0.000 000 000 000 000 000 008 522 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 522 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
0.000 000 000 000 000 000 008 522 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: 0.000 000 000 000 000 000 008 522 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) 0.000 000 000 000 000 000 008 522 1 × 2 = 0 + 0.000 000 000 000 000 000 017 044 2;
  • 2) 0.000 000 000 000 000 000 017 044 2 × 2 = 0 + 0.000 000 000 000 000 000 034 088 4;
  • 3) 0.000 000 000 000 000 000 034 088 4 × 2 = 0 + 0.000 000 000 000 000 000 068 176 8;
  • 4) 0.000 000 000 000 000 000 068 176 8 × 2 = 0 + 0.000 000 000 000 000 000 136 353 6;
  • 5) 0.000 000 000 000 000 000 136 353 6 × 2 = 0 + 0.000 000 000 000 000 000 272 707 2;
  • 6) 0.000 000 000 000 000 000 272 707 2 × 2 = 0 + 0.000 000 000 000 000 000 545 414 4;
  • 7) 0.000 000 000 000 000 000 545 414 4 × 2 = 0 + 0.000 000 000 000 000 001 090 828 8;
  • 8) 0.000 000 000 000 000 001 090 828 8 × 2 = 0 + 0.000 000 000 000 000 002 181 657 6;
  • 9) 0.000 000 000 000 000 002 181 657 6 × 2 = 0 + 0.000 000 000 000 000 004 363 315 2;
  • 10) 0.000 000 000 000 000 004 363 315 2 × 2 = 0 + 0.000 000 000 000 000 008 726 630 4;
  • 11) 0.000 000 000 000 000 008 726 630 4 × 2 = 0 + 0.000 000 000 000 000 017 453 260 8;
  • 12) 0.000 000 000 000 000 017 453 260 8 × 2 = 0 + 0.000 000 000 000 000 034 906 521 6;
  • 13) 0.000 000 000 000 000 034 906 521 6 × 2 = 0 + 0.000 000 000 000 000 069 813 043 2;
  • 14) 0.000 000 000 000 000 069 813 043 2 × 2 = 0 + 0.000 000 000 000 000 139 626 086 4;
  • 15) 0.000 000 000 000 000 139 626 086 4 × 2 = 0 + 0.000 000 000 000 000 279 252 172 8;
  • 16) 0.000 000 000 000 000 279 252 172 8 × 2 = 0 + 0.000 000 000 000 000 558 504 345 6;
  • 17) 0.000 000 000 000 000 558 504 345 6 × 2 = 0 + 0.000 000 000 000 001 117 008 691 2;
  • 18) 0.000 000 000 000 001 117 008 691 2 × 2 = 0 + 0.000 000 000 000 002 234 017 382 4;
  • 19) 0.000 000 000 000 002 234 017 382 4 × 2 = 0 + 0.000 000 000 000 004 468 034 764 8;
  • 20) 0.000 000 000 000 004 468 034 764 8 × 2 = 0 + 0.000 000 000 000 008 936 069 529 6;
  • 21) 0.000 000 000 000 008 936 069 529 6 × 2 = 0 + 0.000 000 000 000 017 872 139 059 2;
  • 22) 0.000 000 000 000 017 872 139 059 2 × 2 = 0 + 0.000 000 000 000 035 744 278 118 4;
  • 23) 0.000 000 000 000 035 744 278 118 4 × 2 = 0 + 0.000 000 000 000 071 488 556 236 8;
  • 24) 0.000 000 000 000 071 488 556 236 8 × 2 = 0 + 0.000 000 000 000 142 977 112 473 6;
  • 25) 0.000 000 000 000 142 977 112 473 6 × 2 = 0 + 0.000 000 000 000 285 954 224 947 2;
  • 26) 0.000 000 000 000 285 954 224 947 2 × 2 = 0 + 0.000 000 000 000 571 908 449 894 4;
  • 27) 0.000 000 000 000 571 908 449 894 4 × 2 = 0 + 0.000 000 000 001 143 816 899 788 8;
  • 28) 0.000 000 000 001 143 816 899 788 8 × 2 = 0 + 0.000 000 000 002 287 633 799 577 6;
  • 29) 0.000 000 000 002 287 633 799 577 6 × 2 = 0 + 0.000 000 000 004 575 267 599 155 2;
  • 30) 0.000 000 000 004 575 267 599 155 2 × 2 = 0 + 0.000 000 000 009 150 535 198 310 4;
  • 31) 0.000 000 000 009 150 535 198 310 4 × 2 = 0 + 0.000 000 000 018 301 070 396 620 8;
  • 32) 0.000 000 000 018 301 070 396 620 8 × 2 = 0 + 0.000 000 000 036 602 140 793 241 6;
  • 33) 0.000 000 000 036 602 140 793 241 6 × 2 = 0 + 0.000 000 000 073 204 281 586 483 2;
  • 34) 0.000 000 000 073 204 281 586 483 2 × 2 = 0 + 0.000 000 000 146 408 563 172 966 4;
  • 35) 0.000 000 000 146 408 563 172 966 4 × 2 = 0 + 0.000 000 000 292 817 126 345 932 8;
  • 36) 0.000 000 000 292 817 126 345 932 8 × 2 = 0 + 0.000 000 000 585 634 252 691 865 6;
  • 37) 0.000 000 000 585 634 252 691 865 6 × 2 = 0 + 0.000 000 001 171 268 505 383 731 2;
  • 38) 0.000 000 001 171 268 505 383 731 2 × 2 = 0 + 0.000 000 002 342 537 010 767 462 4;
  • 39) 0.000 000 002 342 537 010 767 462 4 × 2 = 0 + 0.000 000 004 685 074 021 534 924 8;
  • 40) 0.000 000 004 685 074 021 534 924 8 × 2 = 0 + 0.000 000 009 370 148 043 069 849 6;
  • 41) 0.000 000 009 370 148 043 069 849 6 × 2 = 0 + 0.000 000 018 740 296 086 139 699 2;
  • 42) 0.000 000 018 740 296 086 139 699 2 × 2 = 0 + 0.000 000 037 480 592 172 279 398 4;
  • 43) 0.000 000 037 480 592 172 279 398 4 × 2 = 0 + 0.000 000 074 961 184 344 558 796 8;
  • 44) 0.000 000 074 961 184 344 558 796 8 × 2 = 0 + 0.000 000 149 922 368 689 117 593 6;
  • 45) 0.000 000 149 922 368 689 117 593 6 × 2 = 0 + 0.000 000 299 844 737 378 235 187 2;
  • 46) 0.000 000 299 844 737 378 235 187 2 × 2 = 0 + 0.000 000 599 689 474 756 470 374 4;
  • 47) 0.000 000 599 689 474 756 470 374 4 × 2 = 0 + 0.000 001 199 378 949 512 940 748 8;
  • 48) 0.000 001 199 378 949 512 940 748 8 × 2 = 0 + 0.000 002 398 757 899 025 881 497 6;
  • 49) 0.000 002 398 757 899 025 881 497 6 × 2 = 0 + 0.000 004 797 515 798 051 762 995 2;
  • 50) 0.000 004 797 515 798 051 762 995 2 × 2 = 0 + 0.000 009 595 031 596 103 525 990 4;
  • 51) 0.000 009 595 031 596 103 525 990 4 × 2 = 0 + 0.000 019 190 063 192 207 051 980 8;
  • 52) 0.000 019 190 063 192 207 051 980 8 × 2 = 0 + 0.000 038 380 126 384 414 103 961 6;
  • 53) 0.000 038 380 126 384 414 103 961 6 × 2 = 0 + 0.000 076 760 252 768 828 207 923 2;
  • 54) 0.000 076 760 252 768 828 207 923 2 × 2 = 0 + 0.000 153 520 505 537 656 415 846 4;
  • 55) 0.000 153 520 505 537 656 415 846 4 × 2 = 0 + 0.000 307 041 011 075 312 831 692 8;
  • 56) 0.000 307 041 011 075 312 831 692 8 × 2 = 0 + 0.000 614 082 022 150 625 663 385 6;
  • 57) 0.000 614 082 022 150 625 663 385 6 × 2 = 0 + 0.001 228 164 044 301 251 326 771 2;
  • 58) 0.001 228 164 044 301 251 326 771 2 × 2 = 0 + 0.002 456 328 088 602 502 653 542 4;
  • 59) 0.002 456 328 088 602 502 653 542 4 × 2 = 0 + 0.004 912 656 177 205 005 307 084 8;
  • 60) 0.004 912 656 177 205 005 307 084 8 × 2 = 0 + 0.009 825 312 354 410 010 614 169 6;
  • 61) 0.009 825 312 354 410 010 614 169 6 × 2 = 0 + 0.019 650 624 708 820 021 228 339 2;
  • 62) 0.019 650 624 708 820 021 228 339 2 × 2 = 0 + 0.039 301 249 417 640 042 456 678 4;
  • 63) 0.039 301 249 417 640 042 456 678 4 × 2 = 0 + 0.078 602 498 835 280 084 913 356 8;
  • 64) 0.078 602 498 835 280 084 913 356 8 × 2 = 0 + 0.157 204 997 670 560 169 826 713 6;
  • 65) 0.157 204 997 670 560 169 826 713 6 × 2 = 0 + 0.314 409 995 341 120 339 653 427 2;
  • 66) 0.314 409 995 341 120 339 653 427 2 × 2 = 0 + 0.628 819 990 682 240 679 306 854 4;
  • 67) 0.628 819 990 682 240 679 306 854 4 × 2 = 1 + 0.257 639 981 364 481 358 613 708 8;
  • 68) 0.257 639 981 364 481 358 613 708 8 × 2 = 0 + 0.515 279 962 728 962 717 227 417 6;
  • 69) 0.515 279 962 728 962 717 227 417 6 × 2 = 1 + 0.030 559 925 457 925 434 454 835 2;
  • 70) 0.030 559 925 457 925 434 454 835 2 × 2 = 0 + 0.061 119 850 915 850 868 909 670 4;
  • 71) 0.061 119 850 915 850 868 909 670 4 × 2 = 0 + 0.122 239 701 831 701 737 819 340 8;
  • 72) 0.122 239 701 831 701 737 819 340 8 × 2 = 0 + 0.244 479 403 663 403 475 638 681 6;
  • 73) 0.244 479 403 663 403 475 638 681 6 × 2 = 0 + 0.488 958 807 326 806 951 277 363 2;
  • 74) 0.488 958 807 326 806 951 277 363 2 × 2 = 0 + 0.977 917 614 653 613 902 554 726 4;
  • 75) 0.977 917 614 653 613 902 554 726 4 × 2 = 1 + 0.955 835 229 307 227 805 109 452 8;
  • 76) 0.955 835 229 307 227 805 109 452 8 × 2 = 1 + 0.911 670 458 614 455 610 218 905 6;
  • 77) 0.911 670 458 614 455 610 218 905 6 × 2 = 1 + 0.823 340 917 228 911 220 437 811 2;
  • 78) 0.823 340 917 228 911 220 437 811 2 × 2 = 1 + 0.646 681 834 457 822 440 875 622 4;
  • 79) 0.646 681 834 457 822 440 875 622 4 × 2 = 1 + 0.293 363 668 915 644 881 751 244 8;
  • 80) 0.293 363 668 915 644 881 751 244 8 × 2 = 0 + 0.586 727 337 831 289 763 502 489 6;
  • 81) 0.586 727 337 831 289 763 502 489 6 × 2 = 1 + 0.173 454 675 662 579 527 004 979 2;
  • 82) 0.173 454 675 662 579 527 004 979 2 × 2 = 0 + 0.346 909 351 325 159 054 009 958 4;
  • 83) 0.346 909 351 325 159 054 009 958 4 × 2 = 0 + 0.693 818 702 650 318 108 019 916 8;
  • 84) 0.693 818 702 650 318 108 019 916 8 × 2 = 1 + 0.387 637 405 300 636 216 039 833 6;
  • 85) 0.387 637 405 300 636 216 039 833 6 × 2 = 0 + 0.775 274 810 601 272 432 079 667 2;
  • 86) 0.775 274 810 601 272 432 079 667 2 × 2 = 1 + 0.550 549 621 202 544 864 159 334 4;
  • 87) 0.550 549 621 202 544 864 159 334 4 × 2 = 1 + 0.101 099 242 405 089 728 318 668 8;
  • 88) 0.101 099 242 405 089 728 318 668 8 × 2 = 0 + 0.202 198 484 810 179 456 637 337 6;
  • 89) 0.202 198 484 810 179 456 637 337 6 × 2 = 0 + 0.404 396 969 620 358 913 274 675 2;
  • 90) 0.404 396 969 620 358 913 274 675 2 × 2 = 0 + 0.808 793 939 240 717 826 549 350 4;
  • 91) 0.808 793 939 240 717 826 549 350 4 × 2 = 1 + 0.617 587 878 481 435 653 098 700 8;
  • 92) 0.617 587 878 481 435 653 098 700 8 × 2 = 1 + 0.235 175 756 962 871 306 197 401 6;
  • 93) 0.235 175 756 962 871 306 197 401 6 × 2 = 0 + 0.470 351 513 925 742 612 394 803 2;
  • 94) 0.470 351 513 925 742 612 394 803 2 × 2 = 0 + 0.940 703 027 851 485 224 789 606 4;
  • 95) 0.940 703 027 851 485 224 789 606 4 × 2 = 1 + 0.881 406 055 702 970 449 579 212 8;
  • 96) 0.881 406 055 702 970 449 579 212 8 × 2 = 1 + 0.762 812 111 405 940 899 158 425 6;
  • 97) 0.762 812 111 405 940 899 158 425 6 × 2 = 1 + 0.525 624 222 811 881 798 316 851 2;
  • 98) 0.525 624 222 811 881 798 316 851 2 × 2 = 1 + 0.051 248 445 623 763 596 633 702 4;
  • 99) 0.051 248 445 623 763 596 633 702 4 × 2 = 0 + 0.102 496 891 247 527 193 267 404 8;
  • 100) 0.102 496 891 247 527 193 267 404 8 × 2 = 0 + 0.204 993 782 495 054 386 534 809 6;
  • 101) 0.204 993 782 495 054 386 534 809 6 × 2 = 0 + 0.409 987 564 990 108 773 069 619 2;
  • 102) 0.409 987 564 990 108 773 069 619 2 × 2 = 0 + 0.819 975 129 980 217 546 139 238 4;
  • 103) 0.819 975 129 980 217 546 139 238 4 × 2 = 1 + 0.639 950 259 960 435 092 278 476 8;
  • 104) 0.639 950 259 960 435 092 278 476 8 × 2 = 1 + 0.279 900 519 920 870 184 556 953 6;
  • 105) 0.279 900 519 920 870 184 556 953 6 × 2 = 0 + 0.559 801 039 841 740 369 113 907 2;
  • 106) 0.559 801 039 841 740 369 113 907 2 × 2 = 1 + 0.119 602 079 683 480 738 227 814 4;
  • 107) 0.119 602 079 683 480 738 227 814 4 × 2 = 0 + 0.239 204 159 366 961 476 455 628 8;
  • 108) 0.239 204 159 366 961 476 455 628 8 × 2 = 0 + 0.478 408 318 733 922 952 911 257 6;
  • 109) 0.478 408 318 733 922 952 911 257 6 × 2 = 0 + 0.956 816 637 467 845 905 822 515 2;
  • 110) 0.956 816 637 467 845 905 822 515 2 × 2 = 1 + 0.913 633 274 935 691 811 645 030 4;
  • 111) 0.913 633 274 935 691 811 645 030 4 × 2 = 1 + 0.827 266 549 871 383 623 290 060 8;
  • 112) 0.827 266 549 871 383 623 290 060 8 × 2 = 1 + 0.654 533 099 742 767 246 580 121 6;
  • 113) 0.654 533 099 742 767 246 580 121 6 × 2 = 1 + 0.309 066 199 485 534 493 160 243 2;
  • 114) 0.309 066 199 485 534 493 160 243 2 × 2 = 0 + 0.618 132 398 971 068 986 320 486 4;
  • 115) 0.618 132 398 971 068 986 320 486 4 × 2 = 1 + 0.236 264 797 942 137 972 640 972 8;
  • 116) 0.236 264 797 942 137 972 640 972 8 × 2 = 0 + 0.472 529 595 884 275 945 281 945 6;
  • 117) 0.472 529 595 884 275 945 281 945 6 × 2 = 0 + 0.945 059 191 768 551 890 563 891 2;
  • 118) 0.945 059 191 768 551 890 563 891 2 × 2 = 1 + 0.890 118 383 537 103 781 127 782 4;
  • 119) 0.890 118 383 537 103 781 127 782 4 × 2 = 1 + 0.780 236 767 074 207 562 255 564 8;

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:


0.000 000 000 000 000 000 008 522 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1110 1001 0110 0011 0011 1100 0011 0100 0111 1010 011(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 522 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1110 1001 0110 0011 0011 1100 0011 0100 0111 1010 011(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:


0.000 000 000 000 000 000 008 522 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1110 1001 0110 0011 0011 1100 0011 0100 0111 1010 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1110 1001 0110 0011 0011 1100 0011 0100 0111 1010 011(2) × 20 =

1.0100 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011(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 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011


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 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011 =

0100 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011


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 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011


Decimal number 0.000 000 000 000 000 000 008 522 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0001 1111 0100 1011 0001 1001 1110 0001 1010 0011 1101 0011

Share

» Convert decimal 0.000 000 000 000 000 000 008 524 2 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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