0.000 030 229 39 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 030 229 39(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
0.000 030 229 39(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: 0.000 030 229 39.

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) 0.000 030 229 39 × 2 = 0 + 0.000 060 458 78;
  • 2) 0.000 060 458 78 × 2 = 0 + 0.000 120 917 56;
  • 3) 0.000 120 917 56 × 2 = 0 + 0.000 241 835 12;
  • 4) 0.000 241 835 12 × 2 = 0 + 0.000 483 670 24;
  • 5) 0.000 483 670 24 × 2 = 0 + 0.000 967 340 48;
  • 6) 0.000 967 340 48 × 2 = 0 + 0.001 934 680 96;
  • 7) 0.001 934 680 96 × 2 = 0 + 0.003 869 361 92;
  • 8) 0.003 869 361 92 × 2 = 0 + 0.007 738 723 84;
  • 9) 0.007 738 723 84 × 2 = 0 + 0.015 477 447 68;
  • 10) 0.015 477 447 68 × 2 = 0 + 0.030 954 895 36;
  • 11) 0.030 954 895 36 × 2 = 0 + 0.061 909 790 72;
  • 12) 0.061 909 790 72 × 2 = 0 + 0.123 819 581 44;
  • 13) 0.123 819 581 44 × 2 = 0 + 0.247 639 162 88;
  • 14) 0.247 639 162 88 × 2 = 0 + 0.495 278 325 76;
  • 15) 0.495 278 325 76 × 2 = 0 + 0.990 556 651 52;
  • 16) 0.990 556 651 52 × 2 = 1 + 0.981 113 303 04;
  • 17) 0.981 113 303 04 × 2 = 1 + 0.962 226 606 08;
  • 18) 0.962 226 606 08 × 2 = 1 + 0.924 453 212 16;
  • 19) 0.924 453 212 16 × 2 = 1 + 0.848 906 424 32;
  • 20) 0.848 906 424 32 × 2 = 1 + 0.697 812 848 64;
  • 21) 0.697 812 848 64 × 2 = 1 + 0.395 625 697 28;
  • 22) 0.395 625 697 28 × 2 = 0 + 0.791 251 394 56;
  • 23) 0.791 251 394 56 × 2 = 1 + 0.582 502 789 12;
  • 24) 0.582 502 789 12 × 2 = 1 + 0.165 005 578 24;
  • 25) 0.165 005 578 24 × 2 = 0 + 0.330 011 156 48;
  • 26) 0.330 011 156 48 × 2 = 0 + 0.660 022 312 96;
  • 27) 0.660 022 312 96 × 2 = 1 + 0.320 044 625 92;
  • 28) 0.320 044 625 92 × 2 = 0 + 0.640 089 251 84;
  • 29) 0.640 089 251 84 × 2 = 1 + 0.280 178 503 68;
  • 30) 0.280 178 503 68 × 2 = 0 + 0.560 357 007 36;
  • 31) 0.560 357 007 36 × 2 = 1 + 0.120 714 014 72;
  • 32) 0.120 714 014 72 × 2 = 0 + 0.241 428 029 44;
  • 33) 0.241 428 029 44 × 2 = 0 + 0.482 856 058 88;
  • 34) 0.482 856 058 88 × 2 = 0 + 0.965 712 117 76;
  • 35) 0.965 712 117 76 × 2 = 1 + 0.931 424 235 52;
  • 36) 0.931 424 235 52 × 2 = 1 + 0.862 848 471 04;
  • 37) 0.862 848 471 04 × 2 = 1 + 0.725 696 942 08;
  • 38) 0.725 696 942 08 × 2 = 1 + 0.451 393 884 16;
  • 39) 0.451 393 884 16 × 2 = 0 + 0.902 787 768 32;

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:


0.000 030 229 39(10) =

0.0000 0000 0000 0001 1111 1011 0010 1010 0011 110(2)

5. Positive number before normalization:

0.000 030 229 39(10) =

0.0000 0000 0000 0001 1111 1011 0010 1010 0011 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 030 229 39(10) =

0.0000 0000 0000 0001 1111 1011 0010 1010 0011 110(2) =

0.0000 0000 0000 0001 1111 1011 0010 1010 0011 110(2) × 20 =

1.1111 1011 0010 1010 0011 110(2) × 2-16


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): -16

Mantissa (not normalized):
1.1111 1011 0010 1010 0011 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(8-1) - 1 =

(-16 + 127)(10) =

111(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;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 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) =

111(10) =

0110 1111(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. 111 1101 1001 0101 0001 1110 =

111 1101 1001 0101 0001 1110


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 1111

Mantissa (23 bits) =
111 1101 1001 0101 0001 1110


Decimal number 0.000 030 229 39 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1111 - 111 1101 1001 0101 0001 1110

Share

» Convert decimal 0.000 030 229 4 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 08, 2026 [August]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. Gemini Notebook YouTube Video of 7.50 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