0.000 000 255 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 255(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
0.000 000 255(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: 0.000 000 255.

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) 0.000 000 255 × 2 = 0 + 0.000 000 51;
  • 2) 0.000 000 51 × 2 = 0 + 0.000 001 02;
  • 3) 0.000 001 02 × 2 = 0 + 0.000 002 04;
  • 4) 0.000 002 04 × 2 = 0 + 0.000 004 08;
  • 5) 0.000 004 08 × 2 = 0 + 0.000 008 16;
  • 6) 0.000 008 16 × 2 = 0 + 0.000 016 32;
  • 7) 0.000 016 32 × 2 = 0 + 0.000 032 64;
  • 8) 0.000 032 64 × 2 = 0 + 0.000 065 28;
  • 9) 0.000 065 28 × 2 = 0 + 0.000 130 56;
  • 10) 0.000 130 56 × 2 = 0 + 0.000 261 12;
  • 11) 0.000 261 12 × 2 = 0 + 0.000 522 24;
  • 12) 0.000 522 24 × 2 = 0 + 0.001 044 48;
  • 13) 0.001 044 48 × 2 = 0 + 0.002 088 96;
  • 14) 0.002 088 96 × 2 = 0 + 0.004 177 92;
  • 15) 0.004 177 92 × 2 = 0 + 0.008 355 84;
  • 16) 0.008 355 84 × 2 = 0 + 0.016 711 68;
  • 17) 0.016 711 68 × 2 = 0 + 0.033 423 36;
  • 18) 0.033 423 36 × 2 = 0 + 0.066 846 72;
  • 19) 0.066 846 72 × 2 = 0 + 0.133 693 44;
  • 20) 0.133 693 44 × 2 = 0 + 0.267 386 88;
  • 21) 0.267 386 88 × 2 = 0 + 0.534 773 76;
  • 22) 0.534 773 76 × 2 = 1 + 0.069 547 52;
  • 23) 0.069 547 52 × 2 = 0 + 0.139 095 04;
  • 24) 0.139 095 04 × 2 = 0 + 0.278 190 08;
  • 25) 0.278 190 08 × 2 = 0 + 0.556 380 16;
  • 26) 0.556 380 16 × 2 = 1 + 0.112 760 32;
  • 27) 0.112 760 32 × 2 = 0 + 0.225 520 64;
  • 28) 0.225 520 64 × 2 = 0 + 0.451 041 28;
  • 29) 0.451 041 28 × 2 = 0 + 0.902 082 56;
  • 30) 0.902 082 56 × 2 = 1 + 0.804 165 12;
  • 31) 0.804 165 12 × 2 = 1 + 0.608 330 24;
  • 32) 0.608 330 24 × 2 = 1 + 0.216 660 48;
  • 33) 0.216 660 48 × 2 = 0 + 0.433 320 96;
  • 34) 0.433 320 96 × 2 = 0 + 0.866 641 92;
  • 35) 0.866 641 92 × 2 = 1 + 0.733 283 84;
  • 36) 0.733 283 84 × 2 = 1 + 0.466 567 68;
  • 37) 0.466 567 68 × 2 = 0 + 0.933 135 36;
  • 38) 0.933 135 36 × 2 = 1 + 0.866 270 72;
  • 39) 0.866 270 72 × 2 = 1 + 0.732 541 44;
  • 40) 0.732 541 44 × 2 = 1 + 0.465 082 88;
  • 41) 0.465 082 88 × 2 = 0 + 0.930 165 76;
  • 42) 0.930 165 76 × 2 = 1 + 0.860 331 52;
  • 43) 0.860 331 52 × 2 = 1 + 0.720 663 04;
  • 44) 0.720 663 04 × 2 = 1 + 0.441 326 08;
  • 45) 0.441 326 08 × 2 = 0 + 0.882 652 16;

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:


0.000 000 255(10) =

0.0000 0000 0000 0000 0000 0100 0100 0111 0011 0111 0111 0(2)

5. Positive number before normalization:

0.000 000 255(10) =

0.0000 0000 0000 0000 0000 0100 0100 0111 0011 0111 0111 0(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:


0.000 000 255(10) =

0.0000 0000 0000 0000 0000 0100 0100 0111 0011 0111 0111 0(2) =

0.0000 0000 0000 0000 0000 0100 0100 0111 0011 0111 0111 0(2) × 20 =

1.0001 0001 1100 1101 1101 110(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):
1.0001 0001 1100 1101 1101 110


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. 000 1000 1110 0110 1110 1110 =

000 1000 1110 0110 1110 1110


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) =
000 1000 1110 0110 1110 1110


Decimal number 0.000 000 255 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 1000 1110 0110 1110 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