-0.000 000 000 000 176 557 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 22(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 22(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 22| = 0.000 000 000 000 176 557 22


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 22.

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 22 × 2 = 0 + 0.000 000 000 000 353 114 44;
  • 2) 0.000 000 000 000 353 114 44 × 2 = 0 + 0.000 000 000 000 706 228 88;
  • 3) 0.000 000 000 000 706 228 88 × 2 = 0 + 0.000 000 000 001 412 457 76;
  • 4) 0.000 000 000 001 412 457 76 × 2 = 0 + 0.000 000 000 002 824 915 52;
  • 5) 0.000 000 000 002 824 915 52 × 2 = 0 + 0.000 000 000 005 649 831 04;
  • 6) 0.000 000 000 005 649 831 04 × 2 = 0 + 0.000 000 000 011 299 662 08;
  • 7) 0.000 000 000 011 299 662 08 × 2 = 0 + 0.000 000 000 022 599 324 16;
  • 8) 0.000 000 000 022 599 324 16 × 2 = 0 + 0.000 000 000 045 198 648 32;
  • 9) 0.000 000 000 045 198 648 32 × 2 = 0 + 0.000 000 000 090 397 296 64;
  • 10) 0.000 000 000 090 397 296 64 × 2 = 0 + 0.000 000 000 180 794 593 28;
  • 11) 0.000 000 000 180 794 593 28 × 2 = 0 + 0.000 000 000 361 589 186 56;
  • 12) 0.000 000 000 361 589 186 56 × 2 = 0 + 0.000 000 000 723 178 373 12;
  • 13) 0.000 000 000 723 178 373 12 × 2 = 0 + 0.000 000 001 446 356 746 24;
  • 14) 0.000 000 001 446 356 746 24 × 2 = 0 + 0.000 000 002 892 713 492 48;
  • 15) 0.000 000 002 892 713 492 48 × 2 = 0 + 0.000 000 005 785 426 984 96;
  • 16) 0.000 000 005 785 426 984 96 × 2 = 0 + 0.000 000 011 570 853 969 92;
  • 17) 0.000 000 011 570 853 969 92 × 2 = 0 + 0.000 000 023 141 707 939 84;
  • 18) 0.000 000 023 141 707 939 84 × 2 = 0 + 0.000 000 046 283 415 879 68;
  • 19) 0.000 000 046 283 415 879 68 × 2 = 0 + 0.000 000 092 566 831 759 36;
  • 20) 0.000 000 092 566 831 759 36 × 2 = 0 + 0.000 000 185 133 663 518 72;
  • 21) 0.000 000 185 133 663 518 72 × 2 = 0 + 0.000 000 370 267 327 037 44;
  • 22) 0.000 000 370 267 327 037 44 × 2 = 0 + 0.000 000 740 534 654 074 88;
  • 23) 0.000 000 740 534 654 074 88 × 2 = 0 + 0.000 001 481 069 308 149 76;
  • 24) 0.000 001 481 069 308 149 76 × 2 = 0 + 0.000 002 962 138 616 299 52;
  • 25) 0.000 002 962 138 616 299 52 × 2 = 0 + 0.000 005 924 277 232 599 04;
  • 26) 0.000 005 924 277 232 599 04 × 2 = 0 + 0.000 011 848 554 465 198 08;
  • 27) 0.000 011 848 554 465 198 08 × 2 = 0 + 0.000 023 697 108 930 396 16;
  • 28) 0.000 023 697 108 930 396 16 × 2 = 0 + 0.000 047 394 217 860 792 32;
  • 29) 0.000 047 394 217 860 792 32 × 2 = 0 + 0.000 094 788 435 721 584 64;
  • 30) 0.000 094 788 435 721 584 64 × 2 = 0 + 0.000 189 576 871 443 169 28;
  • 31) 0.000 189 576 871 443 169 28 × 2 = 0 + 0.000 379 153 742 886 338 56;
  • 32) 0.000 379 153 742 886 338 56 × 2 = 0 + 0.000 758 307 485 772 677 12;
  • 33) 0.000 758 307 485 772 677 12 × 2 = 0 + 0.001 516 614 971 545 354 24;
  • 34) 0.001 516 614 971 545 354 24 × 2 = 0 + 0.003 033 229 943 090 708 48;
  • 35) 0.003 033 229 943 090 708 48 × 2 = 0 + 0.006 066 459 886 181 416 96;
  • 36) 0.006 066 459 886 181 416 96 × 2 = 0 + 0.012 132 919 772 362 833 92;
  • 37) 0.012 132 919 772 362 833 92 × 2 = 0 + 0.024 265 839 544 725 667 84;
  • 38) 0.024 265 839 544 725 667 84 × 2 = 0 + 0.048 531 679 089 451 335 68;
  • 39) 0.048 531 679 089 451 335 68 × 2 = 0 + 0.097 063 358 178 902 671 36;
  • 40) 0.097 063 358 178 902 671 36 × 2 = 0 + 0.194 126 716 357 805 342 72;
  • 41) 0.194 126 716 357 805 342 72 × 2 = 0 + 0.388 253 432 715 610 685 44;
  • 42) 0.388 253 432 715 610 685 44 × 2 = 0 + 0.776 506 865 431 221 370 88;
  • 43) 0.776 506 865 431 221 370 88 × 2 = 1 + 0.553 013 730 862 442 741 76;
  • 44) 0.553 013 730 862 442 741 76 × 2 = 1 + 0.106 027 461 724 885 483 52;
  • 45) 0.106 027 461 724 885 483 52 × 2 = 0 + 0.212 054 923 449 770 967 04;
  • 46) 0.212 054 923 449 770 967 04 × 2 = 0 + 0.424 109 846 899 541 934 08;
  • 47) 0.424 109 846 899 541 934 08 × 2 = 0 + 0.848 219 693 799 083 868 16;
  • 48) 0.848 219 693 799 083 868 16 × 2 = 1 + 0.696 439 387 598 167 736 32;
  • 49) 0.696 439 387 598 167 736 32 × 2 = 1 + 0.392 878 775 196 335 472 64;
  • 50) 0.392 878 775 196 335 472 64 × 2 = 0 + 0.785 757 550 392 670 945 28;
  • 51) 0.785 757 550 392 670 945 28 × 2 = 1 + 0.571 515 100 785 341 890 56;
  • 52) 0.571 515 100 785 341 890 56 × 2 = 1 + 0.143 030 201 570 683 781 12;
  • 53) 0.143 030 201 570 683 781 12 × 2 = 0 + 0.286 060 403 141 367 562 24;
  • 54) 0.286 060 403 141 367 562 24 × 2 = 0 + 0.572 120 806 282 735 124 48;
  • 55) 0.572 120 806 282 735 124 48 × 2 = 1 + 0.144 241 612 565 470 248 96;
  • 56) 0.144 241 612 565 470 248 96 × 2 = 0 + 0.288 483 225 130 940 497 92;
  • 57) 0.288 483 225 130 940 497 92 × 2 = 0 + 0.576 966 450 261 880 995 84;
  • 58) 0.576 966 450 261 880 995 84 × 2 = 1 + 0.153 932 900 523 761 991 68;
  • 59) 0.153 932 900 523 761 991 68 × 2 = 0 + 0.307 865 801 047 523 983 36;
  • 60) 0.307 865 801 047 523 983 36 × 2 = 0 + 0.615 731 602 095 047 966 72;
  • 61) 0.615 731 602 095 047 966 72 × 2 = 1 + 0.231 463 204 190 095 933 44;
  • 62) 0.231 463 204 190 095 933 44 × 2 = 0 + 0.462 926 408 380 191 866 88;
  • 63) 0.462 926 408 380 191 866 88 × 2 = 0 + 0.925 852 816 760 383 733 76;
  • 64) 0.925 852 816 760 383 733 76 × 2 = 1 + 0.851 705 633 520 767 467 52;
  • 65) 0.851 705 633 520 767 467 52 × 2 = 1 + 0.703 411 267 041 534 935 04;
  • 66) 0.703 411 267 041 534 935 04 × 2 = 1 + 0.406 822 534 083 069 870 08;
  • 67) 0.406 822 534 083 069 870 08 × 2 = 0 + 0.813 645 068 166 139 740 16;
  • 68) 0.813 645 068 166 139 740 16 × 2 = 1 + 0.627 290 136 332 279 480 32;
  • 69) 0.627 290 136 332 279 480 32 × 2 = 1 + 0.254 580 272 664 558 960 64;
  • 70) 0.254 580 272 664 558 960 64 × 2 = 0 + 0.509 160 545 329 117 921 28;
  • 71) 0.509 160 545 329 117 921 28 × 2 = 1 + 0.018 321 090 658 235 842 56;
  • 72) 0.018 321 090 658 235 842 56 × 2 = 0 + 0.036 642 181 316 471 685 12;
  • 73) 0.036 642 181 316 471 685 12 × 2 = 0 + 0.073 284 362 632 943 370 24;
  • 74) 0.073 284 362 632 943 370 24 × 2 = 0 + 0.146 568 725 265 886 740 48;
  • 75) 0.146 568 725 265 886 740 48 × 2 = 0 + 0.293 137 450 531 773 480 96;
  • 76) 0.293 137 450 531 773 480 96 × 2 = 0 + 0.586 274 901 063 546 961 92;
  • 77) 0.586 274 901 063 546 961 92 × 2 = 1 + 0.172 549 802 127 093 923 84;
  • 78) 0.172 549 802 127 093 923 84 × 2 = 0 + 0.345 099 604 254 187 847 68;
  • 79) 0.345 099 604 254 187 847 68 × 2 = 0 + 0.690 199 208 508 375 695 36;
  • 80) 0.690 199 208 508 375 695 36 × 2 = 1 + 0.380 398 417 016 751 390 72;
  • 81) 0.380 398 417 016 751 390 72 × 2 = 0 + 0.760 796 834 033 502 781 44;
  • 82) 0.760 796 834 033 502 781 44 × 2 = 1 + 0.521 593 668 067 005 562 88;
  • 83) 0.521 593 668 067 005 562 88 × 2 = 1 + 0.043 187 336 134 011 125 76;
  • 84) 0.043 187 336 134 011 125 76 × 2 = 0 + 0.086 374 672 268 022 251 52;
  • 85) 0.086 374 672 268 022 251 52 × 2 = 0 + 0.172 749 344 536 044 503 04;
  • 86) 0.172 749 344 536 044 503 04 × 2 = 0 + 0.345 498 689 072 089 006 08;
  • 87) 0.345 498 689 072 089 006 08 × 2 = 0 + 0.690 997 378 144 178 012 16;
  • 88) 0.690 997 378 144 178 012 16 × 2 = 1 + 0.381 994 756 288 356 024 32;
  • 89) 0.381 994 756 288 356 024 32 × 2 = 0 + 0.763 989 512 576 712 048 64;
  • 90) 0.763 989 512 576 712 048 64 × 2 = 1 + 0.527 979 025 153 424 097 28;
  • 91) 0.527 979 025 153 424 097 28 × 2 = 1 + 0.055 958 050 306 848 194 56;
  • 92) 0.055 958 050 306 848 194 56 × 2 = 0 + 0.111 916 100 613 696 389 12;
  • 93) 0.111 916 100 613 696 389 12 × 2 = 0 + 0.223 832 201 227 392 778 24;
  • 94) 0.223 832 201 227 392 778 24 × 2 = 0 + 0.447 664 402 454 785 556 48;
  • 95) 0.447 664 402 454 785 556 48 × 2 = 0 + 0.895 328 804 909 571 112 96;

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 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 1101 1010 0000 1001 0110 0001 0110 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 1101 1010 0000 1001 0110 0001 0110 000(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 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 1101 1010 0000 1001 0110 0001 0110 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 1101 1010 0000 1001 0110 0001 0110 000(2) × 20 =

1.1000 1101 1001 0010 0100 1110 1101 0000 0100 1011 0000 1011 0000(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 0100 1110 1101 0000 0100 1011 0000 1011 0000


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 0100 1110 1101 0000 0100 1011 0000 1011 0000 =

1000 1101 1001 0010 0100 1110 1101 0000 0100 1011 0000 1011 0000


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 0100 1110 1101 0000 0100 1011 0000 1011 0000


Decimal number -0.000 000 000 000 176 557 22 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0100 1110 1101 0000 0100 1011 0000 1011 0000

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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