Decimal to 64 Bit IEEE 754 Binary: Convert Number 32 913.337 166 009 107 03 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 32 913.337 166 009 107 03(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 32 913.
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;
  • 32 913 ÷ 2 = 16 456 + 1;
  • 16 456 ÷ 2 = 8 228 + 0;
  • 8 228 ÷ 2 = 4 114 + 0;
  • 4 114 ÷ 2 = 2 057 + 0;
  • 2 057 ÷ 2 = 1 028 + 1;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


32 913(10) =


1000 0000 1001 0001(2)


3. Convert to binary (base 2) the fractional part: 0.337 166 009 107 03.

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) 0.337 166 009 107 03 × 2 = 0 + 0.674 332 018 214 06;
  • 2) 0.674 332 018 214 06 × 2 = 1 + 0.348 664 036 428 12;
  • 3) 0.348 664 036 428 12 × 2 = 0 + 0.697 328 072 856 24;
  • 4) 0.697 328 072 856 24 × 2 = 1 + 0.394 656 145 712 48;
  • 5) 0.394 656 145 712 48 × 2 = 0 + 0.789 312 291 424 96;
  • 6) 0.789 312 291 424 96 × 2 = 1 + 0.578 624 582 849 92;
  • 7) 0.578 624 582 849 92 × 2 = 1 + 0.157 249 165 699 84;
  • 8) 0.157 249 165 699 84 × 2 = 0 + 0.314 498 331 399 68;
  • 9) 0.314 498 331 399 68 × 2 = 0 + 0.628 996 662 799 36;
  • 10) 0.628 996 662 799 36 × 2 = 1 + 0.257 993 325 598 72;
  • 11) 0.257 993 325 598 72 × 2 = 0 + 0.515 986 651 197 44;
  • 12) 0.515 986 651 197 44 × 2 = 1 + 0.031 973 302 394 88;
  • 13) 0.031 973 302 394 88 × 2 = 0 + 0.063 946 604 789 76;
  • 14) 0.063 946 604 789 76 × 2 = 0 + 0.127 893 209 579 52;
  • 15) 0.127 893 209 579 52 × 2 = 0 + 0.255 786 419 159 04;
  • 16) 0.255 786 419 159 04 × 2 = 0 + 0.511 572 838 318 08;
  • 17) 0.511 572 838 318 08 × 2 = 1 + 0.023 145 676 636 16;
  • 18) 0.023 145 676 636 16 × 2 = 0 + 0.046 291 353 272 32;
  • 19) 0.046 291 353 272 32 × 2 = 0 + 0.092 582 706 544 64;
  • 20) 0.092 582 706 544 64 × 2 = 0 + 0.185 165 413 089 28;
  • 21) 0.185 165 413 089 28 × 2 = 0 + 0.370 330 826 178 56;
  • 22) 0.370 330 826 178 56 × 2 = 0 + 0.740 661 652 357 12;
  • 23) 0.740 661 652 357 12 × 2 = 1 + 0.481 323 304 714 24;
  • 24) 0.481 323 304 714 24 × 2 = 0 + 0.962 646 609 428 48;
  • 25) 0.962 646 609 428 48 × 2 = 1 + 0.925 293 218 856 96;
  • 26) 0.925 293 218 856 96 × 2 = 1 + 0.850 586 437 713 92;
  • 27) 0.850 586 437 713 92 × 2 = 1 + 0.701 172 875 427 84;
  • 28) 0.701 172 875 427 84 × 2 = 1 + 0.402 345 750 855 68;
  • 29) 0.402 345 750 855 68 × 2 = 0 + 0.804 691 501 711 36;
  • 30) 0.804 691 501 711 36 × 2 = 1 + 0.609 383 003 422 72;
  • 31) 0.609 383 003 422 72 × 2 = 1 + 0.218 766 006 845 44;
  • 32) 0.218 766 006 845 44 × 2 = 0 + 0.437 532 013 690 88;
  • 33) 0.437 532 013 690 88 × 2 = 0 + 0.875 064 027 381 76;
  • 34) 0.875 064 027 381 76 × 2 = 1 + 0.750 128 054 763 52;
  • 35) 0.750 128 054 763 52 × 2 = 1 + 0.500 256 109 527 04;
  • 36) 0.500 256 109 527 04 × 2 = 1 + 0.000 512 219 054 08;
  • 37) 0.000 512 219 054 08 × 2 = 0 + 0.001 024 438 108 16;
  • 38) 0.001 024 438 108 16 × 2 = 0 + 0.002 048 876 216 32;
  • 39) 0.002 048 876 216 32 × 2 = 0 + 0.004 097 752 432 64;
  • 40) 0.004 097 752 432 64 × 2 = 0 + 0.008 195 504 865 28;
  • 41) 0.008 195 504 865 28 × 2 = 0 + 0.016 391 009 730 56;
  • 42) 0.016 391 009 730 56 × 2 = 0 + 0.032 782 019 461 12;
  • 43) 0.032 782 019 461 12 × 2 = 0 + 0.065 564 038 922 24;
  • 44) 0.065 564 038 922 24 × 2 = 0 + 0.131 128 077 844 48;
  • 45) 0.131 128 077 844 48 × 2 = 0 + 0.262 256 155 688 96;
  • 46) 0.262 256 155 688 96 × 2 = 0 + 0.524 512 311 377 92;
  • 47) 0.524 512 311 377 92 × 2 = 1 + 0.049 024 622 755 84;
  • 48) 0.049 024 622 755 84 × 2 = 0 + 0.098 049 245 511 68;
  • 49) 0.098 049 245 511 68 × 2 = 0 + 0.196 098 491 023 36;
  • 50) 0.196 098 491 023 36 × 2 = 0 + 0.392 196 982 046 72;
  • 51) 0.392 196 982 046 72 × 2 = 0 + 0.784 393 964 093 44;
  • 52) 0.784 393 964 093 44 × 2 = 1 + 0.568 787 928 186 88;
  • 53) 0.568 787 928 186 88 × 2 = 1 + 0.137 575 856 373 76;

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:


0.337 166 009 107 03(10) =


0.0101 0110 0101 0000 1000 0010 1111 0110 0111 0000 0000 0010 0001 1(2)


5. Positive number before normalization:

32 913.337 166 009 107 03(10) =


1000 0000 1001 0001.0101 0110 0101 0000 1000 0010 1111 0110 0111 0000 0000 0010 0001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


32 913.337 166 009 107 03(10) =


1000 0000 1001 0001.0101 0110 0101 0000 1000 0010 1111 0110 0111 0000 0000 0010 0001 1(2) =


1000 0000 1001 0001.0101 0110 0101 0000 1000 0010 1111 0110 0111 0000 0000 0010 0001 1(2) × 20 =


1.0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110 0000 0000 0100 0011(2) × 215


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 15


Mantissa (not normalized):
1.0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110 0000 0000 0100 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


15 + 2(11-1) - 1 =


(15 + 1 023)(10) =


1 038(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 038 ÷ 2 = 519 + 0;
  • 519 ÷ 2 = 259 + 1;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1038(10) =


100 0000 1110(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 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110 0000 0000 0100 0011 =


0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
100 0000 1110


Mantissa (52 bits) =
0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110


The base ten decimal number 32 913.337 166 009 107 03 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1110 - 0000 0001 0010 0010 1010 1100 1010 0001 0000 0101 1110 1100 1110

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

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

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100