0.000 020 830 729 321 671 205 134 999 154 509 662 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 509 662 1(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
0.000 020 830 729 321 671 205 134 999 154 509 662 1(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: 0.000 020 830 729 321 671 205 134 999 154 509 662 1.

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) 0.000 020 830 729 321 671 205 134 999 154 509 662 1 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 019 324 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 019 324 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 038 648 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 038 648 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 077 296 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 077 296 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 154 593 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 154 593 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 309 187 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 309 187 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 618 374 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 618 374 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 236 748 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 236 748 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 554 473 497 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 554 473 497 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 108 946 995 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 108 946 995 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 217 893 990 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 217 893 990 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 435 787 980 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 435 787 980 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 871 575 961 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 871 575 961 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 743 151 923 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 743 151 923 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 486 303 846 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 486 303 846 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 972 607 692 8;
  • 16) 0.682 581 338 412 522 049 863 652 294 972 607 692 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 945 215 385 6;
  • 17) 0.365 162 676 825 044 099 727 304 589 945 215 385 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 890 430 771 2;
  • 18) 0.730 325 353 650 088 199 454 609 179 890 430 771 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 359 780 861 542 4;
  • 19) 0.460 650 707 300 176 398 909 218 359 780 861 542 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 719 561 723 084 8;
  • 20) 0.921 301 414 600 352 797 818 436 719 561 723 084 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 439 123 446 169 6;
  • 21) 0.842 602 829 200 705 595 636 873 439 123 446 169 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 878 246 892 339 2;
  • 22) 0.685 205 658 401 411 191 273 746 878 246 892 339 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 756 493 784 678 4;
  • 23) 0.370 411 316 802 822 382 547 493 756 493 784 678 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 512 987 569 356 8;
  • 24) 0.740 822 633 605 644 765 094 987 512 987 569 356 8 × 2 = 1 + 0.481 645 267 211 289 530 189 975 025 975 138 713 6;
  • 25) 0.481 645 267 211 289 530 189 975 025 975 138 713 6 × 2 = 0 + 0.963 290 534 422 579 060 379 950 051 950 277 427 2;
  • 26) 0.963 290 534 422 579 060 379 950 051 950 277 427 2 × 2 = 1 + 0.926 581 068 845 158 120 759 900 103 900 554 854 4;
  • 27) 0.926 581 068 845 158 120 759 900 103 900 554 854 4 × 2 = 1 + 0.853 162 137 690 316 241 519 800 207 801 109 708 8;
  • 28) 0.853 162 137 690 316 241 519 800 207 801 109 708 8 × 2 = 1 + 0.706 324 275 380 632 483 039 600 415 602 219 417 6;
  • 29) 0.706 324 275 380 632 483 039 600 415 602 219 417 6 × 2 = 1 + 0.412 648 550 761 264 966 079 200 831 204 438 835 2;
  • 30) 0.412 648 550 761 264 966 079 200 831 204 438 835 2 × 2 = 0 + 0.825 297 101 522 529 932 158 401 662 408 877 670 4;
  • 31) 0.825 297 101 522 529 932 158 401 662 408 877 670 4 × 2 = 1 + 0.650 594 203 045 059 864 316 803 324 817 755 340 8;
  • 32) 0.650 594 203 045 059 864 316 803 324 817 755 340 8 × 2 = 1 + 0.301 188 406 090 119 728 633 606 649 635 510 681 6;
  • 33) 0.301 188 406 090 119 728 633 606 649 635 510 681 6 × 2 = 0 + 0.602 376 812 180 239 457 267 213 299 271 021 363 2;
  • 34) 0.602 376 812 180 239 457 267 213 299 271 021 363 2 × 2 = 1 + 0.204 753 624 360 478 914 534 426 598 542 042 726 4;
  • 35) 0.204 753 624 360 478 914 534 426 598 542 042 726 4 × 2 = 0 + 0.409 507 248 720 957 829 068 853 197 084 085 452 8;
  • 36) 0.409 507 248 720 957 829 068 853 197 084 085 452 8 × 2 = 0 + 0.819 014 497 441 915 658 137 706 394 168 170 905 6;
  • 37) 0.819 014 497 441 915 658 137 706 394 168 170 905 6 × 2 = 1 + 0.638 028 994 883 831 316 275 412 788 336 341 811 2;
  • 38) 0.638 028 994 883 831 316 275 412 788 336 341 811 2 × 2 = 1 + 0.276 057 989 767 662 632 550 825 576 672 683 622 4;
  • 39) 0.276 057 989 767 662 632 550 825 576 672 683 622 4 × 2 = 0 + 0.552 115 979 535 325 265 101 651 153 345 367 244 8;
  • 40) 0.552 115 979 535 325 265 101 651 153 345 367 244 8 × 2 = 1 + 0.104 231 959 070 650 530 203 302 306 690 734 489 6;
  • 41) 0.104 231 959 070 650 530 203 302 306 690 734 489 6 × 2 = 0 + 0.208 463 918 141 301 060 406 604 613 381 468 979 2;
  • 42) 0.208 463 918 141 301 060 406 604 613 381 468 979 2 × 2 = 0 + 0.416 927 836 282 602 120 813 209 226 762 937 958 4;
  • 43) 0.416 927 836 282 602 120 813 209 226 762 937 958 4 × 2 = 0 + 0.833 855 672 565 204 241 626 418 453 525 875 916 8;
  • 44) 0.833 855 672 565 204 241 626 418 453 525 875 916 8 × 2 = 1 + 0.667 711 345 130 408 483 252 836 907 051 751 833 6;
  • 45) 0.667 711 345 130 408 483 252 836 907 051 751 833 6 × 2 = 1 + 0.335 422 690 260 816 966 505 673 814 103 503 667 2;
  • 46) 0.335 422 690 260 816 966 505 673 814 103 503 667 2 × 2 = 0 + 0.670 845 380 521 633 933 011 347 628 207 007 334 4;
  • 47) 0.670 845 380 521 633 933 011 347 628 207 007 334 4 × 2 = 1 + 0.341 690 761 043 267 866 022 695 256 414 014 668 8;
  • 48) 0.341 690 761 043 267 866 022 695 256 414 014 668 8 × 2 = 0 + 0.683 381 522 086 535 732 045 390 512 828 029 337 6;
  • 49) 0.683 381 522 086 535 732 045 390 512 828 029 337 6 × 2 = 1 + 0.366 763 044 173 071 464 090 781 025 656 058 675 2;
  • 50) 0.366 763 044 173 071 464 090 781 025 656 058 675 2 × 2 = 0 + 0.733 526 088 346 142 928 181 562 051 312 117 350 4;
  • 51) 0.733 526 088 346 142 928 181 562 051 312 117 350 4 × 2 = 1 + 0.467 052 176 692 285 856 363 124 102 624 234 700 8;
  • 52) 0.467 052 176 692 285 856 363 124 102 624 234 700 8 × 2 = 0 + 0.934 104 353 384 571 712 726 248 205 248 469 401 6;
  • 53) 0.934 104 353 384 571 712 726 248 205 248 469 401 6 × 2 = 1 + 0.868 208 706 769 143 425 452 496 410 496 938 803 2;
  • 54) 0.868 208 706 769 143 425 452 496 410 496 938 803 2 × 2 = 1 + 0.736 417 413 538 286 850 904 992 820 993 877 606 4;
  • 55) 0.736 417 413 538 286 850 904 992 820 993 877 606 4 × 2 = 1 + 0.472 834 827 076 573 701 809 985 641 987 755 212 8;
  • 56) 0.472 834 827 076 573 701 809 985 641 987 755 212 8 × 2 = 0 + 0.945 669 654 153 147 403 619 971 283 975 510 425 6;
  • 57) 0.945 669 654 153 147 403 619 971 283 975 510 425 6 × 2 = 1 + 0.891 339 308 306 294 807 239 942 567 951 020 851 2;
  • 58) 0.891 339 308 306 294 807 239 942 567 951 020 851 2 × 2 = 1 + 0.782 678 616 612 589 614 479 885 135 902 041 702 4;
  • 59) 0.782 678 616 612 589 614 479 885 135 902 041 702 4 × 2 = 1 + 0.565 357 233 225 179 228 959 770 271 804 083 404 8;
  • 60) 0.565 357 233 225 179 228 959 770 271 804 083 404 8 × 2 = 1 + 0.130 714 466 450 358 457 919 540 543 608 166 809 6;
  • 61) 0.130 714 466 450 358 457 919 540 543 608 166 809 6 × 2 = 0 + 0.261 428 932 900 716 915 839 081 087 216 333 619 2;
  • 62) 0.261 428 932 900 716 915 839 081 087 216 333 619 2 × 2 = 0 + 0.522 857 865 801 433 831 678 162 174 432 667 238 4;
  • 63) 0.522 857 865 801 433 831 678 162 174 432 667 238 4 × 2 = 1 + 0.045 715 731 602 867 663 356 324 348 865 334 476 8;
  • 64) 0.045 715 731 602 867 663 356 324 348 865 334 476 8 × 2 = 0 + 0.091 431 463 205 735 326 712 648 697 730 668 953 6;
  • 65) 0.091 431 463 205 735 326 712 648 697 730 668 953 6 × 2 = 0 + 0.182 862 926 411 470 653 425 297 395 461 337 907 2;
  • 66) 0.182 862 926 411 470 653 425 297 395 461 337 907 2 × 2 = 0 + 0.365 725 852 822 941 306 850 594 790 922 675 814 4;
  • 67) 0.365 725 852 822 941 306 850 594 790 922 675 814 4 × 2 = 0 + 0.731 451 705 645 882 613 701 189 581 845 351 628 8;
  • 68) 0.731 451 705 645 882 613 701 189 581 845 351 628 8 × 2 = 1 + 0.462 903 411 291 765 227 402 379 163 690 703 257 6;

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:


0.000 020 830 729 321 671 205 134 999 154 509 662 1(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)

5. Positive number before normalization:

0.000 020 830 729 321 671 205 134 999 154 509 662 1(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 020 830 729 321 671 205 134 999 154 509 662 1(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 20 =

1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 2-16


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): -16

Mantissa (not normalized):
1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(11-1) - 1 =

(-16 + 1 023)(10) =

1 007(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;
  • 1 007 ÷ 2 = 503 + 1;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 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) =

1007(10) =

011 1110 1111(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. 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001 =

0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


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 1111

Mantissa (52 bits) =
0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


Decimal number 0.000 020 830 729 321 671 205 134 999 154 509 662 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001

Share

» Convert decimal 0.000 020 830 729 321 671 205 134 999 154 509 669 9 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Show, Don’t Tell, the Strategic Power of Action. Gemini Notebook YouTube Video of 9.21 minutes

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