0.000 000 021 979 552 668 138 406 917 995 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 917 995(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
0.000 000 021 979 552 668 138 406 917 995(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: 0.000 000 021 979 552 668 138 406 917 995.

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) 0.000 000 021 979 552 668 138 406 917 995 × 2 = 0 + 0.000 000 043 959 105 336 276 813 835 99;
  • 2) 0.000 000 043 959 105 336 276 813 835 99 × 2 = 0 + 0.000 000 087 918 210 672 553 627 671 98;
  • 3) 0.000 000 087 918 210 672 553 627 671 98 × 2 = 0 + 0.000 000 175 836 421 345 107 255 343 96;
  • 4) 0.000 000 175 836 421 345 107 255 343 96 × 2 = 0 + 0.000 000 351 672 842 690 214 510 687 92;
  • 5) 0.000 000 351 672 842 690 214 510 687 92 × 2 = 0 + 0.000 000 703 345 685 380 429 021 375 84;
  • 6) 0.000 000 703 345 685 380 429 021 375 84 × 2 = 0 + 0.000 001 406 691 370 760 858 042 751 68;
  • 7) 0.000 001 406 691 370 760 858 042 751 68 × 2 = 0 + 0.000 002 813 382 741 521 716 085 503 36;
  • 8) 0.000 002 813 382 741 521 716 085 503 36 × 2 = 0 + 0.000 005 626 765 483 043 432 171 006 72;
  • 9) 0.000 005 626 765 483 043 432 171 006 72 × 2 = 0 + 0.000 011 253 530 966 086 864 342 013 44;
  • 10) 0.000 011 253 530 966 086 864 342 013 44 × 2 = 0 + 0.000 022 507 061 932 173 728 684 026 88;
  • 11) 0.000 022 507 061 932 173 728 684 026 88 × 2 = 0 + 0.000 045 014 123 864 347 457 368 053 76;
  • 12) 0.000 045 014 123 864 347 457 368 053 76 × 2 = 0 + 0.000 090 028 247 728 694 914 736 107 52;
  • 13) 0.000 090 028 247 728 694 914 736 107 52 × 2 = 0 + 0.000 180 056 495 457 389 829 472 215 04;
  • 14) 0.000 180 056 495 457 389 829 472 215 04 × 2 = 0 + 0.000 360 112 990 914 779 658 944 430 08;
  • 15) 0.000 360 112 990 914 779 658 944 430 08 × 2 = 0 + 0.000 720 225 981 829 559 317 888 860 16;
  • 16) 0.000 720 225 981 829 559 317 888 860 16 × 2 = 0 + 0.001 440 451 963 659 118 635 777 720 32;
  • 17) 0.001 440 451 963 659 118 635 777 720 32 × 2 = 0 + 0.002 880 903 927 318 237 271 555 440 64;
  • 18) 0.002 880 903 927 318 237 271 555 440 64 × 2 = 0 + 0.005 761 807 854 636 474 543 110 881 28;
  • 19) 0.005 761 807 854 636 474 543 110 881 28 × 2 = 0 + 0.011 523 615 709 272 949 086 221 762 56;
  • 20) 0.011 523 615 709 272 949 086 221 762 56 × 2 = 0 + 0.023 047 231 418 545 898 172 443 525 12;
  • 21) 0.023 047 231 418 545 898 172 443 525 12 × 2 = 0 + 0.046 094 462 837 091 796 344 887 050 24;
  • 22) 0.046 094 462 837 091 796 344 887 050 24 × 2 = 0 + 0.092 188 925 674 183 592 689 774 100 48;
  • 23) 0.092 188 925 674 183 592 689 774 100 48 × 2 = 0 + 0.184 377 851 348 367 185 379 548 200 96;
  • 24) 0.184 377 851 348 367 185 379 548 200 96 × 2 = 0 + 0.368 755 702 696 734 370 759 096 401 92;
  • 25) 0.368 755 702 696 734 370 759 096 401 92 × 2 = 0 + 0.737 511 405 393 468 741 518 192 803 84;
  • 26) 0.737 511 405 393 468 741 518 192 803 84 × 2 = 1 + 0.475 022 810 786 937 483 036 385 607 68;
  • 27) 0.475 022 810 786 937 483 036 385 607 68 × 2 = 0 + 0.950 045 621 573 874 966 072 771 215 36;
  • 28) 0.950 045 621 573 874 966 072 771 215 36 × 2 = 1 + 0.900 091 243 147 749 932 145 542 430 72;
  • 29) 0.900 091 243 147 749 932 145 542 430 72 × 2 = 1 + 0.800 182 486 295 499 864 291 084 861 44;
  • 30) 0.800 182 486 295 499 864 291 084 861 44 × 2 = 1 + 0.600 364 972 590 999 728 582 169 722 88;
  • 31) 0.600 364 972 590 999 728 582 169 722 88 × 2 = 1 + 0.200 729 945 181 999 457 164 339 445 76;
  • 32) 0.200 729 945 181 999 457 164 339 445 76 × 2 = 0 + 0.401 459 890 363 998 914 328 678 891 52;
  • 33) 0.401 459 890 363 998 914 328 678 891 52 × 2 = 0 + 0.802 919 780 727 997 828 657 357 783 04;
  • 34) 0.802 919 780 727 997 828 657 357 783 04 × 2 = 1 + 0.605 839 561 455 995 657 314 715 566 08;
  • 35) 0.605 839 561 455 995 657 314 715 566 08 × 2 = 1 + 0.211 679 122 911 991 314 629 431 132 16;
  • 36) 0.211 679 122 911 991 314 629 431 132 16 × 2 = 0 + 0.423 358 245 823 982 629 258 862 264 32;
  • 37) 0.423 358 245 823 982 629 258 862 264 32 × 2 = 0 + 0.846 716 491 647 965 258 517 724 528 64;
  • 38) 0.846 716 491 647 965 258 517 724 528 64 × 2 = 1 + 0.693 432 983 295 930 517 035 449 057 28;
  • 39) 0.693 432 983 295 930 517 035 449 057 28 × 2 = 1 + 0.386 865 966 591 861 034 070 898 114 56;
  • 40) 0.386 865 966 591 861 034 070 898 114 56 × 2 = 0 + 0.773 731 933 183 722 068 141 796 229 12;
  • 41) 0.773 731 933 183 722 068 141 796 229 12 × 2 = 1 + 0.547 463 866 367 444 136 283 592 458 24;
  • 42) 0.547 463 866 367 444 136 283 592 458 24 × 2 = 1 + 0.094 927 732 734 888 272 567 184 916 48;
  • 43) 0.094 927 732 734 888 272 567 184 916 48 × 2 = 0 + 0.189 855 465 469 776 545 134 369 832 96;
  • 44) 0.189 855 465 469 776 545 134 369 832 96 × 2 = 0 + 0.379 710 930 939 553 090 268 739 665 92;
  • 45) 0.379 710 930 939 553 090 268 739 665 92 × 2 = 0 + 0.759 421 861 879 106 180 537 479 331 84;
  • 46) 0.759 421 861 879 106 180 537 479 331 84 × 2 = 1 + 0.518 843 723 758 212 361 074 958 663 68;
  • 47) 0.518 843 723 758 212 361 074 958 663 68 × 2 = 1 + 0.037 687 447 516 424 722 149 917 327 36;
  • 48) 0.037 687 447 516 424 722 149 917 327 36 × 2 = 0 + 0.075 374 895 032 849 444 299 834 654 72;
  • 49) 0.075 374 895 032 849 444 299 834 654 72 × 2 = 0 + 0.150 749 790 065 698 888 599 669 309 44;
  • 50) 0.150 749 790 065 698 888 599 669 309 44 × 2 = 0 + 0.301 499 580 131 397 777 199 338 618 88;
  • 51) 0.301 499 580 131 397 777 199 338 618 88 × 2 = 0 + 0.602 999 160 262 795 554 398 677 237 76;
  • 52) 0.602 999 160 262 795 554 398 677 237 76 × 2 = 1 + 0.205 998 320 525 591 108 797 354 475 52;
  • 53) 0.205 998 320 525 591 108 797 354 475 52 × 2 = 0 + 0.411 996 641 051 182 217 594 708 951 04;
  • 54) 0.411 996 641 051 182 217 594 708 951 04 × 2 = 0 + 0.823 993 282 102 364 435 189 417 902 08;
  • 55) 0.823 993 282 102 364 435 189 417 902 08 × 2 = 1 + 0.647 986 564 204 728 870 378 835 804 16;
  • 56) 0.647 986 564 204 728 870 378 835 804 16 × 2 = 1 + 0.295 973 128 409 457 740 757 671 608 32;
  • 57) 0.295 973 128 409 457 740 757 671 608 32 × 2 = 0 + 0.591 946 256 818 915 481 515 343 216 64;
  • 58) 0.591 946 256 818 915 481 515 343 216 64 × 2 = 1 + 0.183 892 513 637 830 963 030 686 433 28;
  • 59) 0.183 892 513 637 830 963 030 686 433 28 × 2 = 0 + 0.367 785 027 275 661 926 061 372 866 56;
  • 60) 0.367 785 027 275 661 926 061 372 866 56 × 2 = 0 + 0.735 570 054 551 323 852 122 745 733 12;
  • 61) 0.735 570 054 551 323 852 122 745 733 12 × 2 = 1 + 0.471 140 109 102 647 704 245 491 466 24;
  • 62) 0.471 140 109 102 647 704 245 491 466 24 × 2 = 0 + 0.942 280 218 205 295 408 490 982 932 48;
  • 63) 0.942 280 218 205 295 408 490 982 932 48 × 2 = 1 + 0.884 560 436 410 590 816 981 965 864 96;
  • 64) 0.884 560 436 410 590 816 981 965 864 96 × 2 = 1 + 0.769 120 872 821 181 633 963 931 729 92;
  • 65) 0.769 120 872 821 181 633 963 931 729 92 × 2 = 1 + 0.538 241 745 642 363 267 927 863 459 84;
  • 66) 0.538 241 745 642 363 267 927 863 459 84 × 2 = 1 + 0.076 483 491 284 726 535 855 726 919 68;
  • 67) 0.076 483 491 284 726 535 855 726 919 68 × 2 = 0 + 0.152 966 982 569 453 071 711 453 839 36;
  • 68) 0.152 966 982 569 453 071 711 453 839 36 × 2 = 0 + 0.305 933 965 138 906 143 422 907 678 72;
  • 69) 0.305 933 965 138 906 143 422 907 678 72 × 2 = 0 + 0.611 867 930 277 812 286 845 815 357 44;
  • 70) 0.611 867 930 277 812 286 845 815 357 44 × 2 = 1 + 0.223 735 860 555 624 573 691 630 714 88;
  • 71) 0.223 735 860 555 624 573 691 630 714 88 × 2 = 0 + 0.447 471 721 111 249 147 383 261 429 76;
  • 72) 0.447 471 721 111 249 147 383 261 429 76 × 2 = 0 + 0.894 943 442 222 498 294 766 522 859 52;
  • 73) 0.894 943 442 222 498 294 766 522 859 52 × 2 = 1 + 0.789 886 884 444 996 589 533 045 719 04;
  • 74) 0.789 886 884 444 996 589 533 045 719 04 × 2 = 1 + 0.579 773 768 889 993 179 066 091 438 08;
  • 75) 0.579 773 768 889 993 179 066 091 438 08 × 2 = 1 + 0.159 547 537 779 986 358 132 182 876 16;
  • 76) 0.159 547 537 779 986 358 132 182 876 16 × 2 = 0 + 0.319 095 075 559 972 716 264 365 752 32;
  • 77) 0.319 095 075 559 972 716 264 365 752 32 × 2 = 0 + 0.638 190 151 119 945 432 528 731 504 64;
  • 78) 0.638 190 151 119 945 432 528 731 504 64 × 2 = 1 + 0.276 380 302 239 890 865 057 463 009 28;

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:


0.000 000 021 979 552 668 138 406 917 995(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:

0.000 000 021 979 552 668 138 406 917 995(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:


0.000 000 021 979 552 668 138 406 917 995(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 0.000 000 021 979 552 668 138 406 917 995 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