0.000 000 000 000 000 000 008 52 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 008 52(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 52(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 52.

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 52 × 2 = 0 + 0.000 000 000 000 000 000 017 04;
  • 2) 0.000 000 000 000 000 000 017 04 × 2 = 0 + 0.000 000 000 000 000 000 034 08;
  • 3) 0.000 000 000 000 000 000 034 08 × 2 = 0 + 0.000 000 000 000 000 000 068 16;
  • 4) 0.000 000 000 000 000 000 068 16 × 2 = 0 + 0.000 000 000 000 000 000 136 32;
  • 5) 0.000 000 000 000 000 000 136 32 × 2 = 0 + 0.000 000 000 000 000 000 272 64;
  • 6) 0.000 000 000 000 000 000 272 64 × 2 = 0 + 0.000 000 000 000 000 000 545 28;
  • 7) 0.000 000 000 000 000 000 545 28 × 2 = 0 + 0.000 000 000 000 000 001 090 56;
  • 8) 0.000 000 000 000 000 001 090 56 × 2 = 0 + 0.000 000 000 000 000 002 181 12;
  • 9) 0.000 000 000 000 000 002 181 12 × 2 = 0 + 0.000 000 000 000 000 004 362 24;
  • 10) 0.000 000 000 000 000 004 362 24 × 2 = 0 + 0.000 000 000 000 000 008 724 48;
  • 11) 0.000 000 000 000 000 008 724 48 × 2 = 0 + 0.000 000 000 000 000 017 448 96;
  • 12) 0.000 000 000 000 000 017 448 96 × 2 = 0 + 0.000 000 000 000 000 034 897 92;
  • 13) 0.000 000 000 000 000 034 897 92 × 2 = 0 + 0.000 000 000 000 000 069 795 84;
  • 14) 0.000 000 000 000 000 069 795 84 × 2 = 0 + 0.000 000 000 000 000 139 591 68;
  • 15) 0.000 000 000 000 000 139 591 68 × 2 = 0 + 0.000 000 000 000 000 279 183 36;
  • 16) 0.000 000 000 000 000 279 183 36 × 2 = 0 + 0.000 000 000 000 000 558 366 72;
  • 17) 0.000 000 000 000 000 558 366 72 × 2 = 0 + 0.000 000 000 000 001 116 733 44;
  • 18) 0.000 000 000 000 001 116 733 44 × 2 = 0 + 0.000 000 000 000 002 233 466 88;
  • 19) 0.000 000 000 000 002 233 466 88 × 2 = 0 + 0.000 000 000 000 004 466 933 76;
  • 20) 0.000 000 000 000 004 466 933 76 × 2 = 0 + 0.000 000 000 000 008 933 867 52;
  • 21) 0.000 000 000 000 008 933 867 52 × 2 = 0 + 0.000 000 000 000 017 867 735 04;
  • 22) 0.000 000 000 000 017 867 735 04 × 2 = 0 + 0.000 000 000 000 035 735 470 08;
  • 23) 0.000 000 000 000 035 735 470 08 × 2 = 0 + 0.000 000 000 000 071 470 940 16;
  • 24) 0.000 000 000 000 071 470 940 16 × 2 = 0 + 0.000 000 000 000 142 941 880 32;
  • 25) 0.000 000 000 000 142 941 880 32 × 2 = 0 + 0.000 000 000 000 285 883 760 64;
  • 26) 0.000 000 000 000 285 883 760 64 × 2 = 0 + 0.000 000 000 000 571 767 521 28;
  • 27) 0.000 000 000 000 571 767 521 28 × 2 = 0 + 0.000 000 000 001 143 535 042 56;
  • 28) 0.000 000 000 001 143 535 042 56 × 2 = 0 + 0.000 000 000 002 287 070 085 12;
  • 29) 0.000 000 000 002 287 070 085 12 × 2 = 0 + 0.000 000 000 004 574 140 170 24;
  • 30) 0.000 000 000 004 574 140 170 24 × 2 = 0 + 0.000 000 000 009 148 280 340 48;
  • 31) 0.000 000 000 009 148 280 340 48 × 2 = 0 + 0.000 000 000 018 296 560 680 96;
  • 32) 0.000 000 000 018 296 560 680 96 × 2 = 0 + 0.000 000 000 036 593 121 361 92;
  • 33) 0.000 000 000 036 593 121 361 92 × 2 = 0 + 0.000 000 000 073 186 242 723 84;
  • 34) 0.000 000 000 073 186 242 723 84 × 2 = 0 + 0.000 000 000 146 372 485 447 68;
  • 35) 0.000 000 000 146 372 485 447 68 × 2 = 0 + 0.000 000 000 292 744 970 895 36;
  • 36) 0.000 000 000 292 744 970 895 36 × 2 = 0 + 0.000 000 000 585 489 941 790 72;
  • 37) 0.000 000 000 585 489 941 790 72 × 2 = 0 + 0.000 000 001 170 979 883 581 44;
  • 38) 0.000 000 001 170 979 883 581 44 × 2 = 0 + 0.000 000 002 341 959 767 162 88;
  • 39) 0.000 000 002 341 959 767 162 88 × 2 = 0 + 0.000 000 004 683 919 534 325 76;
  • 40) 0.000 000 004 683 919 534 325 76 × 2 = 0 + 0.000 000 009 367 839 068 651 52;
  • 41) 0.000 000 009 367 839 068 651 52 × 2 = 0 + 0.000 000 018 735 678 137 303 04;
  • 42) 0.000 000 018 735 678 137 303 04 × 2 = 0 + 0.000 000 037 471 356 274 606 08;
  • 43) 0.000 000 037 471 356 274 606 08 × 2 = 0 + 0.000 000 074 942 712 549 212 16;
  • 44) 0.000 000 074 942 712 549 212 16 × 2 = 0 + 0.000 000 149 885 425 098 424 32;
  • 45) 0.000 000 149 885 425 098 424 32 × 2 = 0 + 0.000 000 299 770 850 196 848 64;
  • 46) 0.000 000 299 770 850 196 848 64 × 2 = 0 + 0.000 000 599 541 700 393 697 28;
  • 47) 0.000 000 599 541 700 393 697 28 × 2 = 0 + 0.000 001 199 083 400 787 394 56;
  • 48) 0.000 001 199 083 400 787 394 56 × 2 = 0 + 0.000 002 398 166 801 574 789 12;
  • 49) 0.000 002 398 166 801 574 789 12 × 2 = 0 + 0.000 004 796 333 603 149 578 24;
  • 50) 0.000 004 796 333 603 149 578 24 × 2 = 0 + 0.000 009 592 667 206 299 156 48;
  • 51) 0.000 009 592 667 206 299 156 48 × 2 = 0 + 0.000 019 185 334 412 598 312 96;
  • 52) 0.000 019 185 334 412 598 312 96 × 2 = 0 + 0.000 038 370 668 825 196 625 92;
  • 53) 0.000 038 370 668 825 196 625 92 × 2 = 0 + 0.000 076 741 337 650 393 251 84;
  • 54) 0.000 076 741 337 650 393 251 84 × 2 = 0 + 0.000 153 482 675 300 786 503 68;
  • 55) 0.000 153 482 675 300 786 503 68 × 2 = 0 + 0.000 306 965 350 601 573 007 36;
  • 56) 0.000 306 965 350 601 573 007 36 × 2 = 0 + 0.000 613 930 701 203 146 014 72;
  • 57) 0.000 613 930 701 203 146 014 72 × 2 = 0 + 0.001 227 861 402 406 292 029 44;
  • 58) 0.001 227 861 402 406 292 029 44 × 2 = 0 + 0.002 455 722 804 812 584 058 88;
  • 59) 0.002 455 722 804 812 584 058 88 × 2 = 0 + 0.004 911 445 609 625 168 117 76;
  • 60) 0.004 911 445 609 625 168 117 76 × 2 = 0 + 0.009 822 891 219 250 336 235 52;
  • 61) 0.009 822 891 219 250 336 235 52 × 2 = 0 + 0.019 645 782 438 500 672 471 04;
  • 62) 0.019 645 782 438 500 672 471 04 × 2 = 0 + 0.039 291 564 877 001 344 942 08;
  • 63) 0.039 291 564 877 001 344 942 08 × 2 = 0 + 0.078 583 129 754 002 689 884 16;
  • 64) 0.078 583 129 754 002 689 884 16 × 2 = 0 + 0.157 166 259 508 005 379 768 32;
  • 65) 0.157 166 259 508 005 379 768 32 × 2 = 0 + 0.314 332 519 016 010 759 536 64;
  • 66) 0.314 332 519 016 010 759 536 64 × 2 = 0 + 0.628 665 038 032 021 519 073 28;
  • 67) 0.628 665 038 032 021 519 073 28 × 2 = 1 + 0.257 330 076 064 043 038 146 56;
  • 68) 0.257 330 076 064 043 038 146 56 × 2 = 0 + 0.514 660 152 128 086 076 293 12;
  • 69) 0.514 660 152 128 086 076 293 12 × 2 = 1 + 0.029 320 304 256 172 152 586 24;
  • 70) 0.029 320 304 256 172 152 586 24 × 2 = 0 + 0.058 640 608 512 344 305 172 48;
  • 71) 0.058 640 608 512 344 305 172 48 × 2 = 0 + 0.117 281 217 024 688 610 344 96;
  • 72) 0.117 281 217 024 688 610 344 96 × 2 = 0 + 0.234 562 434 049 377 220 689 92;
  • 73) 0.234 562 434 049 377 220 689 92 × 2 = 0 + 0.469 124 868 098 754 441 379 84;
  • 74) 0.469 124 868 098 754 441 379 84 × 2 = 0 + 0.938 249 736 197 508 882 759 68;
  • 75) 0.938 249 736 197 508 882 759 68 × 2 = 1 + 0.876 499 472 395 017 765 519 36;
  • 76) 0.876 499 472 395 017 765 519 36 × 2 = 1 + 0.752 998 944 790 035 531 038 72;
  • 77) 0.752 998 944 790 035 531 038 72 × 2 = 1 + 0.505 997 889 580 071 062 077 44;
  • 78) 0.505 997 889 580 071 062 077 44 × 2 = 1 + 0.011 995 779 160 142 124 154 88;
  • 79) 0.011 995 779 160 142 124 154 88 × 2 = 0 + 0.023 991 558 320 284 248 309 76;
  • 80) 0.023 991 558 320 284 248 309 76 × 2 = 0 + 0.047 983 116 640 568 496 619 52;
  • 81) 0.047 983 116 640 568 496 619 52 × 2 = 0 + 0.095 966 233 281 136 993 239 04;
  • 82) 0.095 966 233 281 136 993 239 04 × 2 = 0 + 0.191 932 466 562 273 986 478 08;
  • 83) 0.191 932 466 562 273 986 478 08 × 2 = 0 + 0.383 864 933 124 547 972 956 16;
  • 84) 0.383 864 933 124 547 972 956 16 × 2 = 0 + 0.767 729 866 249 095 945 912 32;
  • 85) 0.767 729 866 249 095 945 912 32 × 2 = 1 + 0.535 459 732 498 191 891 824 64;
  • 86) 0.535 459 732 498 191 891 824 64 × 2 = 1 + 0.070 919 464 996 383 783 649 28;
  • 87) 0.070 919 464 996 383 783 649 28 × 2 = 0 + 0.141 838 929 992 767 567 298 56;
  • 88) 0.141 838 929 992 767 567 298 56 × 2 = 0 + 0.283 677 859 985 535 134 597 12;
  • 89) 0.283 677 859 985 535 134 597 12 × 2 = 0 + 0.567 355 719 971 070 269 194 24;
  • 90) 0.567 355 719 971 070 269 194 24 × 2 = 1 + 0.134 711 439 942 140 538 388 48;
  • 91) 0.134 711 439 942 140 538 388 48 × 2 = 0 + 0.269 422 879 884 281 076 776 96;
  • 92) 0.269 422 879 884 281 076 776 96 × 2 = 0 + 0.538 845 759 768 562 153 553 92;
  • 93) 0.538 845 759 768 562 153 553 92 × 2 = 1 + 0.077 691 519 537 124 307 107 84;
  • 94) 0.077 691 519 537 124 307 107 84 × 2 = 0 + 0.155 383 039 074 248 614 215 68;
  • 95) 0.155 383 039 074 248 614 215 68 × 2 = 0 + 0.310 766 078 148 497 228 431 36;
  • 96) 0.310 766 078 148 497 228 431 36 × 2 = 0 + 0.621 532 156 296 994 456 862 72;
  • 97) 0.621 532 156 296 994 456 862 72 × 2 = 1 + 0.243 064 312 593 988 913 725 44;
  • 98) 0.243 064 312 593 988 913 725 44 × 2 = 0 + 0.486 128 625 187 977 827 450 88;
  • 99) 0.486 128 625 187 977 827 450 88 × 2 = 0 + 0.972 257 250 375 955 654 901 76;
  • 100) 0.972 257 250 375 955 654 901 76 × 2 = 1 + 0.944 514 500 751 911 309 803 52;
  • 101) 0.944 514 500 751 911 309 803 52 × 2 = 1 + 0.889 029 001 503 822 619 607 04;
  • 102) 0.889 029 001 503 822 619 607 04 × 2 = 1 + 0.778 058 003 007 645 239 214 08;
  • 103) 0.778 058 003 007 645 239 214 08 × 2 = 1 + 0.556 116 006 015 290 478 428 16;
  • 104) 0.556 116 006 015 290 478 428 16 × 2 = 1 + 0.112 232 012 030 580 956 856 32;
  • 105) 0.112 232 012 030 580 956 856 32 × 2 = 0 + 0.224 464 024 061 161 913 712 64;
  • 106) 0.224 464 024 061 161 913 712 64 × 2 = 0 + 0.448 928 048 122 323 827 425 28;
  • 107) 0.448 928 048 122 323 827 425 28 × 2 = 0 + 0.897 856 096 244 647 654 850 56;
  • 108) 0.897 856 096 244 647 654 850 56 × 2 = 1 + 0.795 712 192 489 295 309 701 12;
  • 109) 0.795 712 192 489 295 309 701 12 × 2 = 1 + 0.591 424 384 978 590 619 402 24;
  • 110) 0.591 424 384 978 590 619 402 24 × 2 = 1 + 0.182 848 769 957 181 238 804 48;
  • 111) 0.182 848 769 957 181 238 804 48 × 2 = 0 + 0.365 697 539 914 362 477 608 96;
  • 112) 0.365 697 539 914 362 477 608 96 × 2 = 0 + 0.731 395 079 828 724 955 217 92;
  • 113) 0.731 395 079 828 724 955 217 92 × 2 = 1 + 0.462 790 159 657 449 910 435 84;
  • 114) 0.462 790 159 657 449 910 435 84 × 2 = 0 + 0.925 580 319 314 899 820 871 68;
  • 115) 0.925 580 319 314 899 820 871 68 × 2 = 1 + 0.851 160 638 629 799 641 743 36;
  • 116) 0.851 160 638 629 799 641 743 36 × 2 = 1 + 0.702 321 277 259 599 283 486 72;
  • 117) 0.702 321 277 259 599 283 486 72 × 2 = 1 + 0.404 642 554 519 198 566 973 44;
  • 118) 0.404 642 554 519 198 566 973 44 × 2 = 0 + 0.809 285 109 038 397 133 946 88;
  • 119) 0.809 285 109 038 397 133 946 88 × 2 = 1 + 0.618 570 218 076 794 267 893 76;

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 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1100 0000 1100 0100 1000 1001 1111 0001 1100 1011 101(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 008 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1100 0000 1100 0100 1000 1001 1111 0001 1100 1011 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 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1100 0000 1100 0100 1000 1001 1111 0001 1100 1011 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 0011 1100 0000 1100 0100 1000 1001 1111 0001 1100 1011 101(2) × 20 =

1.0100 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 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 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 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 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 1101 =

0100 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 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 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 1101


Decimal number 0.000 000 000 000 000 000 008 52 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0100 0001 1110 0000 0110 0010 0100 0100 1111 1000 1110 0101 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