Base ten decimal number 0.917 004 043 204 671 231 743 541 594 794 14 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number 0.917 004 043 204 671 231 743 541 594 794 14(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 binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until 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, by taking 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: 0.917 004 043 204 671 231 743 541 594 794 14. Multiply it 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) 0.917 004 043 204 671 231 743 541 594 794 14 × 2 = 1 + 0.834 008 086 409 342 463 487 083 189 588 28;
  • 2) 0.834 008 086 409 342 463 487 083 189 588 28 × 2 = 1 + 0.668 016 172 818 684 926 974 166 379 176 56;
  • 3) 0.668 016 172 818 684 926 974 166 379 176 56 × 2 = 1 + 0.336 032 345 637 369 853 948 332 758 353 12;
  • 4) 0.336 032 345 637 369 853 948 332 758 353 12 × 2 = 0 + 0.672 064 691 274 739 707 896 665 516 706 24;
  • 5) 0.672 064 691 274 739 707 896 665 516 706 24 × 2 = 1 + 0.344 129 382 549 479 415 793 331 033 412 48;
  • 6) 0.344 129 382 549 479 415 793 331 033 412 48 × 2 = 0 + 0.688 258 765 098 958 831 586 662 066 824 96;
  • 7) 0.688 258 765 098 958 831 586 662 066 824 96 × 2 = 1 + 0.376 517 530 197 917 663 173 324 133 649 92;
  • 8) 0.376 517 530 197 917 663 173 324 133 649 92 × 2 = 0 + 0.753 035 060 395 835 326 346 648 267 299 84;
  • 9) 0.753 035 060 395 835 326 346 648 267 299 84 × 2 = 1 + 0.506 070 120 791 670 652 693 296 534 599 68;
  • 10) 0.506 070 120 791 670 652 693 296 534 599 68 × 2 = 1 + 0.012 140 241 583 341 305 386 593 069 199 36;
  • 11) 0.012 140 241 583 341 305 386 593 069 199 36 × 2 = 0 + 0.024 280 483 166 682 610 773 186 138 398 72;
  • 12) 0.024 280 483 166 682 610 773 186 138 398 72 × 2 = 0 + 0.048 560 966 333 365 221 546 372 276 797 44;
  • 13) 0.048 560 966 333 365 221 546 372 276 797 44 × 2 = 0 + 0.097 121 932 666 730 443 092 744 553 594 88;
  • 14) 0.097 121 932 666 730 443 092 744 553 594 88 × 2 = 0 + 0.194 243 865 333 460 886 185 489 107 189 76;
  • 15) 0.194 243 865 333 460 886 185 489 107 189 76 × 2 = 0 + 0.388 487 730 666 921 772 370 978 214 379 52;
  • 16) 0.388 487 730 666 921 772 370 978 214 379 52 × 2 = 0 + 0.776 975 461 333 843 544 741 956 428 759 04;
  • 17) 0.776 975 461 333 843 544 741 956 428 759 04 × 2 = 1 + 0.553 950 922 667 687 089 483 912 857 518 08;
  • 18) 0.553 950 922 667 687 089 483 912 857 518 08 × 2 = 1 + 0.107 901 845 335 374 178 967 825 715 036 16;
  • 19) 0.107 901 845 335 374 178 967 825 715 036 16 × 2 = 0 + 0.215 803 690 670 748 357 935 651 430 072 32;
  • 20) 0.215 803 690 670 748 357 935 651 430 072 32 × 2 = 0 + 0.431 607 381 341 496 715 871 302 860 144 64;
  • 21) 0.431 607 381 341 496 715 871 302 860 144 64 × 2 = 0 + 0.863 214 762 682 993 431 742 605 720 289 28;
  • 22) 0.863 214 762 682 993 431 742 605 720 289 28 × 2 = 1 + 0.726 429 525 365 986 863 485 211 440 578 56;
  • 23) 0.726 429 525 365 986 863 485 211 440 578 56 × 2 = 1 + 0.452 859 050 731 973 726 970 422 881 157 12;
  • 24) 0.452 859 050 731 973 726 970 422 881 157 12 × 2 = 0 + 0.905 718 101 463 947 453 940 845 762 314 24;
  • 25) 0.905 718 101 463 947 453 940 845 762 314 24 × 2 = 1 + 0.811 436 202 927 894 907 881 691 524 628 48;
  • 26) 0.811 436 202 927 894 907 881 691 524 628 48 × 2 = 1 + 0.622 872 405 855 789 815 763 383 049 256 96;
  • 27) 0.622 872 405 855 789 815 763 383 049 256 96 × 2 = 1 + 0.245 744 811 711 579 631 526 766 098 513 92;
  • 28) 0.245 744 811 711 579 631 526 766 098 513 92 × 2 = 0 + 0.491 489 623 423 159 263 053 532 197 027 84;
  • 29) 0.491 489 623 423 159 263 053 532 197 027 84 × 2 = 0 + 0.982 979 246 846 318 526 107 064 394 055 68;
  • 30) 0.982 979 246 846 318 526 107 064 394 055 68 × 2 = 1 + 0.965 958 493 692 637 052 214 128 788 111 36;
  • 31) 0.965 958 493 692 637 052 214 128 788 111 36 × 2 = 1 + 0.931 916 987 385 274 104 428 257 576 222 72;
  • 32) 0.931 916 987 385 274 104 428 257 576 222 72 × 2 = 1 + 0.863 833 974 770 548 208 856 515 152 445 44;
  • 33) 0.863 833 974 770 548 208 856 515 152 445 44 × 2 = 1 + 0.727 667 949 541 096 417 713 030 304 890 88;
  • 34) 0.727 667 949 541 096 417 713 030 304 890 88 × 2 = 1 + 0.455 335 899 082 192 835 426 060 609 781 76;
  • 35) 0.455 335 899 082 192 835 426 060 609 781 76 × 2 = 0 + 0.910 671 798 164 385 670 852 121 219 563 52;
  • 36) 0.910 671 798 164 385 670 852 121 219 563 52 × 2 = 1 + 0.821 343 596 328 771 341 704 242 439 127 04;
  • 37) 0.821 343 596 328 771 341 704 242 439 127 04 × 2 = 1 + 0.642 687 192 657 542 683 408 484 878 254 08;
  • 38) 0.642 687 192 657 542 683 408 484 878 254 08 × 2 = 1 + 0.285 374 385 315 085 366 816 969 756 508 16;
  • 39) 0.285 374 385 315 085 366 816 969 756 508 16 × 2 = 0 + 0.570 748 770 630 170 733 633 939 513 016 32;
  • 40) 0.570 748 770 630 170 733 633 939 513 016 32 × 2 = 1 + 0.141 497 541 260 341 467 267 879 026 032 64;
  • 41) 0.141 497 541 260 341 467 267 879 026 032 64 × 2 = 0 + 0.282 995 082 520 682 934 535 758 052 065 28;
  • 42) 0.282 995 082 520 682 934 535 758 052 065 28 × 2 = 0 + 0.565 990 165 041 365 869 071 516 104 130 56;
  • 43) 0.565 990 165 041 365 869 071 516 104 130 56 × 2 = 1 + 0.131 980 330 082 731 738 143 032 208 261 12;
  • 44) 0.131 980 330 082 731 738 143 032 208 261 12 × 2 = 0 + 0.263 960 660 165 463 476 286 064 416 522 24;
  • 45) 0.263 960 660 165 463 476 286 064 416 522 24 × 2 = 0 + 0.527 921 320 330 926 952 572 128 833 044 48;
  • 46) 0.527 921 320 330 926 952 572 128 833 044 48 × 2 = 1 + 0.055 842 640 661 853 905 144 257 666 088 96;
  • 47) 0.055 842 640 661 853 905 144 257 666 088 96 × 2 = 0 + 0.111 685 281 323 707 810 288 515 332 177 92;
  • 48) 0.111 685 281 323 707 810 288 515 332 177 92 × 2 = 0 + 0.223 370 562 647 415 620 577 030 664 355 84;
  • 49) 0.223 370 562 647 415 620 577 030 664 355 84 × 2 = 0 + 0.446 741 125 294 831 241 154 061 328 711 68;
  • 50) 0.446 741 125 294 831 241 154 061 328 711 68 × 2 = 0 + 0.893 482 250 589 662 482 308 122 657 423 36;
  • 51) 0.893 482 250 589 662 482 308 122 657 423 36 × 2 = 1 + 0.786 964 501 179 324 964 616 245 314 846 72;
  • 52) 0.786 964 501 179 324 964 616 245 314 846 72 × 2 = 1 + 0.573 929 002 358 649 929 232 490 629 693 44;
  • 53) 0.573 929 002 358 649 929 232 490 629 693 44 × 2 = 1 + 0.147 858 004 717 299 858 464 981 259 386 88;

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, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.917 004 043 204 671 231 743 541 594 794 14(10) =


0.1110 1010 1100 0000 1100 0110 1110 0111 1101 1101 0010 0100 0011 1(2)

Positive number before normalization:

0.917 004 043 204 671 231 743 541 594 794 14(10) =


0.1110 1010 1100 0000 1100 0110 1110 0111 1101 1101 0010 0100 0011 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 1 positions to the right so that only one non zero digit remains to the left of it:

0.917 004 043 204 671 231 743 541 594 794 14(10) =


0.1110 1010 1100 0000 1100 0110 1110 0111 1101 1101 0010 0100 0011 1(2) =


0.1110 1010 1100 0000 1100 0110 1110 0111 1101 1101 0010 0100 0011 1(2) × 20 =


1.1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111(2) × 2-1

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): -1


Mantissa (not normalized): 1.1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111

6. 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:

Exponent (adjusted) =


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


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


(-1 + 1 023)(10) =


1 022(10)


  • division = quotient + remainder;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


1022(10) =


011 1111 1110(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 52 bits, only if necessary (not the case here):

Mantissa (normalized) =


1. 1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111 =


1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111

Conclusion:

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 1111 1110


Mantissa (52 bits) =
1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111

Number 0.917 004 043 204 671 231 743 541 594 794 14, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


0 - 011 1111 1110 - 1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 0

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 1

      0

Convert decimal numbers from base ten 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 1 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

0.917 004 043 204 671 231 743 541 594 794 14 = 0 - 011 1111 1110 - 1101 0101 1000 0001 1000 1101 1100 1111 1011 1010 0100 1000 0111 Sep 18 07:00 UTC (GMT)
22.25 = 0 - 100 0000 0011 - 0110 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Sep 18 07:00 UTC (GMT)
0.269 421 = 0 - 011 1111 1101 - 0001 0011 1110 0011 0001 1001 0011 1111 0110 1100 0010 0110 1001 Sep 18 07:00 UTC (GMT)
17.29 = 0 - 100 0000 0011 - 0001 0100 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 Sep 18 06:59 UTC (GMT)
823 575.006 2 = 0 - 100 0001 0010 - 1001 0010 0010 0010 1110 0000 0011 0010 1100 1010 0101 0111 1010 Sep 18 06:59 UTC (GMT)
-0.123 35 = 1 - 011 1111 1011 - 1111 1001 0011 1101 1101 1001 0111 1111 0110 0010 1011 0110 1010 Sep 18 06:59 UTC (GMT)
0.001 931 5 = 0 - 011 1111 0101 - 1111 1010 0101 0100 1100 0101 0101 0100 0011 0010 1000 0111 0011 Sep 18 06:58 UTC (GMT)
-56.654 = 1 - 100 0000 0100 - 1100 0101 0011 1011 0110 0100 0101 1010 0001 1100 1010 1100 0000 Sep 18 06:57 UTC (GMT)
4 294 967 168 = 0 - 100 0001 1110 - 1111 1111 1111 1111 1111 1111 0000 0000 0000 0000 0000 0000 0000 Sep 18 06:56 UTC (GMT)
-170 207 059 = 1 - 100 0001 1010 - 0100 0100 1010 0100 1110 1010 0110 0000 0000 0000 0000 0000 0000 Sep 18 06:56 UTC (GMT)
-3 533 329 530 954 438 969 = 1 - 100 0011 1100 - 1000 1000 0100 0111 0100 0011 1110 0111 0111 0111 0111 1110 1100 Sep 18 06:56 UTC (GMT)
16 020 763 = 0 - 100 0001 0110 - 1110 1000 1110 1010 0011 0110 0000 0000 0000 0000 0000 0000 0000 Sep 18 06:54 UTC (GMT)
4.280 003 580 697 414 7 = 0 - 100 0000 0001 - 0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100 Sep 18 06:54 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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