0.000 000 000 000 000 000 008 615 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 615(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
0.000 000 000 000 000 000 008 615(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: 0.000 000 000 000 000 000 008 615.

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) 0.000 000 000 000 000 000 008 615 × 2 = 0 + 0.000 000 000 000 000 000 017 23;
  • 2) 0.000 000 000 000 000 000 017 23 × 2 = 0 + 0.000 000 000 000 000 000 034 46;
  • 3) 0.000 000 000 000 000 000 034 46 × 2 = 0 + 0.000 000 000 000 000 000 068 92;
  • 4) 0.000 000 000 000 000 000 068 92 × 2 = 0 + 0.000 000 000 000 000 000 137 84;
  • 5) 0.000 000 000 000 000 000 137 84 × 2 = 0 + 0.000 000 000 000 000 000 275 68;
  • 6) 0.000 000 000 000 000 000 275 68 × 2 = 0 + 0.000 000 000 000 000 000 551 36;
  • 7) 0.000 000 000 000 000 000 551 36 × 2 = 0 + 0.000 000 000 000 000 001 102 72;
  • 8) 0.000 000 000 000 000 001 102 72 × 2 = 0 + 0.000 000 000 000 000 002 205 44;
  • 9) 0.000 000 000 000 000 002 205 44 × 2 = 0 + 0.000 000 000 000 000 004 410 88;
  • 10) 0.000 000 000 000 000 004 410 88 × 2 = 0 + 0.000 000 000 000 000 008 821 76;
  • 11) 0.000 000 000 000 000 008 821 76 × 2 = 0 + 0.000 000 000 000 000 017 643 52;
  • 12) 0.000 000 000 000 000 017 643 52 × 2 = 0 + 0.000 000 000 000 000 035 287 04;
  • 13) 0.000 000 000 000 000 035 287 04 × 2 = 0 + 0.000 000 000 000 000 070 574 08;
  • 14) 0.000 000 000 000 000 070 574 08 × 2 = 0 + 0.000 000 000 000 000 141 148 16;
  • 15) 0.000 000 000 000 000 141 148 16 × 2 = 0 + 0.000 000 000 000 000 282 296 32;
  • 16) 0.000 000 000 000 000 282 296 32 × 2 = 0 + 0.000 000 000 000 000 564 592 64;
  • 17) 0.000 000 000 000 000 564 592 64 × 2 = 0 + 0.000 000 000 000 001 129 185 28;
  • 18) 0.000 000 000 000 001 129 185 28 × 2 = 0 + 0.000 000 000 000 002 258 370 56;
  • 19) 0.000 000 000 000 002 258 370 56 × 2 = 0 + 0.000 000 000 000 004 516 741 12;
  • 20) 0.000 000 000 000 004 516 741 12 × 2 = 0 + 0.000 000 000 000 009 033 482 24;
  • 21) 0.000 000 000 000 009 033 482 24 × 2 = 0 + 0.000 000 000 000 018 066 964 48;
  • 22) 0.000 000 000 000 018 066 964 48 × 2 = 0 + 0.000 000 000 000 036 133 928 96;
  • 23) 0.000 000 000 000 036 133 928 96 × 2 = 0 + 0.000 000 000 000 072 267 857 92;
  • 24) 0.000 000 000 000 072 267 857 92 × 2 = 0 + 0.000 000 000 000 144 535 715 84;
  • 25) 0.000 000 000 000 144 535 715 84 × 2 = 0 + 0.000 000 000 000 289 071 431 68;
  • 26) 0.000 000 000 000 289 071 431 68 × 2 = 0 + 0.000 000 000 000 578 142 863 36;
  • 27) 0.000 000 000 000 578 142 863 36 × 2 = 0 + 0.000 000 000 001 156 285 726 72;
  • 28) 0.000 000 000 001 156 285 726 72 × 2 = 0 + 0.000 000 000 002 312 571 453 44;
  • 29) 0.000 000 000 002 312 571 453 44 × 2 = 0 + 0.000 000 000 004 625 142 906 88;
  • 30) 0.000 000 000 004 625 142 906 88 × 2 = 0 + 0.000 000 000 009 250 285 813 76;
  • 31) 0.000 000 000 009 250 285 813 76 × 2 = 0 + 0.000 000 000 018 500 571 627 52;
  • 32) 0.000 000 000 018 500 571 627 52 × 2 = 0 + 0.000 000 000 037 001 143 255 04;
  • 33) 0.000 000 000 037 001 143 255 04 × 2 = 0 + 0.000 000 000 074 002 286 510 08;
  • 34) 0.000 000 000 074 002 286 510 08 × 2 = 0 + 0.000 000 000 148 004 573 020 16;
  • 35) 0.000 000 000 148 004 573 020 16 × 2 = 0 + 0.000 000 000 296 009 146 040 32;
  • 36) 0.000 000 000 296 009 146 040 32 × 2 = 0 + 0.000 000 000 592 018 292 080 64;
  • 37) 0.000 000 000 592 018 292 080 64 × 2 = 0 + 0.000 000 001 184 036 584 161 28;
  • 38) 0.000 000 001 184 036 584 161 28 × 2 = 0 + 0.000 000 002 368 073 168 322 56;
  • 39) 0.000 000 002 368 073 168 322 56 × 2 = 0 + 0.000 000 004 736 146 336 645 12;
  • 40) 0.000 000 004 736 146 336 645 12 × 2 = 0 + 0.000 000 009 472 292 673 290 24;
  • 41) 0.000 000 009 472 292 673 290 24 × 2 = 0 + 0.000 000 018 944 585 346 580 48;
  • 42) 0.000 000 018 944 585 346 580 48 × 2 = 0 + 0.000 000 037 889 170 693 160 96;
  • 43) 0.000 000 037 889 170 693 160 96 × 2 = 0 + 0.000 000 075 778 341 386 321 92;
  • 44) 0.000 000 075 778 341 386 321 92 × 2 = 0 + 0.000 000 151 556 682 772 643 84;
  • 45) 0.000 000 151 556 682 772 643 84 × 2 = 0 + 0.000 000 303 113 365 545 287 68;
  • 46) 0.000 000 303 113 365 545 287 68 × 2 = 0 + 0.000 000 606 226 731 090 575 36;
  • 47) 0.000 000 606 226 731 090 575 36 × 2 = 0 + 0.000 001 212 453 462 181 150 72;
  • 48) 0.000 001 212 453 462 181 150 72 × 2 = 0 + 0.000 002 424 906 924 362 301 44;
  • 49) 0.000 002 424 906 924 362 301 44 × 2 = 0 + 0.000 004 849 813 848 724 602 88;
  • 50) 0.000 004 849 813 848 724 602 88 × 2 = 0 + 0.000 009 699 627 697 449 205 76;
  • 51) 0.000 009 699 627 697 449 205 76 × 2 = 0 + 0.000 019 399 255 394 898 411 52;
  • 52) 0.000 019 399 255 394 898 411 52 × 2 = 0 + 0.000 038 798 510 789 796 823 04;
  • 53) 0.000 038 798 510 789 796 823 04 × 2 = 0 + 0.000 077 597 021 579 593 646 08;
  • 54) 0.000 077 597 021 579 593 646 08 × 2 = 0 + 0.000 155 194 043 159 187 292 16;
  • 55) 0.000 155 194 043 159 187 292 16 × 2 = 0 + 0.000 310 388 086 318 374 584 32;
  • 56) 0.000 310 388 086 318 374 584 32 × 2 = 0 + 0.000 620 776 172 636 749 168 64;
  • 57) 0.000 620 776 172 636 749 168 64 × 2 = 0 + 0.001 241 552 345 273 498 337 28;
  • 58) 0.001 241 552 345 273 498 337 28 × 2 = 0 + 0.002 483 104 690 546 996 674 56;
  • 59) 0.002 483 104 690 546 996 674 56 × 2 = 0 + 0.004 966 209 381 093 993 349 12;
  • 60) 0.004 966 209 381 093 993 349 12 × 2 = 0 + 0.009 932 418 762 187 986 698 24;
  • 61) 0.009 932 418 762 187 986 698 24 × 2 = 0 + 0.019 864 837 524 375 973 396 48;
  • 62) 0.019 864 837 524 375 973 396 48 × 2 = 0 + 0.039 729 675 048 751 946 792 96;
  • 63) 0.039 729 675 048 751 946 792 96 × 2 = 0 + 0.079 459 350 097 503 893 585 92;
  • 64) 0.079 459 350 097 503 893 585 92 × 2 = 0 + 0.158 918 700 195 007 787 171 84;
  • 65) 0.158 918 700 195 007 787 171 84 × 2 = 0 + 0.317 837 400 390 015 574 343 68;
  • 66) 0.317 837 400 390 015 574 343 68 × 2 = 0 + 0.635 674 800 780 031 148 687 36;
  • 67) 0.635 674 800 780 031 148 687 36 × 2 = 1 + 0.271 349 601 560 062 297 374 72;
  • 68) 0.271 349 601 560 062 297 374 72 × 2 = 0 + 0.542 699 203 120 124 594 749 44;
  • 69) 0.542 699 203 120 124 594 749 44 × 2 = 1 + 0.085 398 406 240 249 189 498 88;
  • 70) 0.085 398 406 240 249 189 498 88 × 2 = 0 + 0.170 796 812 480 498 378 997 76;
  • 71) 0.170 796 812 480 498 378 997 76 × 2 = 0 + 0.341 593 624 960 996 757 995 52;
  • 72) 0.341 593 624 960 996 757 995 52 × 2 = 0 + 0.683 187 249 921 993 515 991 04;
  • 73) 0.683 187 249 921 993 515 991 04 × 2 = 1 + 0.366 374 499 843 987 031 982 08;
  • 74) 0.366 374 499 843 987 031 982 08 × 2 = 0 + 0.732 748 999 687 974 063 964 16;
  • 75) 0.732 748 999 687 974 063 964 16 × 2 = 1 + 0.465 497 999 375 948 127 928 32;
  • 76) 0.465 497 999 375 948 127 928 32 × 2 = 0 + 0.930 995 998 751 896 255 856 64;
  • 77) 0.930 995 998 751 896 255 856 64 × 2 = 1 + 0.861 991 997 503 792 511 713 28;
  • 78) 0.861 991 997 503 792 511 713 28 × 2 = 1 + 0.723 983 995 007 585 023 426 56;
  • 79) 0.723 983 995 007 585 023 426 56 × 2 = 1 + 0.447 967 990 015 170 046 853 12;
  • 80) 0.447 967 990 015 170 046 853 12 × 2 = 0 + 0.895 935 980 030 340 093 706 24;
  • 81) 0.895 935 980 030 340 093 706 24 × 2 = 1 + 0.791 871 960 060 680 187 412 48;
  • 82) 0.791 871 960 060 680 187 412 48 × 2 = 1 + 0.583 743 920 121 360 374 824 96;
  • 83) 0.583 743 920 121 360 374 824 96 × 2 = 1 + 0.167 487 840 242 720 749 649 92;
  • 84) 0.167 487 840 242 720 749 649 92 × 2 = 0 + 0.334 975 680 485 441 499 299 84;
  • 85) 0.334 975 680 485 441 499 299 84 × 2 = 0 + 0.669 951 360 970 882 998 599 68;
  • 86) 0.669 951 360 970 882 998 599 68 × 2 = 1 + 0.339 902 721 941 765 997 199 36;
  • 87) 0.339 902 721 941 765 997 199 36 × 2 = 0 + 0.679 805 443 883 531 994 398 72;
  • 88) 0.679 805 443 883 531 994 398 72 × 2 = 1 + 0.359 610 887 767 063 988 797 44;
  • 89) 0.359 610 887 767 063 988 797 44 × 2 = 0 + 0.719 221 775 534 127 977 594 88;
  • 90) 0.719 221 775 534 127 977 594 88 × 2 = 1 + 0.438 443 551 068 255 955 189 76;
  • 91) 0.438 443 551 068 255 955 189 76 × 2 = 0 + 0.876 887 102 136 511 910 379 52;
  • 92) 0.876 887 102 136 511 910 379 52 × 2 = 1 + 0.753 774 204 273 023 820 759 04;
  • 93) 0.753 774 204 273 023 820 759 04 × 2 = 1 + 0.507 548 408 546 047 641 518 08;
  • 94) 0.507 548 408 546 047 641 518 08 × 2 = 1 + 0.015 096 817 092 095 283 036 16;
  • 95) 0.015 096 817 092 095 283 036 16 × 2 = 0 + 0.030 193 634 184 190 566 072 32;
  • 96) 0.030 193 634 184 190 566 072 32 × 2 = 0 + 0.060 387 268 368 381 132 144 64;
  • 97) 0.060 387 268 368 381 132 144 64 × 2 = 0 + 0.120 774 536 736 762 264 289 28;
  • 98) 0.120 774 536 736 762 264 289 28 × 2 = 0 + 0.241 549 073 473 524 528 578 56;
  • 99) 0.241 549 073 473 524 528 578 56 × 2 = 0 + 0.483 098 146 947 049 057 157 12;
  • 100) 0.483 098 146 947 049 057 157 12 × 2 = 0 + 0.966 196 293 894 098 114 314 24;
  • 101) 0.966 196 293 894 098 114 314 24 × 2 = 1 + 0.932 392 587 788 196 228 628 48;
  • 102) 0.932 392 587 788 196 228 628 48 × 2 = 1 + 0.864 785 175 576 392 457 256 96;
  • 103) 0.864 785 175 576 392 457 256 96 × 2 = 1 + 0.729 570 351 152 784 914 513 92;
  • 104) 0.729 570 351 152 784 914 513 92 × 2 = 1 + 0.459 140 702 305 569 829 027 84;
  • 105) 0.459 140 702 305 569 829 027 84 × 2 = 0 + 0.918 281 404 611 139 658 055 68;
  • 106) 0.918 281 404 611 139 658 055 68 × 2 = 1 + 0.836 562 809 222 279 316 111 36;
  • 107) 0.836 562 809 222 279 316 111 36 × 2 = 1 + 0.673 125 618 444 558 632 222 72;
  • 108) 0.673 125 618 444 558 632 222 72 × 2 = 1 + 0.346 251 236 889 117 264 445 44;
  • 109) 0.346 251 236 889 117 264 445 44 × 2 = 0 + 0.692 502 473 778 234 528 890 88;
  • 110) 0.692 502 473 778 234 528 890 88 × 2 = 1 + 0.385 004 947 556 469 057 781 76;
  • 111) 0.385 004 947 556 469 057 781 76 × 2 = 0 + 0.770 009 895 112 938 115 563 52;
  • 112) 0.770 009 895 112 938 115 563 52 × 2 = 1 + 0.540 019 790 225 876 231 127 04;
  • 113) 0.540 019 790 225 876 231 127 04 × 2 = 1 + 0.080 039 580 451 752 462 254 08;
  • 114) 0.080 039 580 451 752 462 254 08 × 2 = 0 + 0.160 079 160 903 504 924 508 16;
  • 115) 0.160 079 160 903 504 924 508 16 × 2 = 0 + 0.320 158 321 807 009 849 016 32;
  • 116) 0.320 158 321 807 009 849 016 32 × 2 = 0 + 0.640 316 643 614 019 698 032 64;
  • 117) 0.640 316 643 614 019 698 032 64 × 2 = 1 + 0.280 633 287 228 039 396 065 28;
  • 118) 0.280 633 287 228 039 396 065 28 × 2 = 0 + 0.561 266 574 456 078 792 130 56;
  • 119) 0.561 266 574 456 078 792 130 56 × 2 = 1 + 0.122 533 148 912 157 584 261 12;

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:


0.000 000 000 000 000 000 008 615(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1010 1110 1110 0101 0101 1100 0000 1111 0111 0101 1000 101(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 615(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1010 1110 1110 0101 0101 1100 0000 1111 0111 0101 1000 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:


0.000 000 000 000 000 000 008 615(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1010 1110 1110 0101 0101 1100 0000 1111 0111 0101 1000 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1010 1110 1110 0101 0101 1100 0000 1111 0111 0101 1000 101(2) × 20 =

1.0100 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101(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 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101


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 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101 =

0100 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101


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 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101


Decimal number 0.000 000 000 000 000 000 008 615 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0101 0111 0111 0010 1010 1110 0000 0111 1011 1010 1100 0101

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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