0.000 000 000 000 000 000 008 545 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 545(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
0.000 000 000 000 000 000 008 545(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: 0.000 000 000 000 000 000 008 545.

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) 0.000 000 000 000 000 000 008 545 × 2 = 0 + 0.000 000 000 000 000 000 017 09;
  • 2) 0.000 000 000 000 000 000 017 09 × 2 = 0 + 0.000 000 000 000 000 000 034 18;
  • 3) 0.000 000 000 000 000 000 034 18 × 2 = 0 + 0.000 000 000 000 000 000 068 36;
  • 4) 0.000 000 000 000 000 000 068 36 × 2 = 0 + 0.000 000 000 000 000 000 136 72;
  • 5) 0.000 000 000 000 000 000 136 72 × 2 = 0 + 0.000 000 000 000 000 000 273 44;
  • 6) 0.000 000 000 000 000 000 273 44 × 2 = 0 + 0.000 000 000 000 000 000 546 88;
  • 7) 0.000 000 000 000 000 000 546 88 × 2 = 0 + 0.000 000 000 000 000 001 093 76;
  • 8) 0.000 000 000 000 000 001 093 76 × 2 = 0 + 0.000 000 000 000 000 002 187 52;
  • 9) 0.000 000 000 000 000 002 187 52 × 2 = 0 + 0.000 000 000 000 000 004 375 04;
  • 10) 0.000 000 000 000 000 004 375 04 × 2 = 0 + 0.000 000 000 000 000 008 750 08;
  • 11) 0.000 000 000 000 000 008 750 08 × 2 = 0 + 0.000 000 000 000 000 017 500 16;
  • 12) 0.000 000 000 000 000 017 500 16 × 2 = 0 + 0.000 000 000 000 000 035 000 32;
  • 13) 0.000 000 000 000 000 035 000 32 × 2 = 0 + 0.000 000 000 000 000 070 000 64;
  • 14) 0.000 000 000 000 000 070 000 64 × 2 = 0 + 0.000 000 000 000 000 140 001 28;
  • 15) 0.000 000 000 000 000 140 001 28 × 2 = 0 + 0.000 000 000 000 000 280 002 56;
  • 16) 0.000 000 000 000 000 280 002 56 × 2 = 0 + 0.000 000 000 000 000 560 005 12;
  • 17) 0.000 000 000 000 000 560 005 12 × 2 = 0 + 0.000 000 000 000 001 120 010 24;
  • 18) 0.000 000 000 000 001 120 010 24 × 2 = 0 + 0.000 000 000 000 002 240 020 48;
  • 19) 0.000 000 000 000 002 240 020 48 × 2 = 0 + 0.000 000 000 000 004 480 040 96;
  • 20) 0.000 000 000 000 004 480 040 96 × 2 = 0 + 0.000 000 000 000 008 960 081 92;
  • 21) 0.000 000 000 000 008 960 081 92 × 2 = 0 + 0.000 000 000 000 017 920 163 84;
  • 22) 0.000 000 000 000 017 920 163 84 × 2 = 0 + 0.000 000 000 000 035 840 327 68;
  • 23) 0.000 000 000 000 035 840 327 68 × 2 = 0 + 0.000 000 000 000 071 680 655 36;
  • 24) 0.000 000 000 000 071 680 655 36 × 2 = 0 + 0.000 000 000 000 143 361 310 72;
  • 25) 0.000 000 000 000 143 361 310 72 × 2 = 0 + 0.000 000 000 000 286 722 621 44;
  • 26) 0.000 000 000 000 286 722 621 44 × 2 = 0 + 0.000 000 000 000 573 445 242 88;
  • 27) 0.000 000 000 000 573 445 242 88 × 2 = 0 + 0.000 000 000 001 146 890 485 76;
  • 28) 0.000 000 000 001 146 890 485 76 × 2 = 0 + 0.000 000 000 002 293 780 971 52;
  • 29) 0.000 000 000 002 293 780 971 52 × 2 = 0 + 0.000 000 000 004 587 561 943 04;
  • 30) 0.000 000 000 004 587 561 943 04 × 2 = 0 + 0.000 000 000 009 175 123 886 08;
  • 31) 0.000 000 000 009 175 123 886 08 × 2 = 0 + 0.000 000 000 018 350 247 772 16;
  • 32) 0.000 000 000 018 350 247 772 16 × 2 = 0 + 0.000 000 000 036 700 495 544 32;
  • 33) 0.000 000 000 036 700 495 544 32 × 2 = 0 + 0.000 000 000 073 400 991 088 64;
  • 34) 0.000 000 000 073 400 991 088 64 × 2 = 0 + 0.000 000 000 146 801 982 177 28;
  • 35) 0.000 000 000 146 801 982 177 28 × 2 = 0 + 0.000 000 000 293 603 964 354 56;
  • 36) 0.000 000 000 293 603 964 354 56 × 2 = 0 + 0.000 000 000 587 207 928 709 12;
  • 37) 0.000 000 000 587 207 928 709 12 × 2 = 0 + 0.000 000 001 174 415 857 418 24;
  • 38) 0.000 000 001 174 415 857 418 24 × 2 = 0 + 0.000 000 002 348 831 714 836 48;
  • 39) 0.000 000 002 348 831 714 836 48 × 2 = 0 + 0.000 000 004 697 663 429 672 96;
  • 40) 0.000 000 004 697 663 429 672 96 × 2 = 0 + 0.000 000 009 395 326 859 345 92;
  • 41) 0.000 000 009 395 326 859 345 92 × 2 = 0 + 0.000 000 018 790 653 718 691 84;
  • 42) 0.000 000 018 790 653 718 691 84 × 2 = 0 + 0.000 000 037 581 307 437 383 68;
  • 43) 0.000 000 037 581 307 437 383 68 × 2 = 0 + 0.000 000 075 162 614 874 767 36;
  • 44) 0.000 000 075 162 614 874 767 36 × 2 = 0 + 0.000 000 150 325 229 749 534 72;
  • 45) 0.000 000 150 325 229 749 534 72 × 2 = 0 + 0.000 000 300 650 459 499 069 44;
  • 46) 0.000 000 300 650 459 499 069 44 × 2 = 0 + 0.000 000 601 300 918 998 138 88;
  • 47) 0.000 000 601 300 918 998 138 88 × 2 = 0 + 0.000 001 202 601 837 996 277 76;
  • 48) 0.000 001 202 601 837 996 277 76 × 2 = 0 + 0.000 002 405 203 675 992 555 52;
  • 49) 0.000 002 405 203 675 992 555 52 × 2 = 0 + 0.000 004 810 407 351 985 111 04;
  • 50) 0.000 004 810 407 351 985 111 04 × 2 = 0 + 0.000 009 620 814 703 970 222 08;
  • 51) 0.000 009 620 814 703 970 222 08 × 2 = 0 + 0.000 019 241 629 407 940 444 16;
  • 52) 0.000 019 241 629 407 940 444 16 × 2 = 0 + 0.000 038 483 258 815 880 888 32;
  • 53) 0.000 038 483 258 815 880 888 32 × 2 = 0 + 0.000 076 966 517 631 761 776 64;
  • 54) 0.000 076 966 517 631 761 776 64 × 2 = 0 + 0.000 153 933 035 263 523 553 28;
  • 55) 0.000 153 933 035 263 523 553 28 × 2 = 0 + 0.000 307 866 070 527 047 106 56;
  • 56) 0.000 307 866 070 527 047 106 56 × 2 = 0 + 0.000 615 732 141 054 094 213 12;
  • 57) 0.000 615 732 141 054 094 213 12 × 2 = 0 + 0.001 231 464 282 108 188 426 24;
  • 58) 0.001 231 464 282 108 188 426 24 × 2 = 0 + 0.002 462 928 564 216 376 852 48;
  • 59) 0.002 462 928 564 216 376 852 48 × 2 = 0 + 0.004 925 857 128 432 753 704 96;
  • 60) 0.004 925 857 128 432 753 704 96 × 2 = 0 + 0.009 851 714 256 865 507 409 92;
  • 61) 0.009 851 714 256 865 507 409 92 × 2 = 0 + 0.019 703 428 513 731 014 819 84;
  • 62) 0.019 703 428 513 731 014 819 84 × 2 = 0 + 0.039 406 857 027 462 029 639 68;
  • 63) 0.039 406 857 027 462 029 639 68 × 2 = 0 + 0.078 813 714 054 924 059 279 36;
  • 64) 0.078 813 714 054 924 059 279 36 × 2 = 0 + 0.157 627 428 109 848 118 558 72;
  • 65) 0.157 627 428 109 848 118 558 72 × 2 = 0 + 0.315 254 856 219 696 237 117 44;
  • 66) 0.315 254 856 219 696 237 117 44 × 2 = 0 + 0.630 509 712 439 392 474 234 88;
  • 67) 0.630 509 712 439 392 474 234 88 × 2 = 1 + 0.261 019 424 878 784 948 469 76;
  • 68) 0.261 019 424 878 784 948 469 76 × 2 = 0 + 0.522 038 849 757 569 896 939 52;
  • 69) 0.522 038 849 757 569 896 939 52 × 2 = 1 + 0.044 077 699 515 139 793 879 04;
  • 70) 0.044 077 699 515 139 793 879 04 × 2 = 0 + 0.088 155 399 030 279 587 758 08;
  • 71) 0.088 155 399 030 279 587 758 08 × 2 = 0 + 0.176 310 798 060 559 175 516 16;
  • 72) 0.176 310 798 060 559 175 516 16 × 2 = 0 + 0.352 621 596 121 118 351 032 32;
  • 73) 0.352 621 596 121 118 351 032 32 × 2 = 0 + 0.705 243 192 242 236 702 064 64;
  • 74) 0.705 243 192 242 236 702 064 64 × 2 = 1 + 0.410 486 384 484 473 404 129 28;
  • 75) 0.410 486 384 484 473 404 129 28 × 2 = 0 + 0.820 972 768 968 946 808 258 56;
  • 76) 0.820 972 768 968 946 808 258 56 × 2 = 1 + 0.641 945 537 937 893 616 517 12;
  • 77) 0.641 945 537 937 893 616 517 12 × 2 = 1 + 0.283 891 075 875 787 233 034 24;
  • 78) 0.283 891 075 875 787 233 034 24 × 2 = 0 + 0.567 782 151 751 574 466 068 48;
  • 79) 0.567 782 151 751 574 466 068 48 × 2 = 1 + 0.135 564 303 503 148 932 136 96;
  • 80) 0.135 564 303 503 148 932 136 96 × 2 = 0 + 0.271 128 607 006 297 864 273 92;
  • 81) 0.271 128 607 006 297 864 273 92 × 2 = 0 + 0.542 257 214 012 595 728 547 84;
  • 82) 0.542 257 214 012 595 728 547 84 × 2 = 1 + 0.084 514 428 025 191 457 095 68;
  • 83) 0.084 514 428 025 191 457 095 68 × 2 = 0 + 0.169 028 856 050 382 914 191 36;
  • 84) 0.169 028 856 050 382 914 191 36 × 2 = 0 + 0.338 057 712 100 765 828 382 72;
  • 85) 0.338 057 712 100 765 828 382 72 × 2 = 0 + 0.676 115 424 201 531 656 765 44;
  • 86) 0.676 115 424 201 531 656 765 44 × 2 = 1 + 0.352 230 848 403 063 313 530 88;
  • 87) 0.352 230 848 403 063 313 530 88 × 2 = 0 + 0.704 461 696 806 126 627 061 76;
  • 88) 0.704 461 696 806 126 627 061 76 × 2 = 1 + 0.408 923 393 612 253 254 123 52;
  • 89) 0.408 923 393 612 253 254 123 52 × 2 = 0 + 0.817 846 787 224 506 508 247 04;
  • 90) 0.817 846 787 224 506 508 247 04 × 2 = 1 + 0.635 693 574 449 013 016 494 08;
  • 91) 0.635 693 574 449 013 016 494 08 × 2 = 1 + 0.271 387 148 898 026 032 988 16;
  • 92) 0.271 387 148 898 026 032 988 16 × 2 = 0 + 0.542 774 297 796 052 065 976 32;
  • 93) 0.542 774 297 796 052 065 976 32 × 2 = 1 + 0.085 548 595 592 104 131 952 64;
  • 94) 0.085 548 595 592 104 131 952 64 × 2 = 0 + 0.171 097 191 184 208 263 905 28;
  • 95) 0.171 097 191 184 208 263 905 28 × 2 = 0 + 0.342 194 382 368 416 527 810 56;
  • 96) 0.342 194 382 368 416 527 810 56 × 2 = 0 + 0.684 388 764 736 833 055 621 12;
  • 97) 0.684 388 764 736 833 055 621 12 × 2 = 1 + 0.368 777 529 473 666 111 242 24;
  • 98) 0.368 777 529 473 666 111 242 24 × 2 = 0 + 0.737 555 058 947 332 222 484 48;
  • 99) 0.737 555 058 947 332 222 484 48 × 2 = 1 + 0.475 110 117 894 664 444 968 96;
  • 100) 0.475 110 117 894 664 444 968 96 × 2 = 0 + 0.950 220 235 789 328 889 937 92;
  • 101) 0.950 220 235 789 328 889 937 92 × 2 = 1 + 0.900 440 471 578 657 779 875 84;
  • 102) 0.900 440 471 578 657 779 875 84 × 2 = 1 + 0.800 880 943 157 315 559 751 68;
  • 103) 0.800 880 943 157 315 559 751 68 × 2 = 1 + 0.601 761 886 314 631 119 503 36;
  • 104) 0.601 761 886 314 631 119 503 36 × 2 = 1 + 0.203 523 772 629 262 239 006 72;
  • 105) 0.203 523 772 629 262 239 006 72 × 2 = 0 + 0.407 047 545 258 524 478 013 44;
  • 106) 0.407 047 545 258 524 478 013 44 × 2 = 0 + 0.814 095 090 517 048 956 026 88;
  • 107) 0.814 095 090 517 048 956 026 88 × 2 = 1 + 0.628 190 181 034 097 912 053 76;
  • 108) 0.628 190 181 034 097 912 053 76 × 2 = 1 + 0.256 380 362 068 195 824 107 52;
  • 109) 0.256 380 362 068 195 824 107 52 × 2 = 0 + 0.512 760 724 136 391 648 215 04;
  • 110) 0.512 760 724 136 391 648 215 04 × 2 = 1 + 0.025 521 448 272 783 296 430 08;
  • 111) 0.025 521 448 272 783 296 430 08 × 2 = 0 + 0.051 042 896 545 566 592 860 16;
  • 112) 0.051 042 896 545 566 592 860 16 × 2 = 0 + 0.102 085 793 091 133 185 720 32;
  • 113) 0.102 085 793 091 133 185 720 32 × 2 = 0 + 0.204 171 586 182 266 371 440 64;
  • 114) 0.204 171 586 182 266 371 440 64 × 2 = 0 + 0.408 343 172 364 532 742 881 28;
  • 115) 0.408 343 172 364 532 742 881 28 × 2 = 0 + 0.816 686 344 729 065 485 762 56;
  • 116) 0.816 686 344 729 065 485 762 56 × 2 = 1 + 0.633 372 689 458 130 971 525 12;
  • 117) 0.633 372 689 458 130 971 525 12 × 2 = 1 + 0.266 745 378 916 261 943 050 24;
  • 118) 0.266 745 378 916 261 943 050 24 × 2 = 0 + 0.533 490 757 832 523 886 100 48;
  • 119) 0.533 490 757 832 523 886 100 48 × 2 = 1 + 0.066 981 515 665 047 772 200 96;

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:


0.000 000 000 000 000 000 008 545(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0101 1010 0100 0101 0110 1000 1010 1111 0011 0100 0001 101(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 545(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0101 1010 0100 0101 0110 1000 1010 1111 0011 0100 0001 101(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:


0.000 000 000 000 000 000 008 545(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0101 1010 0100 0101 0110 1000 1010 1111 0011 0100 0001 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0101 1010 0100 0101 0110 1000 1010 1111 0011 0100 0001 101(2) × 20 =

1.0100 0010 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101(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 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101


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 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101 =

0100 0010 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101


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 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101


Decimal number 0.000 000 000 000 000 000 008 545 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0010 1101 0010 0010 1011 0100 0101 0111 1001 1010 0000 1101

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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