-0.000 204 88 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 204 88(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
-0.000 204 88(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. Start with the positive version of the number:

|-0.000 204 88| = 0.000 204 88


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 204 88.

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) 0.000 204 88 × 2 = 0 + 0.000 409 76;
  • 2) 0.000 409 76 × 2 = 0 + 0.000 819 52;
  • 3) 0.000 819 52 × 2 = 0 + 0.001 639 04;
  • 4) 0.001 639 04 × 2 = 0 + 0.003 278 08;
  • 5) 0.003 278 08 × 2 = 0 + 0.006 556 16;
  • 6) 0.006 556 16 × 2 = 0 + 0.013 112 32;
  • 7) 0.013 112 32 × 2 = 0 + 0.026 224 64;
  • 8) 0.026 224 64 × 2 = 0 + 0.052 449 28;
  • 9) 0.052 449 28 × 2 = 0 + 0.104 898 56;
  • 10) 0.104 898 56 × 2 = 0 + 0.209 797 12;
  • 11) 0.209 797 12 × 2 = 0 + 0.419 594 24;
  • 12) 0.419 594 24 × 2 = 0 + 0.839 188 48;
  • 13) 0.839 188 48 × 2 = 1 + 0.678 376 96;
  • 14) 0.678 376 96 × 2 = 1 + 0.356 753 92;
  • 15) 0.356 753 92 × 2 = 0 + 0.713 507 84;
  • 16) 0.713 507 84 × 2 = 1 + 0.427 015 68;
  • 17) 0.427 015 68 × 2 = 0 + 0.854 031 36;
  • 18) 0.854 031 36 × 2 = 1 + 0.708 062 72;
  • 19) 0.708 062 72 × 2 = 1 + 0.416 125 44;
  • 20) 0.416 125 44 × 2 = 0 + 0.832 250 88;
  • 21) 0.832 250 88 × 2 = 1 + 0.664 501 76;
  • 22) 0.664 501 76 × 2 = 1 + 0.329 003 52;
  • 23) 0.329 003 52 × 2 = 0 + 0.658 007 04;
  • 24) 0.658 007 04 × 2 = 1 + 0.316 014 08;
  • 25) 0.316 014 08 × 2 = 0 + 0.632 028 16;
  • 26) 0.632 028 16 × 2 = 1 + 0.264 056 32;
  • 27) 0.264 056 32 × 2 = 0 + 0.528 112 64;
  • 28) 0.528 112 64 × 2 = 1 + 0.056 225 28;
  • 29) 0.056 225 28 × 2 = 0 + 0.112 450 56;
  • 30) 0.112 450 56 × 2 = 0 + 0.224 901 12;
  • 31) 0.224 901 12 × 2 = 0 + 0.449 802 24;
  • 32) 0.449 802 24 × 2 = 0 + 0.899 604 48;
  • 33) 0.899 604 48 × 2 = 1 + 0.799 208 96;
  • 34) 0.799 208 96 × 2 = 1 + 0.598 417 92;
  • 35) 0.598 417 92 × 2 = 1 + 0.196 835 84;
  • 36) 0.196 835 84 × 2 = 0 + 0.393 671 68;

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).


5. 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:


0.000 204 88(10) =

0.0000 0000 0000 1101 0110 1101 0101 0000 1110(2)

6. Positive number before normalization:

0.000 204 88(10) =

0.0000 0000 0000 1101 0110 1101 0101 0000 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 204 88(10) =

0.0000 0000 0000 1101 0110 1101 0101 0000 1110(2) =

0.0000 0000 0000 1101 0110 1101 0101 0000 1110(2) × 20 =

1.1010 1101 1010 1010 0001 110(2) × 2-13


8. 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 1 (a negative number)

Exponent (unadjusted): -13

Mantissa (not normalized):
1.1010 1101 1010 1010 0001 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(8-1) - 1 =

(-13 + 127)(10) =

114(10)


10. 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;
  • 114 ÷ 2 = 57 + 0;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

114(10) =

0111 0010(2)


12. 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. 101 0110 1101 0101 0000 1110 =

101 0110 1101 0101 0000 1110


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0111 0010

Mantissa (23 bits) =
101 0110 1101 0101 0000 1110


Decimal number -0.000 204 88 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0010 - 101 0110 1101 0101 0000 1110

Share

» Convert decimal -0.000 204 44 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 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 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