0.000 000 021 979 552 668 138 406 902 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 902(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
0.000 000 021 979 552 668 138 406 902(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. 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: 0.000 000 021 979 552 668 138 406 902.

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) 0.000 000 021 979 552 668 138 406 902 × 2 = 0 + 0.000 000 043 959 105 336 276 813 804;
  • 2) 0.000 000 043 959 105 336 276 813 804 × 2 = 0 + 0.000 000 087 918 210 672 553 627 608;
  • 3) 0.000 000 087 918 210 672 553 627 608 × 2 = 0 + 0.000 000 175 836 421 345 107 255 216;
  • 4) 0.000 000 175 836 421 345 107 255 216 × 2 = 0 + 0.000 000 351 672 842 690 214 510 432;
  • 5) 0.000 000 351 672 842 690 214 510 432 × 2 = 0 + 0.000 000 703 345 685 380 429 020 864;
  • 6) 0.000 000 703 345 685 380 429 020 864 × 2 = 0 + 0.000 001 406 691 370 760 858 041 728;
  • 7) 0.000 001 406 691 370 760 858 041 728 × 2 = 0 + 0.000 002 813 382 741 521 716 083 456;
  • 8) 0.000 002 813 382 741 521 716 083 456 × 2 = 0 + 0.000 005 626 765 483 043 432 166 912;
  • 9) 0.000 005 626 765 483 043 432 166 912 × 2 = 0 + 0.000 011 253 530 966 086 864 333 824;
  • 10) 0.000 011 253 530 966 086 864 333 824 × 2 = 0 + 0.000 022 507 061 932 173 728 667 648;
  • 11) 0.000 022 507 061 932 173 728 667 648 × 2 = 0 + 0.000 045 014 123 864 347 457 335 296;
  • 12) 0.000 045 014 123 864 347 457 335 296 × 2 = 0 + 0.000 090 028 247 728 694 914 670 592;
  • 13) 0.000 090 028 247 728 694 914 670 592 × 2 = 0 + 0.000 180 056 495 457 389 829 341 184;
  • 14) 0.000 180 056 495 457 389 829 341 184 × 2 = 0 + 0.000 360 112 990 914 779 658 682 368;
  • 15) 0.000 360 112 990 914 779 658 682 368 × 2 = 0 + 0.000 720 225 981 829 559 317 364 736;
  • 16) 0.000 720 225 981 829 559 317 364 736 × 2 = 0 + 0.001 440 451 963 659 118 634 729 472;
  • 17) 0.001 440 451 963 659 118 634 729 472 × 2 = 0 + 0.002 880 903 927 318 237 269 458 944;
  • 18) 0.002 880 903 927 318 237 269 458 944 × 2 = 0 + 0.005 761 807 854 636 474 538 917 888;
  • 19) 0.005 761 807 854 636 474 538 917 888 × 2 = 0 + 0.011 523 615 709 272 949 077 835 776;
  • 20) 0.011 523 615 709 272 949 077 835 776 × 2 = 0 + 0.023 047 231 418 545 898 155 671 552;
  • 21) 0.023 047 231 418 545 898 155 671 552 × 2 = 0 + 0.046 094 462 837 091 796 311 343 104;
  • 22) 0.046 094 462 837 091 796 311 343 104 × 2 = 0 + 0.092 188 925 674 183 592 622 686 208;
  • 23) 0.092 188 925 674 183 592 622 686 208 × 2 = 0 + 0.184 377 851 348 367 185 245 372 416;
  • 24) 0.184 377 851 348 367 185 245 372 416 × 2 = 0 + 0.368 755 702 696 734 370 490 744 832;
  • 25) 0.368 755 702 696 734 370 490 744 832 × 2 = 0 + 0.737 511 405 393 468 740 981 489 664;
  • 26) 0.737 511 405 393 468 740 981 489 664 × 2 = 1 + 0.475 022 810 786 937 481 962 979 328;
  • 27) 0.475 022 810 786 937 481 962 979 328 × 2 = 0 + 0.950 045 621 573 874 963 925 958 656;
  • 28) 0.950 045 621 573 874 963 925 958 656 × 2 = 1 + 0.900 091 243 147 749 927 851 917 312;
  • 29) 0.900 091 243 147 749 927 851 917 312 × 2 = 1 + 0.800 182 486 295 499 855 703 834 624;
  • 30) 0.800 182 486 295 499 855 703 834 624 × 2 = 1 + 0.600 364 972 590 999 711 407 669 248;
  • 31) 0.600 364 972 590 999 711 407 669 248 × 2 = 1 + 0.200 729 945 181 999 422 815 338 496;
  • 32) 0.200 729 945 181 999 422 815 338 496 × 2 = 0 + 0.401 459 890 363 998 845 630 676 992;
  • 33) 0.401 459 890 363 998 845 630 676 992 × 2 = 0 + 0.802 919 780 727 997 691 261 353 984;
  • 34) 0.802 919 780 727 997 691 261 353 984 × 2 = 1 + 0.605 839 561 455 995 382 522 707 968;
  • 35) 0.605 839 561 455 995 382 522 707 968 × 2 = 1 + 0.211 679 122 911 990 765 045 415 936;
  • 36) 0.211 679 122 911 990 765 045 415 936 × 2 = 0 + 0.423 358 245 823 981 530 090 831 872;
  • 37) 0.423 358 245 823 981 530 090 831 872 × 2 = 0 + 0.846 716 491 647 963 060 181 663 744;
  • 38) 0.846 716 491 647 963 060 181 663 744 × 2 = 1 + 0.693 432 983 295 926 120 363 327 488;
  • 39) 0.693 432 983 295 926 120 363 327 488 × 2 = 1 + 0.386 865 966 591 852 240 726 654 976;
  • 40) 0.386 865 966 591 852 240 726 654 976 × 2 = 0 + 0.773 731 933 183 704 481 453 309 952;
  • 41) 0.773 731 933 183 704 481 453 309 952 × 2 = 1 + 0.547 463 866 367 408 962 906 619 904;
  • 42) 0.547 463 866 367 408 962 906 619 904 × 2 = 1 + 0.094 927 732 734 817 925 813 239 808;
  • 43) 0.094 927 732 734 817 925 813 239 808 × 2 = 0 + 0.189 855 465 469 635 851 626 479 616;
  • 44) 0.189 855 465 469 635 851 626 479 616 × 2 = 0 + 0.379 710 930 939 271 703 252 959 232;
  • 45) 0.379 710 930 939 271 703 252 959 232 × 2 = 0 + 0.759 421 861 878 543 406 505 918 464;
  • 46) 0.759 421 861 878 543 406 505 918 464 × 2 = 1 + 0.518 843 723 757 086 813 011 836 928;
  • 47) 0.518 843 723 757 086 813 011 836 928 × 2 = 1 + 0.037 687 447 514 173 626 023 673 856;
  • 48) 0.037 687 447 514 173 626 023 673 856 × 2 = 0 + 0.075 374 895 028 347 252 047 347 712;
  • 49) 0.075 374 895 028 347 252 047 347 712 × 2 = 0 + 0.150 749 790 056 694 504 094 695 424;
  • 50) 0.150 749 790 056 694 504 094 695 424 × 2 = 0 + 0.301 499 580 113 389 008 189 390 848;
  • 51) 0.301 499 580 113 389 008 189 390 848 × 2 = 0 + 0.602 999 160 226 778 016 378 781 696;
  • 52) 0.602 999 160 226 778 016 378 781 696 × 2 = 1 + 0.205 998 320 453 556 032 757 563 392;
  • 53) 0.205 998 320 453 556 032 757 563 392 × 2 = 0 + 0.411 996 640 907 112 065 515 126 784;
  • 54) 0.411 996 640 907 112 065 515 126 784 × 2 = 0 + 0.823 993 281 814 224 131 030 253 568;
  • 55) 0.823 993 281 814 224 131 030 253 568 × 2 = 1 + 0.647 986 563 628 448 262 060 507 136;
  • 56) 0.647 986 563 628 448 262 060 507 136 × 2 = 1 + 0.295 973 127 256 896 524 121 014 272;
  • 57) 0.295 973 127 256 896 524 121 014 272 × 2 = 0 + 0.591 946 254 513 793 048 242 028 544;
  • 58) 0.591 946 254 513 793 048 242 028 544 × 2 = 1 + 0.183 892 509 027 586 096 484 057 088;
  • 59) 0.183 892 509 027 586 096 484 057 088 × 2 = 0 + 0.367 785 018 055 172 192 968 114 176;
  • 60) 0.367 785 018 055 172 192 968 114 176 × 2 = 0 + 0.735 570 036 110 344 385 936 228 352;
  • 61) 0.735 570 036 110 344 385 936 228 352 × 2 = 1 + 0.471 140 072 220 688 771 872 456 704;
  • 62) 0.471 140 072 220 688 771 872 456 704 × 2 = 0 + 0.942 280 144 441 377 543 744 913 408;
  • 63) 0.942 280 144 441 377 543 744 913 408 × 2 = 1 + 0.884 560 288 882 755 087 489 826 816;
  • 64) 0.884 560 288 882 755 087 489 826 816 × 2 = 1 + 0.769 120 577 765 510 174 979 653 632;
  • 65) 0.769 120 577 765 510 174 979 653 632 × 2 = 1 + 0.538 241 155 531 020 349 959 307 264;
  • 66) 0.538 241 155 531 020 349 959 307 264 × 2 = 1 + 0.076 482 311 062 040 699 918 614 528;
  • 67) 0.076 482 311 062 040 699 918 614 528 × 2 = 0 + 0.152 964 622 124 081 399 837 229 056;
  • 68) 0.152 964 622 124 081 399 837 229 056 × 2 = 0 + 0.305 929 244 248 162 799 674 458 112;
  • 69) 0.305 929 244 248 162 799 674 458 112 × 2 = 0 + 0.611 858 488 496 325 599 348 916 224;
  • 70) 0.611 858 488 496 325 599 348 916 224 × 2 = 1 + 0.223 716 976 992 651 198 697 832 448;
  • 71) 0.223 716 976 992 651 198 697 832 448 × 2 = 0 + 0.447 433 953 985 302 397 395 664 896;
  • 72) 0.447 433 953 985 302 397 395 664 896 × 2 = 0 + 0.894 867 907 970 604 794 791 329 792;
  • 73) 0.894 867 907 970 604 794 791 329 792 × 2 = 1 + 0.789 735 815 941 209 589 582 659 584;
  • 74) 0.789 735 815 941 209 589 582 659 584 × 2 = 1 + 0.579 471 631 882 419 179 165 319 168;
  • 75) 0.579 471 631 882 419 179 165 319 168 × 2 = 1 + 0.158 943 263 764 838 358 330 638 336;
  • 76) 0.158 943 263 764 838 358 330 638 336 × 2 = 0 + 0.317 886 527 529 676 716 661 276 672;
  • 77) 0.317 886 527 529 676 716 661 276 672 × 2 = 0 + 0.635 773 055 059 353 433 322 553 344;
  • 78) 0.635 773 055 059 353 433 322 553 344 × 2 = 1 + 0.271 546 110 118 706 866 645 106 688;

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).


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:


0.000 000 021 979 552 668 138 406 902(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

5. Positive number before normalization:

0.000 000 021 979 552 668 138 406 902(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 979 552 668 138 406 902(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) × 20 =

1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001(2) × 2-26


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): -26

Mantissa (not normalized):
1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

997(10) =

011 1110 0101(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. 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001 =

0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


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 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


Decimal number 0.000 000 021 979 552 668 138 406 902 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001

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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