-0.000 000 000 000 176 557 495 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 495(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
-0.000 000 000 000 176 557 495(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:

|-0.000 000 000 000 176 557 495| = 0.000 000 000 000 176 557 495


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: 0.000 000 000 000 176 557 495.

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) 0.000 000 000 000 176 557 495 × 2 = 0 + 0.000 000 000 000 353 114 99;
  • 2) 0.000 000 000 000 353 114 99 × 2 = 0 + 0.000 000 000 000 706 229 98;
  • 3) 0.000 000 000 000 706 229 98 × 2 = 0 + 0.000 000 000 001 412 459 96;
  • 4) 0.000 000 000 001 412 459 96 × 2 = 0 + 0.000 000 000 002 824 919 92;
  • 5) 0.000 000 000 002 824 919 92 × 2 = 0 + 0.000 000 000 005 649 839 84;
  • 6) 0.000 000 000 005 649 839 84 × 2 = 0 + 0.000 000 000 011 299 679 68;
  • 7) 0.000 000 000 011 299 679 68 × 2 = 0 + 0.000 000 000 022 599 359 36;
  • 8) 0.000 000 000 022 599 359 36 × 2 = 0 + 0.000 000 000 045 198 718 72;
  • 9) 0.000 000 000 045 198 718 72 × 2 = 0 + 0.000 000 000 090 397 437 44;
  • 10) 0.000 000 000 090 397 437 44 × 2 = 0 + 0.000 000 000 180 794 874 88;
  • 11) 0.000 000 000 180 794 874 88 × 2 = 0 + 0.000 000 000 361 589 749 76;
  • 12) 0.000 000 000 361 589 749 76 × 2 = 0 + 0.000 000 000 723 179 499 52;
  • 13) 0.000 000 000 723 179 499 52 × 2 = 0 + 0.000 000 001 446 358 999 04;
  • 14) 0.000 000 001 446 358 999 04 × 2 = 0 + 0.000 000 002 892 717 998 08;
  • 15) 0.000 000 002 892 717 998 08 × 2 = 0 + 0.000 000 005 785 435 996 16;
  • 16) 0.000 000 005 785 435 996 16 × 2 = 0 + 0.000 000 011 570 871 992 32;
  • 17) 0.000 000 011 570 871 992 32 × 2 = 0 + 0.000 000 023 141 743 984 64;
  • 18) 0.000 000 023 141 743 984 64 × 2 = 0 + 0.000 000 046 283 487 969 28;
  • 19) 0.000 000 046 283 487 969 28 × 2 = 0 + 0.000 000 092 566 975 938 56;
  • 20) 0.000 000 092 566 975 938 56 × 2 = 0 + 0.000 000 185 133 951 877 12;
  • 21) 0.000 000 185 133 951 877 12 × 2 = 0 + 0.000 000 370 267 903 754 24;
  • 22) 0.000 000 370 267 903 754 24 × 2 = 0 + 0.000 000 740 535 807 508 48;
  • 23) 0.000 000 740 535 807 508 48 × 2 = 0 + 0.000 001 481 071 615 016 96;
  • 24) 0.000 001 481 071 615 016 96 × 2 = 0 + 0.000 002 962 143 230 033 92;
  • 25) 0.000 002 962 143 230 033 92 × 2 = 0 + 0.000 005 924 286 460 067 84;
  • 26) 0.000 005 924 286 460 067 84 × 2 = 0 + 0.000 011 848 572 920 135 68;
  • 27) 0.000 011 848 572 920 135 68 × 2 = 0 + 0.000 023 697 145 840 271 36;
  • 28) 0.000 023 697 145 840 271 36 × 2 = 0 + 0.000 047 394 291 680 542 72;
  • 29) 0.000 047 394 291 680 542 72 × 2 = 0 + 0.000 094 788 583 361 085 44;
  • 30) 0.000 094 788 583 361 085 44 × 2 = 0 + 0.000 189 577 166 722 170 88;
  • 31) 0.000 189 577 166 722 170 88 × 2 = 0 + 0.000 379 154 333 444 341 76;
  • 32) 0.000 379 154 333 444 341 76 × 2 = 0 + 0.000 758 308 666 888 683 52;
  • 33) 0.000 758 308 666 888 683 52 × 2 = 0 + 0.001 516 617 333 777 367 04;
  • 34) 0.001 516 617 333 777 367 04 × 2 = 0 + 0.003 033 234 667 554 734 08;
  • 35) 0.003 033 234 667 554 734 08 × 2 = 0 + 0.006 066 469 335 109 468 16;
  • 36) 0.006 066 469 335 109 468 16 × 2 = 0 + 0.012 132 938 670 218 936 32;
  • 37) 0.012 132 938 670 218 936 32 × 2 = 0 + 0.024 265 877 340 437 872 64;
  • 38) 0.024 265 877 340 437 872 64 × 2 = 0 + 0.048 531 754 680 875 745 28;
  • 39) 0.048 531 754 680 875 745 28 × 2 = 0 + 0.097 063 509 361 751 490 56;
  • 40) 0.097 063 509 361 751 490 56 × 2 = 0 + 0.194 127 018 723 502 981 12;
  • 41) 0.194 127 018 723 502 981 12 × 2 = 0 + 0.388 254 037 447 005 962 24;
  • 42) 0.388 254 037 447 005 962 24 × 2 = 0 + 0.776 508 074 894 011 924 48;
  • 43) 0.776 508 074 894 011 924 48 × 2 = 1 + 0.553 016 149 788 023 848 96;
  • 44) 0.553 016 149 788 023 848 96 × 2 = 1 + 0.106 032 299 576 047 697 92;
  • 45) 0.106 032 299 576 047 697 92 × 2 = 0 + 0.212 064 599 152 095 395 84;
  • 46) 0.212 064 599 152 095 395 84 × 2 = 0 + 0.424 129 198 304 190 791 68;
  • 47) 0.424 129 198 304 190 791 68 × 2 = 0 + 0.848 258 396 608 381 583 36;
  • 48) 0.848 258 396 608 381 583 36 × 2 = 1 + 0.696 516 793 216 763 166 72;
  • 49) 0.696 516 793 216 763 166 72 × 2 = 1 + 0.393 033 586 433 526 333 44;
  • 50) 0.393 033 586 433 526 333 44 × 2 = 0 + 0.786 067 172 867 052 666 88;
  • 51) 0.786 067 172 867 052 666 88 × 2 = 1 + 0.572 134 345 734 105 333 76;
  • 52) 0.572 134 345 734 105 333 76 × 2 = 1 + 0.144 268 691 468 210 667 52;
  • 53) 0.144 268 691 468 210 667 52 × 2 = 0 + 0.288 537 382 936 421 335 04;
  • 54) 0.288 537 382 936 421 335 04 × 2 = 0 + 0.577 074 765 872 842 670 08;
  • 55) 0.577 074 765 872 842 670 08 × 2 = 1 + 0.154 149 531 745 685 340 16;
  • 56) 0.154 149 531 745 685 340 16 × 2 = 0 + 0.308 299 063 491 370 680 32;
  • 57) 0.308 299 063 491 370 680 32 × 2 = 0 + 0.616 598 126 982 741 360 64;
  • 58) 0.616 598 126 982 741 360 64 × 2 = 1 + 0.233 196 253 965 482 721 28;
  • 59) 0.233 196 253 965 482 721 28 × 2 = 0 + 0.466 392 507 930 965 442 56;
  • 60) 0.466 392 507 930 965 442 56 × 2 = 0 + 0.932 785 015 861 930 885 12;
  • 61) 0.932 785 015 861 930 885 12 × 2 = 1 + 0.865 570 031 723 861 770 24;
  • 62) 0.865 570 031 723 861 770 24 × 2 = 1 + 0.731 140 063 447 723 540 48;
  • 63) 0.731 140 063 447 723 540 48 × 2 = 1 + 0.462 280 126 895 447 080 96;
  • 64) 0.462 280 126 895 447 080 96 × 2 = 0 + 0.924 560 253 790 894 161 92;
  • 65) 0.924 560 253 790 894 161 92 × 2 = 1 + 0.849 120 507 581 788 323 84;
  • 66) 0.849 120 507 581 788 323 84 × 2 = 1 + 0.698 241 015 163 576 647 68;
  • 67) 0.698 241 015 163 576 647 68 × 2 = 1 + 0.396 482 030 327 153 295 36;
  • 68) 0.396 482 030 327 153 295 36 × 2 = 0 + 0.792 964 060 654 306 590 72;
  • 69) 0.792 964 060 654 306 590 72 × 2 = 1 + 0.585 928 121 308 613 181 44;
  • 70) 0.585 928 121 308 613 181 44 × 2 = 1 + 0.171 856 242 617 226 362 88;
  • 71) 0.171 856 242 617 226 362 88 × 2 = 0 + 0.343 712 485 234 452 725 76;
  • 72) 0.343 712 485 234 452 725 76 × 2 = 0 + 0.687 424 970 468 905 451 52;
  • 73) 0.687 424 970 468 905 451 52 × 2 = 1 + 0.374 849 940 937 810 903 04;
  • 74) 0.374 849 940 937 810 903 04 × 2 = 0 + 0.749 699 881 875 621 806 08;
  • 75) 0.749 699 881 875 621 806 08 × 2 = 1 + 0.499 399 763 751 243 612 16;
  • 76) 0.499 399 763 751 243 612 16 × 2 = 0 + 0.998 799 527 502 487 224 32;
  • 77) 0.998 799 527 502 487 224 32 × 2 = 1 + 0.997 599 055 004 974 448 64;
  • 78) 0.997 599 055 004 974 448 64 × 2 = 1 + 0.995 198 110 009 948 897 28;
  • 79) 0.995 198 110 009 948 897 28 × 2 = 1 + 0.990 396 220 019 897 794 56;
  • 80) 0.990 396 220 019 897 794 56 × 2 = 1 + 0.980 792 440 039 795 589 12;
  • 81) 0.980 792 440 039 795 589 12 × 2 = 1 + 0.961 584 880 079 591 178 24;
  • 82) 0.961 584 880 079 591 178 24 × 2 = 1 + 0.923 169 760 159 182 356 48;
  • 83) 0.923 169 760 159 182 356 48 × 2 = 1 + 0.846 339 520 318 364 712 96;
  • 84) 0.846 339 520 318 364 712 96 × 2 = 1 + 0.692 679 040 636 729 425 92;
  • 85) 0.692 679 040 636 729 425 92 × 2 = 1 + 0.385 358 081 273 458 851 84;
  • 86) 0.385 358 081 273 458 851 84 × 2 = 0 + 0.770 716 162 546 917 703 68;
  • 87) 0.770 716 162 546 917 703 68 × 2 = 1 + 0.541 432 325 093 835 407 36;
  • 88) 0.541 432 325 093 835 407 36 × 2 = 1 + 0.082 864 650 187 670 814 72;
  • 89) 0.082 864 650 187 670 814 72 × 2 = 0 + 0.165 729 300 375 341 629 44;
  • 90) 0.165 729 300 375 341 629 44 × 2 = 0 + 0.331 458 600 750 683 258 88;
  • 91) 0.331 458 600 750 683 258 88 × 2 = 0 + 0.662 917 201 501 366 517 76;
  • 92) 0.662 917 201 501 366 517 76 × 2 = 1 + 0.325 834 403 002 733 035 52;
  • 93) 0.325 834 403 002 733 035 52 × 2 = 0 + 0.651 668 806 005 466 071 04;
  • 94) 0.651 668 806 005 466 071 04 × 2 = 1 + 0.303 337 612 010 932 142 08;
  • 95) 0.303 337 612 010 932 142 08 × 2 = 0 + 0.606 675 224 021 864 284 16;

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:


0.000 000 000 000 176 557 495(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1110 1100 1010 1111 1111 1011 0001 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 495(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1110 1100 1010 1111 1111 1011 0001 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 495(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1110 1100 1010 1111 1111 1011 0001 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1110 1100 1010 1111 1111 1011 0001 010(2) × 20 =

1.1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010 =

1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010


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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010


Decimal number -0.000 000 000 000 176 557 495 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0111 0110 0101 0111 1111 1101 1000 1010

Share

» Convert decimal -0.000 000 000 000 176 557 586 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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