64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.805 245 165 974 627 154 089 760 4 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.805 245 165 974 627 154 089 760 4(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. 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: 0.805 245 165 974 627 154 089 760 4.

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) 0.805 245 165 974 627 154 089 760 4 × 2 = 1 + 0.610 490 331 949 254 308 179 520 8;
  • 2) 0.610 490 331 949 254 308 179 520 8 × 2 = 1 + 0.220 980 663 898 508 616 359 041 6;
  • 3) 0.220 980 663 898 508 616 359 041 6 × 2 = 0 + 0.441 961 327 797 017 232 718 083 2;
  • 4) 0.441 961 327 797 017 232 718 083 2 × 2 = 0 + 0.883 922 655 594 034 465 436 166 4;
  • 5) 0.883 922 655 594 034 465 436 166 4 × 2 = 1 + 0.767 845 311 188 068 930 872 332 8;
  • 6) 0.767 845 311 188 068 930 872 332 8 × 2 = 1 + 0.535 690 622 376 137 861 744 665 6;
  • 7) 0.535 690 622 376 137 861 744 665 6 × 2 = 1 + 0.071 381 244 752 275 723 489 331 2;
  • 8) 0.071 381 244 752 275 723 489 331 2 × 2 = 0 + 0.142 762 489 504 551 446 978 662 4;
  • 9) 0.142 762 489 504 551 446 978 662 4 × 2 = 0 + 0.285 524 979 009 102 893 957 324 8;
  • 10) 0.285 524 979 009 102 893 957 324 8 × 2 = 0 + 0.571 049 958 018 205 787 914 649 6;
  • 11) 0.571 049 958 018 205 787 914 649 6 × 2 = 1 + 0.142 099 916 036 411 575 829 299 2;
  • 12) 0.142 099 916 036 411 575 829 299 2 × 2 = 0 + 0.284 199 832 072 823 151 658 598 4;
  • 13) 0.284 199 832 072 823 151 658 598 4 × 2 = 0 + 0.568 399 664 145 646 303 317 196 8;
  • 14) 0.568 399 664 145 646 303 317 196 8 × 2 = 1 + 0.136 799 328 291 292 606 634 393 6;
  • 15) 0.136 799 328 291 292 606 634 393 6 × 2 = 0 + 0.273 598 656 582 585 213 268 787 2;
  • 16) 0.273 598 656 582 585 213 268 787 2 × 2 = 0 + 0.547 197 313 165 170 426 537 574 4;
  • 17) 0.547 197 313 165 170 426 537 574 4 × 2 = 1 + 0.094 394 626 330 340 853 075 148 8;
  • 18) 0.094 394 626 330 340 853 075 148 8 × 2 = 0 + 0.188 789 252 660 681 706 150 297 6;
  • 19) 0.188 789 252 660 681 706 150 297 6 × 2 = 0 + 0.377 578 505 321 363 412 300 595 2;
  • 20) 0.377 578 505 321 363 412 300 595 2 × 2 = 0 + 0.755 157 010 642 726 824 601 190 4;
  • 21) 0.755 157 010 642 726 824 601 190 4 × 2 = 1 + 0.510 314 021 285 453 649 202 380 8;
  • 22) 0.510 314 021 285 453 649 202 380 8 × 2 = 1 + 0.020 628 042 570 907 298 404 761 6;
  • 23) 0.020 628 042 570 907 298 404 761 6 × 2 = 0 + 0.041 256 085 141 814 596 809 523 2;
  • 24) 0.041 256 085 141 814 596 809 523 2 × 2 = 0 + 0.082 512 170 283 629 193 619 046 4;
  • 25) 0.082 512 170 283 629 193 619 046 4 × 2 = 0 + 0.165 024 340 567 258 387 238 092 8;
  • 26) 0.165 024 340 567 258 387 238 092 8 × 2 = 0 + 0.330 048 681 134 516 774 476 185 6;
  • 27) 0.330 048 681 134 516 774 476 185 6 × 2 = 0 + 0.660 097 362 269 033 548 952 371 2;
  • 28) 0.660 097 362 269 033 548 952 371 2 × 2 = 1 + 0.320 194 724 538 067 097 904 742 4;
  • 29) 0.320 194 724 538 067 097 904 742 4 × 2 = 0 + 0.640 389 449 076 134 195 809 484 8;
  • 30) 0.640 389 449 076 134 195 809 484 8 × 2 = 1 + 0.280 778 898 152 268 391 618 969 6;
  • 31) 0.280 778 898 152 268 391 618 969 6 × 2 = 0 + 0.561 557 796 304 536 783 237 939 2;
  • 32) 0.561 557 796 304 536 783 237 939 2 × 2 = 1 + 0.123 115 592 609 073 566 475 878 4;
  • 33) 0.123 115 592 609 073 566 475 878 4 × 2 = 0 + 0.246 231 185 218 147 132 951 756 8;
  • 34) 0.246 231 185 218 147 132 951 756 8 × 2 = 0 + 0.492 462 370 436 294 265 903 513 6;
  • 35) 0.492 462 370 436 294 265 903 513 6 × 2 = 0 + 0.984 924 740 872 588 531 807 027 2;
  • 36) 0.984 924 740 872 588 531 807 027 2 × 2 = 1 + 0.969 849 481 745 177 063 614 054 4;
  • 37) 0.969 849 481 745 177 063 614 054 4 × 2 = 1 + 0.939 698 963 490 354 127 228 108 8;
  • 38) 0.939 698 963 490 354 127 228 108 8 × 2 = 1 + 0.879 397 926 980 708 254 456 217 6;
  • 39) 0.879 397 926 980 708 254 456 217 6 × 2 = 1 + 0.758 795 853 961 416 508 912 435 2;
  • 40) 0.758 795 853 961 416 508 912 435 2 × 2 = 1 + 0.517 591 707 922 833 017 824 870 4;
  • 41) 0.517 591 707 922 833 017 824 870 4 × 2 = 1 + 0.035 183 415 845 666 035 649 740 8;
  • 42) 0.035 183 415 845 666 035 649 740 8 × 2 = 0 + 0.070 366 831 691 332 071 299 481 6;
  • 43) 0.070 366 831 691 332 071 299 481 6 × 2 = 0 + 0.140 733 663 382 664 142 598 963 2;
  • 44) 0.140 733 663 382 664 142 598 963 2 × 2 = 0 + 0.281 467 326 765 328 285 197 926 4;
  • 45) 0.281 467 326 765 328 285 197 926 4 × 2 = 0 + 0.562 934 653 530 656 570 395 852 8;
  • 46) 0.562 934 653 530 656 570 395 852 8 × 2 = 1 + 0.125 869 307 061 313 140 791 705 6;
  • 47) 0.125 869 307 061 313 140 791 705 6 × 2 = 0 + 0.251 738 614 122 626 281 583 411 2;
  • 48) 0.251 738 614 122 626 281 583 411 2 × 2 = 0 + 0.503 477 228 245 252 563 166 822 4;
  • 49) 0.503 477 228 245 252 563 166 822 4 × 2 = 1 + 0.006 954 456 490 505 126 333 644 8;
  • 50) 0.006 954 456 490 505 126 333 644 8 × 2 = 0 + 0.013 908 912 981 010 252 667 289 6;
  • 51) 0.013 908 912 981 010 252 667 289 6 × 2 = 0 + 0.027 817 825 962 020 505 334 579 2;
  • 52) 0.027 817 825 962 020 505 334 579 2 × 2 = 0 + 0.055 635 651 924 041 010 669 158 4;
  • 53) 0.055 635 651 924 041 010 669 158 4 × 2 = 0 + 0.111 271 303 848 082 021 338 316 8;

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...)


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:


0.805 245 165 974 627 154 089 760 4(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2)


5. Positive number before normalization:

0.805 245 165 974 627 154 089 760 4(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.805 245 165 974 627 154 089 760 4(10) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2) =

0.1100 1110 0010 0100 1000 1100 0001 0101 0001 1111 1000 0100 1000 0(2) × 20 =

1.1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000(2) × 2-1


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): -1

Mantissa (not normalized):
1.1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1022(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000 =

1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


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) =
011 1111 1110

Mantissa (52 bits) =
1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000


The base ten decimal number 0.805 245 165 974 627 154 089 760 4 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1110 - 1001 1100 0100 1001 0001 1000 0010 1010 0011 1111 0000 1001 0000

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

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» Number 0.805 245 165 974 627 154 089 760 5 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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