Convert 17 857 142 857 142 857 142 857 142 857 144 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

17 857 142 857 142 857 142 857 142 857 144(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. 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;
  • 17 857 142 857 142 857 142 857 142 857 144 ÷ 2 = 8 928 571 428 571 428 571 428 571 428 572 + 0;
  • 8 928 571 428 571 428 571 428 571 428 572 ÷ 2 = 4 464 285 714 285 714 285 714 285 714 286 + 0;
  • 4 464 285 714 285 714 285 714 285 714 286 ÷ 2 = 2 232 142 857 142 857 142 857 142 857 143 + 0;
  • 2 232 142 857 142 857 142 857 142 857 143 ÷ 2 = 1 116 071 428 571 428 571 428 571 428 571 + 1;
  • 1 116 071 428 571 428 571 428 571 428 571 ÷ 2 = 558 035 714 285 714 285 714 285 714 285 + 1;
  • 558 035 714 285 714 285 714 285 714 285 ÷ 2 = 279 017 857 142 857 142 857 142 857 142 + 1;
  • 279 017 857 142 857 142 857 142 857 142 ÷ 2 = 139 508 928 571 428 571 428 571 428 571 + 0;
  • 139 508 928 571 428 571 428 571 428 571 ÷ 2 = 69 754 464 285 714 285 714 285 714 285 + 1;
  • 69 754 464 285 714 285 714 285 714 285 ÷ 2 = 34 877 232 142 857 142 857 142 857 142 + 1;
  • 34 877 232 142 857 142 857 142 857 142 ÷ 2 = 17 438 616 071 428 571 428 571 428 571 + 0;
  • 17 438 616 071 428 571 428 571 428 571 ÷ 2 = 8 719 308 035 714 285 714 285 714 285 + 1;
  • 8 719 308 035 714 285 714 285 714 285 ÷ 2 = 4 359 654 017 857 142 857 142 857 142 + 1;
  • 4 359 654 017 857 142 857 142 857 142 ÷ 2 = 2 179 827 008 928 571 428 571 428 571 + 0;
  • 2 179 827 008 928 571 428 571 428 571 ÷ 2 = 1 089 913 504 464 285 714 285 714 285 + 1;
  • 1 089 913 504 464 285 714 285 714 285 ÷ 2 = 544 956 752 232 142 857 142 857 142 + 1;
  • 544 956 752 232 142 857 142 857 142 ÷ 2 = 272 478 376 116 071 428 571 428 571 + 0;
  • 272 478 376 116 071 428 571 428 571 ÷ 2 = 136 239 188 058 035 714 285 714 285 + 1;
  • 136 239 188 058 035 714 285 714 285 ÷ 2 = 68 119 594 029 017 857 142 857 142 + 1;
  • 68 119 594 029 017 857 142 857 142 ÷ 2 = 34 059 797 014 508 928 571 428 571 + 0;
  • 34 059 797 014 508 928 571 428 571 ÷ 2 = 17 029 898 507 254 464 285 714 285 + 1;
  • 17 029 898 507 254 464 285 714 285 ÷ 2 = 8 514 949 253 627 232 142 857 142 + 1;
  • 8 514 949 253 627 232 142 857 142 ÷ 2 = 4 257 474 626 813 616 071 428 571 + 0;
  • 4 257 474 626 813 616 071 428 571 ÷ 2 = 2 128 737 313 406 808 035 714 285 + 1;
  • 2 128 737 313 406 808 035 714 285 ÷ 2 = 1 064 368 656 703 404 017 857 142 + 1;
  • 1 064 368 656 703 404 017 857 142 ÷ 2 = 532 184 328 351 702 008 928 571 + 0;
  • 532 184 328 351 702 008 928 571 ÷ 2 = 266 092 164 175 851 004 464 285 + 1;
  • 266 092 164 175 851 004 464 285 ÷ 2 = 133 046 082 087 925 502 232 142 + 1;
  • 133 046 082 087 925 502 232 142 ÷ 2 = 66 523 041 043 962 751 116 071 + 0;
  • 66 523 041 043 962 751 116 071 ÷ 2 = 33 261 520 521 981 375 558 035 + 1;
  • 33 261 520 521 981 375 558 035 ÷ 2 = 16 630 760 260 990 687 779 017 + 1;
  • 16 630 760 260 990 687 779 017 ÷ 2 = 8 315 380 130 495 343 889 508 + 1;
  • 8 315 380 130 495 343 889 508 ÷ 2 = 4 157 690 065 247 671 944 754 + 0;
  • 4 157 690 065 247 671 944 754 ÷ 2 = 2 078 845 032 623 835 972 377 + 0;
  • 2 078 845 032 623 835 972 377 ÷ 2 = 1 039 422 516 311 917 986 188 + 1;
  • 1 039 422 516 311 917 986 188 ÷ 2 = 519 711 258 155 958 993 094 + 0;
  • 519 711 258 155 958 993 094 ÷ 2 = 259 855 629 077 979 496 547 + 0;
  • 259 855 629 077 979 496 547 ÷ 2 = 129 927 814 538 989 748 273 + 1;
  • 129 927 814 538 989 748 273 ÷ 2 = 64 963 907 269 494 874 136 + 1;
  • 64 963 907 269 494 874 136 ÷ 2 = 32 481 953 634 747 437 068 + 0;
  • 32 481 953 634 747 437 068 ÷ 2 = 16 240 976 817 373 718 534 + 0;
  • 16 240 976 817 373 718 534 ÷ 2 = 8 120 488 408 686 859 267 + 0;
  • 8 120 488 408 686 859 267 ÷ 2 = 4 060 244 204 343 429 633 + 1;
  • 4 060 244 204 343 429 633 ÷ 2 = 2 030 122 102 171 714 816 + 1;
  • 2 030 122 102 171 714 816 ÷ 2 = 1 015 061 051 085 857 408 + 0;
  • 1 015 061 051 085 857 408 ÷ 2 = 507 530 525 542 928 704 + 0;
  • 507 530 525 542 928 704 ÷ 2 = 253 765 262 771 464 352 + 0;
  • 253 765 262 771 464 352 ÷ 2 = 126 882 631 385 732 176 + 0;
  • 126 882 631 385 732 176 ÷ 2 = 63 441 315 692 866 088 + 0;
  • 63 441 315 692 866 088 ÷ 2 = 31 720 657 846 433 044 + 0;
  • 31 720 657 846 433 044 ÷ 2 = 15 860 328 923 216 522 + 0;
  • 15 860 328 923 216 522 ÷ 2 = 7 930 164 461 608 261 + 0;
  • 7 930 164 461 608 261 ÷ 2 = 3 965 082 230 804 130 + 1;
  • 3 965 082 230 804 130 ÷ 2 = 1 982 541 115 402 065 + 0;
  • 1 982 541 115 402 065 ÷ 2 = 991 270 557 701 032 + 1;
  • 991 270 557 701 032 ÷ 2 = 495 635 278 850 516 + 0;
  • 495 635 278 850 516 ÷ 2 = 247 817 639 425 258 + 0;
  • 247 817 639 425 258 ÷ 2 = 123 908 819 712 629 + 0;
  • 123 908 819 712 629 ÷ 2 = 61 954 409 856 314 + 1;
  • 61 954 409 856 314 ÷ 2 = 30 977 204 928 157 + 0;
  • 30 977 204 928 157 ÷ 2 = 15 488 602 464 078 + 1;
  • 15 488 602 464 078 ÷ 2 = 7 744 301 232 039 + 0;
  • 7 744 301 232 039 ÷ 2 = 3 872 150 616 019 + 1;
  • 3 872 150 616 019 ÷ 2 = 1 936 075 308 009 + 1;
  • 1 936 075 308 009 ÷ 2 = 968 037 654 004 + 1;
  • 968 037 654 004 ÷ 2 = 484 018 827 002 + 0;
  • 484 018 827 002 ÷ 2 = 242 009 413 501 + 0;
  • 242 009 413 501 ÷ 2 = 121 004 706 750 + 1;
  • 121 004 706 750 ÷ 2 = 60 502 353 375 + 0;
  • 60 502 353 375 ÷ 2 = 30 251 176 687 + 1;
  • 30 251 176 687 ÷ 2 = 15 125 588 343 + 1;
  • 15 125 588 343 ÷ 2 = 7 562 794 171 + 1;
  • 7 562 794 171 ÷ 2 = 3 781 397 085 + 1;
  • 3 781 397 085 ÷ 2 = 1 890 698 542 + 1;
  • 1 890 698 542 ÷ 2 = 945 349 271 + 0;
  • 945 349 271 ÷ 2 = 472 674 635 + 1;
  • 472 674 635 ÷ 2 = 236 337 317 + 1;
  • 236 337 317 ÷ 2 = 118 168 658 + 1;
  • 118 168 658 ÷ 2 = 59 084 329 + 0;
  • 59 084 329 ÷ 2 = 29 542 164 + 1;
  • 29 542 164 ÷ 2 = 14 771 082 + 0;
  • 14 771 082 ÷ 2 = 7 385 541 + 0;
  • 7 385 541 ÷ 2 = 3 692 770 + 1;
  • 3 692 770 ÷ 2 = 1 846 385 + 0;
  • 1 846 385 ÷ 2 = 923 192 + 1;
  • 923 192 ÷ 2 = 461 596 + 0;
  • 461 596 ÷ 2 = 230 798 + 0;
  • 230 798 ÷ 2 = 115 399 + 0;
  • 115 399 ÷ 2 = 57 699 + 1;
  • 57 699 ÷ 2 = 28 849 + 1;
  • 28 849 ÷ 2 = 14 424 + 1;
  • 14 424 ÷ 2 = 7 212 + 0;
  • 7 212 ÷ 2 = 3 606 + 0;
  • 3 606 ÷ 2 = 1 803 + 0;
  • 1 803 ÷ 2 = 901 + 1;
  • 901 ÷ 2 = 450 + 1;
  • 450 ÷ 2 = 225 + 0;
  • 225 ÷ 2 = 112 + 1;
  • 112 ÷ 2 = 56 + 0;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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

17 857 142 857 142 857 142 857 142 857 144(10) =


1110 0001 0110 0011 1000 1010 0101 1101 1111 0100 1110 1010 0010 1000 0000 0110 0011 0010 0111 0110 1101 1011 0110 1101 1011 1000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 103 positions to the left so that only one non zero digit remains to the left of it:

17 857 142 857 142 857 142 857 142 857 144(10) =


1110 0001 0110 0011 1000 1010 0101 1101 1111 0100 1110 1010 0010 1000 0000 0110 0011 0010 0111 0110 1101 1011 0110 1101 1011 1000(2) =


1110 0001 0110 0011 1000 1010 0101 1101 1111 0100 1110 1010 0010 1000 0000 0110 0011 0010 0111 0110 1101 1011 0110 1101 1011 1000(2) × 20 =


1.1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101 0000 0000 1100 0110 0100 1110 1101 1011 0110 1101 1011 0111 000(2) × 2103


4. 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): 103


Mantissa (not normalized):
1.1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101 0000 0000 1100 0110 0100 1110 1101 1011 0110 1101 1011 0111 000


5. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


103 + 2(11-1) - 1 =


(103 + 1 023)(10) =


1 126(10)


6. 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;
  • 1 126 ÷ 2 = 563 + 0;
  • 563 ÷ 2 = 281 + 1;
  • 281 ÷ 2 = 140 + 1;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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

Exponent (adjusted) =


1126(10) =


100 0110 0110(2)


8. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101 000 0000 0110 0011 0010 0111 0110 1101 1011 0110 1101 1011 1000 =


1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101


9. 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) =
100 0110 0110


Mantissa (52 bits) =
1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101


Number 17 857 142 857 142 857 142 857 142 857 144 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0110 0110 - 1100 0010 1100 0111 0001 0100 1011 1011 1110 1001 1101 0100 0101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 1

      58
    • 1

      57
    • 0

      56
    • 0

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 1

      25
    • 1

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

17 857 142 857 142 857 142 857 142 857 143 = ? ... 17 857 142 857 142 857 142 857 142 857 145 = ?


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

17 857 142 857 142 857 142 857 142 857 144 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
1.006 103 515 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
32.42 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
2 312 556.000 000 46 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
201.437 7 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
823 549.664 582 642 758 703 551 187 066 422 004 074 982 775 029 787 740 333 9 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
0.000 000 000 000 000 000 000 013 806 488 12 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
-25.432 1 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:38 UTC (GMT)
964.1 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:37 UTC (GMT)
19 255 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:37 UTC (GMT)
0.025 172 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:37 UTC (GMT)
-142.375 25 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:37 UTC (GMT)
1 438 468 554 to 64 bit double precision IEEE 754 binary floating point = ? Mar 09 09:37 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