64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.569 394 317 378 345 826 851 915 141 920 6 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.569 394 317 378 345 826 851 915 141 920 6(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: 0.569 394 317 378 345 826 851 915 141 920 6.

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) 0.569 394 317 378 345 826 851 915 141 920 6 × 2 = 1 + 0.138 788 634 756 691 653 703 830 283 841 2;
  • 2) 0.138 788 634 756 691 653 703 830 283 841 2 × 2 = 0 + 0.277 577 269 513 383 307 407 660 567 682 4;
  • 3) 0.277 577 269 513 383 307 407 660 567 682 4 × 2 = 0 + 0.555 154 539 026 766 614 815 321 135 364 8;
  • 4) 0.555 154 539 026 766 614 815 321 135 364 8 × 2 = 1 + 0.110 309 078 053 533 229 630 642 270 729 6;
  • 5) 0.110 309 078 053 533 229 630 642 270 729 6 × 2 = 0 + 0.220 618 156 107 066 459 261 284 541 459 2;
  • 6) 0.220 618 156 107 066 459 261 284 541 459 2 × 2 = 0 + 0.441 236 312 214 132 918 522 569 082 918 4;
  • 7) 0.441 236 312 214 132 918 522 569 082 918 4 × 2 = 0 + 0.882 472 624 428 265 837 045 138 165 836 8;
  • 8) 0.882 472 624 428 265 837 045 138 165 836 8 × 2 = 1 + 0.764 945 248 856 531 674 090 276 331 673 6;
  • 9) 0.764 945 248 856 531 674 090 276 331 673 6 × 2 = 1 + 0.529 890 497 713 063 348 180 552 663 347 2;
  • 10) 0.529 890 497 713 063 348 180 552 663 347 2 × 2 = 1 + 0.059 780 995 426 126 696 361 105 326 694 4;
  • 11) 0.059 780 995 426 126 696 361 105 326 694 4 × 2 = 0 + 0.119 561 990 852 253 392 722 210 653 388 8;
  • 12) 0.119 561 990 852 253 392 722 210 653 388 8 × 2 = 0 + 0.239 123 981 704 506 785 444 421 306 777 6;
  • 13) 0.239 123 981 704 506 785 444 421 306 777 6 × 2 = 0 + 0.478 247 963 409 013 570 888 842 613 555 2;
  • 14) 0.478 247 963 409 013 570 888 842 613 555 2 × 2 = 0 + 0.956 495 926 818 027 141 777 685 227 110 4;
  • 15) 0.956 495 926 818 027 141 777 685 227 110 4 × 2 = 1 + 0.912 991 853 636 054 283 555 370 454 220 8;
  • 16) 0.912 991 853 636 054 283 555 370 454 220 8 × 2 = 1 + 0.825 983 707 272 108 567 110 740 908 441 6;
  • 17) 0.825 983 707 272 108 567 110 740 908 441 6 × 2 = 1 + 0.651 967 414 544 217 134 221 481 816 883 2;
  • 18) 0.651 967 414 544 217 134 221 481 816 883 2 × 2 = 1 + 0.303 934 829 088 434 268 442 963 633 766 4;
  • 19) 0.303 934 829 088 434 268 442 963 633 766 4 × 2 = 0 + 0.607 869 658 176 868 536 885 927 267 532 8;
  • 20) 0.607 869 658 176 868 536 885 927 267 532 8 × 2 = 1 + 0.215 739 316 353 737 073 771 854 535 065 6;
  • 21) 0.215 739 316 353 737 073 771 854 535 065 6 × 2 = 0 + 0.431 478 632 707 474 147 543 709 070 131 2;
  • 22) 0.431 478 632 707 474 147 543 709 070 131 2 × 2 = 0 + 0.862 957 265 414 948 295 087 418 140 262 4;
  • 23) 0.862 957 265 414 948 295 087 418 140 262 4 × 2 = 1 + 0.725 914 530 829 896 590 174 836 280 524 8;
  • 24) 0.725 914 530 829 896 590 174 836 280 524 8 × 2 = 1 + 0.451 829 061 659 793 180 349 672 561 049 6;
  • 25) 0.451 829 061 659 793 180 349 672 561 049 6 × 2 = 0 + 0.903 658 123 319 586 360 699 345 122 099 2;
  • 26) 0.903 658 123 319 586 360 699 345 122 099 2 × 2 = 1 + 0.807 316 246 639 172 721 398 690 244 198 4;
  • 27) 0.807 316 246 639 172 721 398 690 244 198 4 × 2 = 1 + 0.614 632 493 278 345 442 797 380 488 396 8;
  • 28) 0.614 632 493 278 345 442 797 380 488 396 8 × 2 = 1 + 0.229 264 986 556 690 885 594 760 976 793 6;
  • 29) 0.229 264 986 556 690 885 594 760 976 793 6 × 2 = 0 + 0.458 529 973 113 381 771 189 521 953 587 2;
  • 30) 0.458 529 973 113 381 771 189 521 953 587 2 × 2 = 0 + 0.917 059 946 226 763 542 379 043 907 174 4;
  • 31) 0.917 059 946 226 763 542 379 043 907 174 4 × 2 = 1 + 0.834 119 892 453 527 084 758 087 814 348 8;
  • 32) 0.834 119 892 453 527 084 758 087 814 348 8 × 2 = 1 + 0.668 239 784 907 054 169 516 175 628 697 6;
  • 33) 0.668 239 784 907 054 169 516 175 628 697 6 × 2 = 1 + 0.336 479 569 814 108 339 032 351 257 395 2;
  • 34) 0.336 479 569 814 108 339 032 351 257 395 2 × 2 = 0 + 0.672 959 139 628 216 678 064 702 514 790 4;
  • 35) 0.672 959 139 628 216 678 064 702 514 790 4 × 2 = 1 + 0.345 918 279 256 433 356 129 405 029 580 8;
  • 36) 0.345 918 279 256 433 356 129 405 029 580 8 × 2 = 0 + 0.691 836 558 512 866 712 258 810 059 161 6;
  • 37) 0.691 836 558 512 866 712 258 810 059 161 6 × 2 = 1 + 0.383 673 117 025 733 424 517 620 118 323 2;
  • 38) 0.383 673 117 025 733 424 517 620 118 323 2 × 2 = 0 + 0.767 346 234 051 466 849 035 240 236 646 4;
  • 39) 0.767 346 234 051 466 849 035 240 236 646 4 × 2 = 1 + 0.534 692 468 102 933 698 070 480 473 292 8;
  • 40) 0.534 692 468 102 933 698 070 480 473 292 8 × 2 = 1 + 0.069 384 936 205 867 396 140 960 946 585 6;
  • 41) 0.069 384 936 205 867 396 140 960 946 585 6 × 2 = 0 + 0.138 769 872 411 734 792 281 921 893 171 2;
  • 42) 0.138 769 872 411 734 792 281 921 893 171 2 × 2 = 0 + 0.277 539 744 823 469 584 563 843 786 342 4;
  • 43) 0.277 539 744 823 469 584 563 843 786 342 4 × 2 = 0 + 0.555 079 489 646 939 169 127 687 572 684 8;
  • 44) 0.555 079 489 646 939 169 127 687 572 684 8 × 2 = 1 + 0.110 158 979 293 878 338 255 375 145 369 6;
  • 45) 0.110 158 979 293 878 338 255 375 145 369 6 × 2 = 0 + 0.220 317 958 587 756 676 510 750 290 739 2;
  • 46) 0.220 317 958 587 756 676 510 750 290 739 2 × 2 = 0 + 0.440 635 917 175 513 353 021 500 581 478 4;
  • 47) 0.440 635 917 175 513 353 021 500 581 478 4 × 2 = 0 + 0.881 271 834 351 026 706 043 001 162 956 8;
  • 48) 0.881 271 834 351 026 706 043 001 162 956 8 × 2 = 1 + 0.762 543 668 702 053 412 086 002 325 913 6;
  • 49) 0.762 543 668 702 053 412 086 002 325 913 6 × 2 = 1 + 0.525 087 337 404 106 824 172 004 651 827 2;
  • 50) 0.525 087 337 404 106 824 172 004 651 827 2 × 2 = 1 + 0.050 174 674 808 213 648 344 009 303 654 4;
  • 51) 0.050 174 674 808 213 648 344 009 303 654 4 × 2 = 0 + 0.100 349 349 616 427 296 688 018 607 308 8;
  • 52) 0.100 349 349 616 427 296 688 018 607 308 8 × 2 = 0 + 0.200 698 699 232 854 593 376 037 214 617 6;
  • 53) 0.200 698 699 232 854 593 376 037 214 617 6 × 2 = 0 + 0.401 397 398 465 709 186 752 074 429 235 2;

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:


0.569 394 317 378 345 826 851 915 141 920 6(10) =


0.1001 0001 1100 0011 1101 0011 0111 0011 1010 1011 0001 0001 1100 0(2)


5. Positive number before normalization:

0.569 394 317 378 345 826 851 915 141 920 6(10) =


0.1001 0001 1100 0011 1101 0011 0111 0011 1010 1011 0001 0001 1100 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:


0.569 394 317 378 345 826 851 915 141 920 6(10) =


0.1001 0001 1100 0011 1101 0011 0111 0011 1010 1011 0001 0001 1100 0(2) =


0.1001 0001 1100 0011 1101 0011 0111 0011 1010 1011 0001 0001 1100 0(2) × 20 =


1.0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 1000(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):
1.0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 1000


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. 0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 1000 =


0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 1000


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) =
0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 1000


The base ten decimal number 0.569 394 317 378 345 826 851 915 141 920 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1110 - 0010 0011 1000 0111 1010 0110 1110 0111 0101 0110 0010 0011 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