0.000 000 000 000 000 000 008 535 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 535 8(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
0.000 000 000 000 000 000 008 535 8(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: 0.000 000 000 000 000 000 008 535 8.

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) 0.000 000 000 000 000 000 008 535 8 × 2 = 0 + 0.000 000 000 000 000 000 017 071 6;
  • 2) 0.000 000 000 000 000 000 017 071 6 × 2 = 0 + 0.000 000 000 000 000 000 034 143 2;
  • 3) 0.000 000 000 000 000 000 034 143 2 × 2 = 0 + 0.000 000 000 000 000 000 068 286 4;
  • 4) 0.000 000 000 000 000 000 068 286 4 × 2 = 0 + 0.000 000 000 000 000 000 136 572 8;
  • 5) 0.000 000 000 000 000 000 136 572 8 × 2 = 0 + 0.000 000 000 000 000 000 273 145 6;
  • 6) 0.000 000 000 000 000 000 273 145 6 × 2 = 0 + 0.000 000 000 000 000 000 546 291 2;
  • 7) 0.000 000 000 000 000 000 546 291 2 × 2 = 0 + 0.000 000 000 000 000 001 092 582 4;
  • 8) 0.000 000 000 000 000 001 092 582 4 × 2 = 0 + 0.000 000 000 000 000 002 185 164 8;
  • 9) 0.000 000 000 000 000 002 185 164 8 × 2 = 0 + 0.000 000 000 000 000 004 370 329 6;
  • 10) 0.000 000 000 000 000 004 370 329 6 × 2 = 0 + 0.000 000 000 000 000 008 740 659 2;
  • 11) 0.000 000 000 000 000 008 740 659 2 × 2 = 0 + 0.000 000 000 000 000 017 481 318 4;
  • 12) 0.000 000 000 000 000 017 481 318 4 × 2 = 0 + 0.000 000 000 000 000 034 962 636 8;
  • 13) 0.000 000 000 000 000 034 962 636 8 × 2 = 0 + 0.000 000 000 000 000 069 925 273 6;
  • 14) 0.000 000 000 000 000 069 925 273 6 × 2 = 0 + 0.000 000 000 000 000 139 850 547 2;
  • 15) 0.000 000 000 000 000 139 850 547 2 × 2 = 0 + 0.000 000 000 000 000 279 701 094 4;
  • 16) 0.000 000 000 000 000 279 701 094 4 × 2 = 0 + 0.000 000 000 000 000 559 402 188 8;
  • 17) 0.000 000 000 000 000 559 402 188 8 × 2 = 0 + 0.000 000 000 000 001 118 804 377 6;
  • 18) 0.000 000 000 000 001 118 804 377 6 × 2 = 0 + 0.000 000 000 000 002 237 608 755 2;
  • 19) 0.000 000 000 000 002 237 608 755 2 × 2 = 0 + 0.000 000 000 000 004 475 217 510 4;
  • 20) 0.000 000 000 000 004 475 217 510 4 × 2 = 0 + 0.000 000 000 000 008 950 435 020 8;
  • 21) 0.000 000 000 000 008 950 435 020 8 × 2 = 0 + 0.000 000 000 000 017 900 870 041 6;
  • 22) 0.000 000 000 000 017 900 870 041 6 × 2 = 0 + 0.000 000 000 000 035 801 740 083 2;
  • 23) 0.000 000 000 000 035 801 740 083 2 × 2 = 0 + 0.000 000 000 000 071 603 480 166 4;
  • 24) 0.000 000 000 000 071 603 480 166 4 × 2 = 0 + 0.000 000 000 000 143 206 960 332 8;
  • 25) 0.000 000 000 000 143 206 960 332 8 × 2 = 0 + 0.000 000 000 000 286 413 920 665 6;
  • 26) 0.000 000 000 000 286 413 920 665 6 × 2 = 0 + 0.000 000 000 000 572 827 841 331 2;
  • 27) 0.000 000 000 000 572 827 841 331 2 × 2 = 0 + 0.000 000 000 001 145 655 682 662 4;
  • 28) 0.000 000 000 001 145 655 682 662 4 × 2 = 0 + 0.000 000 000 002 291 311 365 324 8;
  • 29) 0.000 000 000 002 291 311 365 324 8 × 2 = 0 + 0.000 000 000 004 582 622 730 649 6;
  • 30) 0.000 000 000 004 582 622 730 649 6 × 2 = 0 + 0.000 000 000 009 165 245 461 299 2;
  • 31) 0.000 000 000 009 165 245 461 299 2 × 2 = 0 + 0.000 000 000 018 330 490 922 598 4;
  • 32) 0.000 000 000 018 330 490 922 598 4 × 2 = 0 + 0.000 000 000 036 660 981 845 196 8;
  • 33) 0.000 000 000 036 660 981 845 196 8 × 2 = 0 + 0.000 000 000 073 321 963 690 393 6;
  • 34) 0.000 000 000 073 321 963 690 393 6 × 2 = 0 + 0.000 000 000 146 643 927 380 787 2;
  • 35) 0.000 000 000 146 643 927 380 787 2 × 2 = 0 + 0.000 000 000 293 287 854 761 574 4;
  • 36) 0.000 000 000 293 287 854 761 574 4 × 2 = 0 + 0.000 000 000 586 575 709 523 148 8;
  • 37) 0.000 000 000 586 575 709 523 148 8 × 2 = 0 + 0.000 000 001 173 151 419 046 297 6;
  • 38) 0.000 000 001 173 151 419 046 297 6 × 2 = 0 + 0.000 000 002 346 302 838 092 595 2;
  • 39) 0.000 000 002 346 302 838 092 595 2 × 2 = 0 + 0.000 000 004 692 605 676 185 190 4;
  • 40) 0.000 000 004 692 605 676 185 190 4 × 2 = 0 + 0.000 000 009 385 211 352 370 380 8;
  • 41) 0.000 000 009 385 211 352 370 380 8 × 2 = 0 + 0.000 000 018 770 422 704 740 761 6;
  • 42) 0.000 000 018 770 422 704 740 761 6 × 2 = 0 + 0.000 000 037 540 845 409 481 523 2;
  • 43) 0.000 000 037 540 845 409 481 523 2 × 2 = 0 + 0.000 000 075 081 690 818 963 046 4;
  • 44) 0.000 000 075 081 690 818 963 046 4 × 2 = 0 + 0.000 000 150 163 381 637 926 092 8;
  • 45) 0.000 000 150 163 381 637 926 092 8 × 2 = 0 + 0.000 000 300 326 763 275 852 185 6;
  • 46) 0.000 000 300 326 763 275 852 185 6 × 2 = 0 + 0.000 000 600 653 526 551 704 371 2;
  • 47) 0.000 000 600 653 526 551 704 371 2 × 2 = 0 + 0.000 001 201 307 053 103 408 742 4;
  • 48) 0.000 001 201 307 053 103 408 742 4 × 2 = 0 + 0.000 002 402 614 106 206 817 484 8;
  • 49) 0.000 002 402 614 106 206 817 484 8 × 2 = 0 + 0.000 004 805 228 212 413 634 969 6;
  • 50) 0.000 004 805 228 212 413 634 969 6 × 2 = 0 + 0.000 009 610 456 424 827 269 939 2;
  • 51) 0.000 009 610 456 424 827 269 939 2 × 2 = 0 + 0.000 019 220 912 849 654 539 878 4;
  • 52) 0.000 019 220 912 849 654 539 878 4 × 2 = 0 + 0.000 038 441 825 699 309 079 756 8;
  • 53) 0.000 038 441 825 699 309 079 756 8 × 2 = 0 + 0.000 076 883 651 398 618 159 513 6;
  • 54) 0.000 076 883 651 398 618 159 513 6 × 2 = 0 + 0.000 153 767 302 797 236 319 027 2;
  • 55) 0.000 153 767 302 797 236 319 027 2 × 2 = 0 + 0.000 307 534 605 594 472 638 054 4;
  • 56) 0.000 307 534 605 594 472 638 054 4 × 2 = 0 + 0.000 615 069 211 188 945 276 108 8;
  • 57) 0.000 615 069 211 188 945 276 108 8 × 2 = 0 + 0.001 230 138 422 377 890 552 217 6;
  • 58) 0.001 230 138 422 377 890 552 217 6 × 2 = 0 + 0.002 460 276 844 755 781 104 435 2;
  • 59) 0.002 460 276 844 755 781 104 435 2 × 2 = 0 + 0.004 920 553 689 511 562 208 870 4;
  • 60) 0.004 920 553 689 511 562 208 870 4 × 2 = 0 + 0.009 841 107 379 023 124 417 740 8;
  • 61) 0.009 841 107 379 023 124 417 740 8 × 2 = 0 + 0.019 682 214 758 046 248 835 481 6;
  • 62) 0.019 682 214 758 046 248 835 481 6 × 2 = 0 + 0.039 364 429 516 092 497 670 963 2;
  • 63) 0.039 364 429 516 092 497 670 963 2 × 2 = 0 + 0.078 728 859 032 184 995 341 926 4;
  • 64) 0.078 728 859 032 184 995 341 926 4 × 2 = 0 + 0.157 457 718 064 369 990 683 852 8;
  • 65) 0.157 457 718 064 369 990 683 852 8 × 2 = 0 + 0.314 915 436 128 739 981 367 705 6;
  • 66) 0.314 915 436 128 739 981 367 705 6 × 2 = 0 + 0.629 830 872 257 479 962 735 411 2;
  • 67) 0.629 830 872 257 479 962 735 411 2 × 2 = 1 + 0.259 661 744 514 959 925 470 822 4;
  • 68) 0.259 661 744 514 959 925 470 822 4 × 2 = 0 + 0.519 323 489 029 919 850 941 644 8;
  • 69) 0.519 323 489 029 919 850 941 644 8 × 2 = 1 + 0.038 646 978 059 839 701 883 289 6;
  • 70) 0.038 646 978 059 839 701 883 289 6 × 2 = 0 + 0.077 293 956 119 679 403 766 579 2;
  • 71) 0.077 293 956 119 679 403 766 579 2 × 2 = 0 + 0.154 587 912 239 358 807 533 158 4;
  • 72) 0.154 587 912 239 358 807 533 158 4 × 2 = 0 + 0.309 175 824 478 717 615 066 316 8;
  • 73) 0.309 175 824 478 717 615 066 316 8 × 2 = 0 + 0.618 351 648 957 435 230 132 633 6;
  • 74) 0.618 351 648 957 435 230 132 633 6 × 2 = 1 + 0.236 703 297 914 870 460 265 267 2;
  • 75) 0.236 703 297 914 870 460 265 267 2 × 2 = 0 + 0.473 406 595 829 740 920 530 534 4;
  • 76) 0.473 406 595 829 740 920 530 534 4 × 2 = 0 + 0.946 813 191 659 481 841 061 068 8;
  • 77) 0.946 813 191 659 481 841 061 068 8 × 2 = 1 + 0.893 626 383 318 963 682 122 137 6;
  • 78) 0.893 626 383 318 963 682 122 137 6 × 2 = 1 + 0.787 252 766 637 927 364 244 275 2;
  • 79) 0.787 252 766 637 927 364 244 275 2 × 2 = 1 + 0.574 505 533 275 854 728 488 550 4;
  • 80) 0.574 505 533 275 854 728 488 550 4 × 2 = 1 + 0.149 011 066 551 709 456 977 100 8;
  • 81) 0.149 011 066 551 709 456 977 100 8 × 2 = 0 + 0.298 022 133 103 418 913 954 201 6;
  • 82) 0.298 022 133 103 418 913 954 201 6 × 2 = 0 + 0.596 044 266 206 837 827 908 403 2;
  • 83) 0.596 044 266 206 837 827 908 403 2 × 2 = 1 + 0.192 088 532 413 675 655 816 806 4;
  • 84) 0.192 088 532 413 675 655 816 806 4 × 2 = 0 + 0.384 177 064 827 351 311 633 612 8;
  • 85) 0.384 177 064 827 351 311 633 612 8 × 2 = 0 + 0.768 354 129 654 702 623 267 225 6;
  • 86) 0.768 354 129 654 702 623 267 225 6 × 2 = 1 + 0.536 708 259 309 405 246 534 451 2;
  • 87) 0.536 708 259 309 405 246 534 451 2 × 2 = 1 + 0.073 416 518 618 810 493 068 902 4;
  • 88) 0.073 416 518 618 810 493 068 902 4 × 2 = 0 + 0.146 833 037 237 620 986 137 804 8;
  • 89) 0.146 833 037 237 620 986 137 804 8 × 2 = 0 + 0.293 666 074 475 241 972 275 609 6;
  • 90) 0.293 666 074 475 241 972 275 609 6 × 2 = 0 + 0.587 332 148 950 483 944 551 219 2;
  • 91) 0.587 332 148 950 483 944 551 219 2 × 2 = 1 + 0.174 664 297 900 967 889 102 438 4;
  • 92) 0.174 664 297 900 967 889 102 438 4 × 2 = 0 + 0.349 328 595 801 935 778 204 876 8;
  • 93) 0.349 328 595 801 935 778 204 876 8 × 2 = 0 + 0.698 657 191 603 871 556 409 753 6;
  • 94) 0.698 657 191 603 871 556 409 753 6 × 2 = 1 + 0.397 314 383 207 743 112 819 507 2;
  • 95) 0.397 314 383 207 743 112 819 507 2 × 2 = 0 + 0.794 628 766 415 486 225 639 014 4;
  • 96) 0.794 628 766 415 486 225 639 014 4 × 2 = 1 + 0.589 257 532 830 972 451 278 028 8;
  • 97) 0.589 257 532 830 972 451 278 028 8 × 2 = 1 + 0.178 515 065 661 944 902 556 057 6;
  • 98) 0.178 515 065 661 944 902 556 057 6 × 2 = 0 + 0.357 030 131 323 889 805 112 115 2;
  • 99) 0.357 030 131 323 889 805 112 115 2 × 2 = 0 + 0.714 060 262 647 779 610 224 230 4;
  • 100) 0.714 060 262 647 779 610 224 230 4 × 2 = 1 + 0.428 120 525 295 559 220 448 460 8;
  • 101) 0.428 120 525 295 559 220 448 460 8 × 2 = 0 + 0.856 241 050 591 118 440 896 921 6;
  • 102) 0.856 241 050 591 118 440 896 921 6 × 2 = 1 + 0.712 482 101 182 236 881 793 843 2;
  • 103) 0.712 482 101 182 236 881 793 843 2 × 2 = 1 + 0.424 964 202 364 473 763 587 686 4;
  • 104) 0.424 964 202 364 473 763 587 686 4 × 2 = 0 + 0.849 928 404 728 947 527 175 372 8;
  • 105) 0.849 928 404 728 947 527 175 372 8 × 2 = 1 + 0.699 856 809 457 895 054 350 745 6;
  • 106) 0.699 856 809 457 895 054 350 745 6 × 2 = 1 + 0.399 713 618 915 790 108 701 491 2;
  • 107) 0.399 713 618 915 790 108 701 491 2 × 2 = 0 + 0.799 427 237 831 580 217 402 982 4;
  • 108) 0.799 427 237 831 580 217 402 982 4 × 2 = 1 + 0.598 854 475 663 160 434 805 964 8;
  • 109) 0.598 854 475 663 160 434 805 964 8 × 2 = 1 + 0.197 708 951 326 320 869 611 929 6;
  • 110) 0.197 708 951 326 320 869 611 929 6 × 2 = 0 + 0.395 417 902 652 641 739 223 859 2;
  • 111) 0.395 417 902 652 641 739 223 859 2 × 2 = 0 + 0.790 835 805 305 283 478 447 718 4;
  • 112) 0.790 835 805 305 283 478 447 718 4 × 2 = 1 + 0.581 671 610 610 566 956 895 436 8;
  • 113) 0.581 671 610 610 566 956 895 436 8 × 2 = 1 + 0.163 343 221 221 133 913 790 873 6;
  • 114) 0.163 343 221 221 133 913 790 873 6 × 2 = 0 + 0.326 686 442 442 267 827 581 747 2;
  • 115) 0.326 686 442 442 267 827 581 747 2 × 2 = 0 + 0.653 372 884 884 535 655 163 494 4;
  • 116) 0.653 372 884 884 535 655 163 494 4 × 2 = 1 + 0.306 745 769 769 071 310 326 988 8;
  • 117) 0.306 745 769 769 071 310 326 988 8 × 2 = 0 + 0.613 491 539 538 142 620 653 977 6;
  • 118) 0.613 491 539 538 142 620 653 977 6 × 2 = 1 + 0.226 983 079 076 285 241 307 955 2;
  • 119) 0.226 983 079 076 285 241 307 955 2 × 2 = 0 + 0.453 966 158 152 570 482 615 910 4;

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:


0.000 000 000 000 000 000 008 535 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1111 0010 0110 0010 0101 1001 0110 1101 1001 1001 010(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 535 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1111 0010 0110 0010 0101 1001 0110 1101 1001 1001 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 67 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 008 535 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1111 0010 0110 0010 0101 1001 0110 1101 1001 1001 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1111 0010 0110 0010 0101 1001 0110 1101 1001 1001 010(2) × 20 =

1.0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010(2) × 2-67


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): -67

Mantissa (not normalized):
1.0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-67 + 2(11-1) - 1 =

(-67 + 1 023)(10) =

956(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;
  • 956 ÷ 2 = 478 + 0;
  • 478 ÷ 2 = 239 + 0;
  • 239 ÷ 2 = 119 + 1;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 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) =

956(10) =

011 1011 1100(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. 0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010 =

0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010


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 1011 1100

Mantissa (52 bits) =
0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010


Decimal number 0.000 000 000 000 000 000 008 535 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0010 0111 1001 0011 0001 0010 1100 1011 0110 1100 1100 1010

Share

» Convert decimal 0.000 000 000 000 000 000 008 545 4 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. NotebookLM YouTube Video of 7.54 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