-0.000 000 000 000 014 366 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 366 4(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
-0.000 000 000 000 014 366 4(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:

|-0.000 000 000 000 014 366 4| = 0.000 000 000 000 014 366 4


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: 0.000 000 000 000 014 366 4.

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) 0.000 000 000 000 014 366 4 × 2 = 0 + 0.000 000 000 000 028 732 8;
  • 2) 0.000 000 000 000 028 732 8 × 2 = 0 + 0.000 000 000 000 057 465 6;
  • 3) 0.000 000 000 000 057 465 6 × 2 = 0 + 0.000 000 000 000 114 931 2;
  • 4) 0.000 000 000 000 114 931 2 × 2 = 0 + 0.000 000 000 000 229 862 4;
  • 5) 0.000 000 000 000 229 862 4 × 2 = 0 + 0.000 000 000 000 459 724 8;
  • 6) 0.000 000 000 000 459 724 8 × 2 = 0 + 0.000 000 000 000 919 449 6;
  • 7) 0.000 000 000 000 919 449 6 × 2 = 0 + 0.000 000 000 001 838 899 2;
  • 8) 0.000 000 000 001 838 899 2 × 2 = 0 + 0.000 000 000 003 677 798 4;
  • 9) 0.000 000 000 003 677 798 4 × 2 = 0 + 0.000 000 000 007 355 596 8;
  • 10) 0.000 000 000 007 355 596 8 × 2 = 0 + 0.000 000 000 014 711 193 6;
  • 11) 0.000 000 000 014 711 193 6 × 2 = 0 + 0.000 000 000 029 422 387 2;
  • 12) 0.000 000 000 029 422 387 2 × 2 = 0 + 0.000 000 000 058 844 774 4;
  • 13) 0.000 000 000 058 844 774 4 × 2 = 0 + 0.000 000 000 117 689 548 8;
  • 14) 0.000 000 000 117 689 548 8 × 2 = 0 + 0.000 000 000 235 379 097 6;
  • 15) 0.000 000 000 235 379 097 6 × 2 = 0 + 0.000 000 000 470 758 195 2;
  • 16) 0.000 000 000 470 758 195 2 × 2 = 0 + 0.000 000 000 941 516 390 4;
  • 17) 0.000 000 000 941 516 390 4 × 2 = 0 + 0.000 000 001 883 032 780 8;
  • 18) 0.000 000 001 883 032 780 8 × 2 = 0 + 0.000 000 003 766 065 561 6;
  • 19) 0.000 000 003 766 065 561 6 × 2 = 0 + 0.000 000 007 532 131 123 2;
  • 20) 0.000 000 007 532 131 123 2 × 2 = 0 + 0.000 000 015 064 262 246 4;
  • 21) 0.000 000 015 064 262 246 4 × 2 = 0 + 0.000 000 030 128 524 492 8;
  • 22) 0.000 000 030 128 524 492 8 × 2 = 0 + 0.000 000 060 257 048 985 6;
  • 23) 0.000 000 060 257 048 985 6 × 2 = 0 + 0.000 000 120 514 097 971 2;
  • 24) 0.000 000 120 514 097 971 2 × 2 = 0 + 0.000 000 241 028 195 942 4;
  • 25) 0.000 000 241 028 195 942 4 × 2 = 0 + 0.000 000 482 056 391 884 8;
  • 26) 0.000 000 482 056 391 884 8 × 2 = 0 + 0.000 000 964 112 783 769 6;
  • 27) 0.000 000 964 112 783 769 6 × 2 = 0 + 0.000 001 928 225 567 539 2;
  • 28) 0.000 001 928 225 567 539 2 × 2 = 0 + 0.000 003 856 451 135 078 4;
  • 29) 0.000 003 856 451 135 078 4 × 2 = 0 + 0.000 007 712 902 270 156 8;
  • 30) 0.000 007 712 902 270 156 8 × 2 = 0 + 0.000 015 425 804 540 313 6;
  • 31) 0.000 015 425 804 540 313 6 × 2 = 0 + 0.000 030 851 609 080 627 2;
  • 32) 0.000 030 851 609 080 627 2 × 2 = 0 + 0.000 061 703 218 161 254 4;
  • 33) 0.000 061 703 218 161 254 4 × 2 = 0 + 0.000 123 406 436 322 508 8;
  • 34) 0.000 123 406 436 322 508 8 × 2 = 0 + 0.000 246 812 872 645 017 6;
  • 35) 0.000 246 812 872 645 017 6 × 2 = 0 + 0.000 493 625 745 290 035 2;
  • 36) 0.000 493 625 745 290 035 2 × 2 = 0 + 0.000 987 251 490 580 070 4;
  • 37) 0.000 987 251 490 580 070 4 × 2 = 0 + 0.001 974 502 981 160 140 8;
  • 38) 0.001 974 502 981 160 140 8 × 2 = 0 + 0.003 949 005 962 320 281 6;
  • 39) 0.003 949 005 962 320 281 6 × 2 = 0 + 0.007 898 011 924 640 563 2;
  • 40) 0.007 898 011 924 640 563 2 × 2 = 0 + 0.015 796 023 849 281 126 4;
  • 41) 0.015 796 023 849 281 126 4 × 2 = 0 + 0.031 592 047 698 562 252 8;
  • 42) 0.031 592 047 698 562 252 8 × 2 = 0 + 0.063 184 095 397 124 505 6;
  • 43) 0.063 184 095 397 124 505 6 × 2 = 0 + 0.126 368 190 794 249 011 2;
  • 44) 0.126 368 190 794 249 011 2 × 2 = 0 + 0.252 736 381 588 498 022 4;
  • 45) 0.252 736 381 588 498 022 4 × 2 = 0 + 0.505 472 763 176 996 044 8;
  • 46) 0.505 472 763 176 996 044 8 × 2 = 1 + 0.010 945 526 353 992 089 6;
  • 47) 0.010 945 526 353 992 089 6 × 2 = 0 + 0.021 891 052 707 984 179 2;
  • 48) 0.021 891 052 707 984 179 2 × 2 = 0 + 0.043 782 105 415 968 358 4;
  • 49) 0.043 782 105 415 968 358 4 × 2 = 0 + 0.087 564 210 831 936 716 8;
  • 50) 0.087 564 210 831 936 716 8 × 2 = 0 + 0.175 128 421 663 873 433 6;
  • 51) 0.175 128 421 663 873 433 6 × 2 = 0 + 0.350 256 843 327 746 867 2;
  • 52) 0.350 256 843 327 746 867 2 × 2 = 0 + 0.700 513 686 655 493 734 4;
  • 53) 0.700 513 686 655 493 734 4 × 2 = 1 + 0.401 027 373 310 987 468 8;
  • 54) 0.401 027 373 310 987 468 8 × 2 = 0 + 0.802 054 746 621 974 937 6;
  • 55) 0.802 054 746 621 974 937 6 × 2 = 1 + 0.604 109 493 243 949 875 2;
  • 56) 0.604 109 493 243 949 875 2 × 2 = 1 + 0.208 218 986 487 899 750 4;
  • 57) 0.208 218 986 487 899 750 4 × 2 = 0 + 0.416 437 972 975 799 500 8;
  • 58) 0.416 437 972 975 799 500 8 × 2 = 0 + 0.832 875 945 951 599 001 6;
  • 59) 0.832 875 945 951 599 001 6 × 2 = 1 + 0.665 751 891 903 198 003 2;
  • 60) 0.665 751 891 903 198 003 2 × 2 = 1 + 0.331 503 783 806 396 006 4;
  • 61) 0.331 503 783 806 396 006 4 × 2 = 0 + 0.663 007 567 612 792 012 8;
  • 62) 0.663 007 567 612 792 012 8 × 2 = 1 + 0.326 015 135 225 584 025 6;
  • 63) 0.326 015 135 225 584 025 6 × 2 = 0 + 0.652 030 270 451 168 051 2;
  • 64) 0.652 030 270 451 168 051 2 × 2 = 1 + 0.304 060 540 902 336 102 4;
  • 65) 0.304 060 540 902 336 102 4 × 2 = 0 + 0.608 121 081 804 672 204 8;
  • 66) 0.608 121 081 804 672 204 8 × 2 = 1 + 0.216 242 163 609 344 409 6;
  • 67) 0.216 242 163 609 344 409 6 × 2 = 0 + 0.432 484 327 218 688 819 2;
  • 68) 0.432 484 327 218 688 819 2 × 2 = 0 + 0.864 968 654 437 377 638 4;
  • 69) 0.864 968 654 437 377 638 4 × 2 = 1 + 0.729 937 308 874 755 276 8;
  • 70) 0.729 937 308 874 755 276 8 × 2 = 1 + 0.459 874 617 749 510 553 6;
  • 71) 0.459 874 617 749 510 553 6 × 2 = 0 + 0.919 749 235 499 021 107 2;
  • 72) 0.919 749 235 499 021 107 2 × 2 = 1 + 0.839 498 470 998 042 214 4;
  • 73) 0.839 498 470 998 042 214 4 × 2 = 1 + 0.678 996 941 996 084 428 8;
  • 74) 0.678 996 941 996 084 428 8 × 2 = 1 + 0.357 993 883 992 168 857 6;
  • 75) 0.357 993 883 992 168 857 6 × 2 = 0 + 0.715 987 767 984 337 715 2;
  • 76) 0.715 987 767 984 337 715 2 × 2 = 1 + 0.431 975 535 968 675 430 4;
  • 77) 0.431 975 535 968 675 430 4 × 2 = 0 + 0.863 951 071 937 350 860 8;
  • 78) 0.863 951 071 937 350 860 8 × 2 = 1 + 0.727 902 143 874 701 721 6;
  • 79) 0.727 902 143 874 701 721 6 × 2 = 1 + 0.455 804 287 749 403 443 2;
  • 80) 0.455 804 287 749 403 443 2 × 2 = 0 + 0.911 608 575 498 806 886 4;
  • 81) 0.911 608 575 498 806 886 4 × 2 = 1 + 0.823 217 150 997 613 772 8;
  • 82) 0.823 217 150 997 613 772 8 × 2 = 1 + 0.646 434 301 995 227 545 6;
  • 83) 0.646 434 301 995 227 545 6 × 2 = 1 + 0.292 868 603 990 455 091 2;
  • 84) 0.292 868 603 990 455 091 2 × 2 = 0 + 0.585 737 207 980 910 182 4;
  • 85) 0.585 737 207 980 910 182 4 × 2 = 1 + 0.171 474 415 961 820 364 8;
  • 86) 0.171 474 415 961 820 364 8 × 2 = 0 + 0.342 948 831 923 640 729 6;
  • 87) 0.342 948 831 923 640 729 6 × 2 = 0 + 0.685 897 663 847 281 459 2;
  • 88) 0.685 897 663 847 281 459 2 × 2 = 1 + 0.371 795 327 694 562 918 4;
  • 89) 0.371 795 327 694 562 918 4 × 2 = 0 + 0.743 590 655 389 125 836 8;
  • 90) 0.743 590 655 389 125 836 8 × 2 = 1 + 0.487 181 310 778 251 673 6;
  • 91) 0.487 181 310 778 251 673 6 × 2 = 0 + 0.974 362 621 556 503 347 2;
  • 92) 0.974 362 621 556 503 347 2 × 2 = 1 + 0.948 725 243 113 006 694 4;
  • 93) 0.948 725 243 113 006 694 4 × 2 = 1 + 0.897 450 486 226 013 388 8;
  • 94) 0.897 450 486 226 013 388 8 × 2 = 1 + 0.794 900 972 452 026 777 6;
  • 95) 0.794 900 972 452 026 777 6 × 2 = 1 + 0.589 801 944 904 053 555 2;
  • 96) 0.589 801 944 904 053 555 2 × 2 = 1 + 0.179 603 889 808 107 110 4;
  • 97) 0.179 603 889 808 107 110 4 × 2 = 0 + 0.359 207 779 616 214 220 8;
  • 98) 0.359 207 779 616 214 220 8 × 2 = 0 + 0.718 415 559 232 428 441 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 - 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:


0.000 000 000 000 014 366 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0011 0101 0100 1101 1101 0110 1110 1001 0101 1111 00(2)

6. Positive number before normalization:

0.000 000 000 000 014 366 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0011 0101 0100 1101 1101 0110 1110 1001 0101 1111 00(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:


0.000 000 000 000 014 366 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0011 0101 0100 1101 1101 0110 1110 1001 0101 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0011 0101 0100 1101 1101 0110 1110 1001 0101 1111 00(2) × 20 =

1.0000 0010 1100 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100(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 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100


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 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100 =

0000 0010 1100 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100


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 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100


Decimal number -0.000 000 000 000 014 366 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1100 1101 0101 0011 0111 0101 1011 1010 0101 0111 1100

» Convert decimal -0.000 000 000 000 014 356 9 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