0.000 000 020 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 020 2(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 2(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 2.

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 2 × 2 = 0 + 0.000 000 040 4;
  • 2) 0.000 000 040 4 × 2 = 0 + 0.000 000 080 8;
  • 3) 0.000 000 080 8 × 2 = 0 + 0.000 000 161 6;
  • 4) 0.000 000 161 6 × 2 = 0 + 0.000 000 323 2;
  • 5) 0.000 000 323 2 × 2 = 0 + 0.000 000 646 4;
  • 6) 0.000 000 646 4 × 2 = 0 + 0.000 001 292 8;
  • 7) 0.000 001 292 8 × 2 = 0 + 0.000 002 585 6;
  • 8) 0.000 002 585 6 × 2 = 0 + 0.000 005 171 2;
  • 9) 0.000 005 171 2 × 2 = 0 + 0.000 010 342 4;
  • 10) 0.000 010 342 4 × 2 = 0 + 0.000 020 684 8;
  • 11) 0.000 020 684 8 × 2 = 0 + 0.000 041 369 6;
  • 12) 0.000 041 369 6 × 2 = 0 + 0.000 082 739 2;
  • 13) 0.000 082 739 2 × 2 = 0 + 0.000 165 478 4;
  • 14) 0.000 165 478 4 × 2 = 0 + 0.000 330 956 8;
  • 15) 0.000 330 956 8 × 2 = 0 + 0.000 661 913 6;
  • 16) 0.000 661 913 6 × 2 = 0 + 0.001 323 827 2;
  • 17) 0.001 323 827 2 × 2 = 0 + 0.002 647 654 4;
  • 18) 0.002 647 654 4 × 2 = 0 + 0.005 295 308 8;
  • 19) 0.005 295 308 8 × 2 = 0 + 0.010 590 617 6;
  • 20) 0.010 590 617 6 × 2 = 0 + 0.021 181 235 2;
  • 21) 0.021 181 235 2 × 2 = 0 + 0.042 362 470 4;
  • 22) 0.042 362 470 4 × 2 = 0 + 0.084 724 940 8;
  • 23) 0.084 724 940 8 × 2 = 0 + 0.169 449 881 6;
  • 24) 0.169 449 881 6 × 2 = 0 + 0.338 899 763 2;
  • 25) 0.338 899 763 2 × 2 = 0 + 0.677 799 526 4;
  • 26) 0.677 799 526 4 × 2 = 1 + 0.355 599 052 8;
  • 27) 0.355 599 052 8 × 2 = 0 + 0.711 198 105 6;
  • 28) 0.711 198 105 6 × 2 = 1 + 0.422 396 211 2;
  • 29) 0.422 396 211 2 × 2 = 0 + 0.844 792 422 4;
  • 30) 0.844 792 422 4 × 2 = 1 + 0.689 584 844 8;
  • 31) 0.689 584 844 8 × 2 = 1 + 0.379 169 689 6;
  • 32) 0.379 169 689 6 × 2 = 0 + 0.758 339 379 2;
  • 33) 0.758 339 379 2 × 2 = 1 + 0.516 678 758 4;
  • 34) 0.516 678 758 4 × 2 = 1 + 0.033 357 516 8;
  • 35) 0.033 357 516 8 × 2 = 0 + 0.066 715 033 6;
  • 36) 0.066 715 033 6 × 2 = 0 + 0.133 430 067 2;
  • 37) 0.133 430 067 2 × 2 = 0 + 0.266 860 134 4;
  • 38) 0.266 860 134 4 × 2 = 0 + 0.533 720 268 8;
  • 39) 0.533 720 268 8 × 2 = 1 + 0.067 440 537 6;
  • 40) 0.067 440 537 6 × 2 = 0 + 0.134 881 075 2;
  • 41) 0.134 881 075 2 × 2 = 0 + 0.269 762 150 4;
  • 42) 0.269 762 150 4 × 2 = 0 + 0.539 524 300 8;
  • 43) 0.539 524 300 8 × 2 = 1 + 0.079 048 601 6;
  • 44) 0.079 048 601 6 × 2 = 0 + 0.158 097 203 2;
  • 45) 0.158 097 203 2 × 2 = 0 + 0.316 194 406 4;
  • 46) 0.316 194 406 4 × 2 = 0 + 0.632 388 812 8;
  • 47) 0.632 388 812 8 × 2 = 1 + 0.264 777 625 6;
  • 48) 0.264 777 625 6 × 2 = 0 + 0.529 555 251 2;
  • 49) 0.529 555 251 2 × 2 = 1 + 0.059 110 502 4;

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 2(10) =

0.0000 0000 0000 0000 0000 0000 0101 0110 1100 0010 0010 0010 1(2)

5. Positive number before normalization:

0.000 000 020 2(10) =

0.0000 0000 0000 0000 0000 0000 0101 0110 1100 0010 0010 0010 1(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 2(10) =

0.0000 0000 0000 0000 0000 0000 0101 0110 1100 0010 0010 0010 1(2) =

0.0000 0000 0000 0000 0000 0000 0101 0110 1100 0010 0010 0010 1(2) × 20 =

1.0101 1011 0000 1000 1000 101(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):
1.0101 1011 0000 1000 1000 101


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. 010 1101 1000 0100 0100 0101 =

010 1101 1000 0100 0100 0101


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) =
010 1101 1000 0100 0100 0101


Decimal number 0.000 000 020 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 010 1101 1000 0100 0100 0101

Share

» Convert decimal 0.000 000 012 3 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: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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