-0.000 000 000 000 014 364 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 364 6(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
-0.000 000 000 000 014 364 6(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:

|-0.000 000 000 000 014 364 6| = 0.000 000 000 000 014 364 6


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: 0.000 000 000 000 014 364 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) 0.000 000 000 000 014 364 6 × 2 = 0 + 0.000 000 000 000 028 729 2;
  • 2) 0.000 000 000 000 028 729 2 × 2 = 0 + 0.000 000 000 000 057 458 4;
  • 3) 0.000 000 000 000 057 458 4 × 2 = 0 + 0.000 000 000 000 114 916 8;
  • 4) 0.000 000 000 000 114 916 8 × 2 = 0 + 0.000 000 000 000 229 833 6;
  • 5) 0.000 000 000 000 229 833 6 × 2 = 0 + 0.000 000 000 000 459 667 2;
  • 6) 0.000 000 000 000 459 667 2 × 2 = 0 + 0.000 000 000 000 919 334 4;
  • 7) 0.000 000 000 000 919 334 4 × 2 = 0 + 0.000 000 000 001 838 668 8;
  • 8) 0.000 000 000 001 838 668 8 × 2 = 0 + 0.000 000 000 003 677 337 6;
  • 9) 0.000 000 000 003 677 337 6 × 2 = 0 + 0.000 000 000 007 354 675 2;
  • 10) 0.000 000 000 007 354 675 2 × 2 = 0 + 0.000 000 000 014 709 350 4;
  • 11) 0.000 000 000 014 709 350 4 × 2 = 0 + 0.000 000 000 029 418 700 8;
  • 12) 0.000 000 000 029 418 700 8 × 2 = 0 + 0.000 000 000 058 837 401 6;
  • 13) 0.000 000 000 058 837 401 6 × 2 = 0 + 0.000 000 000 117 674 803 2;
  • 14) 0.000 000 000 117 674 803 2 × 2 = 0 + 0.000 000 000 235 349 606 4;
  • 15) 0.000 000 000 235 349 606 4 × 2 = 0 + 0.000 000 000 470 699 212 8;
  • 16) 0.000 000 000 470 699 212 8 × 2 = 0 + 0.000 000 000 941 398 425 6;
  • 17) 0.000 000 000 941 398 425 6 × 2 = 0 + 0.000 000 001 882 796 851 2;
  • 18) 0.000 000 001 882 796 851 2 × 2 = 0 + 0.000 000 003 765 593 702 4;
  • 19) 0.000 000 003 765 593 702 4 × 2 = 0 + 0.000 000 007 531 187 404 8;
  • 20) 0.000 000 007 531 187 404 8 × 2 = 0 + 0.000 000 015 062 374 809 6;
  • 21) 0.000 000 015 062 374 809 6 × 2 = 0 + 0.000 000 030 124 749 619 2;
  • 22) 0.000 000 030 124 749 619 2 × 2 = 0 + 0.000 000 060 249 499 238 4;
  • 23) 0.000 000 060 249 499 238 4 × 2 = 0 + 0.000 000 120 498 998 476 8;
  • 24) 0.000 000 120 498 998 476 8 × 2 = 0 + 0.000 000 240 997 996 953 6;
  • 25) 0.000 000 240 997 996 953 6 × 2 = 0 + 0.000 000 481 995 993 907 2;
  • 26) 0.000 000 481 995 993 907 2 × 2 = 0 + 0.000 000 963 991 987 814 4;
  • 27) 0.000 000 963 991 987 814 4 × 2 = 0 + 0.000 001 927 983 975 628 8;
  • 28) 0.000 001 927 983 975 628 8 × 2 = 0 + 0.000 003 855 967 951 257 6;
  • 29) 0.000 003 855 967 951 257 6 × 2 = 0 + 0.000 007 711 935 902 515 2;
  • 30) 0.000 007 711 935 902 515 2 × 2 = 0 + 0.000 015 423 871 805 030 4;
  • 31) 0.000 015 423 871 805 030 4 × 2 = 0 + 0.000 030 847 743 610 060 8;
  • 32) 0.000 030 847 743 610 060 8 × 2 = 0 + 0.000 061 695 487 220 121 6;
  • 33) 0.000 061 695 487 220 121 6 × 2 = 0 + 0.000 123 390 974 440 243 2;
  • 34) 0.000 123 390 974 440 243 2 × 2 = 0 + 0.000 246 781 948 880 486 4;
  • 35) 0.000 246 781 948 880 486 4 × 2 = 0 + 0.000 493 563 897 760 972 8;
  • 36) 0.000 493 563 897 760 972 8 × 2 = 0 + 0.000 987 127 795 521 945 6;
  • 37) 0.000 987 127 795 521 945 6 × 2 = 0 + 0.001 974 255 591 043 891 2;
  • 38) 0.001 974 255 591 043 891 2 × 2 = 0 + 0.003 948 511 182 087 782 4;
  • 39) 0.003 948 511 182 087 782 4 × 2 = 0 + 0.007 897 022 364 175 564 8;
  • 40) 0.007 897 022 364 175 564 8 × 2 = 0 + 0.015 794 044 728 351 129 6;
  • 41) 0.015 794 044 728 351 129 6 × 2 = 0 + 0.031 588 089 456 702 259 2;
  • 42) 0.031 588 089 456 702 259 2 × 2 = 0 + 0.063 176 178 913 404 518 4;
  • 43) 0.063 176 178 913 404 518 4 × 2 = 0 + 0.126 352 357 826 809 036 8;
  • 44) 0.126 352 357 826 809 036 8 × 2 = 0 + 0.252 704 715 653 618 073 6;
  • 45) 0.252 704 715 653 618 073 6 × 2 = 0 + 0.505 409 431 307 236 147 2;
  • 46) 0.505 409 431 307 236 147 2 × 2 = 1 + 0.010 818 862 614 472 294 4;
  • 47) 0.010 818 862 614 472 294 4 × 2 = 0 + 0.021 637 725 228 944 588 8;
  • 48) 0.021 637 725 228 944 588 8 × 2 = 0 + 0.043 275 450 457 889 177 6;
  • 49) 0.043 275 450 457 889 177 6 × 2 = 0 + 0.086 550 900 915 778 355 2;
  • 50) 0.086 550 900 915 778 355 2 × 2 = 0 + 0.173 101 801 831 556 710 4;
  • 51) 0.173 101 801 831 556 710 4 × 2 = 0 + 0.346 203 603 663 113 420 8;
  • 52) 0.346 203 603 663 113 420 8 × 2 = 0 + 0.692 407 207 326 226 841 6;
  • 53) 0.692 407 207 326 226 841 6 × 2 = 1 + 0.384 814 414 652 453 683 2;
  • 54) 0.384 814 414 652 453 683 2 × 2 = 0 + 0.769 628 829 304 907 366 4;
  • 55) 0.769 628 829 304 907 366 4 × 2 = 1 + 0.539 257 658 609 814 732 8;
  • 56) 0.539 257 658 609 814 732 8 × 2 = 1 + 0.078 515 317 219 629 465 6;
  • 57) 0.078 515 317 219 629 465 6 × 2 = 0 + 0.157 030 634 439 258 931 2;
  • 58) 0.157 030 634 439 258 931 2 × 2 = 0 + 0.314 061 268 878 517 862 4;
  • 59) 0.314 061 268 878 517 862 4 × 2 = 0 + 0.628 122 537 757 035 724 8;
  • 60) 0.628 122 537 757 035 724 8 × 2 = 1 + 0.256 245 075 514 071 449 6;
  • 61) 0.256 245 075 514 071 449 6 × 2 = 0 + 0.512 490 151 028 142 899 2;
  • 62) 0.512 490 151 028 142 899 2 × 2 = 1 + 0.024 980 302 056 285 798 4;
  • 63) 0.024 980 302 056 285 798 4 × 2 = 0 + 0.049 960 604 112 571 596 8;
  • 64) 0.049 960 604 112 571 596 8 × 2 = 0 + 0.099 921 208 225 143 193 6;
  • 65) 0.099 921 208 225 143 193 6 × 2 = 0 + 0.199 842 416 450 286 387 2;
  • 66) 0.199 842 416 450 286 387 2 × 2 = 0 + 0.399 684 832 900 572 774 4;
  • 67) 0.399 684 832 900 572 774 4 × 2 = 0 + 0.799 369 665 801 145 548 8;
  • 68) 0.799 369 665 801 145 548 8 × 2 = 1 + 0.598 739 331 602 291 097 6;
  • 69) 0.598 739 331 602 291 097 6 × 2 = 1 + 0.197 478 663 204 582 195 2;
  • 70) 0.197 478 663 204 582 195 2 × 2 = 0 + 0.394 957 326 409 164 390 4;
  • 71) 0.394 957 326 409 164 390 4 × 2 = 0 + 0.789 914 652 818 328 780 8;
  • 72) 0.789 914 652 818 328 780 8 × 2 = 1 + 0.579 829 305 636 657 561 6;
  • 73) 0.579 829 305 636 657 561 6 × 2 = 1 + 0.159 658 611 273 315 123 2;
  • 74) 0.159 658 611 273 315 123 2 × 2 = 0 + 0.319 317 222 546 630 246 4;
  • 75) 0.319 317 222 546 630 246 4 × 2 = 0 + 0.638 634 445 093 260 492 8;
  • 76) 0.638 634 445 093 260 492 8 × 2 = 1 + 0.277 268 890 186 520 985 6;
  • 77) 0.277 268 890 186 520 985 6 × 2 = 0 + 0.554 537 780 373 041 971 2;
  • 78) 0.554 537 780 373 041 971 2 × 2 = 1 + 0.109 075 560 746 083 942 4;
  • 79) 0.109 075 560 746 083 942 4 × 2 = 0 + 0.218 151 121 492 167 884 8;
  • 80) 0.218 151 121 492 167 884 8 × 2 = 0 + 0.436 302 242 984 335 769 6;
  • 81) 0.436 302 242 984 335 769 6 × 2 = 0 + 0.872 604 485 968 671 539 2;
  • 82) 0.872 604 485 968 671 539 2 × 2 = 1 + 0.745 208 971 937 343 078 4;
  • 83) 0.745 208 971 937 343 078 4 × 2 = 1 + 0.490 417 943 874 686 156 8;
  • 84) 0.490 417 943 874 686 156 8 × 2 = 0 + 0.980 835 887 749 372 313 6;
  • 85) 0.980 835 887 749 372 313 6 × 2 = 1 + 0.961 671 775 498 744 627 2;
  • 86) 0.961 671 775 498 744 627 2 × 2 = 1 + 0.923 343 550 997 489 254 4;
  • 87) 0.923 343 550 997 489 254 4 × 2 = 1 + 0.846 687 101 994 978 508 8;
  • 88) 0.846 687 101 994 978 508 8 × 2 = 1 + 0.693 374 203 989 957 017 6;
  • 89) 0.693 374 203 989 957 017 6 × 2 = 1 + 0.386 748 407 979 914 035 2;
  • 90) 0.386 748 407 979 914 035 2 × 2 = 0 + 0.773 496 815 959 828 070 4;
  • 91) 0.773 496 815 959 828 070 4 × 2 = 1 + 0.546 993 631 919 656 140 8;
  • 92) 0.546 993 631 919 656 140 8 × 2 = 1 + 0.093 987 263 839 312 281 6;
  • 93) 0.093 987 263 839 312 281 6 × 2 = 0 + 0.187 974 527 678 624 563 2;
  • 94) 0.187 974 527 678 624 563 2 × 2 = 0 + 0.375 949 055 357 249 126 4;
  • 95) 0.375 949 055 357 249 126 4 × 2 = 0 + 0.751 898 110 714 498 252 8;
  • 96) 0.751 898 110 714 498 252 8 × 2 = 1 + 0.503 796 221 428 996 505 6;
  • 97) 0.503 796 221 428 996 505 6 × 2 = 1 + 0.007 592 442 857 993 011 2;
  • 98) 0.007 592 442 857 993 011 2 × 2 = 0 + 0.015 184 885 715 986 022 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:


0.000 000 000 000 014 364 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0001 0100 0001 1001 1001 0100 0110 1111 1011 0001 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 364 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0001 0100 0001 1001 1001 0100 0110 1111 1011 0001 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 364 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0001 0100 0001 1001 1001 0100 0110 1111 1011 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0001 0100 0001 1001 1001 0100 0110 1111 1011 0001 10(2) × 20 =

1.0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110 =

0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110


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 1101 0001

Mantissa (52 bits) =
0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110


Decimal number -0.000 000 000 000 014 364 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1100 0101 0000 0110 0110 0101 0001 1011 1110 1100 0110

» Convert decimal -0.000 000 000 000 014 373 3 to 64 bit double precision IEEE 754 binary floating point representation standard

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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