0.000 003 669 408 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 669 408 63(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
0.000 003 669 408 63(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: 0.000 003 669 408 63.

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) 0.000 003 669 408 63 × 2 = 0 + 0.000 007 338 817 26;
  • 2) 0.000 007 338 817 26 × 2 = 0 + 0.000 014 677 634 52;
  • 3) 0.000 014 677 634 52 × 2 = 0 + 0.000 029 355 269 04;
  • 4) 0.000 029 355 269 04 × 2 = 0 + 0.000 058 710 538 08;
  • 5) 0.000 058 710 538 08 × 2 = 0 + 0.000 117 421 076 16;
  • 6) 0.000 117 421 076 16 × 2 = 0 + 0.000 234 842 152 32;
  • 7) 0.000 234 842 152 32 × 2 = 0 + 0.000 469 684 304 64;
  • 8) 0.000 469 684 304 64 × 2 = 0 + 0.000 939 368 609 28;
  • 9) 0.000 939 368 609 28 × 2 = 0 + 0.001 878 737 218 56;
  • 10) 0.001 878 737 218 56 × 2 = 0 + 0.003 757 474 437 12;
  • 11) 0.003 757 474 437 12 × 2 = 0 + 0.007 514 948 874 24;
  • 12) 0.007 514 948 874 24 × 2 = 0 + 0.015 029 897 748 48;
  • 13) 0.015 029 897 748 48 × 2 = 0 + 0.030 059 795 496 96;
  • 14) 0.030 059 795 496 96 × 2 = 0 + 0.060 119 590 993 92;
  • 15) 0.060 119 590 993 92 × 2 = 0 + 0.120 239 181 987 84;
  • 16) 0.120 239 181 987 84 × 2 = 0 + 0.240 478 363 975 68;
  • 17) 0.240 478 363 975 68 × 2 = 0 + 0.480 956 727 951 36;
  • 18) 0.480 956 727 951 36 × 2 = 0 + 0.961 913 455 902 72;
  • 19) 0.961 913 455 902 72 × 2 = 1 + 0.923 826 911 805 44;
  • 20) 0.923 826 911 805 44 × 2 = 1 + 0.847 653 823 610 88;
  • 21) 0.847 653 823 610 88 × 2 = 1 + 0.695 307 647 221 76;
  • 22) 0.695 307 647 221 76 × 2 = 1 + 0.390 615 294 443 52;
  • 23) 0.390 615 294 443 52 × 2 = 0 + 0.781 230 588 887 04;
  • 24) 0.781 230 588 887 04 × 2 = 1 + 0.562 461 177 774 08;
  • 25) 0.562 461 177 774 08 × 2 = 1 + 0.124 922 355 548 16;
  • 26) 0.124 922 355 548 16 × 2 = 0 + 0.249 844 711 096 32;
  • 27) 0.249 844 711 096 32 × 2 = 0 + 0.499 689 422 192 64;
  • 28) 0.499 689 422 192 64 × 2 = 0 + 0.999 378 844 385 28;
  • 29) 0.999 378 844 385 28 × 2 = 1 + 0.998 757 688 770 56;
  • 30) 0.998 757 688 770 56 × 2 = 1 + 0.997 515 377 541 12;
  • 31) 0.997 515 377 541 12 × 2 = 1 + 0.995 030 755 082 24;
  • 32) 0.995 030 755 082 24 × 2 = 1 + 0.990 061 510 164 48;
  • 33) 0.990 061 510 164 48 × 2 = 1 + 0.980 123 020 328 96;
  • 34) 0.980 123 020 328 96 × 2 = 1 + 0.960 246 040 657 92;
  • 35) 0.960 246 040 657 92 × 2 = 1 + 0.920 492 081 315 84;
  • 36) 0.920 492 081 315 84 × 2 = 1 + 0.840 984 162 631 68;
  • 37) 0.840 984 162 631 68 × 2 = 1 + 0.681 968 325 263 36;
  • 38) 0.681 968 325 263 36 × 2 = 1 + 0.363 936 650 526 72;
  • 39) 0.363 936 650 526 72 × 2 = 0 + 0.727 873 301 053 44;
  • 40) 0.727 873 301 053 44 × 2 = 1 + 0.455 746 602 106 88;
  • 41) 0.455 746 602 106 88 × 2 = 0 + 0.911 493 204 213 76;
  • 42) 0.911 493 204 213 76 × 2 = 1 + 0.822 986 408 427 52;

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:


0.000 003 669 408 63(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1101 01(2)

5. Positive number before normalization:

0.000 003 669 408 63(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1101 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 669 408 63(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1101 01(2) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1101 01(2) × 20 =

1.1110 1100 0111 1111 1110 101(2) × 2-19


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): -19

Mantissa (not normalized):
1.1110 1100 0111 1111 1110 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(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;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 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) =

108(10) =

0110 1100(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 0110 0011 1111 1111 0101 =

111 0110 0011 1111 1111 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 1100

Mantissa (23 bits) =
111 0110 0011 1111 1111 0101


Decimal number 0.000 003 669 408 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 111 0110 0011 1111 1111 0101

Share

» Convert decimal 0.000 003 669 409 13 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: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 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