Convert 0.000 020 830 729 321 671 205 134 999 154 509 660 6 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number 0.000 020 830 729 321 671 205 134 999 154 509 660 6(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 020 830 729 321 671 205 134 999 154 509 660 6.

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) 0.000 020 830 729 321 671 205 134 999 154 509 660 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 019 321 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 019 321 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 038 642 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 038 642 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 077 284 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 077 284 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 154 569 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 154 569 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 309 139 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 309 139 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 618 278 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 618 278 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 236 556 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 236 556 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 554 473 113 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 554 473 113 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 108 946 227 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 108 946 227 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 217 892 454 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 217 892 454 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 435 784 908 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 435 784 908 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 871 569 817 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 871 569 817 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 743 139 635 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 743 139 635 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 486 279 270 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 486 279 270 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 972 558 540 8;
  • 16) 0.682 581 338 412 522 049 863 652 294 972 558 540 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 945 117 081 6;
  • 17) 0.365 162 676 825 044 099 727 304 589 945 117 081 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 890 234 163 2;
  • 18) 0.730 325 353 650 088 199 454 609 179 890 234 163 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 359 780 468 326 4;
  • 19) 0.460 650 707 300 176 398 909 218 359 780 468 326 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 719 560 936 652 8;
  • 20) 0.921 301 414 600 352 797 818 436 719 560 936 652 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 439 121 873 305 6;
  • 21) 0.842 602 829 200 705 595 636 873 439 121 873 305 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 878 243 746 611 2;
  • 22) 0.685 205 658 401 411 191 273 746 878 243 746 611 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 756 487 493 222 4;
  • 23) 0.370 411 316 802 822 382 547 493 756 487 493 222 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 512 974 986 444 8;
  • 24) 0.740 822 633 605 644 765 094 987 512 974 986 444 8 × 2 = 1 + 0.481 645 267 211 289 530 189 975 025 949 972 889 6;
  • 25) 0.481 645 267 211 289 530 189 975 025 949 972 889 6 × 2 = 0 + 0.963 290 534 422 579 060 379 950 051 899 945 779 2;
  • 26) 0.963 290 534 422 579 060 379 950 051 899 945 779 2 × 2 = 1 + 0.926 581 068 845 158 120 759 900 103 799 891 558 4;
  • 27) 0.926 581 068 845 158 120 759 900 103 799 891 558 4 × 2 = 1 + 0.853 162 137 690 316 241 519 800 207 599 783 116 8;
  • 28) 0.853 162 137 690 316 241 519 800 207 599 783 116 8 × 2 = 1 + 0.706 324 275 380 632 483 039 600 415 199 566 233 6;
  • 29) 0.706 324 275 380 632 483 039 600 415 199 566 233 6 × 2 = 1 + 0.412 648 550 761 264 966 079 200 830 399 132 467 2;
  • 30) 0.412 648 550 761 264 966 079 200 830 399 132 467 2 × 2 = 0 + 0.825 297 101 522 529 932 158 401 660 798 264 934 4;
  • 31) 0.825 297 101 522 529 932 158 401 660 798 264 934 4 × 2 = 1 + 0.650 594 203 045 059 864 316 803 321 596 529 868 8;
  • 32) 0.650 594 203 045 059 864 316 803 321 596 529 868 8 × 2 = 1 + 0.301 188 406 090 119 728 633 606 643 193 059 737 6;
  • 33) 0.301 188 406 090 119 728 633 606 643 193 059 737 6 × 2 = 0 + 0.602 376 812 180 239 457 267 213 286 386 119 475 2;
  • 34) 0.602 376 812 180 239 457 267 213 286 386 119 475 2 × 2 = 1 + 0.204 753 624 360 478 914 534 426 572 772 238 950 4;
  • 35) 0.204 753 624 360 478 914 534 426 572 772 238 950 4 × 2 = 0 + 0.409 507 248 720 957 829 068 853 145 544 477 900 8;
  • 36) 0.409 507 248 720 957 829 068 853 145 544 477 900 8 × 2 = 0 + 0.819 014 497 441 915 658 137 706 291 088 955 801 6;
  • 37) 0.819 014 497 441 915 658 137 706 291 088 955 801 6 × 2 = 1 + 0.638 028 994 883 831 316 275 412 582 177 911 603 2;
  • 38) 0.638 028 994 883 831 316 275 412 582 177 911 603 2 × 2 = 1 + 0.276 057 989 767 662 632 550 825 164 355 823 206 4;
  • 39) 0.276 057 989 767 662 632 550 825 164 355 823 206 4 × 2 = 0 + 0.552 115 979 535 325 265 101 650 328 711 646 412 8;
  • 40) 0.552 115 979 535 325 265 101 650 328 711 646 412 8 × 2 = 1 + 0.104 231 959 070 650 530 203 300 657 423 292 825 6;
  • 41) 0.104 231 959 070 650 530 203 300 657 423 292 825 6 × 2 = 0 + 0.208 463 918 141 301 060 406 601 314 846 585 651 2;
  • 42) 0.208 463 918 141 301 060 406 601 314 846 585 651 2 × 2 = 0 + 0.416 927 836 282 602 120 813 202 629 693 171 302 4;
  • 43) 0.416 927 836 282 602 120 813 202 629 693 171 302 4 × 2 = 0 + 0.833 855 672 565 204 241 626 405 259 386 342 604 8;
  • 44) 0.833 855 672 565 204 241 626 405 259 386 342 604 8 × 2 = 1 + 0.667 711 345 130 408 483 252 810 518 772 685 209 6;
  • 45) 0.667 711 345 130 408 483 252 810 518 772 685 209 6 × 2 = 1 + 0.335 422 690 260 816 966 505 621 037 545 370 419 2;
  • 46) 0.335 422 690 260 816 966 505 621 037 545 370 419 2 × 2 = 0 + 0.670 845 380 521 633 933 011 242 075 090 740 838 4;
  • 47) 0.670 845 380 521 633 933 011 242 075 090 740 838 4 × 2 = 1 + 0.341 690 761 043 267 866 022 484 150 181 481 676 8;
  • 48) 0.341 690 761 043 267 866 022 484 150 181 481 676 8 × 2 = 0 + 0.683 381 522 086 535 732 044 968 300 362 963 353 6;
  • 49) 0.683 381 522 086 535 732 044 968 300 362 963 353 6 × 2 = 1 + 0.366 763 044 173 071 464 089 936 600 725 926 707 2;
  • 50) 0.366 763 044 173 071 464 089 936 600 725 926 707 2 × 2 = 0 + 0.733 526 088 346 142 928 179 873 201 451 853 414 4;
  • 51) 0.733 526 088 346 142 928 179 873 201 451 853 414 4 × 2 = 1 + 0.467 052 176 692 285 856 359 746 402 903 706 828 8;
  • 52) 0.467 052 176 692 285 856 359 746 402 903 706 828 8 × 2 = 0 + 0.934 104 353 384 571 712 719 492 805 807 413 657 6;
  • 53) 0.934 104 353 384 571 712 719 492 805 807 413 657 6 × 2 = 1 + 0.868 208 706 769 143 425 438 985 611 614 827 315 2;
  • 54) 0.868 208 706 769 143 425 438 985 611 614 827 315 2 × 2 = 1 + 0.736 417 413 538 286 850 877 971 223 229 654 630 4;
  • 55) 0.736 417 413 538 286 850 877 971 223 229 654 630 4 × 2 = 1 + 0.472 834 827 076 573 701 755 942 446 459 309 260 8;
  • 56) 0.472 834 827 076 573 701 755 942 446 459 309 260 8 × 2 = 0 + 0.945 669 654 153 147 403 511 884 892 918 618 521 6;
  • 57) 0.945 669 654 153 147 403 511 884 892 918 618 521 6 × 2 = 1 + 0.891 339 308 306 294 807 023 769 785 837 237 043 2;
  • 58) 0.891 339 308 306 294 807 023 769 785 837 237 043 2 × 2 = 1 + 0.782 678 616 612 589 614 047 539 571 674 474 086 4;
  • 59) 0.782 678 616 612 589 614 047 539 571 674 474 086 4 × 2 = 1 + 0.565 357 233 225 179 228 095 079 143 348 948 172 8;
  • 60) 0.565 357 233 225 179 228 095 079 143 348 948 172 8 × 2 = 1 + 0.130 714 466 450 358 456 190 158 286 697 896 345 6;
  • 61) 0.130 714 466 450 358 456 190 158 286 697 896 345 6 × 2 = 0 + 0.261 428 932 900 716 912 380 316 573 395 792 691 2;
  • 62) 0.261 428 932 900 716 912 380 316 573 395 792 691 2 × 2 = 0 + 0.522 857 865 801 433 824 760 633 146 791 585 382 4;
  • 63) 0.522 857 865 801 433 824 760 633 146 791 585 382 4 × 2 = 1 + 0.045 715 731 602 867 649 521 266 293 583 170 764 8;
  • 64) 0.045 715 731 602 867 649 521 266 293 583 170 764 8 × 2 = 0 + 0.091 431 463 205 735 299 042 532 587 166 341 529 6;
  • 65) 0.091 431 463 205 735 299 042 532 587 166 341 529 6 × 2 = 0 + 0.182 862 926 411 470 598 085 065 174 332 683 059 2;
  • 66) 0.182 862 926 411 470 598 085 065 174 332 683 059 2 × 2 = 0 + 0.365 725 852 822 941 196 170 130 348 665 366 118 4;
  • 67) 0.365 725 852 822 941 196 170 130 348 665 366 118 4 × 2 = 0 + 0.731 451 705 645 882 392 340 260 697 330 732 236 8;
  • 68) 0.731 451 705 645 882 392 340 260 697 330 732 236 8 × 2 = 1 + 0.462 903 411 291 764 784 680 521 394 661 464 473 6;

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...)


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:

0.000 020 830 729 321 671 205 134 999 154 509 660 6(10) =


0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)


5. Positive number before normalization:

0.000 020 830 729 321 671 205 134 999 154 509 660 6(10) =


0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right so that only one non zero digit remains to the left of it:

0.000 020 830 729 321 671 205 134 999 154 509 660 6(10) =


0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) =


0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 20 =


1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 2-16


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): -16


Mantissa (not normalized):
1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-16 + 2(11-1) - 1 =


(-16 + 1 023)(10) =


1 007(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;
  • 1 007 ÷ 2 = 503 + 1;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 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) =


1007(10) =


011 1110 1111(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. 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001 =


0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


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 1110 1111


Mantissa (52 bits) =
0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


Conclusion:

Number 0.000 020 830 729 321 671 205 134 999 154 509 660 6 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 011 1110 1111 - 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 0

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 1

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

0.000 020 830 729 321 671 205 134 999 154 509 660 5 = ? ... 0.000 020 830 729 321 671 205 134 999 154 509 660 7 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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