0.000 000 021 979 552 668 138 406 910 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 910 3(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
0.000 000 021 979 552 668 138 406 910 3(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: 0.000 000 021 979 552 668 138 406 910 3.

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) 0.000 000 021 979 552 668 138 406 910 3 × 2 = 0 + 0.000 000 043 959 105 336 276 813 820 6;
  • 2) 0.000 000 043 959 105 336 276 813 820 6 × 2 = 0 + 0.000 000 087 918 210 672 553 627 641 2;
  • 3) 0.000 000 087 918 210 672 553 627 641 2 × 2 = 0 + 0.000 000 175 836 421 345 107 255 282 4;
  • 4) 0.000 000 175 836 421 345 107 255 282 4 × 2 = 0 + 0.000 000 351 672 842 690 214 510 564 8;
  • 5) 0.000 000 351 672 842 690 214 510 564 8 × 2 = 0 + 0.000 000 703 345 685 380 429 021 129 6;
  • 6) 0.000 000 703 345 685 380 429 021 129 6 × 2 = 0 + 0.000 001 406 691 370 760 858 042 259 2;
  • 7) 0.000 001 406 691 370 760 858 042 259 2 × 2 = 0 + 0.000 002 813 382 741 521 716 084 518 4;
  • 8) 0.000 002 813 382 741 521 716 084 518 4 × 2 = 0 + 0.000 005 626 765 483 043 432 169 036 8;
  • 9) 0.000 005 626 765 483 043 432 169 036 8 × 2 = 0 + 0.000 011 253 530 966 086 864 338 073 6;
  • 10) 0.000 011 253 530 966 086 864 338 073 6 × 2 = 0 + 0.000 022 507 061 932 173 728 676 147 2;
  • 11) 0.000 022 507 061 932 173 728 676 147 2 × 2 = 0 + 0.000 045 014 123 864 347 457 352 294 4;
  • 12) 0.000 045 014 123 864 347 457 352 294 4 × 2 = 0 + 0.000 090 028 247 728 694 914 704 588 8;
  • 13) 0.000 090 028 247 728 694 914 704 588 8 × 2 = 0 + 0.000 180 056 495 457 389 829 409 177 6;
  • 14) 0.000 180 056 495 457 389 829 409 177 6 × 2 = 0 + 0.000 360 112 990 914 779 658 818 355 2;
  • 15) 0.000 360 112 990 914 779 658 818 355 2 × 2 = 0 + 0.000 720 225 981 829 559 317 636 710 4;
  • 16) 0.000 720 225 981 829 559 317 636 710 4 × 2 = 0 + 0.001 440 451 963 659 118 635 273 420 8;
  • 17) 0.001 440 451 963 659 118 635 273 420 8 × 2 = 0 + 0.002 880 903 927 318 237 270 546 841 6;
  • 18) 0.002 880 903 927 318 237 270 546 841 6 × 2 = 0 + 0.005 761 807 854 636 474 541 093 683 2;
  • 19) 0.005 761 807 854 636 474 541 093 683 2 × 2 = 0 + 0.011 523 615 709 272 949 082 187 366 4;
  • 20) 0.011 523 615 709 272 949 082 187 366 4 × 2 = 0 + 0.023 047 231 418 545 898 164 374 732 8;
  • 21) 0.023 047 231 418 545 898 164 374 732 8 × 2 = 0 + 0.046 094 462 837 091 796 328 749 465 6;
  • 22) 0.046 094 462 837 091 796 328 749 465 6 × 2 = 0 + 0.092 188 925 674 183 592 657 498 931 2;
  • 23) 0.092 188 925 674 183 592 657 498 931 2 × 2 = 0 + 0.184 377 851 348 367 185 314 997 862 4;
  • 24) 0.184 377 851 348 367 185 314 997 862 4 × 2 = 0 + 0.368 755 702 696 734 370 629 995 724 8;
  • 25) 0.368 755 702 696 734 370 629 995 724 8 × 2 = 0 + 0.737 511 405 393 468 741 259 991 449 6;
  • 26) 0.737 511 405 393 468 741 259 991 449 6 × 2 = 1 + 0.475 022 810 786 937 482 519 982 899 2;
  • 27) 0.475 022 810 786 937 482 519 982 899 2 × 2 = 0 + 0.950 045 621 573 874 965 039 965 798 4;
  • 28) 0.950 045 621 573 874 965 039 965 798 4 × 2 = 1 + 0.900 091 243 147 749 930 079 931 596 8;
  • 29) 0.900 091 243 147 749 930 079 931 596 8 × 2 = 1 + 0.800 182 486 295 499 860 159 863 193 6;
  • 30) 0.800 182 486 295 499 860 159 863 193 6 × 2 = 1 + 0.600 364 972 590 999 720 319 726 387 2;
  • 31) 0.600 364 972 590 999 720 319 726 387 2 × 2 = 1 + 0.200 729 945 181 999 440 639 452 774 4;
  • 32) 0.200 729 945 181 999 440 639 452 774 4 × 2 = 0 + 0.401 459 890 363 998 881 278 905 548 8;
  • 33) 0.401 459 890 363 998 881 278 905 548 8 × 2 = 0 + 0.802 919 780 727 997 762 557 811 097 6;
  • 34) 0.802 919 780 727 997 762 557 811 097 6 × 2 = 1 + 0.605 839 561 455 995 525 115 622 195 2;
  • 35) 0.605 839 561 455 995 525 115 622 195 2 × 2 = 1 + 0.211 679 122 911 991 050 231 244 390 4;
  • 36) 0.211 679 122 911 991 050 231 244 390 4 × 2 = 0 + 0.423 358 245 823 982 100 462 488 780 8;
  • 37) 0.423 358 245 823 982 100 462 488 780 8 × 2 = 0 + 0.846 716 491 647 964 200 924 977 561 6;
  • 38) 0.846 716 491 647 964 200 924 977 561 6 × 2 = 1 + 0.693 432 983 295 928 401 849 955 123 2;
  • 39) 0.693 432 983 295 928 401 849 955 123 2 × 2 = 1 + 0.386 865 966 591 856 803 699 910 246 4;
  • 40) 0.386 865 966 591 856 803 699 910 246 4 × 2 = 0 + 0.773 731 933 183 713 607 399 820 492 8;
  • 41) 0.773 731 933 183 713 607 399 820 492 8 × 2 = 1 + 0.547 463 866 367 427 214 799 640 985 6;
  • 42) 0.547 463 866 367 427 214 799 640 985 6 × 2 = 1 + 0.094 927 732 734 854 429 599 281 971 2;
  • 43) 0.094 927 732 734 854 429 599 281 971 2 × 2 = 0 + 0.189 855 465 469 708 859 198 563 942 4;
  • 44) 0.189 855 465 469 708 859 198 563 942 4 × 2 = 0 + 0.379 710 930 939 417 718 397 127 884 8;
  • 45) 0.379 710 930 939 417 718 397 127 884 8 × 2 = 0 + 0.759 421 861 878 835 436 794 255 769 6;
  • 46) 0.759 421 861 878 835 436 794 255 769 6 × 2 = 1 + 0.518 843 723 757 670 873 588 511 539 2;
  • 47) 0.518 843 723 757 670 873 588 511 539 2 × 2 = 1 + 0.037 687 447 515 341 747 177 023 078 4;
  • 48) 0.037 687 447 515 341 747 177 023 078 4 × 2 = 0 + 0.075 374 895 030 683 494 354 046 156 8;
  • 49) 0.075 374 895 030 683 494 354 046 156 8 × 2 = 0 + 0.150 749 790 061 366 988 708 092 313 6;
  • 50) 0.150 749 790 061 366 988 708 092 313 6 × 2 = 0 + 0.301 499 580 122 733 977 416 184 627 2;
  • 51) 0.301 499 580 122 733 977 416 184 627 2 × 2 = 0 + 0.602 999 160 245 467 954 832 369 254 4;
  • 52) 0.602 999 160 245 467 954 832 369 254 4 × 2 = 1 + 0.205 998 320 490 935 909 664 738 508 8;
  • 53) 0.205 998 320 490 935 909 664 738 508 8 × 2 = 0 + 0.411 996 640 981 871 819 329 477 017 6;
  • 54) 0.411 996 640 981 871 819 329 477 017 6 × 2 = 0 + 0.823 993 281 963 743 638 658 954 035 2;
  • 55) 0.823 993 281 963 743 638 658 954 035 2 × 2 = 1 + 0.647 986 563 927 487 277 317 908 070 4;
  • 56) 0.647 986 563 927 487 277 317 908 070 4 × 2 = 1 + 0.295 973 127 854 974 554 635 816 140 8;
  • 57) 0.295 973 127 854 974 554 635 816 140 8 × 2 = 0 + 0.591 946 255 709 949 109 271 632 281 6;
  • 58) 0.591 946 255 709 949 109 271 632 281 6 × 2 = 1 + 0.183 892 511 419 898 218 543 264 563 2;
  • 59) 0.183 892 511 419 898 218 543 264 563 2 × 2 = 0 + 0.367 785 022 839 796 437 086 529 126 4;
  • 60) 0.367 785 022 839 796 437 086 529 126 4 × 2 = 0 + 0.735 570 045 679 592 874 173 058 252 8;
  • 61) 0.735 570 045 679 592 874 173 058 252 8 × 2 = 1 + 0.471 140 091 359 185 748 346 116 505 6;
  • 62) 0.471 140 091 359 185 748 346 116 505 6 × 2 = 0 + 0.942 280 182 718 371 496 692 233 011 2;
  • 63) 0.942 280 182 718 371 496 692 233 011 2 × 2 = 1 + 0.884 560 365 436 742 993 384 466 022 4;
  • 64) 0.884 560 365 436 742 993 384 466 022 4 × 2 = 1 + 0.769 120 730 873 485 986 768 932 044 8;
  • 65) 0.769 120 730 873 485 986 768 932 044 8 × 2 = 1 + 0.538 241 461 746 971 973 537 864 089 6;
  • 66) 0.538 241 461 746 971 973 537 864 089 6 × 2 = 1 + 0.076 482 923 493 943 947 075 728 179 2;
  • 67) 0.076 482 923 493 943 947 075 728 179 2 × 2 = 0 + 0.152 965 846 987 887 894 151 456 358 4;
  • 68) 0.152 965 846 987 887 894 151 456 358 4 × 2 = 0 + 0.305 931 693 975 775 788 302 912 716 8;
  • 69) 0.305 931 693 975 775 788 302 912 716 8 × 2 = 0 + 0.611 863 387 951 551 576 605 825 433 6;
  • 70) 0.611 863 387 951 551 576 605 825 433 6 × 2 = 1 + 0.223 726 775 903 103 153 211 650 867 2;
  • 71) 0.223 726 775 903 103 153 211 650 867 2 × 2 = 0 + 0.447 453 551 806 206 306 423 301 734 4;
  • 72) 0.447 453 551 806 206 306 423 301 734 4 × 2 = 0 + 0.894 907 103 612 412 612 846 603 468 8;
  • 73) 0.894 907 103 612 412 612 846 603 468 8 × 2 = 1 + 0.789 814 207 224 825 225 693 206 937 6;
  • 74) 0.789 814 207 224 825 225 693 206 937 6 × 2 = 1 + 0.579 628 414 449 650 451 386 413 875 2;
  • 75) 0.579 628 414 449 650 451 386 413 875 2 × 2 = 1 + 0.159 256 828 899 300 902 772 827 750 4;
  • 76) 0.159 256 828 899 300 902 772 827 750 4 × 2 = 0 + 0.318 513 657 798 601 805 545 655 500 8;
  • 77) 0.318 513 657 798 601 805 545 655 500 8 × 2 = 0 + 0.637 027 315 597 203 611 091 311 001 6;
  • 78) 0.637 027 315 597 203 611 091 311 001 6 × 2 = 1 + 0.274 054 631 194 407 222 182 622 003 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 - 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:


0.000 000 021 979 552 668 138 406 910 3(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:

0.000 000 021 979 552 668 138 406 910 3(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:


0.000 000 021 979 552 668 138 406 910 3(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 0.000 000 021 979 552 668 138 406 910 3 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