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

Convert decimal -0.000 000 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 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 81| = 0.000 000 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 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 81 × 2 = 0 + 0.000 001 62;
  • 2) 0.000 001 62 × 2 = 0 + 0.000 003 24;
  • 3) 0.000 003 24 × 2 = 0 + 0.000 006 48;
  • 4) 0.000 006 48 × 2 = 0 + 0.000 012 96;
  • 5) 0.000 012 96 × 2 = 0 + 0.000 025 92;
  • 6) 0.000 025 92 × 2 = 0 + 0.000 051 84;
  • 7) 0.000 051 84 × 2 = 0 + 0.000 103 68;
  • 8) 0.000 103 68 × 2 = 0 + 0.000 207 36;
  • 9) 0.000 207 36 × 2 = 0 + 0.000 414 72;
  • 10) 0.000 414 72 × 2 = 0 + 0.000 829 44;
  • 11) 0.000 829 44 × 2 = 0 + 0.001 658 88;
  • 12) 0.001 658 88 × 2 = 0 + 0.003 317 76;
  • 13) 0.003 317 76 × 2 = 0 + 0.006 635 52;
  • 14) 0.006 635 52 × 2 = 0 + 0.013 271 04;
  • 15) 0.013 271 04 × 2 = 0 + 0.026 542 08;
  • 16) 0.026 542 08 × 2 = 0 + 0.053 084 16;
  • 17) 0.053 084 16 × 2 = 0 + 0.106 168 32;
  • 18) 0.106 168 32 × 2 = 0 + 0.212 336 64;
  • 19) 0.212 336 64 × 2 = 0 + 0.424 673 28;
  • 20) 0.424 673 28 × 2 = 0 + 0.849 346 56;
  • 21) 0.849 346 56 × 2 = 1 + 0.698 693 12;
  • 22) 0.698 693 12 × 2 = 1 + 0.397 386 24;
  • 23) 0.397 386 24 × 2 = 0 + 0.794 772 48;
  • 24) 0.794 772 48 × 2 = 1 + 0.589 544 96;
  • 25) 0.589 544 96 × 2 = 1 + 0.179 089 92;
  • 26) 0.179 089 92 × 2 = 0 + 0.358 179 84;
  • 27) 0.358 179 84 × 2 = 0 + 0.716 359 68;
  • 28) 0.716 359 68 × 2 = 1 + 0.432 719 36;
  • 29) 0.432 719 36 × 2 = 0 + 0.865 438 72;
  • 30) 0.865 438 72 × 2 = 1 + 0.730 877 44;
  • 31) 0.730 877 44 × 2 = 1 + 0.461 754 88;
  • 32) 0.461 754 88 × 2 = 0 + 0.923 509 76;
  • 33) 0.923 509 76 × 2 = 1 + 0.847 019 52;
  • 34) 0.847 019 52 × 2 = 1 + 0.694 039 04;
  • 35) 0.694 039 04 × 2 = 1 + 0.388 078 08;
  • 36) 0.388 078 08 × 2 = 0 + 0.776 156 16;
  • 37) 0.776 156 16 × 2 = 1 + 0.552 312 32;
  • 38) 0.552 312 32 × 2 = 1 + 0.104 624 64;
  • 39) 0.104 624 64 × 2 = 0 + 0.209 249 28;
  • 40) 0.209 249 28 × 2 = 0 + 0.418 498 56;
  • 41) 0.418 498 56 × 2 = 0 + 0.836 997 12;
  • 42) 0.836 997 12 × 2 = 1 + 0.673 994 24;
  • 43) 0.673 994 24 × 2 = 1 + 0.347 988 48;
  • 44) 0.347 988 48 × 2 = 0 + 0.695 976 96;

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 81(10) =

0.0000 0000 0000 0000 0000 1101 1001 0110 1110 1100 0110(2)

6. Positive number before normalization:

0.000 000 81(10) =

0.0000 0000 0000 0000 0000 1101 1001 0110 1110 1100 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 81(10) =

0.0000 0000 0000 0000 0000 1101 1001 0110 1110 1100 0110(2) =

0.0000 0000 0000 0000 0000 1101 1001 0110 1110 1100 0110(2) × 20 =

1.1011 0010 1101 1101 1000 110(2) × 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): -21

Mantissa (not normalized):
1.1011 0010 1101 1101 1000 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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 1001 0110 1110 1100 0110 =

101 1001 0110 1110 1100 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 1010

Mantissa (23 bits) =
101 1001 0110 1110 1100 0110


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

1 - 0110 1010 - 101 1001 0110 1110 1100 0110

Share

» Convert decimal -0.000 000 45 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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