0.000 000 021 979 552 668 138 406 878 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 878(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
0.000 000 021 979 552 668 138 406 878(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: 0.000 000 021 979 552 668 138 406 878.

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) 0.000 000 021 979 552 668 138 406 878 × 2 = 0 + 0.000 000 043 959 105 336 276 813 756;
  • 2) 0.000 000 043 959 105 336 276 813 756 × 2 = 0 + 0.000 000 087 918 210 672 553 627 512;
  • 3) 0.000 000 087 918 210 672 553 627 512 × 2 = 0 + 0.000 000 175 836 421 345 107 255 024;
  • 4) 0.000 000 175 836 421 345 107 255 024 × 2 = 0 + 0.000 000 351 672 842 690 214 510 048;
  • 5) 0.000 000 351 672 842 690 214 510 048 × 2 = 0 + 0.000 000 703 345 685 380 429 020 096;
  • 6) 0.000 000 703 345 685 380 429 020 096 × 2 = 0 + 0.000 001 406 691 370 760 858 040 192;
  • 7) 0.000 001 406 691 370 760 858 040 192 × 2 = 0 + 0.000 002 813 382 741 521 716 080 384;
  • 8) 0.000 002 813 382 741 521 716 080 384 × 2 = 0 + 0.000 005 626 765 483 043 432 160 768;
  • 9) 0.000 005 626 765 483 043 432 160 768 × 2 = 0 + 0.000 011 253 530 966 086 864 321 536;
  • 10) 0.000 011 253 530 966 086 864 321 536 × 2 = 0 + 0.000 022 507 061 932 173 728 643 072;
  • 11) 0.000 022 507 061 932 173 728 643 072 × 2 = 0 + 0.000 045 014 123 864 347 457 286 144;
  • 12) 0.000 045 014 123 864 347 457 286 144 × 2 = 0 + 0.000 090 028 247 728 694 914 572 288;
  • 13) 0.000 090 028 247 728 694 914 572 288 × 2 = 0 + 0.000 180 056 495 457 389 829 144 576;
  • 14) 0.000 180 056 495 457 389 829 144 576 × 2 = 0 + 0.000 360 112 990 914 779 658 289 152;
  • 15) 0.000 360 112 990 914 779 658 289 152 × 2 = 0 + 0.000 720 225 981 829 559 316 578 304;
  • 16) 0.000 720 225 981 829 559 316 578 304 × 2 = 0 + 0.001 440 451 963 659 118 633 156 608;
  • 17) 0.001 440 451 963 659 118 633 156 608 × 2 = 0 + 0.002 880 903 927 318 237 266 313 216;
  • 18) 0.002 880 903 927 318 237 266 313 216 × 2 = 0 + 0.005 761 807 854 636 474 532 626 432;
  • 19) 0.005 761 807 854 636 474 532 626 432 × 2 = 0 + 0.011 523 615 709 272 949 065 252 864;
  • 20) 0.011 523 615 709 272 949 065 252 864 × 2 = 0 + 0.023 047 231 418 545 898 130 505 728;
  • 21) 0.023 047 231 418 545 898 130 505 728 × 2 = 0 + 0.046 094 462 837 091 796 261 011 456;
  • 22) 0.046 094 462 837 091 796 261 011 456 × 2 = 0 + 0.092 188 925 674 183 592 522 022 912;
  • 23) 0.092 188 925 674 183 592 522 022 912 × 2 = 0 + 0.184 377 851 348 367 185 044 045 824;
  • 24) 0.184 377 851 348 367 185 044 045 824 × 2 = 0 + 0.368 755 702 696 734 370 088 091 648;
  • 25) 0.368 755 702 696 734 370 088 091 648 × 2 = 0 + 0.737 511 405 393 468 740 176 183 296;
  • 26) 0.737 511 405 393 468 740 176 183 296 × 2 = 1 + 0.475 022 810 786 937 480 352 366 592;
  • 27) 0.475 022 810 786 937 480 352 366 592 × 2 = 0 + 0.950 045 621 573 874 960 704 733 184;
  • 28) 0.950 045 621 573 874 960 704 733 184 × 2 = 1 + 0.900 091 243 147 749 921 409 466 368;
  • 29) 0.900 091 243 147 749 921 409 466 368 × 2 = 1 + 0.800 182 486 295 499 842 818 932 736;
  • 30) 0.800 182 486 295 499 842 818 932 736 × 2 = 1 + 0.600 364 972 590 999 685 637 865 472;
  • 31) 0.600 364 972 590 999 685 637 865 472 × 2 = 1 + 0.200 729 945 181 999 371 275 730 944;
  • 32) 0.200 729 945 181 999 371 275 730 944 × 2 = 0 + 0.401 459 890 363 998 742 551 461 888;
  • 33) 0.401 459 890 363 998 742 551 461 888 × 2 = 0 + 0.802 919 780 727 997 485 102 923 776;
  • 34) 0.802 919 780 727 997 485 102 923 776 × 2 = 1 + 0.605 839 561 455 994 970 205 847 552;
  • 35) 0.605 839 561 455 994 970 205 847 552 × 2 = 1 + 0.211 679 122 911 989 940 411 695 104;
  • 36) 0.211 679 122 911 989 940 411 695 104 × 2 = 0 + 0.423 358 245 823 979 880 823 390 208;
  • 37) 0.423 358 245 823 979 880 823 390 208 × 2 = 0 + 0.846 716 491 647 959 761 646 780 416;
  • 38) 0.846 716 491 647 959 761 646 780 416 × 2 = 1 + 0.693 432 983 295 919 523 293 560 832;
  • 39) 0.693 432 983 295 919 523 293 560 832 × 2 = 1 + 0.386 865 966 591 839 046 587 121 664;
  • 40) 0.386 865 966 591 839 046 587 121 664 × 2 = 0 + 0.773 731 933 183 678 093 174 243 328;
  • 41) 0.773 731 933 183 678 093 174 243 328 × 2 = 1 + 0.547 463 866 367 356 186 348 486 656;
  • 42) 0.547 463 866 367 356 186 348 486 656 × 2 = 1 + 0.094 927 732 734 712 372 696 973 312;
  • 43) 0.094 927 732 734 712 372 696 973 312 × 2 = 0 + 0.189 855 465 469 424 745 393 946 624;
  • 44) 0.189 855 465 469 424 745 393 946 624 × 2 = 0 + 0.379 710 930 938 849 490 787 893 248;
  • 45) 0.379 710 930 938 849 490 787 893 248 × 2 = 0 + 0.759 421 861 877 698 981 575 786 496;
  • 46) 0.759 421 861 877 698 981 575 786 496 × 2 = 1 + 0.518 843 723 755 397 963 151 572 992;
  • 47) 0.518 843 723 755 397 963 151 572 992 × 2 = 1 + 0.037 687 447 510 795 926 303 145 984;
  • 48) 0.037 687 447 510 795 926 303 145 984 × 2 = 0 + 0.075 374 895 021 591 852 606 291 968;
  • 49) 0.075 374 895 021 591 852 606 291 968 × 2 = 0 + 0.150 749 790 043 183 705 212 583 936;
  • 50) 0.150 749 790 043 183 705 212 583 936 × 2 = 0 + 0.301 499 580 086 367 410 425 167 872;
  • 51) 0.301 499 580 086 367 410 425 167 872 × 2 = 0 + 0.602 999 160 172 734 820 850 335 744;
  • 52) 0.602 999 160 172 734 820 850 335 744 × 2 = 1 + 0.205 998 320 345 469 641 700 671 488;
  • 53) 0.205 998 320 345 469 641 700 671 488 × 2 = 0 + 0.411 996 640 690 939 283 401 342 976;
  • 54) 0.411 996 640 690 939 283 401 342 976 × 2 = 0 + 0.823 993 281 381 878 566 802 685 952;
  • 55) 0.823 993 281 381 878 566 802 685 952 × 2 = 1 + 0.647 986 562 763 757 133 605 371 904;
  • 56) 0.647 986 562 763 757 133 605 371 904 × 2 = 1 + 0.295 973 125 527 514 267 210 743 808;
  • 57) 0.295 973 125 527 514 267 210 743 808 × 2 = 0 + 0.591 946 251 055 028 534 421 487 616;
  • 58) 0.591 946 251 055 028 534 421 487 616 × 2 = 1 + 0.183 892 502 110 057 068 842 975 232;
  • 59) 0.183 892 502 110 057 068 842 975 232 × 2 = 0 + 0.367 785 004 220 114 137 685 950 464;
  • 60) 0.367 785 004 220 114 137 685 950 464 × 2 = 0 + 0.735 570 008 440 228 275 371 900 928;
  • 61) 0.735 570 008 440 228 275 371 900 928 × 2 = 1 + 0.471 140 016 880 456 550 743 801 856;
  • 62) 0.471 140 016 880 456 550 743 801 856 × 2 = 0 + 0.942 280 033 760 913 101 487 603 712;
  • 63) 0.942 280 033 760 913 101 487 603 712 × 2 = 1 + 0.884 560 067 521 826 202 975 207 424;
  • 64) 0.884 560 067 521 826 202 975 207 424 × 2 = 1 + 0.769 120 135 043 652 405 950 414 848;
  • 65) 0.769 120 135 043 652 405 950 414 848 × 2 = 1 + 0.538 240 270 087 304 811 900 829 696;
  • 66) 0.538 240 270 087 304 811 900 829 696 × 2 = 1 + 0.076 480 540 174 609 623 801 659 392;
  • 67) 0.076 480 540 174 609 623 801 659 392 × 2 = 0 + 0.152 961 080 349 219 247 603 318 784;
  • 68) 0.152 961 080 349 219 247 603 318 784 × 2 = 0 + 0.305 922 160 698 438 495 206 637 568;
  • 69) 0.305 922 160 698 438 495 206 637 568 × 2 = 0 + 0.611 844 321 396 876 990 413 275 136;
  • 70) 0.611 844 321 396 876 990 413 275 136 × 2 = 1 + 0.223 688 642 793 753 980 826 550 272;
  • 71) 0.223 688 642 793 753 980 826 550 272 × 2 = 0 + 0.447 377 285 587 507 961 653 100 544;
  • 72) 0.447 377 285 587 507 961 653 100 544 × 2 = 0 + 0.894 754 571 175 015 923 306 201 088;
  • 73) 0.894 754 571 175 015 923 306 201 088 × 2 = 1 + 0.789 509 142 350 031 846 612 402 176;
  • 74) 0.789 509 142 350 031 846 612 402 176 × 2 = 1 + 0.579 018 284 700 063 693 224 804 352;
  • 75) 0.579 018 284 700 063 693 224 804 352 × 2 = 1 + 0.158 036 569 400 127 386 449 608 704;
  • 76) 0.158 036 569 400 127 386 449 608 704 × 2 = 0 + 0.316 073 138 800 254 772 899 217 408;
  • 77) 0.316 073 138 800 254 772 899 217 408 × 2 = 0 + 0.632 146 277 600 509 545 798 434 816;
  • 78) 0.632 146 277 600 509 545 798 434 816 × 2 = 1 + 0.264 292 555 201 019 091 596 869 632;

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:


0.000 000 021 979 552 668 138 406 878(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:

0.000 000 021 979 552 668 138 406 878(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:


0.000 000 021 979 552 668 138 406 878(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 0.000 000 021 979 552 668 138 406 878 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