-2.211 829 052 383 358 300 119 548 661 699 713 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 713(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
-2.211 829 052 383 358 300 119 548 661 699 713(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:

|-2.211 829 052 383 358 300 119 548 661 699 713| = 2.211 829 052 383 358 300 119 548 661 699 713


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: 0.211 829 052 383 358 300 119 548 661 699 713.

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) 0.211 829 052 383 358 300 119 548 661 699 713 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 426;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 426 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 852;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 852 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 597 704;
  • 4) 0.694 632 419 066 866 400 956 389 293 597 704 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 195 408;
  • 5) 0.389 264 838 133 732 801 912 778 587 195 408 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 390 816;
  • 6) 0.778 529 676 267 465 603 825 557 174 390 816 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 781 632;
  • 7) 0.557 059 352 534 931 207 651 114 348 781 632 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 563 264;
  • 8) 0.114 118 705 069 862 415 302 228 697 563 264 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 126 528;
  • 9) 0.228 237 410 139 724 830 604 457 395 126 528 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 253 056;
  • 10) 0.456 474 820 279 449 661 208 914 790 253 056 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 506 112;
  • 11) 0.912 949 640 558 899 322 417 829 580 506 112 × 2 = 1 + 0.825 899 281 117 798 644 835 659 161 012 224;
  • 12) 0.825 899 281 117 798 644 835 659 161 012 224 × 2 = 1 + 0.651 798 562 235 597 289 671 318 322 024 448;
  • 13) 0.651 798 562 235 597 289 671 318 322 024 448 × 2 = 1 + 0.303 597 124 471 194 579 342 636 644 048 896;
  • 14) 0.303 597 124 471 194 579 342 636 644 048 896 × 2 = 0 + 0.607 194 248 942 389 158 685 273 288 097 792;
  • 15) 0.607 194 248 942 389 158 685 273 288 097 792 × 2 = 1 + 0.214 388 497 884 778 317 370 546 576 195 584;
  • 16) 0.214 388 497 884 778 317 370 546 576 195 584 × 2 = 0 + 0.428 776 995 769 556 634 741 093 152 391 168;
  • 17) 0.428 776 995 769 556 634 741 093 152 391 168 × 2 = 0 + 0.857 553 991 539 113 269 482 186 304 782 336;
  • 18) 0.857 553 991 539 113 269 482 186 304 782 336 × 2 = 1 + 0.715 107 983 078 226 538 964 372 609 564 672;
  • 19) 0.715 107 983 078 226 538 964 372 609 564 672 × 2 = 1 + 0.430 215 966 156 453 077 928 745 219 129 344;
  • 20) 0.430 215 966 156 453 077 928 745 219 129 344 × 2 = 0 + 0.860 431 932 312 906 155 857 490 438 258 688;
  • 21) 0.860 431 932 312 906 155 857 490 438 258 688 × 2 = 1 + 0.720 863 864 625 812 311 714 980 876 517 376;
  • 22) 0.720 863 864 625 812 311 714 980 876 517 376 × 2 = 1 + 0.441 727 729 251 624 623 429 961 753 034 752;
  • 23) 0.441 727 729 251 624 623 429 961 753 034 752 × 2 = 0 + 0.883 455 458 503 249 246 859 923 506 069 504;
  • 24) 0.883 455 458 503 249 246 859 923 506 069 504 × 2 = 1 + 0.766 910 917 006 498 493 719 847 012 139 008;
  • 25) 0.766 910 917 006 498 493 719 847 012 139 008 × 2 = 1 + 0.533 821 834 012 996 987 439 694 024 278 016;
  • 26) 0.533 821 834 012 996 987 439 694 024 278 016 × 2 = 1 + 0.067 643 668 025 993 974 879 388 048 556 032;
  • 27) 0.067 643 668 025 993 974 879 388 048 556 032 × 2 = 0 + 0.135 287 336 051 987 949 758 776 097 112 064;
  • 28) 0.135 287 336 051 987 949 758 776 097 112 064 × 2 = 0 + 0.270 574 672 103 975 899 517 552 194 224 128;
  • 29) 0.270 574 672 103 975 899 517 552 194 224 128 × 2 = 0 + 0.541 149 344 207 951 799 035 104 388 448 256;
  • 30) 0.541 149 344 207 951 799 035 104 388 448 256 × 2 = 1 + 0.082 298 688 415 903 598 070 208 776 896 512;
  • 31) 0.082 298 688 415 903 598 070 208 776 896 512 × 2 = 0 + 0.164 597 376 831 807 196 140 417 553 793 024;
  • 32) 0.164 597 376 831 807 196 140 417 553 793 024 × 2 = 0 + 0.329 194 753 663 614 392 280 835 107 586 048;
  • 33) 0.329 194 753 663 614 392 280 835 107 586 048 × 2 = 0 + 0.658 389 507 327 228 784 561 670 215 172 096;
  • 34) 0.658 389 507 327 228 784 561 670 215 172 096 × 2 = 1 + 0.316 779 014 654 457 569 123 340 430 344 192;
  • 35) 0.316 779 014 654 457 569 123 340 430 344 192 × 2 = 0 + 0.633 558 029 308 915 138 246 680 860 688 384;
  • 36) 0.633 558 029 308 915 138 246 680 860 688 384 × 2 = 1 + 0.267 116 058 617 830 276 493 361 721 376 768;
  • 37) 0.267 116 058 617 830 276 493 361 721 376 768 × 2 = 0 + 0.534 232 117 235 660 552 986 723 442 753 536;
  • 38) 0.534 232 117 235 660 552 986 723 442 753 536 × 2 = 1 + 0.068 464 234 471 321 105 973 446 885 507 072;
  • 39) 0.068 464 234 471 321 105 973 446 885 507 072 × 2 = 0 + 0.136 928 468 942 642 211 946 893 771 014 144;
  • 40) 0.136 928 468 942 642 211 946 893 771 014 144 × 2 = 0 + 0.273 856 937 885 284 423 893 787 542 028 288;
  • 41) 0.273 856 937 885 284 423 893 787 542 028 288 × 2 = 0 + 0.547 713 875 770 568 847 787 575 084 056 576;
  • 42) 0.547 713 875 770 568 847 787 575 084 056 576 × 2 = 1 + 0.095 427 751 541 137 695 575 150 168 113 152;
  • 43) 0.095 427 751 541 137 695 575 150 168 113 152 × 2 = 0 + 0.190 855 503 082 275 391 150 300 336 226 304;
  • 44) 0.190 855 503 082 275 391 150 300 336 226 304 × 2 = 0 + 0.381 711 006 164 550 782 300 600 672 452 608;
  • 45) 0.381 711 006 164 550 782 300 600 672 452 608 × 2 = 0 + 0.763 422 012 329 101 564 601 201 344 905 216;
  • 46) 0.763 422 012 329 101 564 601 201 344 905 216 × 2 = 1 + 0.526 844 024 658 203 129 202 402 689 810 432;
  • 47) 0.526 844 024 658 203 129 202 402 689 810 432 × 2 = 1 + 0.053 688 049 316 406 258 404 805 379 620 864;
  • 48) 0.053 688 049 316 406 258 404 805 379 620 864 × 2 = 0 + 0.107 376 098 632 812 516 809 610 759 241 728;
  • 49) 0.107 376 098 632 812 516 809 610 759 241 728 × 2 = 0 + 0.214 752 197 265 625 033 619 221 518 483 456;
  • 50) 0.214 752 197 265 625 033 619 221 518 483 456 × 2 = 0 + 0.429 504 394 531 250 067 238 443 036 966 912;
  • 51) 0.429 504 394 531 250 067 238 443 036 966 912 × 2 = 0 + 0.859 008 789 062 500 134 476 886 073 933 824;
  • 52) 0.859 008 789 062 500 134 476 886 073 933 824 × 2 = 1 + 0.718 017 578 125 000 268 953 772 147 867 648;
  • 53) 0.718 017 578 125 000 268 953 772 147 867 648 × 2 = 1 + 0.436 035 156 250 000 537 907 544 295 735 296;

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:


0.211 829 052 383 358 300 119 548 661 699 713(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 713(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:


2.211 829 052 383 358 300 119 548 661 699 713(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 -2.211 829 052 383 358 300 119 548 661 699 713 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

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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