0.000 000 000 000 000 000 008 07 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 07(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 07(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 07.

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 07 × 2 = 0 + 0.000 000 000 000 000 000 016 14;
  • 2) 0.000 000 000 000 000 000 016 14 × 2 = 0 + 0.000 000 000 000 000 000 032 28;
  • 3) 0.000 000 000 000 000 000 032 28 × 2 = 0 + 0.000 000 000 000 000 000 064 56;
  • 4) 0.000 000 000 000 000 000 064 56 × 2 = 0 + 0.000 000 000 000 000 000 129 12;
  • 5) 0.000 000 000 000 000 000 129 12 × 2 = 0 + 0.000 000 000 000 000 000 258 24;
  • 6) 0.000 000 000 000 000 000 258 24 × 2 = 0 + 0.000 000 000 000 000 000 516 48;
  • 7) 0.000 000 000 000 000 000 516 48 × 2 = 0 + 0.000 000 000 000 000 001 032 96;
  • 8) 0.000 000 000 000 000 001 032 96 × 2 = 0 + 0.000 000 000 000 000 002 065 92;
  • 9) 0.000 000 000 000 000 002 065 92 × 2 = 0 + 0.000 000 000 000 000 004 131 84;
  • 10) 0.000 000 000 000 000 004 131 84 × 2 = 0 + 0.000 000 000 000 000 008 263 68;
  • 11) 0.000 000 000 000 000 008 263 68 × 2 = 0 + 0.000 000 000 000 000 016 527 36;
  • 12) 0.000 000 000 000 000 016 527 36 × 2 = 0 + 0.000 000 000 000 000 033 054 72;
  • 13) 0.000 000 000 000 000 033 054 72 × 2 = 0 + 0.000 000 000 000 000 066 109 44;
  • 14) 0.000 000 000 000 000 066 109 44 × 2 = 0 + 0.000 000 000 000 000 132 218 88;
  • 15) 0.000 000 000 000 000 132 218 88 × 2 = 0 + 0.000 000 000 000 000 264 437 76;
  • 16) 0.000 000 000 000 000 264 437 76 × 2 = 0 + 0.000 000 000 000 000 528 875 52;
  • 17) 0.000 000 000 000 000 528 875 52 × 2 = 0 + 0.000 000 000 000 001 057 751 04;
  • 18) 0.000 000 000 000 001 057 751 04 × 2 = 0 + 0.000 000 000 000 002 115 502 08;
  • 19) 0.000 000 000 000 002 115 502 08 × 2 = 0 + 0.000 000 000 000 004 231 004 16;
  • 20) 0.000 000 000 000 004 231 004 16 × 2 = 0 + 0.000 000 000 000 008 462 008 32;
  • 21) 0.000 000 000 000 008 462 008 32 × 2 = 0 + 0.000 000 000 000 016 924 016 64;
  • 22) 0.000 000 000 000 016 924 016 64 × 2 = 0 + 0.000 000 000 000 033 848 033 28;
  • 23) 0.000 000 000 000 033 848 033 28 × 2 = 0 + 0.000 000 000 000 067 696 066 56;
  • 24) 0.000 000 000 000 067 696 066 56 × 2 = 0 + 0.000 000 000 000 135 392 133 12;
  • 25) 0.000 000 000 000 135 392 133 12 × 2 = 0 + 0.000 000 000 000 270 784 266 24;
  • 26) 0.000 000 000 000 270 784 266 24 × 2 = 0 + 0.000 000 000 000 541 568 532 48;
  • 27) 0.000 000 000 000 541 568 532 48 × 2 = 0 + 0.000 000 000 001 083 137 064 96;
  • 28) 0.000 000 000 001 083 137 064 96 × 2 = 0 + 0.000 000 000 002 166 274 129 92;
  • 29) 0.000 000 000 002 166 274 129 92 × 2 = 0 + 0.000 000 000 004 332 548 259 84;
  • 30) 0.000 000 000 004 332 548 259 84 × 2 = 0 + 0.000 000 000 008 665 096 519 68;
  • 31) 0.000 000 000 008 665 096 519 68 × 2 = 0 + 0.000 000 000 017 330 193 039 36;
  • 32) 0.000 000 000 017 330 193 039 36 × 2 = 0 + 0.000 000 000 034 660 386 078 72;
  • 33) 0.000 000 000 034 660 386 078 72 × 2 = 0 + 0.000 000 000 069 320 772 157 44;
  • 34) 0.000 000 000 069 320 772 157 44 × 2 = 0 + 0.000 000 000 138 641 544 314 88;
  • 35) 0.000 000 000 138 641 544 314 88 × 2 = 0 + 0.000 000 000 277 283 088 629 76;
  • 36) 0.000 000 000 277 283 088 629 76 × 2 = 0 + 0.000 000 000 554 566 177 259 52;
  • 37) 0.000 000 000 554 566 177 259 52 × 2 = 0 + 0.000 000 001 109 132 354 519 04;
  • 38) 0.000 000 001 109 132 354 519 04 × 2 = 0 + 0.000 000 002 218 264 709 038 08;
  • 39) 0.000 000 002 218 264 709 038 08 × 2 = 0 + 0.000 000 004 436 529 418 076 16;
  • 40) 0.000 000 004 436 529 418 076 16 × 2 = 0 + 0.000 000 008 873 058 836 152 32;
  • 41) 0.000 000 008 873 058 836 152 32 × 2 = 0 + 0.000 000 017 746 117 672 304 64;
  • 42) 0.000 000 017 746 117 672 304 64 × 2 = 0 + 0.000 000 035 492 235 344 609 28;
  • 43) 0.000 000 035 492 235 344 609 28 × 2 = 0 + 0.000 000 070 984 470 689 218 56;
  • 44) 0.000 000 070 984 470 689 218 56 × 2 = 0 + 0.000 000 141 968 941 378 437 12;
  • 45) 0.000 000 141 968 941 378 437 12 × 2 = 0 + 0.000 000 283 937 882 756 874 24;
  • 46) 0.000 000 283 937 882 756 874 24 × 2 = 0 + 0.000 000 567 875 765 513 748 48;
  • 47) 0.000 000 567 875 765 513 748 48 × 2 = 0 + 0.000 001 135 751 531 027 496 96;
  • 48) 0.000 001 135 751 531 027 496 96 × 2 = 0 + 0.000 002 271 503 062 054 993 92;
  • 49) 0.000 002 271 503 062 054 993 92 × 2 = 0 + 0.000 004 543 006 124 109 987 84;
  • 50) 0.000 004 543 006 124 109 987 84 × 2 = 0 + 0.000 009 086 012 248 219 975 68;
  • 51) 0.000 009 086 012 248 219 975 68 × 2 = 0 + 0.000 018 172 024 496 439 951 36;
  • 52) 0.000 018 172 024 496 439 951 36 × 2 = 0 + 0.000 036 344 048 992 879 902 72;
  • 53) 0.000 036 344 048 992 879 902 72 × 2 = 0 + 0.000 072 688 097 985 759 805 44;
  • 54) 0.000 072 688 097 985 759 805 44 × 2 = 0 + 0.000 145 376 195 971 519 610 88;
  • 55) 0.000 145 376 195 971 519 610 88 × 2 = 0 + 0.000 290 752 391 943 039 221 76;
  • 56) 0.000 290 752 391 943 039 221 76 × 2 = 0 + 0.000 581 504 783 886 078 443 52;
  • 57) 0.000 581 504 783 886 078 443 52 × 2 = 0 + 0.001 163 009 567 772 156 887 04;
  • 58) 0.001 163 009 567 772 156 887 04 × 2 = 0 + 0.002 326 019 135 544 313 774 08;
  • 59) 0.002 326 019 135 544 313 774 08 × 2 = 0 + 0.004 652 038 271 088 627 548 16;
  • 60) 0.004 652 038 271 088 627 548 16 × 2 = 0 + 0.009 304 076 542 177 255 096 32;
  • 61) 0.009 304 076 542 177 255 096 32 × 2 = 0 + 0.018 608 153 084 354 510 192 64;
  • 62) 0.018 608 153 084 354 510 192 64 × 2 = 0 + 0.037 216 306 168 709 020 385 28;
  • 63) 0.037 216 306 168 709 020 385 28 × 2 = 0 + 0.074 432 612 337 418 040 770 56;
  • 64) 0.074 432 612 337 418 040 770 56 × 2 = 0 + 0.148 865 224 674 836 081 541 12;
  • 65) 0.148 865 224 674 836 081 541 12 × 2 = 0 + 0.297 730 449 349 672 163 082 24;
  • 66) 0.297 730 449 349 672 163 082 24 × 2 = 0 + 0.595 460 898 699 344 326 164 48;
  • 67) 0.595 460 898 699 344 326 164 48 × 2 = 1 + 0.190 921 797 398 688 652 328 96;
  • 68) 0.190 921 797 398 688 652 328 96 × 2 = 0 + 0.381 843 594 797 377 304 657 92;
  • 69) 0.381 843 594 797 377 304 657 92 × 2 = 0 + 0.763 687 189 594 754 609 315 84;
  • 70) 0.763 687 189 594 754 609 315 84 × 2 = 1 + 0.527 374 379 189 509 218 631 68;
  • 71) 0.527 374 379 189 509 218 631 68 × 2 = 1 + 0.054 748 758 379 018 437 263 36;
  • 72) 0.054 748 758 379 018 437 263 36 × 2 = 0 + 0.109 497 516 758 036 874 526 72;
  • 73) 0.109 497 516 758 036 874 526 72 × 2 = 0 + 0.218 995 033 516 073 749 053 44;
  • 74) 0.218 995 033 516 073 749 053 44 × 2 = 0 + 0.437 990 067 032 147 498 106 88;
  • 75) 0.437 990 067 032 147 498 106 88 × 2 = 0 + 0.875 980 134 064 294 996 213 76;
  • 76) 0.875 980 134 064 294 996 213 76 × 2 = 1 + 0.751 960 268 128 589 992 427 52;
  • 77) 0.751 960 268 128 589 992 427 52 × 2 = 1 + 0.503 920 536 257 179 984 855 04;
  • 78) 0.503 920 536 257 179 984 855 04 × 2 = 1 + 0.007 841 072 514 359 969 710 08;
  • 79) 0.007 841 072 514 359 969 710 08 × 2 = 0 + 0.015 682 145 028 719 939 420 16;
  • 80) 0.015 682 145 028 719 939 420 16 × 2 = 0 + 0.031 364 290 057 439 878 840 32;
  • 81) 0.031 364 290 057 439 878 840 32 × 2 = 0 + 0.062 728 580 114 879 757 680 64;
  • 82) 0.062 728 580 114 879 757 680 64 × 2 = 0 + 0.125 457 160 229 759 515 361 28;
  • 83) 0.125 457 160 229 759 515 361 28 × 2 = 0 + 0.250 914 320 459 519 030 722 56;
  • 84) 0.250 914 320 459 519 030 722 56 × 2 = 0 + 0.501 828 640 919 038 061 445 12;
  • 85) 0.501 828 640 919 038 061 445 12 × 2 = 1 + 0.003 657 281 838 076 122 890 24;
  • 86) 0.003 657 281 838 076 122 890 24 × 2 = 0 + 0.007 314 563 676 152 245 780 48;
  • 87) 0.007 314 563 676 152 245 780 48 × 2 = 0 + 0.014 629 127 352 304 491 560 96;
  • 88) 0.014 629 127 352 304 491 560 96 × 2 = 0 + 0.029 258 254 704 608 983 121 92;
  • 89) 0.029 258 254 704 608 983 121 92 × 2 = 0 + 0.058 516 509 409 217 966 243 84;
  • 90) 0.058 516 509 409 217 966 243 84 × 2 = 0 + 0.117 033 018 818 435 932 487 68;
  • 91) 0.117 033 018 818 435 932 487 68 × 2 = 0 + 0.234 066 037 636 871 864 975 36;
  • 92) 0.234 066 037 636 871 864 975 36 × 2 = 0 + 0.468 132 075 273 743 729 950 72;
  • 93) 0.468 132 075 273 743 729 950 72 × 2 = 0 + 0.936 264 150 547 487 459 901 44;
  • 94) 0.936 264 150 547 487 459 901 44 × 2 = 1 + 0.872 528 301 094 974 919 802 88;
  • 95) 0.872 528 301 094 974 919 802 88 × 2 = 1 + 0.745 056 602 189 949 839 605 76;
  • 96) 0.745 056 602 189 949 839 605 76 × 2 = 1 + 0.490 113 204 379 899 679 211 52;
  • 97) 0.490 113 204 379 899 679 211 52 × 2 = 0 + 0.980 226 408 759 799 358 423 04;
  • 98) 0.980 226 408 759 799 358 423 04 × 2 = 1 + 0.960 452 817 519 598 716 846 08;
  • 99) 0.960 452 817 519 598 716 846 08 × 2 = 1 + 0.920 905 635 039 197 433 692 16;
  • 100) 0.920 905 635 039 197 433 692 16 × 2 = 1 + 0.841 811 270 078 394 867 384 32;
  • 101) 0.841 811 270 078 394 867 384 32 × 2 = 1 + 0.683 622 540 156 789 734 768 64;
  • 102) 0.683 622 540 156 789 734 768 64 × 2 = 1 + 0.367 245 080 313 579 469 537 28;
  • 103) 0.367 245 080 313 579 469 537 28 × 2 = 0 + 0.734 490 160 627 158 939 074 56;
  • 104) 0.734 490 160 627 158 939 074 56 × 2 = 1 + 0.468 980 321 254 317 878 149 12;
  • 105) 0.468 980 321 254 317 878 149 12 × 2 = 0 + 0.937 960 642 508 635 756 298 24;
  • 106) 0.937 960 642 508 635 756 298 24 × 2 = 1 + 0.875 921 285 017 271 512 596 48;
  • 107) 0.875 921 285 017 271 512 596 48 × 2 = 1 + 0.751 842 570 034 543 025 192 96;
  • 108) 0.751 842 570 034 543 025 192 96 × 2 = 1 + 0.503 685 140 069 086 050 385 92;
  • 109) 0.503 685 140 069 086 050 385 92 × 2 = 1 + 0.007 370 280 138 172 100 771 84;
  • 110) 0.007 370 280 138 172 100 771 84 × 2 = 0 + 0.014 740 560 276 344 201 543 68;
  • 111) 0.014 740 560 276 344 201 543 68 × 2 = 0 + 0.029 481 120 552 688 403 087 36;
  • 112) 0.029 481 120 552 688 403 087 36 × 2 = 0 + 0.058 962 241 105 376 806 174 72;
  • 113) 0.058 962 241 105 376 806 174 72 × 2 = 0 + 0.117 924 482 210 753 612 349 44;
  • 114) 0.117 924 482 210 753 612 349 44 × 2 = 0 + 0.235 848 964 421 507 224 698 88;
  • 115) 0.235 848 964 421 507 224 698 88 × 2 = 0 + 0.471 697 928 843 014 449 397 76;
  • 116) 0.471 697 928 843 014 449 397 76 × 2 = 0 + 0.943 395 857 686 028 898 795 52;
  • 117) 0.943 395 857 686 028 898 795 52 × 2 = 1 + 0.886 791 715 372 057 797 591 04;
  • 118) 0.886 791 715 372 057 797 591 04 × 2 = 1 + 0.773 583 430 744 115 595 182 08;
  • 119) 0.773 583 430 744 115 595 182 08 × 2 = 1 + 0.547 166 861 488 231 190 364 16;

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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 0001 1100 0000 1000 0000 0111 0111 1101 0111 1000 0000 111(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 0001 1100 0000 1000 0000 0111 0111 1101 0111 1000 0000 111(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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 0001 1100 0000 1000 0000 0111 0111 1101 0111 1000 0000 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 0001 1100 0000 1000 0000 0111 0111 1101 0111 1000 0000 111(2) × 20 =

1.0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111(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):
1.0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111


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. 0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111 =

0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111


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) =
0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111


Decimal number 0.000 000 000 000 000 000 008 07 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0011 0000 1110 0000 0100 0000 0011 1011 1110 1011 1100 0000 0111

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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