-0.000 000 003 31 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 003 31(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
-0.000 000 003 31(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:

|-0.000 000 003 31| = 0.000 000 003 31


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: 0.000 000 003 31.

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) 0.000 000 003 31 × 2 = 0 + 0.000 000 006 62;
  • 2) 0.000 000 006 62 × 2 = 0 + 0.000 000 013 24;
  • 3) 0.000 000 013 24 × 2 = 0 + 0.000 000 026 48;
  • 4) 0.000 000 026 48 × 2 = 0 + 0.000 000 052 96;
  • 5) 0.000 000 052 96 × 2 = 0 + 0.000 000 105 92;
  • 6) 0.000 000 105 92 × 2 = 0 + 0.000 000 211 84;
  • 7) 0.000 000 211 84 × 2 = 0 + 0.000 000 423 68;
  • 8) 0.000 000 423 68 × 2 = 0 + 0.000 000 847 36;
  • 9) 0.000 000 847 36 × 2 = 0 + 0.000 001 694 72;
  • 10) 0.000 001 694 72 × 2 = 0 + 0.000 003 389 44;
  • 11) 0.000 003 389 44 × 2 = 0 + 0.000 006 778 88;
  • 12) 0.000 006 778 88 × 2 = 0 + 0.000 013 557 76;
  • 13) 0.000 013 557 76 × 2 = 0 + 0.000 027 115 52;
  • 14) 0.000 027 115 52 × 2 = 0 + 0.000 054 231 04;
  • 15) 0.000 054 231 04 × 2 = 0 + 0.000 108 462 08;
  • 16) 0.000 108 462 08 × 2 = 0 + 0.000 216 924 16;
  • 17) 0.000 216 924 16 × 2 = 0 + 0.000 433 848 32;
  • 18) 0.000 433 848 32 × 2 = 0 + 0.000 867 696 64;
  • 19) 0.000 867 696 64 × 2 = 0 + 0.001 735 393 28;
  • 20) 0.001 735 393 28 × 2 = 0 + 0.003 470 786 56;
  • 21) 0.003 470 786 56 × 2 = 0 + 0.006 941 573 12;
  • 22) 0.006 941 573 12 × 2 = 0 + 0.013 883 146 24;
  • 23) 0.013 883 146 24 × 2 = 0 + 0.027 766 292 48;
  • 24) 0.027 766 292 48 × 2 = 0 + 0.055 532 584 96;
  • 25) 0.055 532 584 96 × 2 = 0 + 0.111 065 169 92;
  • 26) 0.111 065 169 92 × 2 = 0 + 0.222 130 339 84;
  • 27) 0.222 130 339 84 × 2 = 0 + 0.444 260 679 68;
  • 28) 0.444 260 679 68 × 2 = 0 + 0.888 521 359 36;
  • 29) 0.888 521 359 36 × 2 = 1 + 0.777 042 718 72;
  • 30) 0.777 042 718 72 × 2 = 1 + 0.554 085 437 44;
  • 31) 0.554 085 437 44 × 2 = 1 + 0.108 170 874 88;
  • 32) 0.108 170 874 88 × 2 = 0 + 0.216 341 749 76;
  • 33) 0.216 341 749 76 × 2 = 0 + 0.432 683 499 52;
  • 34) 0.432 683 499 52 × 2 = 0 + 0.865 366 999 04;
  • 35) 0.865 366 999 04 × 2 = 1 + 0.730 733 998 08;
  • 36) 0.730 733 998 08 × 2 = 1 + 0.461 467 996 16;
  • 37) 0.461 467 996 16 × 2 = 0 + 0.922 935 992 32;
  • 38) 0.922 935 992 32 × 2 = 1 + 0.845 871 984 64;
  • 39) 0.845 871 984 64 × 2 = 1 + 0.691 743 969 28;
  • 40) 0.691 743 969 28 × 2 = 1 + 0.383 487 938 56;
  • 41) 0.383 487 938 56 × 2 = 0 + 0.766 975 877 12;
  • 42) 0.766 975 877 12 × 2 = 1 + 0.533 951 754 24;
  • 43) 0.533 951 754 24 × 2 = 1 + 0.067 903 508 48;
  • 44) 0.067 903 508 48 × 2 = 0 + 0.135 807 016 96;
  • 45) 0.135 807 016 96 × 2 = 0 + 0.271 614 033 92;
  • 46) 0.271 614 033 92 × 2 = 0 + 0.543 228 067 84;
  • 47) 0.543 228 067 84 × 2 = 1 + 0.086 456 135 68;
  • 48) 0.086 456 135 68 × 2 = 0 + 0.172 912 271 36;
  • 49) 0.172 912 271 36 × 2 = 0 + 0.345 824 542 72;
  • 50) 0.345 824 542 72 × 2 = 0 + 0.691 649 085 44;
  • 51) 0.691 649 085 44 × 2 = 1 + 0.383 298 170 88;
  • 52) 0.383 298 170 88 × 2 = 0 + 0.766 596 341 76;

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:


0.000 000 003 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0011 0111 0110 0010 0010(2)

6. Positive number before normalization:

0.000 000 003 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0011 0111 0110 0010 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 003 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0011 0111 0110 0010 0010(2) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0011 0111 0110 0010 0010(2) × 20 =

1.1100 0110 1110 1100 0100 010(2) × 2-29


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): -29

Mantissa (not normalized):
1.1100 0110 1110 1100 0100 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 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) =

98(10) =

0110 0010(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 0011 0111 0110 0010 0010 =

110 0011 0111 0110 0010 0010


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 0010

Mantissa (23 bits) =
110 0011 0111 0110 0010 0010


Decimal number -0.000 000 003 31 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 110 0011 0111 0110 0010 0010

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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