-2.211 829 052 383 358 300 119 548 661 699 653 279 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 653 279 76(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-2.211 829 052 383 358 300 119 548 661 699 653 279 76(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-2.211 829 052 383 358 300 119 548 661 699 653 279 76| = 2.211 829 052 383 358 300 119 548 661 699 653 279 76


2. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

2(10) =


10(2)


4. Convert to binary (base 2) the fractional part: 0.211 829 052 383 358 300 119 548 661 699 653 279 76.

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) 0.211 829 052 383 358 300 119 548 661 699 653 279 76 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 306 559 52;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 306 559 52 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 613 119 04;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 613 119 04 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 597 226 238 08;
  • 4) 0.694 632 419 066 866 400 956 389 293 597 226 238 08 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 194 452 476 16;
  • 5) 0.389 264 838 133 732 801 912 778 587 194 452 476 16 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 388 904 952 32;
  • 6) 0.778 529 676 267 465 603 825 557 174 388 904 952 32 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 777 809 904 64;
  • 7) 0.557 059 352 534 931 207 651 114 348 777 809 904 64 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 555 619 809 28;
  • 8) 0.114 118 705 069 862 415 302 228 697 555 619 809 28 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 111 239 618 56;
  • 9) 0.228 237 410 139 724 830 604 457 395 111 239 618 56 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 222 479 237 12;
  • 10) 0.456 474 820 279 449 661 208 914 790 222 479 237 12 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 444 958 474 24;
  • 11) 0.912 949 640 558 899 322 417 829 580 444 958 474 24 × 2 = 1 + 0.825 899 281 117 798 644 835 659 160 889 916 948 48;
  • 12) 0.825 899 281 117 798 644 835 659 160 889 916 948 48 × 2 = 1 + 0.651 798 562 235 597 289 671 318 321 779 833 896 96;
  • 13) 0.651 798 562 235 597 289 671 318 321 779 833 896 96 × 2 = 1 + 0.303 597 124 471 194 579 342 636 643 559 667 793 92;
  • 14) 0.303 597 124 471 194 579 342 636 643 559 667 793 92 × 2 = 0 + 0.607 194 248 942 389 158 685 273 287 119 335 587 84;
  • 15) 0.607 194 248 942 389 158 685 273 287 119 335 587 84 × 2 = 1 + 0.214 388 497 884 778 317 370 546 574 238 671 175 68;
  • 16) 0.214 388 497 884 778 317 370 546 574 238 671 175 68 × 2 = 0 + 0.428 776 995 769 556 634 741 093 148 477 342 351 36;
  • 17) 0.428 776 995 769 556 634 741 093 148 477 342 351 36 × 2 = 0 + 0.857 553 991 539 113 269 482 186 296 954 684 702 72;
  • 18) 0.857 553 991 539 113 269 482 186 296 954 684 702 72 × 2 = 1 + 0.715 107 983 078 226 538 964 372 593 909 369 405 44;
  • 19) 0.715 107 983 078 226 538 964 372 593 909 369 405 44 × 2 = 1 + 0.430 215 966 156 453 077 928 745 187 818 738 810 88;
  • 20) 0.430 215 966 156 453 077 928 745 187 818 738 810 88 × 2 = 0 + 0.860 431 932 312 906 155 857 490 375 637 477 621 76;
  • 21) 0.860 431 932 312 906 155 857 490 375 637 477 621 76 × 2 = 1 + 0.720 863 864 625 812 311 714 980 751 274 955 243 52;
  • 22) 0.720 863 864 625 812 311 714 980 751 274 955 243 52 × 2 = 1 + 0.441 727 729 251 624 623 429 961 502 549 910 487 04;
  • 23) 0.441 727 729 251 624 623 429 961 502 549 910 487 04 × 2 = 0 + 0.883 455 458 503 249 246 859 923 005 099 820 974 08;
  • 24) 0.883 455 458 503 249 246 859 923 005 099 820 974 08 × 2 = 1 + 0.766 910 917 006 498 493 719 846 010 199 641 948 16;
  • 25) 0.766 910 917 006 498 493 719 846 010 199 641 948 16 × 2 = 1 + 0.533 821 834 012 996 987 439 692 020 399 283 896 32;
  • 26) 0.533 821 834 012 996 987 439 692 020 399 283 896 32 × 2 = 1 + 0.067 643 668 025 993 974 879 384 040 798 567 792 64;
  • 27) 0.067 643 668 025 993 974 879 384 040 798 567 792 64 × 2 = 0 + 0.135 287 336 051 987 949 758 768 081 597 135 585 28;
  • 28) 0.135 287 336 051 987 949 758 768 081 597 135 585 28 × 2 = 0 + 0.270 574 672 103 975 899 517 536 163 194 271 170 56;
  • 29) 0.270 574 672 103 975 899 517 536 163 194 271 170 56 × 2 = 0 + 0.541 149 344 207 951 799 035 072 326 388 542 341 12;
  • 30) 0.541 149 344 207 951 799 035 072 326 388 542 341 12 × 2 = 1 + 0.082 298 688 415 903 598 070 144 652 777 084 682 24;
  • 31) 0.082 298 688 415 903 598 070 144 652 777 084 682 24 × 2 = 0 + 0.164 597 376 831 807 196 140 289 305 554 169 364 48;
  • 32) 0.164 597 376 831 807 196 140 289 305 554 169 364 48 × 2 = 0 + 0.329 194 753 663 614 392 280 578 611 108 338 728 96;
  • 33) 0.329 194 753 663 614 392 280 578 611 108 338 728 96 × 2 = 0 + 0.658 389 507 327 228 784 561 157 222 216 677 457 92;
  • 34) 0.658 389 507 327 228 784 561 157 222 216 677 457 92 × 2 = 1 + 0.316 779 014 654 457 569 122 314 444 433 354 915 84;
  • 35) 0.316 779 014 654 457 569 122 314 444 433 354 915 84 × 2 = 0 + 0.633 558 029 308 915 138 244 628 888 866 709 831 68;
  • 36) 0.633 558 029 308 915 138 244 628 888 866 709 831 68 × 2 = 1 + 0.267 116 058 617 830 276 489 257 777 733 419 663 36;
  • 37) 0.267 116 058 617 830 276 489 257 777 733 419 663 36 × 2 = 0 + 0.534 232 117 235 660 552 978 515 555 466 839 326 72;
  • 38) 0.534 232 117 235 660 552 978 515 555 466 839 326 72 × 2 = 1 + 0.068 464 234 471 321 105 957 031 110 933 678 653 44;
  • 39) 0.068 464 234 471 321 105 957 031 110 933 678 653 44 × 2 = 0 + 0.136 928 468 942 642 211 914 062 221 867 357 306 88;
  • 40) 0.136 928 468 942 642 211 914 062 221 867 357 306 88 × 2 = 0 + 0.273 856 937 885 284 423 828 124 443 734 714 613 76;
  • 41) 0.273 856 937 885 284 423 828 124 443 734 714 613 76 × 2 = 0 + 0.547 713 875 770 568 847 656 248 887 469 429 227 52;
  • 42) 0.547 713 875 770 568 847 656 248 887 469 429 227 52 × 2 = 1 + 0.095 427 751 541 137 695 312 497 774 938 858 455 04;
  • 43) 0.095 427 751 541 137 695 312 497 774 938 858 455 04 × 2 = 0 + 0.190 855 503 082 275 390 624 995 549 877 716 910 08;
  • 44) 0.190 855 503 082 275 390 624 995 549 877 716 910 08 × 2 = 0 + 0.381 711 006 164 550 781 249 991 099 755 433 820 16;
  • 45) 0.381 711 006 164 550 781 249 991 099 755 433 820 16 × 2 = 0 + 0.763 422 012 329 101 562 499 982 199 510 867 640 32;
  • 46) 0.763 422 012 329 101 562 499 982 199 510 867 640 32 × 2 = 1 + 0.526 844 024 658 203 124 999 964 399 021 735 280 64;
  • 47) 0.526 844 024 658 203 124 999 964 399 021 735 280 64 × 2 = 1 + 0.053 688 049 316 406 249 999 928 798 043 470 561 28;
  • 48) 0.053 688 049 316 406 249 999 928 798 043 470 561 28 × 2 = 0 + 0.107 376 098 632 812 499 999 857 596 086 941 122 56;
  • 49) 0.107 376 098 632 812 499 999 857 596 086 941 122 56 × 2 = 0 + 0.214 752 197 265 624 999 999 715 192 173 882 245 12;
  • 50) 0.214 752 197 265 624 999 999 715 192 173 882 245 12 × 2 = 0 + 0.429 504 394 531 249 999 999 430 384 347 764 490 24;
  • 51) 0.429 504 394 531 249 999 999 430 384 347 764 490 24 × 2 = 0 + 0.859 008 789 062 499 999 998 860 768 695 528 980 48;
  • 52) 0.859 008 789 062 499 999 998 860 768 695 528 980 48 × 2 = 1 + 0.718 017 578 124 999 999 997 721 537 391 057 960 96;
  • 53) 0.718 017 578 124 999 999 997 721 537 391 057 960 96 × 2 = 1 + 0.436 035 156 249 999 999 995 443 074 782 115 921 92;

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).


5. 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:


0.211 829 052 383 358 300 119 548 661 699 653 279 76(10) =


0.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 653 279 76(10) =


10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

7. Normalize the binary representation of the number.

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


2.211 829 052 383 358 300 119 548 661 699 653 279 76(10) =


10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) =


10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) × 20 =


1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11(2) × 21


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): 1


Mantissa (not normalized):
1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11


9. 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) =


1 024(10)


10. 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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 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;

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

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


Exponent (adjusted) =


1024(10) =


100 0000 0000(2)


12. 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. 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11 =


0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


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

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


Exponent (11 bits) =
100 0000 0000


Mantissa (52 bits) =
0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


Decimal number -2.211 829 052 383 358 300 119 548 661 699 653 279 76 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


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