-0.000 000 000 000 000 000 001 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 001 1(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
-0.000 000 000 000 000 000 001 1(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. Start with the positive version of the number:

|-0.000 000 000 000 000 000 001 1| = 0.000 000 000 000 000 000 001 1


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 001 1.

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) 0.000 000 000 000 000 000 001 1 × 2 = 0 + 0.000 000 000 000 000 000 002 2;
  • 2) 0.000 000 000 000 000 000 002 2 × 2 = 0 + 0.000 000 000 000 000 000 004 4;
  • 3) 0.000 000 000 000 000 000 004 4 × 2 = 0 + 0.000 000 000 000 000 000 008 8;
  • 4) 0.000 000 000 000 000 000 008 8 × 2 = 0 + 0.000 000 000 000 000 000 017 6;
  • 5) 0.000 000 000 000 000 000 017 6 × 2 = 0 + 0.000 000 000 000 000 000 035 2;
  • 6) 0.000 000 000 000 000 000 035 2 × 2 = 0 + 0.000 000 000 000 000 000 070 4;
  • 7) 0.000 000 000 000 000 000 070 4 × 2 = 0 + 0.000 000 000 000 000 000 140 8;
  • 8) 0.000 000 000 000 000 000 140 8 × 2 = 0 + 0.000 000 000 000 000 000 281 6;
  • 9) 0.000 000 000 000 000 000 281 6 × 2 = 0 + 0.000 000 000 000 000 000 563 2;
  • 10) 0.000 000 000 000 000 000 563 2 × 2 = 0 + 0.000 000 000 000 000 001 126 4;
  • 11) 0.000 000 000 000 000 001 126 4 × 2 = 0 + 0.000 000 000 000 000 002 252 8;
  • 12) 0.000 000 000 000 000 002 252 8 × 2 = 0 + 0.000 000 000 000 000 004 505 6;
  • 13) 0.000 000 000 000 000 004 505 6 × 2 = 0 + 0.000 000 000 000 000 009 011 2;
  • 14) 0.000 000 000 000 000 009 011 2 × 2 = 0 + 0.000 000 000 000 000 018 022 4;
  • 15) 0.000 000 000 000 000 018 022 4 × 2 = 0 + 0.000 000 000 000 000 036 044 8;
  • 16) 0.000 000 000 000 000 036 044 8 × 2 = 0 + 0.000 000 000 000 000 072 089 6;
  • 17) 0.000 000 000 000 000 072 089 6 × 2 = 0 + 0.000 000 000 000 000 144 179 2;
  • 18) 0.000 000 000 000 000 144 179 2 × 2 = 0 + 0.000 000 000 000 000 288 358 4;
  • 19) 0.000 000 000 000 000 288 358 4 × 2 = 0 + 0.000 000 000 000 000 576 716 8;
  • 20) 0.000 000 000 000 000 576 716 8 × 2 = 0 + 0.000 000 000 000 001 153 433 6;
  • 21) 0.000 000 000 000 001 153 433 6 × 2 = 0 + 0.000 000 000 000 002 306 867 2;
  • 22) 0.000 000 000 000 002 306 867 2 × 2 = 0 + 0.000 000 000 000 004 613 734 4;
  • 23) 0.000 000 000 000 004 613 734 4 × 2 = 0 + 0.000 000 000 000 009 227 468 8;
  • 24) 0.000 000 000 000 009 227 468 8 × 2 = 0 + 0.000 000 000 000 018 454 937 6;
  • 25) 0.000 000 000 000 018 454 937 6 × 2 = 0 + 0.000 000 000 000 036 909 875 2;
  • 26) 0.000 000 000 000 036 909 875 2 × 2 = 0 + 0.000 000 000 000 073 819 750 4;
  • 27) 0.000 000 000 000 073 819 750 4 × 2 = 0 + 0.000 000 000 000 147 639 500 8;
  • 28) 0.000 000 000 000 147 639 500 8 × 2 = 0 + 0.000 000 000 000 295 279 001 6;
  • 29) 0.000 000 000 000 295 279 001 6 × 2 = 0 + 0.000 000 000 000 590 558 003 2;
  • 30) 0.000 000 000 000 590 558 003 2 × 2 = 0 + 0.000 000 000 001 181 116 006 4;
  • 31) 0.000 000 000 001 181 116 006 4 × 2 = 0 + 0.000 000 000 002 362 232 012 8;
  • 32) 0.000 000 000 002 362 232 012 8 × 2 = 0 + 0.000 000 000 004 724 464 025 6;
  • 33) 0.000 000 000 004 724 464 025 6 × 2 = 0 + 0.000 000 000 009 448 928 051 2;
  • 34) 0.000 000 000 009 448 928 051 2 × 2 = 0 + 0.000 000 000 018 897 856 102 4;
  • 35) 0.000 000 000 018 897 856 102 4 × 2 = 0 + 0.000 000 000 037 795 712 204 8;
  • 36) 0.000 000 000 037 795 712 204 8 × 2 = 0 + 0.000 000 000 075 591 424 409 6;
  • 37) 0.000 000 000 075 591 424 409 6 × 2 = 0 + 0.000 000 000 151 182 848 819 2;
  • 38) 0.000 000 000 151 182 848 819 2 × 2 = 0 + 0.000 000 000 302 365 697 638 4;
  • 39) 0.000 000 000 302 365 697 638 4 × 2 = 0 + 0.000 000 000 604 731 395 276 8;
  • 40) 0.000 000 000 604 731 395 276 8 × 2 = 0 + 0.000 000 001 209 462 790 553 6;
  • 41) 0.000 000 001 209 462 790 553 6 × 2 = 0 + 0.000 000 002 418 925 581 107 2;
  • 42) 0.000 000 002 418 925 581 107 2 × 2 = 0 + 0.000 000 004 837 851 162 214 4;
  • 43) 0.000 000 004 837 851 162 214 4 × 2 = 0 + 0.000 000 009 675 702 324 428 8;
  • 44) 0.000 000 009 675 702 324 428 8 × 2 = 0 + 0.000 000 019 351 404 648 857 6;
  • 45) 0.000 000 019 351 404 648 857 6 × 2 = 0 + 0.000 000 038 702 809 297 715 2;
  • 46) 0.000 000 038 702 809 297 715 2 × 2 = 0 + 0.000 000 077 405 618 595 430 4;
  • 47) 0.000 000 077 405 618 595 430 4 × 2 = 0 + 0.000 000 154 811 237 190 860 8;
  • 48) 0.000 000 154 811 237 190 860 8 × 2 = 0 + 0.000 000 309 622 474 381 721 6;
  • 49) 0.000 000 309 622 474 381 721 6 × 2 = 0 + 0.000 000 619 244 948 763 443 2;
  • 50) 0.000 000 619 244 948 763 443 2 × 2 = 0 + 0.000 001 238 489 897 526 886 4;
  • 51) 0.000 001 238 489 897 526 886 4 × 2 = 0 + 0.000 002 476 979 795 053 772 8;
  • 52) 0.000 002 476 979 795 053 772 8 × 2 = 0 + 0.000 004 953 959 590 107 545 6;
  • 53) 0.000 004 953 959 590 107 545 6 × 2 = 0 + 0.000 009 907 919 180 215 091 2;
  • 54) 0.000 009 907 919 180 215 091 2 × 2 = 0 + 0.000 019 815 838 360 430 182 4;
  • 55) 0.000 019 815 838 360 430 182 4 × 2 = 0 + 0.000 039 631 676 720 860 364 8;
  • 56) 0.000 039 631 676 720 860 364 8 × 2 = 0 + 0.000 079 263 353 441 720 729 6;
  • 57) 0.000 079 263 353 441 720 729 6 × 2 = 0 + 0.000 158 526 706 883 441 459 2;
  • 58) 0.000 158 526 706 883 441 459 2 × 2 = 0 + 0.000 317 053 413 766 882 918 4;
  • 59) 0.000 317 053 413 766 882 918 4 × 2 = 0 + 0.000 634 106 827 533 765 836 8;
  • 60) 0.000 634 106 827 533 765 836 8 × 2 = 0 + 0.001 268 213 655 067 531 673 6;
  • 61) 0.001 268 213 655 067 531 673 6 × 2 = 0 + 0.002 536 427 310 135 063 347 2;
  • 62) 0.002 536 427 310 135 063 347 2 × 2 = 0 + 0.005 072 854 620 270 126 694 4;
  • 63) 0.005 072 854 620 270 126 694 4 × 2 = 0 + 0.010 145 709 240 540 253 388 8;
  • 64) 0.010 145 709 240 540 253 388 8 × 2 = 0 + 0.020 291 418 481 080 506 777 6;
  • 65) 0.020 291 418 481 080 506 777 6 × 2 = 0 + 0.040 582 836 962 161 013 555 2;
  • 66) 0.040 582 836 962 161 013 555 2 × 2 = 0 + 0.081 165 673 924 322 027 110 4;
  • 67) 0.081 165 673 924 322 027 110 4 × 2 = 0 + 0.162 331 347 848 644 054 220 8;
  • 68) 0.162 331 347 848 644 054 220 8 × 2 = 0 + 0.324 662 695 697 288 108 441 6;
  • 69) 0.324 662 695 697 288 108 441 6 × 2 = 0 + 0.649 325 391 394 576 216 883 2;
  • 70) 0.649 325 391 394 576 216 883 2 × 2 = 1 + 0.298 650 782 789 152 433 766 4;
  • 71) 0.298 650 782 789 152 433 766 4 × 2 = 0 + 0.597 301 565 578 304 867 532 8;
  • 72) 0.597 301 565 578 304 867 532 8 × 2 = 1 + 0.194 603 131 156 609 735 065 6;
  • 73) 0.194 603 131 156 609 735 065 6 × 2 = 0 + 0.389 206 262 313 219 470 131 2;
  • 74) 0.389 206 262 313 219 470 131 2 × 2 = 0 + 0.778 412 524 626 438 940 262 4;
  • 75) 0.778 412 524 626 438 940 262 4 × 2 = 1 + 0.556 825 049 252 877 880 524 8;
  • 76) 0.556 825 049 252 877 880 524 8 × 2 = 1 + 0.113 650 098 505 755 761 049 6;
  • 77) 0.113 650 098 505 755 761 049 6 × 2 = 0 + 0.227 300 197 011 511 522 099 2;
  • 78) 0.227 300 197 011 511 522 099 2 × 2 = 0 + 0.454 600 394 023 023 044 198 4;
  • 79) 0.454 600 394 023 023 044 198 4 × 2 = 0 + 0.909 200 788 046 046 088 396 8;
  • 80) 0.909 200 788 046 046 088 396 8 × 2 = 1 + 0.818 401 576 092 092 176 793 6;
  • 81) 0.818 401 576 092 092 176 793 6 × 2 = 1 + 0.636 803 152 184 184 353 587 2;
  • 82) 0.636 803 152 184 184 353 587 2 × 2 = 1 + 0.273 606 304 368 368 707 174 4;
  • 83) 0.273 606 304 368 368 707 174 4 × 2 = 0 + 0.547 212 608 736 737 414 348 8;
  • 84) 0.547 212 608 736 737 414 348 8 × 2 = 1 + 0.094 425 217 473 474 828 697 6;
  • 85) 0.094 425 217 473 474 828 697 6 × 2 = 0 + 0.188 850 434 946 949 657 395 2;
  • 86) 0.188 850 434 946 949 657 395 2 × 2 = 0 + 0.377 700 869 893 899 314 790 4;
  • 87) 0.377 700 869 893 899 314 790 4 × 2 = 0 + 0.755 401 739 787 798 629 580 8;
  • 88) 0.755 401 739 787 798 629 580 8 × 2 = 1 + 0.510 803 479 575 597 259 161 6;
  • 89) 0.510 803 479 575 597 259 161 6 × 2 = 1 + 0.021 606 959 151 194 518 323 2;
  • 90) 0.021 606 959 151 194 518 323 2 × 2 = 0 + 0.043 213 918 302 389 036 646 4;
  • 91) 0.043 213 918 302 389 036 646 4 × 2 = 0 + 0.086 427 836 604 778 073 292 8;
  • 92) 0.086 427 836 604 778 073 292 8 × 2 = 0 + 0.172 855 673 209 556 146 585 6;
  • 93) 0.172 855 673 209 556 146 585 6 × 2 = 0 + 0.345 711 346 419 112 293 171 2;
  • 94) 0.345 711 346 419 112 293 171 2 × 2 = 0 + 0.691 422 692 838 224 586 342 4;
  • 95) 0.691 422 692 838 224 586 342 4 × 2 = 1 + 0.382 845 385 676 449 172 684 8;
  • 96) 0.382 845 385 676 449 172 684 8 × 2 = 0 + 0.765 690 771 352 898 345 369 6;
  • 97) 0.765 690 771 352 898 345 369 6 × 2 = 1 + 0.531 381 542 705 796 690 739 2;
  • 98) 0.531 381 542 705 796 690 739 2 × 2 = 1 + 0.062 763 085 411 593 381 478 4;
  • 99) 0.062 763 085 411 593 381 478 4 × 2 = 0 + 0.125 526 170 823 186 762 956 8;
  • 100) 0.125 526 170 823 186 762 956 8 × 2 = 0 + 0.251 052 341 646 373 525 913 6;
  • 101) 0.251 052 341 646 373 525 913 6 × 2 = 0 + 0.502 104 683 292 747 051 827 2;
  • 102) 0.502 104 683 292 747 051 827 2 × 2 = 1 + 0.004 209 366 585 494 103 654 4;
  • 103) 0.004 209 366 585 494 103 654 4 × 2 = 0 + 0.008 418 733 170 988 207 308 8;
  • 104) 0.008 418 733 170 988 207 308 8 × 2 = 0 + 0.016 837 466 341 976 414 617 6;
  • 105) 0.016 837 466 341 976 414 617 6 × 2 = 0 + 0.033 674 932 683 952 829 235 2;
  • 106) 0.033 674 932 683 952 829 235 2 × 2 = 0 + 0.067 349 865 367 905 658 470 4;
  • 107) 0.067 349 865 367 905 658 470 4 × 2 = 0 + 0.134 699 730 735 811 316 940 8;
  • 108) 0.134 699 730 735 811 316 940 8 × 2 = 0 + 0.269 399 461 471 622 633 881 6;
  • 109) 0.269 399 461 471 622 633 881 6 × 2 = 0 + 0.538 798 922 943 245 267 763 2;
  • 110) 0.538 798 922 943 245 267 763 2 × 2 = 1 + 0.077 597 845 886 490 535 526 4;
  • 111) 0.077 597 845 886 490 535 526 4 × 2 = 0 + 0.155 195 691 772 981 071 052 8;
  • 112) 0.155 195 691 772 981 071 052 8 × 2 = 0 + 0.310 391 383 545 962 142 105 6;
  • 113) 0.310 391 383 545 962 142 105 6 × 2 = 0 + 0.620 782 767 091 924 284 211 2;
  • 114) 0.620 782 767 091 924 284 211 2 × 2 = 1 + 0.241 565 534 183 848 568 422 4;
  • 115) 0.241 565 534 183 848 568 422 4 × 2 = 0 + 0.483 131 068 367 697 136 844 8;
  • 116) 0.483 131 068 367 697 136 844 8 × 2 = 0 + 0.966 262 136 735 394 273 689 6;
  • 117) 0.966 262 136 735 394 273 689 6 × 2 = 1 + 0.932 524 273 470 788 547 379 2;
  • 118) 0.932 524 273 470 788 547 379 2 × 2 = 1 + 0.865 048 546 941 577 094 758 4;
  • 119) 0.865 048 546 941 577 094 758 4 × 2 = 1 + 0.730 097 093 883 154 189 516 8;
  • 120) 0.730 097 093 883 154 189 516 8 × 2 = 1 + 0.460 194 187 766 308 379 033 6;
  • 121) 0.460 194 187 766 308 379 033 6 × 2 = 0 + 0.920 388 375 532 616 758 067 2;
  • 122) 0.920 388 375 532 616 758 067 2 × 2 = 1 + 0.840 776 751 065 233 516 134 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).


5. 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:


0.000 000 000 000 000 000 001 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0011 0001 1101 0001 1000 0010 1100 0100 0000 0100 0100 1111 01(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 001 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0011 0001 1101 0001 1000 0010 1100 0100 0000 0100 0100 1111 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 70 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 001 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0011 0001 1101 0001 1000 0010 1100 0100 0000 0100 0100 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0011 0001 1101 0001 1000 0010 1100 0100 0000 0100 0100 1111 01(2) × 20 =

1.0100 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101(2) × 2-70


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): -70

Mantissa (not normalized):
1.0100 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-70 + 2(11-1) - 1 =

(-70 + 1 023)(10) =

953(10)


10. 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;
  • 953 ÷ 2 = 476 + 1;
  • 476 ÷ 2 = 238 + 0;
  • 238 ÷ 2 = 119 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

953(10) =

011 1011 1001(2)


12. 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 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101 =

0100 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1011 1001

Mantissa (52 bits) =
0100 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101


Decimal number -0.000 000 000 000 000 000 001 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1011 1001 - 0100 1100 0111 0100 0110 0000 1011 0001 0000 0001 0001 0011 1101

Share

» Convert decimal -0.000 000 000 000 000 000 009 7 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Pull vs. Push Leadership: The Invisible Power. NotebookLM YouTube Video of 11.31 minutes

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