-0.000 000 000 000 000 004 573 696 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 004 573 696(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
-0.000 000 000 000 000 004 573 696(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. Start with the positive version of the number:

|-0.000 000 000 000 000 004 573 696| = 0.000 000 000 000 000 004 573 696


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 004 573 696.

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) 0.000 000 000 000 000 004 573 696 × 2 = 0 + 0.000 000 000 000 000 009 147 392;
  • 2) 0.000 000 000 000 000 009 147 392 × 2 = 0 + 0.000 000 000 000 000 018 294 784;
  • 3) 0.000 000 000 000 000 018 294 784 × 2 = 0 + 0.000 000 000 000 000 036 589 568;
  • 4) 0.000 000 000 000 000 036 589 568 × 2 = 0 + 0.000 000 000 000 000 073 179 136;
  • 5) 0.000 000 000 000 000 073 179 136 × 2 = 0 + 0.000 000 000 000 000 146 358 272;
  • 6) 0.000 000 000 000 000 146 358 272 × 2 = 0 + 0.000 000 000 000 000 292 716 544;
  • 7) 0.000 000 000 000 000 292 716 544 × 2 = 0 + 0.000 000 000 000 000 585 433 088;
  • 8) 0.000 000 000 000 000 585 433 088 × 2 = 0 + 0.000 000 000 000 001 170 866 176;
  • 9) 0.000 000 000 000 001 170 866 176 × 2 = 0 + 0.000 000 000 000 002 341 732 352;
  • 10) 0.000 000 000 000 002 341 732 352 × 2 = 0 + 0.000 000 000 000 004 683 464 704;
  • 11) 0.000 000 000 000 004 683 464 704 × 2 = 0 + 0.000 000 000 000 009 366 929 408;
  • 12) 0.000 000 000 000 009 366 929 408 × 2 = 0 + 0.000 000 000 000 018 733 858 816;
  • 13) 0.000 000 000 000 018 733 858 816 × 2 = 0 + 0.000 000 000 000 037 467 717 632;
  • 14) 0.000 000 000 000 037 467 717 632 × 2 = 0 + 0.000 000 000 000 074 935 435 264;
  • 15) 0.000 000 000 000 074 935 435 264 × 2 = 0 + 0.000 000 000 000 149 870 870 528;
  • 16) 0.000 000 000 000 149 870 870 528 × 2 = 0 + 0.000 000 000 000 299 741 741 056;
  • 17) 0.000 000 000 000 299 741 741 056 × 2 = 0 + 0.000 000 000 000 599 483 482 112;
  • 18) 0.000 000 000 000 599 483 482 112 × 2 = 0 + 0.000 000 000 001 198 966 964 224;
  • 19) 0.000 000 000 001 198 966 964 224 × 2 = 0 + 0.000 000 000 002 397 933 928 448;
  • 20) 0.000 000 000 002 397 933 928 448 × 2 = 0 + 0.000 000 000 004 795 867 856 896;
  • 21) 0.000 000 000 004 795 867 856 896 × 2 = 0 + 0.000 000 000 009 591 735 713 792;
  • 22) 0.000 000 000 009 591 735 713 792 × 2 = 0 + 0.000 000 000 019 183 471 427 584;
  • 23) 0.000 000 000 019 183 471 427 584 × 2 = 0 + 0.000 000 000 038 366 942 855 168;
  • 24) 0.000 000 000 038 366 942 855 168 × 2 = 0 + 0.000 000 000 076 733 885 710 336;
  • 25) 0.000 000 000 076 733 885 710 336 × 2 = 0 + 0.000 000 000 153 467 771 420 672;
  • 26) 0.000 000 000 153 467 771 420 672 × 2 = 0 + 0.000 000 000 306 935 542 841 344;
  • 27) 0.000 000 000 306 935 542 841 344 × 2 = 0 + 0.000 000 000 613 871 085 682 688;
  • 28) 0.000 000 000 613 871 085 682 688 × 2 = 0 + 0.000 000 001 227 742 171 365 376;
  • 29) 0.000 000 001 227 742 171 365 376 × 2 = 0 + 0.000 000 002 455 484 342 730 752;
  • 30) 0.000 000 002 455 484 342 730 752 × 2 = 0 + 0.000 000 004 910 968 685 461 504;
  • 31) 0.000 000 004 910 968 685 461 504 × 2 = 0 + 0.000 000 009 821 937 370 923 008;
  • 32) 0.000 000 009 821 937 370 923 008 × 2 = 0 + 0.000 000 019 643 874 741 846 016;
  • 33) 0.000 000 019 643 874 741 846 016 × 2 = 0 + 0.000 000 039 287 749 483 692 032;
  • 34) 0.000 000 039 287 749 483 692 032 × 2 = 0 + 0.000 000 078 575 498 967 384 064;
  • 35) 0.000 000 078 575 498 967 384 064 × 2 = 0 + 0.000 000 157 150 997 934 768 128;
  • 36) 0.000 000 157 150 997 934 768 128 × 2 = 0 + 0.000 000 314 301 995 869 536 256;
  • 37) 0.000 000 314 301 995 869 536 256 × 2 = 0 + 0.000 000 628 603 991 739 072 512;
  • 38) 0.000 000 628 603 991 739 072 512 × 2 = 0 + 0.000 001 257 207 983 478 145 024;
  • 39) 0.000 001 257 207 983 478 145 024 × 2 = 0 + 0.000 002 514 415 966 956 290 048;
  • 40) 0.000 002 514 415 966 956 290 048 × 2 = 0 + 0.000 005 028 831 933 912 580 096;
  • 41) 0.000 005 028 831 933 912 580 096 × 2 = 0 + 0.000 010 057 663 867 825 160 192;
  • 42) 0.000 010 057 663 867 825 160 192 × 2 = 0 + 0.000 020 115 327 735 650 320 384;
  • 43) 0.000 020 115 327 735 650 320 384 × 2 = 0 + 0.000 040 230 655 471 300 640 768;
  • 44) 0.000 040 230 655 471 300 640 768 × 2 = 0 + 0.000 080 461 310 942 601 281 536;
  • 45) 0.000 080 461 310 942 601 281 536 × 2 = 0 + 0.000 160 922 621 885 202 563 072;
  • 46) 0.000 160 922 621 885 202 563 072 × 2 = 0 + 0.000 321 845 243 770 405 126 144;
  • 47) 0.000 321 845 243 770 405 126 144 × 2 = 0 + 0.000 643 690 487 540 810 252 288;
  • 48) 0.000 643 690 487 540 810 252 288 × 2 = 0 + 0.001 287 380 975 081 620 504 576;
  • 49) 0.001 287 380 975 081 620 504 576 × 2 = 0 + 0.002 574 761 950 163 241 009 152;
  • 50) 0.002 574 761 950 163 241 009 152 × 2 = 0 + 0.005 149 523 900 326 482 018 304;
  • 51) 0.005 149 523 900 326 482 018 304 × 2 = 0 + 0.010 299 047 800 652 964 036 608;
  • 52) 0.010 299 047 800 652 964 036 608 × 2 = 0 + 0.020 598 095 601 305 928 073 216;
  • 53) 0.020 598 095 601 305 928 073 216 × 2 = 0 + 0.041 196 191 202 611 856 146 432;
  • 54) 0.041 196 191 202 611 856 146 432 × 2 = 0 + 0.082 392 382 405 223 712 292 864;
  • 55) 0.082 392 382 405 223 712 292 864 × 2 = 0 + 0.164 784 764 810 447 424 585 728;
  • 56) 0.164 784 764 810 447 424 585 728 × 2 = 0 + 0.329 569 529 620 894 849 171 456;
  • 57) 0.329 569 529 620 894 849 171 456 × 2 = 0 + 0.659 139 059 241 789 698 342 912;
  • 58) 0.659 139 059 241 789 698 342 912 × 2 = 1 + 0.318 278 118 483 579 396 685 824;
  • 59) 0.318 278 118 483 579 396 685 824 × 2 = 0 + 0.636 556 236 967 158 793 371 648;
  • 60) 0.636 556 236 967 158 793 371 648 × 2 = 1 + 0.273 112 473 934 317 586 743 296;
  • 61) 0.273 112 473 934 317 586 743 296 × 2 = 0 + 0.546 224 947 868 635 173 486 592;
  • 62) 0.546 224 947 868 635 173 486 592 × 2 = 1 + 0.092 449 895 737 270 346 973 184;
  • 63) 0.092 449 895 737 270 346 973 184 × 2 = 0 + 0.184 899 791 474 540 693 946 368;
  • 64) 0.184 899 791 474 540 693 946 368 × 2 = 0 + 0.369 799 582 949 081 387 892 736;
  • 65) 0.369 799 582 949 081 387 892 736 × 2 = 0 + 0.739 599 165 898 162 775 785 472;
  • 66) 0.739 599 165 898 162 775 785 472 × 2 = 1 + 0.479 198 331 796 325 551 570 944;
  • 67) 0.479 198 331 796 325 551 570 944 × 2 = 0 + 0.958 396 663 592 651 103 141 888;
  • 68) 0.958 396 663 592 651 103 141 888 × 2 = 1 + 0.916 793 327 185 302 206 283 776;
  • 69) 0.916 793 327 185 302 206 283 776 × 2 = 1 + 0.833 586 654 370 604 412 567 552;
  • 70) 0.833 586 654 370 604 412 567 552 × 2 = 1 + 0.667 173 308 741 208 825 135 104;
  • 71) 0.667 173 308 741 208 825 135 104 × 2 = 1 + 0.334 346 617 482 417 650 270 208;
  • 72) 0.334 346 617 482 417 650 270 208 × 2 = 0 + 0.668 693 234 964 835 300 540 416;
  • 73) 0.668 693 234 964 835 300 540 416 × 2 = 1 + 0.337 386 469 929 670 601 080 832;
  • 74) 0.337 386 469 929 670 601 080 832 × 2 = 0 + 0.674 772 939 859 341 202 161 664;
  • 75) 0.674 772 939 859 341 202 161 664 × 2 = 1 + 0.349 545 879 718 682 404 323 328;
  • 76) 0.349 545 879 718 682 404 323 328 × 2 = 0 + 0.699 091 759 437 364 808 646 656;
  • 77) 0.699 091 759 437 364 808 646 656 × 2 = 1 + 0.398 183 518 874 729 617 293 312;
  • 78) 0.398 183 518 874 729 617 293 312 × 2 = 0 + 0.796 367 037 749 459 234 586 624;
  • 79) 0.796 367 037 749 459 234 586 624 × 2 = 1 + 0.592 734 075 498 918 469 173 248;
  • 80) 0.592 734 075 498 918 469 173 248 × 2 = 1 + 0.185 468 150 997 836 938 346 496;
  • 81) 0.185 468 150 997 836 938 346 496 × 2 = 0 + 0.370 936 301 995 673 876 692 992;
  • 82) 0.370 936 301 995 673 876 692 992 × 2 = 0 + 0.741 872 603 991 347 753 385 984;
  • 83) 0.741 872 603 991 347 753 385 984 × 2 = 1 + 0.483 745 207 982 695 506 771 968;
  • 84) 0.483 745 207 982 695 506 771 968 × 2 = 0 + 0.967 490 415 965 391 013 543 936;
  • 85) 0.967 490 415 965 391 013 543 936 × 2 = 1 + 0.934 980 831 930 782 027 087 872;
  • 86) 0.934 980 831 930 782 027 087 872 × 2 = 1 + 0.869 961 663 861 564 054 175 744;
  • 87) 0.869 961 663 861 564 054 175 744 × 2 = 1 + 0.739 923 327 723 128 108 351 488;
  • 88) 0.739 923 327 723 128 108 351 488 × 2 = 1 + 0.479 846 655 446 256 216 702 976;
  • 89) 0.479 846 655 446 256 216 702 976 × 2 = 0 + 0.959 693 310 892 512 433 405 952;
  • 90) 0.959 693 310 892 512 433 405 952 × 2 = 1 + 0.919 386 621 785 024 866 811 904;
  • 91) 0.919 386 621 785 024 866 811 904 × 2 = 1 + 0.838 773 243 570 049 733 623 808;
  • 92) 0.838 773 243 570 049 733 623 808 × 2 = 1 + 0.677 546 487 140 099 467 247 616;
  • 93) 0.677 546 487 140 099 467 247 616 × 2 = 1 + 0.355 092 974 280 198 934 495 232;
  • 94) 0.355 092 974 280 198 934 495 232 × 2 = 0 + 0.710 185 948 560 397 868 990 464;
  • 95) 0.710 185 948 560 397 868 990 464 × 2 = 1 + 0.420 371 897 120 795 737 980 928;
  • 96) 0.420 371 897 120 795 737 980 928 × 2 = 0 + 0.840 743 794 241 591 475 961 856;
  • 97) 0.840 743 794 241 591 475 961 856 × 2 = 1 + 0.681 487 588 483 182 951 923 712;
  • 98) 0.681 487 588 483 182 951 923 712 × 2 = 1 + 0.362 975 176 966 365 903 847 424;
  • 99) 0.362 975 176 966 365 903 847 424 × 2 = 0 + 0.725 950 353 932 731 807 694 848;
  • 100) 0.725 950 353 932 731 807 694 848 × 2 = 1 + 0.451 900 707 865 463 615 389 696;
  • 101) 0.451 900 707 865 463 615 389 696 × 2 = 0 + 0.903 801 415 730 927 230 779 392;
  • 102) 0.903 801 415 730 927 230 779 392 × 2 = 1 + 0.807 602 831 461 854 461 558 784;
  • 103) 0.807 602 831 461 854 461 558 784 × 2 = 1 + 0.615 205 662 923 708 923 117 568;
  • 104) 0.615 205 662 923 708 923 117 568 × 2 = 1 + 0.230 411 325 847 417 846 235 136;
  • 105) 0.230 411 325 847 417 846 235 136 × 2 = 0 + 0.460 822 651 694 835 692 470 272;
  • 106) 0.460 822 651 694 835 692 470 272 × 2 = 0 + 0.921 645 303 389 671 384 940 544;
  • 107) 0.921 645 303 389 671 384 940 544 × 2 = 1 + 0.843 290 606 779 342 769 881 088;
  • 108) 0.843 290 606 779 342 769 881 088 × 2 = 1 + 0.686 581 213 558 685 539 762 176;
  • 109) 0.686 581 213 558 685 539 762 176 × 2 = 1 + 0.373 162 427 117 371 079 524 352;
  • 110) 0.373 162 427 117 371 079 524 352 × 2 = 0 + 0.746 324 854 234 742 159 048 704;

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).


5. 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:


0.000 000 000 000 000 004 573 696(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0100 0101 1110 1010 1011 0010 1111 0111 1010 1101 0111 0011 10(2)

6. Positive number before normalization:

0.000 000 000 000 000 004 573 696(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0100 0101 1110 1010 1011 0010 1111 0111 1010 1101 0111 0011 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 58 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 004 573 696(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0100 0101 1110 1010 1011 0010 1111 0111 1010 1101 0111 0011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0100 0101 1110 1010 1011 0010 1111 0111 1010 1101 0111 0011 10(2) × 20 =

1.0101 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110(2) × 2-58


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): -58

Mantissa (not normalized):
1.0101 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-58 + 2(11-1) - 1 =

(-58 + 1 023)(10) =

965(10)


10. 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;
  • 965 ÷ 2 = 482 + 1;
  • 482 ÷ 2 = 241 + 0;
  • 241 ÷ 2 = 120 + 1;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

965(10) =

011 1100 0101(2)


12. 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 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110 =

0101 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1100 0101

Mantissa (52 bits) =
0101 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110


Decimal number -0.000 000 000 000 000 004 573 696 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0101 - 0101 0001 0111 1010 1010 1100 1011 1101 1110 1011 0101 1100 1110

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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