0.000 000 000 000 000 000 008 535 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 535 2(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 2(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 2.

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 2 × 2 = 0 + 0.000 000 000 000 000 000 017 070 4;
  • 2) 0.000 000 000 000 000 000 017 070 4 × 2 = 0 + 0.000 000 000 000 000 000 034 140 8;
  • 3) 0.000 000 000 000 000 000 034 140 8 × 2 = 0 + 0.000 000 000 000 000 000 068 281 6;
  • 4) 0.000 000 000 000 000 000 068 281 6 × 2 = 0 + 0.000 000 000 000 000 000 136 563 2;
  • 5) 0.000 000 000 000 000 000 136 563 2 × 2 = 0 + 0.000 000 000 000 000 000 273 126 4;
  • 6) 0.000 000 000 000 000 000 273 126 4 × 2 = 0 + 0.000 000 000 000 000 000 546 252 8;
  • 7) 0.000 000 000 000 000 000 546 252 8 × 2 = 0 + 0.000 000 000 000 000 001 092 505 6;
  • 8) 0.000 000 000 000 000 001 092 505 6 × 2 = 0 + 0.000 000 000 000 000 002 185 011 2;
  • 9) 0.000 000 000 000 000 002 185 011 2 × 2 = 0 + 0.000 000 000 000 000 004 370 022 4;
  • 10) 0.000 000 000 000 000 004 370 022 4 × 2 = 0 + 0.000 000 000 000 000 008 740 044 8;
  • 11) 0.000 000 000 000 000 008 740 044 8 × 2 = 0 + 0.000 000 000 000 000 017 480 089 6;
  • 12) 0.000 000 000 000 000 017 480 089 6 × 2 = 0 + 0.000 000 000 000 000 034 960 179 2;
  • 13) 0.000 000 000 000 000 034 960 179 2 × 2 = 0 + 0.000 000 000 000 000 069 920 358 4;
  • 14) 0.000 000 000 000 000 069 920 358 4 × 2 = 0 + 0.000 000 000 000 000 139 840 716 8;
  • 15) 0.000 000 000 000 000 139 840 716 8 × 2 = 0 + 0.000 000 000 000 000 279 681 433 6;
  • 16) 0.000 000 000 000 000 279 681 433 6 × 2 = 0 + 0.000 000 000 000 000 559 362 867 2;
  • 17) 0.000 000 000 000 000 559 362 867 2 × 2 = 0 + 0.000 000 000 000 001 118 725 734 4;
  • 18) 0.000 000 000 000 001 118 725 734 4 × 2 = 0 + 0.000 000 000 000 002 237 451 468 8;
  • 19) 0.000 000 000 000 002 237 451 468 8 × 2 = 0 + 0.000 000 000 000 004 474 902 937 6;
  • 20) 0.000 000 000 000 004 474 902 937 6 × 2 = 0 + 0.000 000 000 000 008 949 805 875 2;
  • 21) 0.000 000 000 000 008 949 805 875 2 × 2 = 0 + 0.000 000 000 000 017 899 611 750 4;
  • 22) 0.000 000 000 000 017 899 611 750 4 × 2 = 0 + 0.000 000 000 000 035 799 223 500 8;
  • 23) 0.000 000 000 000 035 799 223 500 8 × 2 = 0 + 0.000 000 000 000 071 598 447 001 6;
  • 24) 0.000 000 000 000 071 598 447 001 6 × 2 = 0 + 0.000 000 000 000 143 196 894 003 2;
  • 25) 0.000 000 000 000 143 196 894 003 2 × 2 = 0 + 0.000 000 000 000 286 393 788 006 4;
  • 26) 0.000 000 000 000 286 393 788 006 4 × 2 = 0 + 0.000 000 000 000 572 787 576 012 8;
  • 27) 0.000 000 000 000 572 787 576 012 8 × 2 = 0 + 0.000 000 000 001 145 575 152 025 6;
  • 28) 0.000 000 000 001 145 575 152 025 6 × 2 = 0 + 0.000 000 000 002 291 150 304 051 2;
  • 29) 0.000 000 000 002 291 150 304 051 2 × 2 = 0 + 0.000 000 000 004 582 300 608 102 4;
  • 30) 0.000 000 000 004 582 300 608 102 4 × 2 = 0 + 0.000 000 000 009 164 601 216 204 8;
  • 31) 0.000 000 000 009 164 601 216 204 8 × 2 = 0 + 0.000 000 000 018 329 202 432 409 6;
  • 32) 0.000 000 000 018 329 202 432 409 6 × 2 = 0 + 0.000 000 000 036 658 404 864 819 2;
  • 33) 0.000 000 000 036 658 404 864 819 2 × 2 = 0 + 0.000 000 000 073 316 809 729 638 4;
  • 34) 0.000 000 000 073 316 809 729 638 4 × 2 = 0 + 0.000 000 000 146 633 619 459 276 8;
  • 35) 0.000 000 000 146 633 619 459 276 8 × 2 = 0 + 0.000 000 000 293 267 238 918 553 6;
  • 36) 0.000 000 000 293 267 238 918 553 6 × 2 = 0 + 0.000 000 000 586 534 477 837 107 2;
  • 37) 0.000 000 000 586 534 477 837 107 2 × 2 = 0 + 0.000 000 001 173 068 955 674 214 4;
  • 38) 0.000 000 001 173 068 955 674 214 4 × 2 = 0 + 0.000 000 002 346 137 911 348 428 8;
  • 39) 0.000 000 002 346 137 911 348 428 8 × 2 = 0 + 0.000 000 004 692 275 822 696 857 6;
  • 40) 0.000 000 004 692 275 822 696 857 6 × 2 = 0 + 0.000 000 009 384 551 645 393 715 2;
  • 41) 0.000 000 009 384 551 645 393 715 2 × 2 = 0 + 0.000 000 018 769 103 290 787 430 4;
  • 42) 0.000 000 018 769 103 290 787 430 4 × 2 = 0 + 0.000 000 037 538 206 581 574 860 8;
  • 43) 0.000 000 037 538 206 581 574 860 8 × 2 = 0 + 0.000 000 075 076 413 163 149 721 6;
  • 44) 0.000 000 075 076 413 163 149 721 6 × 2 = 0 + 0.000 000 150 152 826 326 299 443 2;
  • 45) 0.000 000 150 152 826 326 299 443 2 × 2 = 0 + 0.000 000 300 305 652 652 598 886 4;
  • 46) 0.000 000 300 305 652 652 598 886 4 × 2 = 0 + 0.000 000 600 611 305 305 197 772 8;
  • 47) 0.000 000 600 611 305 305 197 772 8 × 2 = 0 + 0.000 001 201 222 610 610 395 545 6;
  • 48) 0.000 001 201 222 610 610 395 545 6 × 2 = 0 + 0.000 002 402 445 221 220 791 091 2;
  • 49) 0.000 002 402 445 221 220 791 091 2 × 2 = 0 + 0.000 004 804 890 442 441 582 182 4;
  • 50) 0.000 004 804 890 442 441 582 182 4 × 2 = 0 + 0.000 009 609 780 884 883 164 364 8;
  • 51) 0.000 009 609 780 884 883 164 364 8 × 2 = 0 + 0.000 019 219 561 769 766 328 729 6;
  • 52) 0.000 019 219 561 769 766 328 729 6 × 2 = 0 + 0.000 038 439 123 539 532 657 459 2;
  • 53) 0.000 038 439 123 539 532 657 459 2 × 2 = 0 + 0.000 076 878 247 079 065 314 918 4;
  • 54) 0.000 076 878 247 079 065 314 918 4 × 2 = 0 + 0.000 153 756 494 158 130 629 836 8;
  • 55) 0.000 153 756 494 158 130 629 836 8 × 2 = 0 + 0.000 307 512 988 316 261 259 673 6;
  • 56) 0.000 307 512 988 316 261 259 673 6 × 2 = 0 + 0.000 615 025 976 632 522 519 347 2;
  • 57) 0.000 615 025 976 632 522 519 347 2 × 2 = 0 + 0.001 230 051 953 265 045 038 694 4;
  • 58) 0.001 230 051 953 265 045 038 694 4 × 2 = 0 + 0.002 460 103 906 530 090 077 388 8;
  • 59) 0.002 460 103 906 530 090 077 388 8 × 2 = 0 + 0.004 920 207 813 060 180 154 777 6;
  • 60) 0.004 920 207 813 060 180 154 777 6 × 2 = 0 + 0.009 840 415 626 120 360 309 555 2;
  • 61) 0.009 840 415 626 120 360 309 555 2 × 2 = 0 + 0.019 680 831 252 240 720 619 110 4;
  • 62) 0.019 680 831 252 240 720 619 110 4 × 2 = 0 + 0.039 361 662 504 481 441 238 220 8;
  • 63) 0.039 361 662 504 481 441 238 220 8 × 2 = 0 + 0.078 723 325 008 962 882 476 441 6;
  • 64) 0.078 723 325 008 962 882 476 441 6 × 2 = 0 + 0.157 446 650 017 925 764 952 883 2;
  • 65) 0.157 446 650 017 925 764 952 883 2 × 2 = 0 + 0.314 893 300 035 851 529 905 766 4;
  • 66) 0.314 893 300 035 851 529 905 766 4 × 2 = 0 + 0.629 786 600 071 703 059 811 532 8;
  • 67) 0.629 786 600 071 703 059 811 532 8 × 2 = 1 + 0.259 573 200 143 406 119 623 065 6;
  • 68) 0.259 573 200 143 406 119 623 065 6 × 2 = 0 + 0.519 146 400 286 812 239 246 131 2;
  • 69) 0.519 146 400 286 812 239 246 131 2 × 2 = 1 + 0.038 292 800 573 624 478 492 262 4;
  • 70) 0.038 292 800 573 624 478 492 262 4 × 2 = 0 + 0.076 585 601 147 248 956 984 524 8;
  • 71) 0.076 585 601 147 248 956 984 524 8 × 2 = 0 + 0.153 171 202 294 497 913 969 049 6;
  • 72) 0.153 171 202 294 497 913 969 049 6 × 2 = 0 + 0.306 342 404 588 995 827 938 099 2;
  • 73) 0.306 342 404 588 995 827 938 099 2 × 2 = 0 + 0.612 684 809 177 991 655 876 198 4;
  • 74) 0.612 684 809 177 991 655 876 198 4 × 2 = 1 + 0.225 369 618 355 983 311 752 396 8;
  • 75) 0.225 369 618 355 983 311 752 396 8 × 2 = 0 + 0.450 739 236 711 966 623 504 793 6;
  • 76) 0.450 739 236 711 966 623 504 793 6 × 2 = 0 + 0.901 478 473 423 933 247 009 587 2;
  • 77) 0.901 478 473 423 933 247 009 587 2 × 2 = 1 + 0.802 956 946 847 866 494 019 174 4;
  • 78) 0.802 956 946 847 866 494 019 174 4 × 2 = 1 + 0.605 913 893 695 732 988 038 348 8;
  • 79) 0.605 913 893 695 732 988 038 348 8 × 2 = 1 + 0.211 827 787 391 465 976 076 697 6;
  • 80) 0.211 827 787 391 465 976 076 697 6 × 2 = 0 + 0.423 655 574 782 931 952 153 395 2;
  • 81) 0.423 655 574 782 931 952 153 395 2 × 2 = 0 + 0.847 311 149 565 863 904 306 790 4;
  • 82) 0.847 311 149 565 863 904 306 790 4 × 2 = 1 + 0.694 622 299 131 727 808 613 580 8;
  • 83) 0.694 622 299 131 727 808 613 580 8 × 2 = 1 + 0.389 244 598 263 455 617 227 161 6;
  • 84) 0.389 244 598 263 455 617 227 161 6 × 2 = 0 + 0.778 489 196 526 911 234 454 323 2;
  • 85) 0.778 489 196 526 911 234 454 323 2 × 2 = 1 + 0.556 978 393 053 822 468 908 646 4;
  • 86) 0.556 978 393 053 822 468 908 646 4 × 2 = 1 + 0.113 956 786 107 644 937 817 292 8;
  • 87) 0.113 956 786 107 644 937 817 292 8 × 2 = 0 + 0.227 913 572 215 289 875 634 585 6;
  • 88) 0.227 913 572 215 289 875 634 585 6 × 2 = 0 + 0.455 827 144 430 579 751 269 171 2;
  • 89) 0.455 827 144 430 579 751 269 171 2 × 2 = 0 + 0.911 654 288 861 159 502 538 342 4;
  • 90) 0.911 654 288 861 159 502 538 342 4 × 2 = 1 + 0.823 308 577 722 319 005 076 684 8;
  • 91) 0.823 308 577 722 319 005 076 684 8 × 2 = 1 + 0.646 617 155 444 638 010 153 369 6;
  • 92) 0.646 617 155 444 638 010 153 369 6 × 2 = 1 + 0.293 234 310 889 276 020 306 739 2;
  • 93) 0.293 234 310 889 276 020 306 739 2 × 2 = 0 + 0.586 468 621 778 552 040 613 478 4;
  • 94) 0.586 468 621 778 552 040 613 478 4 × 2 = 1 + 0.172 937 243 557 104 081 226 956 8;
  • 95) 0.172 937 243 557 104 081 226 956 8 × 2 = 0 + 0.345 874 487 114 208 162 453 913 6;
  • 96) 0.345 874 487 114 208 162 453 913 6 × 2 = 0 + 0.691 748 974 228 416 324 907 827 2;
  • 97) 0.691 748 974 228 416 324 907 827 2 × 2 = 1 + 0.383 497 948 456 832 649 815 654 4;
  • 98) 0.383 497 948 456 832 649 815 654 4 × 2 = 0 + 0.766 995 896 913 665 299 631 308 8;
  • 99) 0.766 995 896 913 665 299 631 308 8 × 2 = 1 + 0.533 991 793 827 330 599 262 617 6;
  • 100) 0.533 991 793 827 330 599 262 617 6 × 2 = 1 + 0.067 983 587 654 661 198 525 235 2;
  • 101) 0.067 983 587 654 661 198 525 235 2 × 2 = 0 + 0.135 967 175 309 322 397 050 470 4;
  • 102) 0.135 967 175 309 322 397 050 470 4 × 2 = 0 + 0.271 934 350 618 644 794 100 940 8;
  • 103) 0.271 934 350 618 644 794 100 940 8 × 2 = 0 + 0.543 868 701 237 289 588 201 881 6;
  • 104) 0.543 868 701 237 289 588 201 881 6 × 2 = 1 + 0.087 737 402 474 579 176 403 763 2;
  • 105) 0.087 737 402 474 579 176 403 763 2 × 2 = 0 + 0.175 474 804 949 158 352 807 526 4;
  • 106) 0.175 474 804 949 158 352 807 526 4 × 2 = 0 + 0.350 949 609 898 316 705 615 052 8;
  • 107) 0.350 949 609 898 316 705 615 052 8 × 2 = 0 + 0.701 899 219 796 633 411 230 105 6;
  • 108) 0.701 899 219 796 633 411 230 105 6 × 2 = 1 + 0.403 798 439 593 266 822 460 211 2;
  • 109) 0.403 798 439 593 266 822 460 211 2 × 2 = 0 + 0.807 596 879 186 533 644 920 422 4;
  • 110) 0.807 596 879 186 533 644 920 422 4 × 2 = 1 + 0.615 193 758 373 067 289 840 844 8;
  • 111) 0.615 193 758 373 067 289 840 844 8 × 2 = 1 + 0.230 387 516 746 134 579 681 689 6;
  • 112) 0.230 387 516 746 134 579 681 689 6 × 2 = 0 + 0.460 775 033 492 269 159 363 379 2;
  • 113) 0.460 775 033 492 269 159 363 379 2 × 2 = 0 + 0.921 550 066 984 538 318 726 758 4;
  • 114) 0.921 550 066 984 538 318 726 758 4 × 2 = 1 + 0.843 100 133 969 076 637 453 516 8;
  • 115) 0.843 100 133 969 076 637 453 516 8 × 2 = 1 + 0.686 200 267 938 153 274 907 033 6;
  • 116) 0.686 200 267 938 153 274 907 033 6 × 2 = 1 + 0.372 400 535 876 306 549 814 067 2;
  • 117) 0.372 400 535 876 306 549 814 067 2 × 2 = 0 + 0.744 801 071 752 613 099 628 134 4;
  • 118) 0.744 801 071 752 613 099 628 134 4 × 2 = 1 + 0.489 602 143 505 226 199 256 268 8;
  • 119) 0.489 602 143 505 226 199 256 268 8 × 2 = 0 + 0.979 204 287 010 452 398 512 537 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:


0.000 000 000 000 000 000 008 535 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1110 0110 1100 0111 0100 1011 0001 0001 0110 0111 010(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 535 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1110 0110 1100 0111 0100 1011 0001 0001 0110 0111 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 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1110 0110 1100 0111 0100 1011 0001 0001 0110 0111 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1110 0110 1100 0111 0100 1011 0001 0001 0110 0111 010(2) × 20 =

1.0100 0010 0111 0011 0110 0011 1010 0101 1000 1000 1011 0011 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 0011 0110 0011 1010 0101 1000 1000 1011 0011 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 0011 0110 0011 1010 0101 1000 1000 1011 0011 1010 =

0100 0010 0111 0011 0110 0011 1010 0101 1000 1000 1011 0011 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 0011 0110 0011 1010 0101 1000 1000 1011 0011 1010


Decimal number 0.000 000 000 000 000 000 008 535 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0010 0111 0011 0110 0011 1010 0101 1000 1000 1011 0011 1010

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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