64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 6 670 903 752 021 072 936 893 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 6 670 903 752 021 072 936 893(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;
  • 6 670 903 752 021 072 936 893 ÷ 2 = 3 335 451 876 010 536 468 446 + 1;
  • 3 335 451 876 010 536 468 446 ÷ 2 = 1 667 725 938 005 268 234 223 + 0;
  • 1 667 725 938 005 268 234 223 ÷ 2 = 833 862 969 002 634 117 111 + 1;
  • 833 862 969 002 634 117 111 ÷ 2 = 416 931 484 501 317 058 555 + 1;
  • 416 931 484 501 317 058 555 ÷ 2 = 208 465 742 250 658 529 277 + 1;
  • 208 465 742 250 658 529 277 ÷ 2 = 104 232 871 125 329 264 638 + 1;
  • 104 232 871 125 329 264 638 ÷ 2 = 52 116 435 562 664 632 319 + 0;
  • 52 116 435 562 664 632 319 ÷ 2 = 26 058 217 781 332 316 159 + 1;
  • 26 058 217 781 332 316 159 ÷ 2 = 13 029 108 890 666 158 079 + 1;
  • 13 029 108 890 666 158 079 ÷ 2 = 6 514 554 445 333 079 039 + 1;
  • 6 514 554 445 333 079 039 ÷ 2 = 3 257 277 222 666 539 519 + 1;
  • 3 257 277 222 666 539 519 ÷ 2 = 1 628 638 611 333 269 759 + 1;
  • 1 628 638 611 333 269 759 ÷ 2 = 814 319 305 666 634 879 + 1;
  • 814 319 305 666 634 879 ÷ 2 = 407 159 652 833 317 439 + 1;
  • 407 159 652 833 317 439 ÷ 2 = 203 579 826 416 658 719 + 1;
  • 203 579 826 416 658 719 ÷ 2 = 101 789 913 208 329 359 + 1;
  • 101 789 913 208 329 359 ÷ 2 = 50 894 956 604 164 679 + 1;
  • 50 894 956 604 164 679 ÷ 2 = 25 447 478 302 082 339 + 1;
  • 25 447 478 302 082 339 ÷ 2 = 12 723 739 151 041 169 + 1;
  • 12 723 739 151 041 169 ÷ 2 = 6 361 869 575 520 584 + 1;
  • 6 361 869 575 520 584 ÷ 2 = 3 180 934 787 760 292 + 0;
  • 3 180 934 787 760 292 ÷ 2 = 1 590 467 393 880 146 + 0;
  • 1 590 467 393 880 146 ÷ 2 = 795 233 696 940 073 + 0;
  • 795 233 696 940 073 ÷ 2 = 397 616 848 470 036 + 1;
  • 397 616 848 470 036 ÷ 2 = 198 808 424 235 018 + 0;
  • 198 808 424 235 018 ÷ 2 = 99 404 212 117 509 + 0;
  • 99 404 212 117 509 ÷ 2 = 49 702 106 058 754 + 1;
  • 49 702 106 058 754 ÷ 2 = 24 851 053 029 377 + 0;
  • 24 851 053 029 377 ÷ 2 = 12 425 526 514 688 + 1;
  • 12 425 526 514 688 ÷ 2 = 6 212 763 257 344 + 0;
  • 6 212 763 257 344 ÷ 2 = 3 106 381 628 672 + 0;
  • 3 106 381 628 672 ÷ 2 = 1 553 190 814 336 + 0;
  • 1 553 190 814 336 ÷ 2 = 776 595 407 168 + 0;
  • 776 595 407 168 ÷ 2 = 388 297 703 584 + 0;
  • 388 297 703 584 ÷ 2 = 194 148 851 792 + 0;
  • 194 148 851 792 ÷ 2 = 97 074 425 896 + 0;
  • 97 074 425 896 ÷ 2 = 48 537 212 948 + 0;
  • 48 537 212 948 ÷ 2 = 24 268 606 474 + 0;
  • 24 268 606 474 ÷ 2 = 12 134 303 237 + 0;
  • 12 134 303 237 ÷ 2 = 6 067 151 618 + 1;
  • 6 067 151 618 ÷ 2 = 3 033 575 809 + 0;
  • 3 033 575 809 ÷ 2 = 1 516 787 904 + 1;
  • 1 516 787 904 ÷ 2 = 758 393 952 + 0;
  • 758 393 952 ÷ 2 = 379 196 976 + 0;
  • 379 196 976 ÷ 2 = 189 598 488 + 0;
  • 189 598 488 ÷ 2 = 94 799 244 + 0;
  • 94 799 244 ÷ 2 = 47 399 622 + 0;
  • 47 399 622 ÷ 2 = 23 699 811 + 0;
  • 23 699 811 ÷ 2 = 11 849 905 + 1;
  • 11 849 905 ÷ 2 = 5 924 952 + 1;
  • 5 924 952 ÷ 2 = 2 962 476 + 0;
  • 2 962 476 ÷ 2 = 1 481 238 + 0;
  • 1 481 238 ÷ 2 = 740 619 + 0;
  • 740 619 ÷ 2 = 370 309 + 1;
  • 370 309 ÷ 2 = 185 154 + 1;
  • 185 154 ÷ 2 = 92 577 + 0;
  • 92 577 ÷ 2 = 46 288 + 1;
  • 46 288 ÷ 2 = 23 144 + 0;
  • 23 144 ÷ 2 = 11 572 + 0;
  • 11 572 ÷ 2 = 5 786 + 0;
  • 5 786 ÷ 2 = 2 893 + 0;
  • 2 893 ÷ 2 = 1 446 + 1;
  • 1 446 ÷ 2 = 723 + 0;
  • 723 ÷ 2 = 361 + 1;
  • 361 ÷ 2 = 180 + 1;
  • 180 ÷ 2 = 90 + 0;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 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.


6 670 903 752 021 072 936 893(10) =


1 0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 72 positions to the left, so that only one non zero digit remains to the left of it:


6 670 903 752 021 072 936 893(10) =


1 0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101(2) =


1 0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101(2) × 20 =


1.0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101(2) × 272


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): 72


Mantissa (not normalized):
1.0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


72 + 2(11-1) - 1 =


(72 + 1 023)(10) =


1 095(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;
  • 1 095 ÷ 2 = 547 + 1;
  • 547 ÷ 2 = 273 + 1;
  • 273 ÷ 2 = 136 + 1;
  • 136 ÷ 2 = 68 + 0;
  • 68 ÷ 2 = 34 + 0;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 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) =


1095(10) =


100 0100 0111(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. 0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000 1111 1111 1111 1011 1101 =


0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000


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 0100 0111


Mantissa (52 bits) =
0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000


The base ten decimal number 6 670 903 752 021 072 936 893 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0100 0111 - 0110 1001 1010 0001 0110 0011 0000 0010 1000 0000 0001 0100 1000

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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