0.000 000 331 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 331(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
0.000 000 331(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. 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: 0.000 000 331.

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) 0.000 000 331 × 2 = 0 + 0.000 000 662;
  • 2) 0.000 000 662 × 2 = 0 + 0.000 001 324;
  • 3) 0.000 001 324 × 2 = 0 + 0.000 002 648;
  • 4) 0.000 002 648 × 2 = 0 + 0.000 005 296;
  • 5) 0.000 005 296 × 2 = 0 + 0.000 010 592;
  • 6) 0.000 010 592 × 2 = 0 + 0.000 021 184;
  • 7) 0.000 021 184 × 2 = 0 + 0.000 042 368;
  • 8) 0.000 042 368 × 2 = 0 + 0.000 084 736;
  • 9) 0.000 084 736 × 2 = 0 + 0.000 169 472;
  • 10) 0.000 169 472 × 2 = 0 + 0.000 338 944;
  • 11) 0.000 338 944 × 2 = 0 + 0.000 677 888;
  • 12) 0.000 677 888 × 2 = 0 + 0.001 355 776;
  • 13) 0.001 355 776 × 2 = 0 + 0.002 711 552;
  • 14) 0.002 711 552 × 2 = 0 + 0.005 423 104;
  • 15) 0.005 423 104 × 2 = 0 + 0.010 846 208;
  • 16) 0.010 846 208 × 2 = 0 + 0.021 692 416;
  • 17) 0.021 692 416 × 2 = 0 + 0.043 384 832;
  • 18) 0.043 384 832 × 2 = 0 + 0.086 769 664;
  • 19) 0.086 769 664 × 2 = 0 + 0.173 539 328;
  • 20) 0.173 539 328 × 2 = 0 + 0.347 078 656;
  • 21) 0.347 078 656 × 2 = 0 + 0.694 157 312;
  • 22) 0.694 157 312 × 2 = 1 + 0.388 314 624;
  • 23) 0.388 314 624 × 2 = 0 + 0.776 629 248;
  • 24) 0.776 629 248 × 2 = 1 + 0.553 258 496;
  • 25) 0.553 258 496 × 2 = 1 + 0.106 516 992;
  • 26) 0.106 516 992 × 2 = 0 + 0.213 033 984;
  • 27) 0.213 033 984 × 2 = 0 + 0.426 067 968;
  • 28) 0.426 067 968 × 2 = 0 + 0.852 135 936;
  • 29) 0.852 135 936 × 2 = 1 + 0.704 271 872;
  • 30) 0.704 271 872 × 2 = 1 + 0.408 543 744;
  • 31) 0.408 543 744 × 2 = 0 + 0.817 087 488;
  • 32) 0.817 087 488 × 2 = 1 + 0.634 174 976;
  • 33) 0.634 174 976 × 2 = 1 + 0.268 349 952;
  • 34) 0.268 349 952 × 2 = 0 + 0.536 699 904;
  • 35) 0.536 699 904 × 2 = 1 + 0.073 399 808;
  • 36) 0.073 399 808 × 2 = 0 + 0.146 799 616;
  • 37) 0.146 799 616 × 2 = 0 + 0.293 599 232;
  • 38) 0.293 599 232 × 2 = 0 + 0.587 198 464;
  • 39) 0.587 198 464 × 2 = 1 + 0.174 396 928;
  • 40) 0.174 396 928 × 2 = 0 + 0.348 793 856;
  • 41) 0.348 793 856 × 2 = 0 + 0.697 587 712;
  • 42) 0.697 587 712 × 2 = 1 + 0.395 175 424;
  • 43) 0.395 175 424 × 2 = 0 + 0.790 350 848;
  • 44) 0.790 350 848 × 2 = 1 + 0.580 701 696;
  • 45) 0.580 701 696 × 2 = 1 + 0.161 403 392;

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).


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:


0.000 000 331(10) =

0.0000 0000 0000 0000 0000 0101 1000 1101 1010 0010 0101 1(2)

5. Positive number before normalization:

0.000 000 331(10) =

0.0000 0000 0000 0000 0000 0101 1000 1101 1010 0010 0101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 331(10) =

0.0000 0000 0000 0000 0000 0101 1000 1101 1010 0010 0101 1(2) =

0.0000 0000 0000 0000 0000 0101 1000 1101 1010 0010 0101 1(2) × 20 =

1.0110 0011 0110 1000 1001 011(2) × 2-22


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): -22

Mantissa (not normalized):
1.0110 0011 0110 1000 1001 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

105(10) =

0110 1001(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. 011 0001 1011 0100 0100 1011 =

011 0001 1011 0100 0100 1011


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) =
0110 1001

Mantissa (23 bits) =
011 0001 1011 0100 0100 1011


Decimal number 0.000 000 331 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 011 0001 1011 0100 0100 1011

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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