0.000 000 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 66(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 66(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 66.

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 66 × 2 = 0 + 0.000 001 32;
  • 2) 0.000 001 32 × 2 = 0 + 0.000 002 64;
  • 3) 0.000 002 64 × 2 = 0 + 0.000 005 28;
  • 4) 0.000 005 28 × 2 = 0 + 0.000 010 56;
  • 5) 0.000 010 56 × 2 = 0 + 0.000 021 12;
  • 6) 0.000 021 12 × 2 = 0 + 0.000 042 24;
  • 7) 0.000 042 24 × 2 = 0 + 0.000 084 48;
  • 8) 0.000 084 48 × 2 = 0 + 0.000 168 96;
  • 9) 0.000 168 96 × 2 = 0 + 0.000 337 92;
  • 10) 0.000 337 92 × 2 = 0 + 0.000 675 84;
  • 11) 0.000 675 84 × 2 = 0 + 0.001 351 68;
  • 12) 0.001 351 68 × 2 = 0 + 0.002 703 36;
  • 13) 0.002 703 36 × 2 = 0 + 0.005 406 72;
  • 14) 0.005 406 72 × 2 = 0 + 0.010 813 44;
  • 15) 0.010 813 44 × 2 = 0 + 0.021 626 88;
  • 16) 0.021 626 88 × 2 = 0 + 0.043 253 76;
  • 17) 0.043 253 76 × 2 = 0 + 0.086 507 52;
  • 18) 0.086 507 52 × 2 = 0 + 0.173 015 04;
  • 19) 0.173 015 04 × 2 = 0 + 0.346 030 08;
  • 20) 0.346 030 08 × 2 = 0 + 0.692 060 16;
  • 21) 0.692 060 16 × 2 = 1 + 0.384 120 32;
  • 22) 0.384 120 32 × 2 = 0 + 0.768 240 64;
  • 23) 0.768 240 64 × 2 = 1 + 0.536 481 28;
  • 24) 0.536 481 28 × 2 = 1 + 0.072 962 56;
  • 25) 0.072 962 56 × 2 = 0 + 0.145 925 12;
  • 26) 0.145 925 12 × 2 = 0 + 0.291 850 24;
  • 27) 0.291 850 24 × 2 = 0 + 0.583 700 48;
  • 28) 0.583 700 48 × 2 = 1 + 0.167 400 96;
  • 29) 0.167 400 96 × 2 = 0 + 0.334 801 92;
  • 30) 0.334 801 92 × 2 = 0 + 0.669 603 84;
  • 31) 0.669 603 84 × 2 = 1 + 0.339 207 68;
  • 32) 0.339 207 68 × 2 = 0 + 0.678 415 36;
  • 33) 0.678 415 36 × 2 = 1 + 0.356 830 72;
  • 34) 0.356 830 72 × 2 = 0 + 0.713 661 44;
  • 35) 0.713 661 44 × 2 = 1 + 0.427 322 88;
  • 36) 0.427 322 88 × 2 = 0 + 0.854 645 76;
  • 37) 0.854 645 76 × 2 = 1 + 0.709 291 52;
  • 38) 0.709 291 52 × 2 = 1 + 0.418 583 04;
  • 39) 0.418 583 04 × 2 = 0 + 0.837 166 08;
  • 40) 0.837 166 08 × 2 = 1 + 0.674 332 16;
  • 41) 0.674 332 16 × 2 = 1 + 0.348 664 32;
  • 42) 0.348 664 32 × 2 = 0 + 0.697 328 64;
  • 43) 0.697 328 64 × 2 = 1 + 0.394 657 28;
  • 44) 0.394 657 28 × 2 = 0 + 0.789 314 56;

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 66(10) =

0.0000 0000 0000 0000 0000 1011 0001 0010 1010 1101 1010(2)

5. Positive number before normalization:

0.000 000 66(10) =

0.0000 0000 0000 0000 0000 1011 0001 0010 1010 1101 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 66(10) =

0.0000 0000 0000 0000 0000 1011 0001 0010 1010 1101 1010(2) =

0.0000 0000 0000 0000 0000 1011 0001 0010 1010 1101 1010(2) × 20 =

1.0110 0010 0101 0101 1011 010(2) × 2-21


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): -21

Mantissa (not normalized):
1.0110 0010 0101 0101 1011 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 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) =

106(10) =

0110 1010(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 0010 1010 1101 1010 =

011 0001 0010 1010 1101 1010


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 1010

Mantissa (23 bits) =
011 0001 0010 1010 1101 1010


Decimal number 0.000 000 66 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 011 0001 0010 1010 1101 1010

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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