0.000 000 000 000 000 000 008 531 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 531 3(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
0.000 000 000 000 000 000 008 531 3(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: 0.000 000 000 000 000 000 008 531 3.

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) 0.000 000 000 000 000 000 008 531 3 × 2 = 0 + 0.000 000 000 000 000 000 017 062 6;
  • 2) 0.000 000 000 000 000 000 017 062 6 × 2 = 0 + 0.000 000 000 000 000 000 034 125 2;
  • 3) 0.000 000 000 000 000 000 034 125 2 × 2 = 0 + 0.000 000 000 000 000 000 068 250 4;
  • 4) 0.000 000 000 000 000 000 068 250 4 × 2 = 0 + 0.000 000 000 000 000 000 136 500 8;
  • 5) 0.000 000 000 000 000 000 136 500 8 × 2 = 0 + 0.000 000 000 000 000 000 273 001 6;
  • 6) 0.000 000 000 000 000 000 273 001 6 × 2 = 0 + 0.000 000 000 000 000 000 546 003 2;
  • 7) 0.000 000 000 000 000 000 546 003 2 × 2 = 0 + 0.000 000 000 000 000 001 092 006 4;
  • 8) 0.000 000 000 000 000 001 092 006 4 × 2 = 0 + 0.000 000 000 000 000 002 184 012 8;
  • 9) 0.000 000 000 000 000 002 184 012 8 × 2 = 0 + 0.000 000 000 000 000 004 368 025 6;
  • 10) 0.000 000 000 000 000 004 368 025 6 × 2 = 0 + 0.000 000 000 000 000 008 736 051 2;
  • 11) 0.000 000 000 000 000 008 736 051 2 × 2 = 0 + 0.000 000 000 000 000 017 472 102 4;
  • 12) 0.000 000 000 000 000 017 472 102 4 × 2 = 0 + 0.000 000 000 000 000 034 944 204 8;
  • 13) 0.000 000 000 000 000 034 944 204 8 × 2 = 0 + 0.000 000 000 000 000 069 888 409 6;
  • 14) 0.000 000 000 000 000 069 888 409 6 × 2 = 0 + 0.000 000 000 000 000 139 776 819 2;
  • 15) 0.000 000 000 000 000 139 776 819 2 × 2 = 0 + 0.000 000 000 000 000 279 553 638 4;
  • 16) 0.000 000 000 000 000 279 553 638 4 × 2 = 0 + 0.000 000 000 000 000 559 107 276 8;
  • 17) 0.000 000 000 000 000 559 107 276 8 × 2 = 0 + 0.000 000 000 000 001 118 214 553 6;
  • 18) 0.000 000 000 000 001 118 214 553 6 × 2 = 0 + 0.000 000 000 000 002 236 429 107 2;
  • 19) 0.000 000 000 000 002 236 429 107 2 × 2 = 0 + 0.000 000 000 000 004 472 858 214 4;
  • 20) 0.000 000 000 000 004 472 858 214 4 × 2 = 0 + 0.000 000 000 000 008 945 716 428 8;
  • 21) 0.000 000 000 000 008 945 716 428 8 × 2 = 0 + 0.000 000 000 000 017 891 432 857 6;
  • 22) 0.000 000 000 000 017 891 432 857 6 × 2 = 0 + 0.000 000 000 000 035 782 865 715 2;
  • 23) 0.000 000 000 000 035 782 865 715 2 × 2 = 0 + 0.000 000 000 000 071 565 731 430 4;
  • 24) 0.000 000 000 000 071 565 731 430 4 × 2 = 0 + 0.000 000 000 000 143 131 462 860 8;
  • 25) 0.000 000 000 000 143 131 462 860 8 × 2 = 0 + 0.000 000 000 000 286 262 925 721 6;
  • 26) 0.000 000 000 000 286 262 925 721 6 × 2 = 0 + 0.000 000 000 000 572 525 851 443 2;
  • 27) 0.000 000 000 000 572 525 851 443 2 × 2 = 0 + 0.000 000 000 001 145 051 702 886 4;
  • 28) 0.000 000 000 001 145 051 702 886 4 × 2 = 0 + 0.000 000 000 002 290 103 405 772 8;
  • 29) 0.000 000 000 002 290 103 405 772 8 × 2 = 0 + 0.000 000 000 004 580 206 811 545 6;
  • 30) 0.000 000 000 004 580 206 811 545 6 × 2 = 0 + 0.000 000 000 009 160 413 623 091 2;
  • 31) 0.000 000 000 009 160 413 623 091 2 × 2 = 0 + 0.000 000 000 018 320 827 246 182 4;
  • 32) 0.000 000 000 018 320 827 246 182 4 × 2 = 0 + 0.000 000 000 036 641 654 492 364 8;
  • 33) 0.000 000 000 036 641 654 492 364 8 × 2 = 0 + 0.000 000 000 073 283 308 984 729 6;
  • 34) 0.000 000 000 073 283 308 984 729 6 × 2 = 0 + 0.000 000 000 146 566 617 969 459 2;
  • 35) 0.000 000 000 146 566 617 969 459 2 × 2 = 0 + 0.000 000 000 293 133 235 938 918 4;
  • 36) 0.000 000 000 293 133 235 938 918 4 × 2 = 0 + 0.000 000 000 586 266 471 877 836 8;
  • 37) 0.000 000 000 586 266 471 877 836 8 × 2 = 0 + 0.000 000 001 172 532 943 755 673 6;
  • 38) 0.000 000 001 172 532 943 755 673 6 × 2 = 0 + 0.000 000 002 345 065 887 511 347 2;
  • 39) 0.000 000 002 345 065 887 511 347 2 × 2 = 0 + 0.000 000 004 690 131 775 022 694 4;
  • 40) 0.000 000 004 690 131 775 022 694 4 × 2 = 0 + 0.000 000 009 380 263 550 045 388 8;
  • 41) 0.000 000 009 380 263 550 045 388 8 × 2 = 0 + 0.000 000 018 760 527 100 090 777 6;
  • 42) 0.000 000 018 760 527 100 090 777 6 × 2 = 0 + 0.000 000 037 521 054 200 181 555 2;
  • 43) 0.000 000 037 521 054 200 181 555 2 × 2 = 0 + 0.000 000 075 042 108 400 363 110 4;
  • 44) 0.000 000 075 042 108 400 363 110 4 × 2 = 0 + 0.000 000 150 084 216 800 726 220 8;
  • 45) 0.000 000 150 084 216 800 726 220 8 × 2 = 0 + 0.000 000 300 168 433 601 452 441 6;
  • 46) 0.000 000 300 168 433 601 452 441 6 × 2 = 0 + 0.000 000 600 336 867 202 904 883 2;
  • 47) 0.000 000 600 336 867 202 904 883 2 × 2 = 0 + 0.000 001 200 673 734 405 809 766 4;
  • 48) 0.000 001 200 673 734 405 809 766 4 × 2 = 0 + 0.000 002 401 347 468 811 619 532 8;
  • 49) 0.000 002 401 347 468 811 619 532 8 × 2 = 0 + 0.000 004 802 694 937 623 239 065 6;
  • 50) 0.000 004 802 694 937 623 239 065 6 × 2 = 0 + 0.000 009 605 389 875 246 478 131 2;
  • 51) 0.000 009 605 389 875 246 478 131 2 × 2 = 0 + 0.000 019 210 779 750 492 956 262 4;
  • 52) 0.000 019 210 779 750 492 956 262 4 × 2 = 0 + 0.000 038 421 559 500 985 912 524 8;
  • 53) 0.000 038 421 559 500 985 912 524 8 × 2 = 0 + 0.000 076 843 119 001 971 825 049 6;
  • 54) 0.000 076 843 119 001 971 825 049 6 × 2 = 0 + 0.000 153 686 238 003 943 650 099 2;
  • 55) 0.000 153 686 238 003 943 650 099 2 × 2 = 0 + 0.000 307 372 476 007 887 300 198 4;
  • 56) 0.000 307 372 476 007 887 300 198 4 × 2 = 0 + 0.000 614 744 952 015 774 600 396 8;
  • 57) 0.000 614 744 952 015 774 600 396 8 × 2 = 0 + 0.001 229 489 904 031 549 200 793 6;
  • 58) 0.001 229 489 904 031 549 200 793 6 × 2 = 0 + 0.002 458 979 808 063 098 401 587 2;
  • 59) 0.002 458 979 808 063 098 401 587 2 × 2 = 0 + 0.004 917 959 616 126 196 803 174 4;
  • 60) 0.004 917 959 616 126 196 803 174 4 × 2 = 0 + 0.009 835 919 232 252 393 606 348 8;
  • 61) 0.009 835 919 232 252 393 606 348 8 × 2 = 0 + 0.019 671 838 464 504 787 212 697 6;
  • 62) 0.019 671 838 464 504 787 212 697 6 × 2 = 0 + 0.039 343 676 929 009 574 425 395 2;
  • 63) 0.039 343 676 929 009 574 425 395 2 × 2 = 0 + 0.078 687 353 858 019 148 850 790 4;
  • 64) 0.078 687 353 858 019 148 850 790 4 × 2 = 0 + 0.157 374 707 716 038 297 701 580 8;
  • 65) 0.157 374 707 716 038 297 701 580 8 × 2 = 0 + 0.314 749 415 432 076 595 403 161 6;
  • 66) 0.314 749 415 432 076 595 403 161 6 × 2 = 0 + 0.629 498 830 864 153 190 806 323 2;
  • 67) 0.629 498 830 864 153 190 806 323 2 × 2 = 1 + 0.258 997 661 728 306 381 612 646 4;
  • 68) 0.258 997 661 728 306 381 612 646 4 × 2 = 0 + 0.517 995 323 456 612 763 225 292 8;
  • 69) 0.517 995 323 456 612 763 225 292 8 × 2 = 1 + 0.035 990 646 913 225 526 450 585 6;
  • 70) 0.035 990 646 913 225 526 450 585 6 × 2 = 0 + 0.071 981 293 826 451 052 901 171 2;
  • 71) 0.071 981 293 826 451 052 901 171 2 × 2 = 0 + 0.143 962 587 652 902 105 802 342 4;
  • 72) 0.143 962 587 652 902 105 802 342 4 × 2 = 0 + 0.287 925 175 305 804 211 604 684 8;
  • 73) 0.287 925 175 305 804 211 604 684 8 × 2 = 0 + 0.575 850 350 611 608 423 209 369 6;
  • 74) 0.575 850 350 611 608 423 209 369 6 × 2 = 1 + 0.151 700 701 223 216 846 418 739 2;
  • 75) 0.151 700 701 223 216 846 418 739 2 × 2 = 0 + 0.303 401 402 446 433 692 837 478 4;
  • 76) 0.303 401 402 446 433 692 837 478 4 × 2 = 0 + 0.606 802 804 892 867 385 674 956 8;
  • 77) 0.606 802 804 892 867 385 674 956 8 × 2 = 1 + 0.213 605 609 785 734 771 349 913 6;
  • 78) 0.213 605 609 785 734 771 349 913 6 × 2 = 0 + 0.427 211 219 571 469 542 699 827 2;
  • 79) 0.427 211 219 571 469 542 699 827 2 × 2 = 0 + 0.854 422 439 142 939 085 399 654 4;
  • 80) 0.854 422 439 142 939 085 399 654 4 × 2 = 1 + 0.708 844 878 285 878 170 799 308 8;
  • 81) 0.708 844 878 285 878 170 799 308 8 × 2 = 1 + 0.417 689 756 571 756 341 598 617 6;
  • 82) 0.417 689 756 571 756 341 598 617 6 × 2 = 0 + 0.835 379 513 143 512 683 197 235 2;
  • 83) 0.835 379 513 143 512 683 197 235 2 × 2 = 1 + 0.670 759 026 287 025 366 394 470 4;
  • 84) 0.670 759 026 287 025 366 394 470 4 × 2 = 1 + 0.341 518 052 574 050 732 788 940 8;
  • 85) 0.341 518 052 574 050 732 788 940 8 × 2 = 0 + 0.683 036 105 148 101 465 577 881 6;
  • 86) 0.683 036 105 148 101 465 577 881 6 × 2 = 1 + 0.366 072 210 296 202 931 155 763 2;
  • 87) 0.366 072 210 296 202 931 155 763 2 × 2 = 0 + 0.732 144 420 592 405 862 311 526 4;
  • 88) 0.732 144 420 592 405 862 311 526 4 × 2 = 1 + 0.464 288 841 184 811 724 623 052 8;
  • 89) 0.464 288 841 184 811 724 623 052 8 × 2 = 0 + 0.928 577 682 369 623 449 246 105 6;
  • 90) 0.928 577 682 369 623 449 246 105 6 × 2 = 1 + 0.857 155 364 739 246 898 492 211 2;
  • 91) 0.857 155 364 739 246 898 492 211 2 × 2 = 1 + 0.714 310 729 478 493 796 984 422 4;
  • 92) 0.714 310 729 478 493 796 984 422 4 × 2 = 1 + 0.428 621 458 956 987 593 968 844 8;
  • 93) 0.428 621 458 956 987 593 968 844 8 × 2 = 0 + 0.857 242 917 913 975 187 937 689 6;
  • 94) 0.857 242 917 913 975 187 937 689 6 × 2 = 1 + 0.714 485 835 827 950 375 875 379 2;
  • 95) 0.714 485 835 827 950 375 875 379 2 × 2 = 1 + 0.428 971 671 655 900 751 750 758 4;
  • 96) 0.428 971 671 655 900 751 750 758 4 × 2 = 0 + 0.857 943 343 311 801 503 501 516 8;
  • 97) 0.857 943 343 311 801 503 501 516 8 × 2 = 1 + 0.715 886 686 623 603 007 003 033 6;
  • 98) 0.715 886 686 623 603 007 003 033 6 × 2 = 1 + 0.431 773 373 247 206 014 006 067 2;
  • 99) 0.431 773 373 247 206 014 006 067 2 × 2 = 0 + 0.863 546 746 494 412 028 012 134 4;
  • 100) 0.863 546 746 494 412 028 012 134 4 × 2 = 1 + 0.727 093 492 988 824 056 024 268 8;
  • 101) 0.727 093 492 988 824 056 024 268 8 × 2 = 1 + 0.454 186 985 977 648 112 048 537 6;
  • 102) 0.454 186 985 977 648 112 048 537 6 × 2 = 0 + 0.908 373 971 955 296 224 097 075 2;
  • 103) 0.908 373 971 955 296 224 097 075 2 × 2 = 1 + 0.816 747 943 910 592 448 194 150 4;
  • 104) 0.816 747 943 910 592 448 194 150 4 × 2 = 1 + 0.633 495 887 821 184 896 388 300 8;
  • 105) 0.633 495 887 821 184 896 388 300 8 × 2 = 1 + 0.266 991 775 642 369 792 776 601 6;
  • 106) 0.266 991 775 642 369 792 776 601 6 × 2 = 0 + 0.533 983 551 284 739 585 553 203 2;
  • 107) 0.533 983 551 284 739 585 553 203 2 × 2 = 1 + 0.067 967 102 569 479 171 106 406 4;
  • 108) 0.067 967 102 569 479 171 106 406 4 × 2 = 0 + 0.135 934 205 138 958 342 212 812 8;
  • 109) 0.135 934 205 138 958 342 212 812 8 × 2 = 0 + 0.271 868 410 277 916 684 425 625 6;
  • 110) 0.271 868 410 277 916 684 425 625 6 × 2 = 0 + 0.543 736 820 555 833 368 851 251 2;
  • 111) 0.543 736 820 555 833 368 851 251 2 × 2 = 1 + 0.087 473 641 111 666 737 702 502 4;
  • 112) 0.087 473 641 111 666 737 702 502 4 × 2 = 0 + 0.174 947 282 223 333 475 405 004 8;
  • 113) 0.174 947 282 223 333 475 405 004 8 × 2 = 0 + 0.349 894 564 446 666 950 810 009 6;
  • 114) 0.349 894 564 446 666 950 810 009 6 × 2 = 0 + 0.699 789 128 893 333 901 620 019 2;
  • 115) 0.699 789 128 893 333 901 620 019 2 × 2 = 1 + 0.399 578 257 786 667 803 240 038 4;
  • 116) 0.399 578 257 786 667 803 240 038 4 × 2 = 0 + 0.799 156 515 573 335 606 480 076 8;
  • 117) 0.799 156 515 573 335 606 480 076 8 × 2 = 1 + 0.598 313 031 146 671 212 960 153 6;
  • 118) 0.598 313 031 146 671 212 960 153 6 × 2 = 1 + 0.196 626 062 293 342 425 920 307 2;
  • 119) 0.196 626 062 293 342 425 920 307 2 × 2 = 0 + 0.393 252 124 586 684 851 840 614 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:


0.000 000 000 000 000 000 008 531 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1001 1011 0101 0111 0110 1101 1011 1010 0010 0010 110(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 531 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1001 1011 0101 0111 0110 1101 1011 1010 0010 0010 110(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:


0.000 000 000 000 000 000 008 531 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1001 1011 0101 0111 0110 1101 1011 1010 0010 0010 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0100 1001 1011 0101 0111 0110 1101 1011 1010 0010 0010 110(2) × 20 =

1.0100 0010 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110(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 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110


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 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110 =

0100 0010 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110


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 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110


Decimal number 0.000 000 000 000 000 000 008 531 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0010 0100 1101 1010 1011 1011 0110 1101 1101 0001 0001 0110

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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