 0.682 783 907 84;
  • 47) 0.682 783 907 84 × 2 = 1 + 0.365 567 815 68;
  • 48) 0.365 567 815 68 × 2 = 0 + 0.731 135 631 36;
  • 49) 0.731 135 631 36 × 2 = 1 + 0.462 271 262 72;
  • 50) 0.462 271 262 72 × 2 = 0 + 0.924 542 525 44;
  • 51) 0.924 542 525 44 × 2 = 1 + 0.849 085 050 88;
  • 52) 0.849 085 050 88 × 2 = 1 + 0.698 170 101 76;
  • 53) 0.698 170 101 76 × 2 = 1 + 0.396 340 203 52;
  • 54) 0.396 340 203 52 × 2 = 0 + 0.792 680 407 04;

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 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1010 1001 1010 1011 10(2)

6. Positive number before normalization:

0.000 000 000 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1010 1001 1010 1011 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 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1010 1001 1010 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1010 1001 1010 1011 10(2) × 20 =

1.1011 1101 0100 1101 0101 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):
1.1011 1101 0100 1101 0101 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. 101 1110 1010 0110 1010 1110 =

101 1110 1010 0110 1010 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) =
0110 0000

Mantissa (23 bits) =
101 1110 1010 0110 1010 1110


Decimal number -0.000 000 000 81 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1110 1010 0110 1010 1110

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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