Decimal to 32 Bit IEEE 754 Binary: Convert Number 0.000 525 932 8 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.000 525 932 8(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (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: 0.000 525 932 8.

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) 0.000 525 932 8 × 2 = 0 + 0.001 051 865 6;
  • 2) 0.001 051 865 6 × 2 = 0 + 0.002 103 731 2;
  • 3) 0.002 103 731 2 × 2 = 0 + 0.004 207 462 4;
  • 4) 0.004 207 462 4 × 2 = 0 + 0.008 414 924 8;
  • 5) 0.008 414 924 8 × 2 = 0 + 0.016 829 849 6;
  • 6) 0.016 829 849 6 × 2 = 0 + 0.033 659 699 2;
  • 7) 0.033 659 699 2 × 2 = 0 + 0.067 319 398 4;
  • 8) 0.067 319 398 4 × 2 = 0 + 0.134 638 796 8;
  • 9) 0.134 638 796 8 × 2 = 0 + 0.269 277 593 6;
  • 10) 0.269 277 593 6 × 2 = 0 + 0.538 555 187 2;
  • 11) 0.538 555 187 2 × 2 = 1 + 0.077 110 374 4;
  • 12) 0.077 110 374 4 × 2 = 0 + 0.154 220 748 8;
  • 13) 0.154 220 748 8 × 2 = 0 + 0.308 441 497 6;
  • 14) 0.308 441 497 6 × 2 = 0 + 0.616 882 995 2;
  • 15) 0.616 882 995 2 × 2 = 1 + 0.233 765 990 4;
  • 16) 0.233 765 990 4 × 2 = 0 + 0.467 531 980 8;
  • 17) 0.467 531 980 8 × 2 = 0 + 0.935 063 961 6;
  • 18) 0.935 063 961 6 × 2 = 1 + 0.870 127 923 2;
  • 19) 0.870 127 923 2 × 2 = 1 + 0.740 255 846 4;
  • 20) 0.740 255 846 4 × 2 = 1 + 0.480 511 692 8;
  • 21) 0.480 511 692 8 × 2 = 0 + 0.961 023 385 6;
  • 22) 0.961 023 385 6 × 2 = 1 + 0.922 046 771 2;
  • 23) 0.922 046 771 2 × 2 = 1 + 0.844 093 542 4;
  • 24) 0.844 093 542 4 × 2 = 1 + 0.688 187 084 8;
  • 25) 0.688 187 084 8 × 2 = 1 + 0.376 374 169 6;
  • 26) 0.376 374 169 6 × 2 = 0 + 0.752 748 339 2;
  • 27) 0.752 748 339 2 × 2 = 1 + 0.505 496 678 4;
  • 28) 0.505 496 678 4 × 2 = 1 + 0.010 993 356 8;
  • 29) 0.010 993 356 8 × 2 = 0 + 0.021 986 713 6;
  • 30) 0.021 986 713 6 × 2 = 0 + 0.043 973 427 2;
  • 31) 0.043 973 427 2 × 2 = 0 + 0.087 946 854 4;
  • 32) 0.087 946 854 4 × 2 = 0 + 0.175 893 708 8;
  • 33) 0.175 893 708 8 × 2 = 0 + 0.351 787 417 6;
  • 34) 0.351 787 417 6 × 2 = 0 + 0.703 574 835 2;

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:


0.000 525 932 8(10) =

0.0000 0000 0010 0010 0111 0111 1011 0000 00(2)

5. Positive number before normalization:

0.000 525 932 8(10) =

0.0000 0000 0010 0010 0111 0111 1011 0000 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 525 932 8(10) =

0.0000 0000 0010 0010 0111 0111 1011 0000 00(2) =

0.0000 0000 0010 0010 0111 0111 1011 0000 00(2) × 20 =

1.0001 0011 1011 1101 1000 000(2) × 2-11


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): -11

Mantissa (not normalized):
1.0001 0011 1011 1101 1000 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-11 + 2(8-1) - 1 =

(-11 + 127)(10) =

116(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;
  • 116 ÷ 2 = 58 + 0;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

116(10) =

0111 0100(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 1001 1101 1110 1100 0000 =

000 1001 1101 1110 1100 0000


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) =
0111 0100

Mantissa (23 bits) =
000 1001 1101 1110 1100 0000


The base ten decimal number 0.000 525 932 8 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0100 - 000 1001 1101 1110 1100 0000

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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