64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 11 665 660 768 132 694 792 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 11 665 660 768 132 694 792(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 11 665 660 768 132 694 792 ÷ 2 = 5 832 830 384 066 347 396 + 0;
  • 5 832 830 384 066 347 396 ÷ 2 = 2 916 415 192 033 173 698 + 0;
  • 2 916 415 192 033 173 698 ÷ 2 = 1 458 207 596 016 586 849 + 0;
  • 1 458 207 596 016 586 849 ÷ 2 = 729 103 798 008 293 424 + 1;
  • 729 103 798 008 293 424 ÷ 2 = 364 551 899 004 146 712 + 0;
  • 364 551 899 004 146 712 ÷ 2 = 182 275 949 502 073 356 + 0;
  • 182 275 949 502 073 356 ÷ 2 = 91 137 974 751 036 678 + 0;
  • 91 137 974 751 036 678 ÷ 2 = 45 568 987 375 518 339 + 0;
  • 45 568 987 375 518 339 ÷ 2 = 22 784 493 687 759 169 + 1;
  • 22 784 493 687 759 169 ÷ 2 = 11 392 246 843 879 584 + 1;
  • 11 392 246 843 879 584 ÷ 2 = 5 696 123 421 939 792 + 0;
  • 5 696 123 421 939 792 ÷ 2 = 2 848 061 710 969 896 + 0;
  • 2 848 061 710 969 896 ÷ 2 = 1 424 030 855 484 948 + 0;
  • 1 424 030 855 484 948 ÷ 2 = 712 015 427 742 474 + 0;
  • 712 015 427 742 474 ÷ 2 = 356 007 713 871 237 + 0;
  • 356 007 713 871 237 ÷ 2 = 178 003 856 935 618 + 1;
  • 178 003 856 935 618 ÷ 2 = 89 001 928 467 809 + 0;
  • 89 001 928 467 809 ÷ 2 = 44 500 964 233 904 + 1;
  • 44 500 964 233 904 ÷ 2 = 22 250 482 116 952 + 0;
  • 22 250 482 116 952 ÷ 2 = 11 125 241 058 476 + 0;
  • 11 125 241 058 476 ÷ 2 = 5 562 620 529 238 + 0;
  • 5 562 620 529 238 ÷ 2 = 2 781 310 264 619 + 0;
  • 2 781 310 264 619 ÷ 2 = 1 390 655 132 309 + 1;
  • 1 390 655 132 309 ÷ 2 = 695 327 566 154 + 1;
  • 695 327 566 154 ÷ 2 = 347 663 783 077 + 0;
  • 347 663 783 077 ÷ 2 = 173 831 891 538 + 1;
  • 173 831 891 538 ÷ 2 = 86 915 945 769 + 0;
  • 86 915 945 769 ÷ 2 = 43 457 972 884 + 1;
  • 43 457 972 884 ÷ 2 = 21 728 986 442 + 0;
  • 21 728 986 442 ÷ 2 = 10 864 493 221 + 0;
  • 10 864 493 221 ÷ 2 = 5 432 246 610 + 1;
  • 5 432 246 610 ÷ 2 = 2 716 123 305 + 0;
  • 2 716 123 305 ÷ 2 = 1 358 061 652 + 1;
  • 1 358 061 652 ÷ 2 = 679 030 826 + 0;
  • 679 030 826 ÷ 2 = 339 515 413 + 0;
  • 339 515 413 ÷ 2 = 169 757 706 + 1;
  • 169 757 706 ÷ 2 = 84 878 853 + 0;
  • 84 878 853 ÷ 2 = 42 439 426 + 1;
  • 42 439 426 ÷ 2 = 21 219 713 + 0;
  • 21 219 713 ÷ 2 = 10 609 856 + 1;
  • 10 609 856 ÷ 2 = 5 304 928 + 0;
  • 5 304 928 ÷ 2 = 2 652 464 + 0;
  • 2 652 464 ÷ 2 = 1 326 232 + 0;
  • 1 326 232 ÷ 2 = 663 116 + 0;
  • 663 116 ÷ 2 = 331 558 + 0;
  • 331 558 ÷ 2 = 165 779 + 0;
  • 165 779 ÷ 2 = 82 889 + 1;
  • 82 889 ÷ 2 = 41 444 + 1;
  • 41 444 ÷ 2 = 20 722 + 0;
  • 20 722 ÷ 2 = 10 361 + 0;
  • 10 361 ÷ 2 = 5 180 + 1;
  • 5 180 ÷ 2 = 2 590 + 0;
  • 2 590 ÷ 2 = 1 295 + 0;
  • 1 295 ÷ 2 = 647 + 1;
  • 647 ÷ 2 = 323 + 1;
  • 323 ÷ 2 = 161 + 1;
  • 161 ÷ 2 = 80 + 1;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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


11 665 660 768 132 694 792(10) =


1010 0001 1110 0100 1100 0000 1010 1001 0100 1010 1100 0010 1000 0011 0000 1000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the left, so that only one non zero digit remains to the left of it:


11 665 660 768 132 694 792(10) =


1010 0001 1110 0100 1100 0000 1010 1001 0100 1010 1100 0010 1000 0011 0000 1000(2) =


1010 0001 1110 0100 1100 0000 1010 1001 0100 1010 1100 0010 1000 0011 0000 1000(2) × 20 =


1.0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000 0110 0001 000(2) × 263


4. 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): 63


Mantissa (not normalized):
1.0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000 0110 0001 000


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


63 + 2(11-1) - 1 =


(63 + 1 023)(10) =


1 086(10)


6. 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;
  • 1 086 ÷ 2 = 543 + 0;
  • 543 ÷ 2 = 271 + 1;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1086(10) =


100 0011 1110(2)


8. 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. 0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000 011 0000 1000 =


0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000


9. 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 0011 1110


Mantissa (52 bits) =
0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000


The base ten decimal number 11 665 660 768 132 694 792 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0011 1110 - 0100 0011 1100 1001 1000 0001 0101 0010 1001 0101 1000 0101 0000

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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