0.000 000 020 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 020 9(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
0.000 000 020 9(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: 0.000 000 020 9.

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) 0.000 000 020 9 × 2 = 0 + 0.000 000 041 8;
  • 2) 0.000 000 041 8 × 2 = 0 + 0.000 000 083 6;
  • 3) 0.000 000 083 6 × 2 = 0 + 0.000 000 167 2;
  • 4) 0.000 000 167 2 × 2 = 0 + 0.000 000 334 4;
  • 5) 0.000 000 334 4 × 2 = 0 + 0.000 000 668 8;
  • 6) 0.000 000 668 8 × 2 = 0 + 0.000 001 337 6;
  • 7) 0.000 001 337 6 × 2 = 0 + 0.000 002 675 2;
  • 8) 0.000 002 675 2 × 2 = 0 + 0.000 005 350 4;
  • 9) 0.000 005 350 4 × 2 = 0 + 0.000 010 700 8;
  • 10) 0.000 010 700 8 × 2 = 0 + 0.000 021 401 6;
  • 11) 0.000 021 401 6 × 2 = 0 + 0.000 042 803 2;
  • 12) 0.000 042 803 2 × 2 = 0 + 0.000 085 606 4;
  • 13) 0.000 085 606 4 × 2 = 0 + 0.000 171 212 8;
  • 14) 0.000 171 212 8 × 2 = 0 + 0.000 342 425 6;
  • 15) 0.000 342 425 6 × 2 = 0 + 0.000 684 851 2;
  • 16) 0.000 684 851 2 × 2 = 0 + 0.001 369 702 4;
  • 17) 0.001 369 702 4 × 2 = 0 + 0.002 739 404 8;
  • 18) 0.002 739 404 8 × 2 = 0 + 0.005 478 809 6;
  • 19) 0.005 478 809 6 × 2 = 0 + 0.010 957 619 2;
  • 20) 0.010 957 619 2 × 2 = 0 + 0.021 915 238 4;
  • 21) 0.021 915 238 4 × 2 = 0 + 0.043 830 476 8;
  • 22) 0.043 830 476 8 × 2 = 0 + 0.087 660 953 6;
  • 23) 0.087 660 953 6 × 2 = 0 + 0.175 321 907 2;
  • 24) 0.175 321 907 2 × 2 = 0 + 0.350 643 814 4;
  • 25) 0.350 643 814 4 × 2 = 0 + 0.701 287 628 8;
  • 26) 0.701 287 628 8 × 2 = 1 + 0.402 575 257 6;
  • 27) 0.402 575 257 6 × 2 = 0 + 0.805 150 515 2;
  • 28) 0.805 150 515 2 × 2 = 1 + 0.610 301 030 4;
  • 29) 0.610 301 030 4 × 2 = 1 + 0.220 602 060 8;
  • 30) 0.220 602 060 8 × 2 = 0 + 0.441 204 121 6;
  • 31) 0.441 204 121 6 × 2 = 0 + 0.882 408 243 2;
  • 32) 0.882 408 243 2 × 2 = 1 + 0.764 816 486 4;
  • 33) 0.764 816 486 4 × 2 = 1 + 0.529 632 972 8;
  • 34) 0.529 632 972 8 × 2 = 1 + 0.059 265 945 6;
  • 35) 0.059 265 945 6 × 2 = 0 + 0.118 531 891 2;
  • 36) 0.118 531 891 2 × 2 = 0 + 0.237 063 782 4;
  • 37) 0.237 063 782 4 × 2 = 0 + 0.474 127 564 8;
  • 38) 0.474 127 564 8 × 2 = 0 + 0.948 255 129 6;
  • 39) 0.948 255 129 6 × 2 = 1 + 0.896 510 259 2;
  • 40) 0.896 510 259 2 × 2 = 1 + 0.793 020 518 4;
  • 41) 0.793 020 518 4 × 2 = 1 + 0.586 041 036 8;
  • 42) 0.586 041 036 8 × 2 = 1 + 0.172 082 073 6;
  • 43) 0.172 082 073 6 × 2 = 0 + 0.344 164 147 2;
  • 44) 0.344 164 147 2 × 2 = 0 + 0.688 328 294 4;
  • 45) 0.688 328 294 4 × 2 = 1 + 0.376 656 588 8;
  • 46) 0.376 656 588 8 × 2 = 0 + 0.753 313 177 6;
  • 47) 0.753 313 177 6 × 2 = 1 + 0.506 626 355 2;
  • 48) 0.506 626 355 2 × 2 = 1 + 0.013 252 710 4;
  • 49) 0.013 252 710 4 × 2 = 0 + 0.026 505 420 8;

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:


0.000 000 020 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1001 1100 0011 1100 1011 0(2)

5. Positive number before normalization:

0.000 000 020 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1001 1100 0011 1100 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 020 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1001 1100 0011 1100 1011 0(2) =

0.0000 0000 0000 0000 0000 0000 0101 1001 1100 0011 1100 1011 0(2) × 20 =

1.0110 0111 0000 1111 0010 110(2) × 2-26


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): -26

Mantissa (not normalized):
1.0110 0111 0000 1111 0010 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

101(10) =

0110 0101(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 0011 1000 0111 1001 0110 =

011 0011 1000 0111 1001 0110


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 0101

Mantissa (23 bits) =
011 0011 1000 0111 1001 0110


Decimal number 0.000 000 020 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 011 0011 1000 0111 1001 0110

Share

» Convert decimal 0.000 000 029 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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