0.000 000 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 99(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 99(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 99.

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 99 × 2 = 0 + 0.000 001 98;
  • 2) 0.000 001 98 × 2 = 0 + 0.000 003 96;
  • 3) 0.000 003 96 × 2 = 0 + 0.000 007 92;
  • 4) 0.000 007 92 × 2 = 0 + 0.000 015 84;
  • 5) 0.000 015 84 × 2 = 0 + 0.000 031 68;
  • 6) 0.000 031 68 × 2 = 0 + 0.000 063 36;
  • 7) 0.000 063 36 × 2 = 0 + 0.000 126 72;
  • 8) 0.000 126 72 × 2 = 0 + 0.000 253 44;
  • 9) 0.000 253 44 × 2 = 0 + 0.000 506 88;
  • 10) 0.000 506 88 × 2 = 0 + 0.001 013 76;
  • 11) 0.001 013 76 × 2 = 0 + 0.002 027 52;
  • 12) 0.002 027 52 × 2 = 0 + 0.004 055 04;
  • 13) 0.004 055 04 × 2 = 0 + 0.008 110 08;
  • 14) 0.008 110 08 × 2 = 0 + 0.016 220 16;
  • 15) 0.016 220 16 × 2 = 0 + 0.032 440 32;
  • 16) 0.032 440 32 × 2 = 0 + 0.064 880 64;
  • 17) 0.064 880 64 × 2 = 0 + 0.129 761 28;
  • 18) 0.129 761 28 × 2 = 0 + 0.259 522 56;
  • 19) 0.259 522 56 × 2 = 0 + 0.519 045 12;
  • 20) 0.519 045 12 × 2 = 1 + 0.038 090 24;
  • 21) 0.038 090 24 × 2 = 0 + 0.076 180 48;
  • 22) 0.076 180 48 × 2 = 0 + 0.152 360 96;
  • 23) 0.152 360 96 × 2 = 0 + 0.304 721 92;
  • 24) 0.304 721 92 × 2 = 0 + 0.609 443 84;
  • 25) 0.609 443 84 × 2 = 1 + 0.218 887 68;
  • 26) 0.218 887 68 × 2 = 0 + 0.437 775 36;
  • 27) 0.437 775 36 × 2 = 0 + 0.875 550 72;
  • 28) 0.875 550 72 × 2 = 1 + 0.751 101 44;
  • 29) 0.751 101 44 × 2 = 1 + 0.502 202 88;
  • 30) 0.502 202 88 × 2 = 1 + 0.004 405 76;
  • 31) 0.004 405 76 × 2 = 0 + 0.008 811 52;
  • 32) 0.008 811 52 × 2 = 0 + 0.017 623 04;
  • 33) 0.017 623 04 × 2 = 0 + 0.035 246 08;
  • 34) 0.035 246 08 × 2 = 0 + 0.070 492 16;
  • 35) 0.070 492 16 × 2 = 0 + 0.140 984 32;
  • 36) 0.140 984 32 × 2 = 0 + 0.281 968 64;
  • 37) 0.281 968 64 × 2 = 0 + 0.563 937 28;
  • 38) 0.563 937 28 × 2 = 1 + 0.127 874 56;
  • 39) 0.127 874 56 × 2 = 0 + 0.255 749 12;
  • 40) 0.255 749 12 × 2 = 0 + 0.511 498 24;
  • 41) 0.511 498 24 × 2 = 1 + 0.022 996 48;
  • 42) 0.022 996 48 × 2 = 0 + 0.045 992 96;
  • 43) 0.045 992 96 × 2 = 0 + 0.091 985 92;

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 99(10) =

0.0000 0000 0000 0000 0001 0000 1001 1100 0000 0100 100(2)

5. Positive number before normalization:

0.000 000 99(10) =

0.0000 0000 0000 0000 0001 0000 1001 1100 0000 0100 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 99(10) =

0.0000 0000 0000 0000 0001 0000 1001 1100 0000 0100 100(2) =

0.0000 0000 0000 0000 0001 0000 1001 1100 0000 0100 100(2) × 20 =

1.0000 1001 1100 0000 0100 100(2) × 2-20


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): -20

Mantissa (not normalized):
1.0000 1001 1100 0000 0100 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 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) =

107(10) =

0110 1011(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 0100 1110 0000 0010 0100 =

000 0100 1110 0000 0010 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 1011

Mantissa (23 bits) =
000 0100 1110 0000 0010 0100


Decimal number 0.000 000 99 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 000 0100 1110 0000 0010 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