0.000 000 443 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 443(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
0.000 000 443(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: 0.000 000 443.

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) 0.000 000 443 × 2 = 0 + 0.000 000 886;
  • 2) 0.000 000 886 × 2 = 0 + 0.000 001 772;
  • 3) 0.000 001 772 × 2 = 0 + 0.000 003 544;
  • 4) 0.000 003 544 × 2 = 0 + 0.000 007 088;
  • 5) 0.000 007 088 × 2 = 0 + 0.000 014 176;
  • 6) 0.000 014 176 × 2 = 0 + 0.000 028 352;
  • 7) 0.000 028 352 × 2 = 0 + 0.000 056 704;
  • 8) 0.000 056 704 × 2 = 0 + 0.000 113 408;
  • 9) 0.000 113 408 × 2 = 0 + 0.000 226 816;
  • 10) 0.000 226 816 × 2 = 0 + 0.000 453 632;
  • 11) 0.000 453 632 × 2 = 0 + 0.000 907 264;
  • 12) 0.000 907 264 × 2 = 0 + 0.001 814 528;
  • 13) 0.001 814 528 × 2 = 0 + 0.003 629 056;
  • 14) 0.003 629 056 × 2 = 0 + 0.007 258 112;
  • 15) 0.007 258 112 × 2 = 0 + 0.014 516 224;
  • 16) 0.014 516 224 × 2 = 0 + 0.029 032 448;
  • 17) 0.029 032 448 × 2 = 0 + 0.058 064 896;
  • 18) 0.058 064 896 × 2 = 0 + 0.116 129 792;
  • 19) 0.116 129 792 × 2 = 0 + 0.232 259 584;
  • 20) 0.232 259 584 × 2 = 0 + 0.464 519 168;
  • 21) 0.464 519 168 × 2 = 0 + 0.929 038 336;
  • 22) 0.929 038 336 × 2 = 1 + 0.858 076 672;
  • 23) 0.858 076 672 × 2 = 1 + 0.716 153 344;
  • 24) 0.716 153 344 × 2 = 1 + 0.432 306 688;
  • 25) 0.432 306 688 × 2 = 0 + 0.864 613 376;
  • 26) 0.864 613 376 × 2 = 1 + 0.729 226 752;
  • 27) 0.729 226 752 × 2 = 1 + 0.458 453 504;
  • 28) 0.458 453 504 × 2 = 0 + 0.916 907 008;
  • 29) 0.916 907 008 × 2 = 1 + 0.833 814 016;
  • 30) 0.833 814 016 × 2 = 1 + 0.667 628 032;
  • 31) 0.667 628 032 × 2 = 1 + 0.335 256 064;
  • 32) 0.335 256 064 × 2 = 0 + 0.670 512 128;
  • 33) 0.670 512 128 × 2 = 1 + 0.341 024 256;
  • 34) 0.341 024 256 × 2 = 0 + 0.682 048 512;
  • 35) 0.682 048 512 × 2 = 1 + 0.364 097 024;
  • 36) 0.364 097 024 × 2 = 0 + 0.728 194 048;
  • 37) 0.728 194 048 × 2 = 1 + 0.456 388 096;
  • 38) 0.456 388 096 × 2 = 0 + 0.912 776 192;
  • 39) 0.912 776 192 × 2 = 1 + 0.825 552 384;
  • 40) 0.825 552 384 × 2 = 1 + 0.651 104 768;
  • 41) 0.651 104 768 × 2 = 1 + 0.302 209 536;
  • 42) 0.302 209 536 × 2 = 0 + 0.604 419 072;
  • 43) 0.604 419 072 × 2 = 1 + 0.208 838 144;
  • 44) 0.208 838 144 × 2 = 0 + 0.417 676 288;
  • 45) 0.417 676 288 × 2 = 0 + 0.835 352 576;

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:


0.000 000 443(10) =

0.0000 0000 0000 0000 0000 0111 0110 1110 1010 1011 1010 0(2)

5. Positive number before normalization:

0.000 000 443(10) =

0.0000 0000 0000 0000 0000 0111 0110 1110 1010 1011 1010 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:


0.000 000 443(10) =

0.0000 0000 0000 0000 0000 0111 0110 1110 1010 1011 1010 0(2) =

0.0000 0000 0000 0000 0000 0111 0110 1110 1010 1011 1010 0(2) × 20 =

1.1101 1011 1010 1010 1110 100(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):
1.1101 1011 1010 1010 1110 100


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. 110 1101 1101 0101 0111 0100 =

110 1101 1101 0101 0111 0100


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) =
110 1101 1101 0101 0111 0100


Decimal number 0.000 000 443 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 110 1101 1101 0101 0111 0100

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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