-0.000 000 000 000 014 389 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 389 8(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
-0.000 000 000 000 014 389 8(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:

|-0.000 000 000 000 014 389 8| = 0.000 000 000 000 014 389 8


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: 0.000 000 000 000 014 389 8.

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) 0.000 000 000 000 014 389 8 × 2 = 0 + 0.000 000 000 000 028 779 6;
  • 2) 0.000 000 000 000 028 779 6 × 2 = 0 + 0.000 000 000 000 057 559 2;
  • 3) 0.000 000 000 000 057 559 2 × 2 = 0 + 0.000 000 000 000 115 118 4;
  • 4) 0.000 000 000 000 115 118 4 × 2 = 0 + 0.000 000 000 000 230 236 8;
  • 5) 0.000 000 000 000 230 236 8 × 2 = 0 + 0.000 000 000 000 460 473 6;
  • 6) 0.000 000 000 000 460 473 6 × 2 = 0 + 0.000 000 000 000 920 947 2;
  • 7) 0.000 000 000 000 920 947 2 × 2 = 0 + 0.000 000 000 001 841 894 4;
  • 8) 0.000 000 000 001 841 894 4 × 2 = 0 + 0.000 000 000 003 683 788 8;
  • 9) 0.000 000 000 003 683 788 8 × 2 = 0 + 0.000 000 000 007 367 577 6;
  • 10) 0.000 000 000 007 367 577 6 × 2 = 0 + 0.000 000 000 014 735 155 2;
  • 11) 0.000 000 000 014 735 155 2 × 2 = 0 + 0.000 000 000 029 470 310 4;
  • 12) 0.000 000 000 029 470 310 4 × 2 = 0 + 0.000 000 000 058 940 620 8;
  • 13) 0.000 000 000 058 940 620 8 × 2 = 0 + 0.000 000 000 117 881 241 6;
  • 14) 0.000 000 000 117 881 241 6 × 2 = 0 + 0.000 000 000 235 762 483 2;
  • 15) 0.000 000 000 235 762 483 2 × 2 = 0 + 0.000 000 000 471 524 966 4;
  • 16) 0.000 000 000 471 524 966 4 × 2 = 0 + 0.000 000 000 943 049 932 8;
  • 17) 0.000 000 000 943 049 932 8 × 2 = 0 + 0.000 000 001 886 099 865 6;
  • 18) 0.000 000 001 886 099 865 6 × 2 = 0 + 0.000 000 003 772 199 731 2;
  • 19) 0.000 000 003 772 199 731 2 × 2 = 0 + 0.000 000 007 544 399 462 4;
  • 20) 0.000 000 007 544 399 462 4 × 2 = 0 + 0.000 000 015 088 798 924 8;
  • 21) 0.000 000 015 088 798 924 8 × 2 = 0 + 0.000 000 030 177 597 849 6;
  • 22) 0.000 000 030 177 597 849 6 × 2 = 0 + 0.000 000 060 355 195 699 2;
  • 23) 0.000 000 060 355 195 699 2 × 2 = 0 + 0.000 000 120 710 391 398 4;
  • 24) 0.000 000 120 710 391 398 4 × 2 = 0 + 0.000 000 241 420 782 796 8;
  • 25) 0.000 000 241 420 782 796 8 × 2 = 0 + 0.000 000 482 841 565 593 6;
  • 26) 0.000 000 482 841 565 593 6 × 2 = 0 + 0.000 000 965 683 131 187 2;
  • 27) 0.000 000 965 683 131 187 2 × 2 = 0 + 0.000 001 931 366 262 374 4;
  • 28) 0.000 001 931 366 262 374 4 × 2 = 0 + 0.000 003 862 732 524 748 8;
  • 29) 0.000 003 862 732 524 748 8 × 2 = 0 + 0.000 007 725 465 049 497 6;
  • 30) 0.000 007 725 465 049 497 6 × 2 = 0 + 0.000 015 450 930 098 995 2;
  • 31) 0.000 015 450 930 098 995 2 × 2 = 0 + 0.000 030 901 860 197 990 4;
  • 32) 0.000 030 901 860 197 990 4 × 2 = 0 + 0.000 061 803 720 395 980 8;
  • 33) 0.000 061 803 720 395 980 8 × 2 = 0 + 0.000 123 607 440 791 961 6;
  • 34) 0.000 123 607 440 791 961 6 × 2 = 0 + 0.000 247 214 881 583 923 2;
  • 35) 0.000 247 214 881 583 923 2 × 2 = 0 + 0.000 494 429 763 167 846 4;
  • 36) 0.000 494 429 763 167 846 4 × 2 = 0 + 0.000 988 859 526 335 692 8;
  • 37) 0.000 988 859 526 335 692 8 × 2 = 0 + 0.001 977 719 052 671 385 6;
  • 38) 0.001 977 719 052 671 385 6 × 2 = 0 + 0.003 955 438 105 342 771 2;
  • 39) 0.003 955 438 105 342 771 2 × 2 = 0 + 0.007 910 876 210 685 542 4;
  • 40) 0.007 910 876 210 685 542 4 × 2 = 0 + 0.015 821 752 421 371 084 8;
  • 41) 0.015 821 752 421 371 084 8 × 2 = 0 + 0.031 643 504 842 742 169 6;
  • 42) 0.031 643 504 842 742 169 6 × 2 = 0 + 0.063 287 009 685 484 339 2;
  • 43) 0.063 287 009 685 484 339 2 × 2 = 0 + 0.126 574 019 370 968 678 4;
  • 44) 0.126 574 019 370 968 678 4 × 2 = 0 + 0.253 148 038 741 937 356 8;
  • 45) 0.253 148 038 741 937 356 8 × 2 = 0 + 0.506 296 077 483 874 713 6;
  • 46) 0.506 296 077 483 874 713 6 × 2 = 1 + 0.012 592 154 967 749 427 2;
  • 47) 0.012 592 154 967 749 427 2 × 2 = 0 + 0.025 184 309 935 498 854 4;
  • 48) 0.025 184 309 935 498 854 4 × 2 = 0 + 0.050 368 619 870 997 708 8;
  • 49) 0.050 368 619 870 997 708 8 × 2 = 0 + 0.100 737 239 741 995 417 6;
  • 50) 0.100 737 239 741 995 417 6 × 2 = 0 + 0.201 474 479 483 990 835 2;
  • 51) 0.201 474 479 483 990 835 2 × 2 = 0 + 0.402 948 958 967 981 670 4;
  • 52) 0.402 948 958 967 981 670 4 × 2 = 0 + 0.805 897 917 935 963 340 8;
  • 53) 0.805 897 917 935 963 340 8 × 2 = 1 + 0.611 795 835 871 926 681 6;
  • 54) 0.611 795 835 871 926 681 6 × 2 = 1 + 0.223 591 671 743 853 363 2;
  • 55) 0.223 591 671 743 853 363 2 × 2 = 0 + 0.447 183 343 487 706 726 4;
  • 56) 0.447 183 343 487 706 726 4 × 2 = 0 + 0.894 366 686 975 413 452 8;
  • 57) 0.894 366 686 975 413 452 8 × 2 = 1 + 0.788 733 373 950 826 905 6;
  • 58) 0.788 733 373 950 826 905 6 × 2 = 1 + 0.577 466 747 901 653 811 2;
  • 59) 0.577 466 747 901 653 811 2 × 2 = 1 + 0.154 933 495 803 307 622 4;
  • 60) 0.154 933 495 803 307 622 4 × 2 = 0 + 0.309 866 991 606 615 244 8;
  • 61) 0.309 866 991 606 615 244 8 × 2 = 0 + 0.619 733 983 213 230 489 6;
  • 62) 0.619 733 983 213 230 489 6 × 2 = 1 + 0.239 467 966 426 460 979 2;
  • 63) 0.239 467 966 426 460 979 2 × 2 = 0 + 0.478 935 932 852 921 958 4;
  • 64) 0.478 935 932 852 921 958 4 × 2 = 0 + 0.957 871 865 705 843 916 8;
  • 65) 0.957 871 865 705 843 916 8 × 2 = 1 + 0.915 743 731 411 687 833 6;
  • 66) 0.915 743 731 411 687 833 6 × 2 = 1 + 0.831 487 462 823 375 667 2;
  • 67) 0.831 487 462 823 375 667 2 × 2 = 1 + 0.662 974 925 646 751 334 4;
  • 68) 0.662 974 925 646 751 334 4 × 2 = 1 + 0.325 949 851 293 502 668 8;
  • 69) 0.325 949 851 293 502 668 8 × 2 = 0 + 0.651 899 702 587 005 337 6;
  • 70) 0.651 899 702 587 005 337 6 × 2 = 1 + 0.303 799 405 174 010 675 2;
  • 71) 0.303 799 405 174 010 675 2 × 2 = 0 + 0.607 598 810 348 021 350 4;
  • 72) 0.607 598 810 348 021 350 4 × 2 = 1 + 0.215 197 620 696 042 700 8;
  • 73) 0.215 197 620 696 042 700 8 × 2 = 0 + 0.430 395 241 392 085 401 6;
  • 74) 0.430 395 241 392 085 401 6 × 2 = 0 + 0.860 790 482 784 170 803 2;
  • 75) 0.860 790 482 784 170 803 2 × 2 = 1 + 0.721 580 965 568 341 606 4;
  • 76) 0.721 580 965 568 341 606 4 × 2 = 1 + 0.443 161 931 136 683 212 8;
  • 77) 0.443 161 931 136 683 212 8 × 2 = 0 + 0.886 323 862 273 366 425 6;
  • 78) 0.886 323 862 273 366 425 6 × 2 = 1 + 0.772 647 724 546 732 851 2;
  • 79) 0.772 647 724 546 732 851 2 × 2 = 1 + 0.545 295 449 093 465 702 4;
  • 80) 0.545 295 449 093 465 702 4 × 2 = 1 + 0.090 590 898 186 931 404 8;
  • 81) 0.090 590 898 186 931 404 8 × 2 = 0 + 0.181 181 796 373 862 809 6;
  • 82) 0.181 181 796 373 862 809 6 × 2 = 0 + 0.362 363 592 747 725 619 2;
  • 83) 0.362 363 592 747 725 619 2 × 2 = 0 + 0.724 727 185 495 451 238 4;
  • 84) 0.724 727 185 495 451 238 4 × 2 = 1 + 0.449 454 370 990 902 476 8;
  • 85) 0.449 454 370 990 902 476 8 × 2 = 0 + 0.898 908 741 981 804 953 6;
  • 86) 0.898 908 741 981 804 953 6 × 2 = 1 + 0.797 817 483 963 609 907 2;
  • 87) 0.797 817 483 963 609 907 2 × 2 = 1 + 0.595 634 967 927 219 814 4;
  • 88) 0.595 634 967 927 219 814 4 × 2 = 1 + 0.191 269 935 854 439 628 8;
  • 89) 0.191 269 935 854 439 628 8 × 2 = 0 + 0.382 539 871 708 879 257 6;
  • 90) 0.382 539 871 708 879 257 6 × 2 = 0 + 0.765 079 743 417 758 515 2;
  • 91) 0.765 079 743 417 758 515 2 × 2 = 1 + 0.530 159 486 835 517 030 4;
  • 92) 0.530 159 486 835 517 030 4 × 2 = 1 + 0.060 318 973 671 034 060 8;
  • 93) 0.060 318 973 671 034 060 8 × 2 = 0 + 0.120 637 947 342 068 121 6;
  • 94) 0.120 637 947 342 068 121 6 × 2 = 0 + 0.241 275 894 684 136 243 2;
  • 95) 0.241 275 894 684 136 243 2 × 2 = 0 + 0.482 551 789 368 272 486 4;
  • 96) 0.482 551 789 368 272 486 4 × 2 = 0 + 0.965 103 578 736 544 972 8;
  • 97) 0.965 103 578 736 544 972 8 × 2 = 1 + 0.930 207 157 473 089 945 6;
  • 98) 0.930 207 157 473 089 945 6 × 2 = 1 + 0.860 414 314 946 179 891 2;

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:


0.000 000 000 000 014 389 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1110 0100 1111 0101 0011 0111 0001 0111 0011 0000 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 389 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1110 0100 1111 0101 0011 0111 0001 0111 0011 0000 11(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:


0.000 000 000 000 014 389 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1110 0100 1111 0101 0011 0111 0001 0111 0011 0000 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1110 0100 1111 0101 0011 0111 0001 0111 0011 0000 11(2) × 20 =

1.0000 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011(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 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011


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 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011 =

0000 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011


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 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011


Decimal number -0.000 000 000 000 014 389 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0011 1001 0011 1101 0100 1101 1100 0101 1100 1100 0011

» Convert decimal -0.000 000 000 000 014 386 2 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