-0.000 000 000 000 176 544 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 544 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
-0.000 000 000 000 176 544 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:

|-0.000 000 000 000 176 544 8| = 0.000 000 000 000 176 544 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: 0.000 000 000 000 176 544 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) 0.000 000 000 000 176 544 8 × 2 = 0 + 0.000 000 000 000 353 089 6;
  • 2) 0.000 000 000 000 353 089 6 × 2 = 0 + 0.000 000 000 000 706 179 2;
  • 3) 0.000 000 000 000 706 179 2 × 2 = 0 + 0.000 000 000 001 412 358 4;
  • 4) 0.000 000 000 001 412 358 4 × 2 = 0 + 0.000 000 000 002 824 716 8;
  • 5) 0.000 000 000 002 824 716 8 × 2 = 0 + 0.000 000 000 005 649 433 6;
  • 6) 0.000 000 000 005 649 433 6 × 2 = 0 + 0.000 000 000 011 298 867 2;
  • 7) 0.000 000 000 011 298 867 2 × 2 = 0 + 0.000 000 000 022 597 734 4;
  • 8) 0.000 000 000 022 597 734 4 × 2 = 0 + 0.000 000 000 045 195 468 8;
  • 9) 0.000 000 000 045 195 468 8 × 2 = 0 + 0.000 000 000 090 390 937 6;
  • 10) 0.000 000 000 090 390 937 6 × 2 = 0 + 0.000 000 000 180 781 875 2;
  • 11) 0.000 000 000 180 781 875 2 × 2 = 0 + 0.000 000 000 361 563 750 4;
  • 12) 0.000 000 000 361 563 750 4 × 2 = 0 + 0.000 000 000 723 127 500 8;
  • 13) 0.000 000 000 723 127 500 8 × 2 = 0 + 0.000 000 001 446 255 001 6;
  • 14) 0.000 000 001 446 255 001 6 × 2 = 0 + 0.000 000 002 892 510 003 2;
  • 15) 0.000 000 002 892 510 003 2 × 2 = 0 + 0.000 000 005 785 020 006 4;
  • 16) 0.000 000 005 785 020 006 4 × 2 = 0 + 0.000 000 011 570 040 012 8;
  • 17) 0.000 000 011 570 040 012 8 × 2 = 0 + 0.000 000 023 140 080 025 6;
  • 18) 0.000 000 023 140 080 025 6 × 2 = 0 + 0.000 000 046 280 160 051 2;
  • 19) 0.000 000 046 280 160 051 2 × 2 = 0 + 0.000 000 092 560 320 102 4;
  • 20) 0.000 000 092 560 320 102 4 × 2 = 0 + 0.000 000 185 120 640 204 8;
  • 21) 0.000 000 185 120 640 204 8 × 2 = 0 + 0.000 000 370 241 280 409 6;
  • 22) 0.000 000 370 241 280 409 6 × 2 = 0 + 0.000 000 740 482 560 819 2;
  • 23) 0.000 000 740 482 560 819 2 × 2 = 0 + 0.000 001 480 965 121 638 4;
  • 24) 0.000 001 480 965 121 638 4 × 2 = 0 + 0.000 002 961 930 243 276 8;
  • 25) 0.000 002 961 930 243 276 8 × 2 = 0 + 0.000 005 923 860 486 553 6;
  • 26) 0.000 005 923 860 486 553 6 × 2 = 0 + 0.000 011 847 720 973 107 2;
  • 27) 0.000 011 847 720 973 107 2 × 2 = 0 + 0.000 023 695 441 946 214 4;
  • 28) 0.000 023 695 441 946 214 4 × 2 = 0 + 0.000 047 390 883 892 428 8;
  • 29) 0.000 047 390 883 892 428 8 × 2 = 0 + 0.000 094 781 767 784 857 6;
  • 30) 0.000 094 781 767 784 857 6 × 2 = 0 + 0.000 189 563 535 569 715 2;
  • 31) 0.000 189 563 535 569 715 2 × 2 = 0 + 0.000 379 127 071 139 430 4;
  • 32) 0.000 379 127 071 139 430 4 × 2 = 0 + 0.000 758 254 142 278 860 8;
  • 33) 0.000 758 254 142 278 860 8 × 2 = 0 + 0.001 516 508 284 557 721 6;
  • 34) 0.001 516 508 284 557 721 6 × 2 = 0 + 0.003 033 016 569 115 443 2;
  • 35) 0.003 033 016 569 115 443 2 × 2 = 0 + 0.006 066 033 138 230 886 4;
  • 36) 0.006 066 033 138 230 886 4 × 2 = 0 + 0.012 132 066 276 461 772 8;
  • 37) 0.012 132 066 276 461 772 8 × 2 = 0 + 0.024 264 132 552 923 545 6;
  • 38) 0.024 264 132 552 923 545 6 × 2 = 0 + 0.048 528 265 105 847 091 2;
  • 39) 0.048 528 265 105 847 091 2 × 2 = 0 + 0.097 056 530 211 694 182 4;
  • 40) 0.097 056 530 211 694 182 4 × 2 = 0 + 0.194 113 060 423 388 364 8;
  • 41) 0.194 113 060 423 388 364 8 × 2 = 0 + 0.388 226 120 846 776 729 6;
  • 42) 0.388 226 120 846 776 729 6 × 2 = 0 + 0.776 452 241 693 553 459 2;
  • 43) 0.776 452 241 693 553 459 2 × 2 = 1 + 0.552 904 483 387 106 918 4;
  • 44) 0.552 904 483 387 106 918 4 × 2 = 1 + 0.105 808 966 774 213 836 8;
  • 45) 0.105 808 966 774 213 836 8 × 2 = 0 + 0.211 617 933 548 427 673 6;
  • 46) 0.211 617 933 548 427 673 6 × 2 = 0 + 0.423 235 867 096 855 347 2;
  • 47) 0.423 235 867 096 855 347 2 × 2 = 0 + 0.846 471 734 193 710 694 4;
  • 48) 0.846 471 734 193 710 694 4 × 2 = 1 + 0.692 943 468 387 421 388 8;
  • 49) 0.692 943 468 387 421 388 8 × 2 = 1 + 0.385 886 936 774 842 777 6;
  • 50) 0.385 886 936 774 842 777 6 × 2 = 0 + 0.771 773 873 549 685 555 2;
  • 51) 0.771 773 873 549 685 555 2 × 2 = 1 + 0.543 547 747 099 371 110 4;
  • 52) 0.543 547 747 099 371 110 4 × 2 = 1 + 0.087 095 494 198 742 220 8;
  • 53) 0.087 095 494 198 742 220 8 × 2 = 0 + 0.174 190 988 397 484 441 6;
  • 54) 0.174 190 988 397 484 441 6 × 2 = 0 + 0.348 381 976 794 968 883 2;
  • 55) 0.348 381 976 794 968 883 2 × 2 = 0 + 0.696 763 953 589 937 766 4;
  • 56) 0.696 763 953 589 937 766 4 × 2 = 1 + 0.393 527 907 179 875 532 8;
  • 57) 0.393 527 907 179 875 532 8 × 2 = 0 + 0.787 055 814 359 751 065 6;
  • 58) 0.787 055 814 359 751 065 6 × 2 = 1 + 0.574 111 628 719 502 131 2;
  • 59) 0.574 111 628 719 502 131 2 × 2 = 1 + 0.148 223 257 439 004 262 4;
  • 60) 0.148 223 257 439 004 262 4 × 2 = 0 + 0.296 446 514 878 008 524 8;
  • 61) 0.296 446 514 878 008 524 8 × 2 = 0 + 0.592 893 029 756 017 049 6;
  • 62) 0.592 893 029 756 017 049 6 × 2 = 1 + 0.185 786 059 512 034 099 2;
  • 63) 0.185 786 059 512 034 099 2 × 2 = 0 + 0.371 572 119 024 068 198 4;
  • 64) 0.371 572 119 024 068 198 4 × 2 = 0 + 0.743 144 238 048 136 396 8;
  • 65) 0.743 144 238 048 136 396 8 × 2 = 1 + 0.486 288 476 096 272 793 6;
  • 66) 0.486 288 476 096 272 793 6 × 2 = 0 + 0.972 576 952 192 545 587 2;
  • 67) 0.972 576 952 192 545 587 2 × 2 = 1 + 0.945 153 904 385 091 174 4;
  • 68) 0.945 153 904 385 091 174 4 × 2 = 1 + 0.890 307 808 770 182 348 8;
  • 69) 0.890 307 808 770 182 348 8 × 2 = 1 + 0.780 615 617 540 364 697 6;
  • 70) 0.780 615 617 540 364 697 6 × 2 = 1 + 0.561 231 235 080 729 395 2;
  • 71) 0.561 231 235 080 729 395 2 × 2 = 1 + 0.122 462 470 161 458 790 4;
  • 72) 0.122 462 470 161 458 790 4 × 2 = 0 + 0.244 924 940 322 917 580 8;
  • 73) 0.244 924 940 322 917 580 8 × 2 = 0 + 0.489 849 880 645 835 161 6;
  • 74) 0.489 849 880 645 835 161 6 × 2 = 0 + 0.979 699 761 291 670 323 2;
  • 75) 0.979 699 761 291 670 323 2 × 2 = 1 + 0.959 399 522 583 340 646 4;
  • 76) 0.959 399 522 583 340 646 4 × 2 = 1 + 0.918 799 045 166 681 292 8;
  • 77) 0.918 799 045 166 681 292 8 × 2 = 1 + 0.837 598 090 333 362 585 6;
  • 78) 0.837 598 090 333 362 585 6 × 2 = 1 + 0.675 196 180 666 725 171 2;
  • 79) 0.675 196 180 666 725 171 2 × 2 = 1 + 0.350 392 361 333 450 342 4;
  • 80) 0.350 392 361 333 450 342 4 × 2 = 0 + 0.700 784 722 666 900 684 8;
  • 81) 0.700 784 722 666 900 684 8 × 2 = 1 + 0.401 569 445 333 801 369 6;
  • 82) 0.401 569 445 333 801 369 6 × 2 = 0 + 0.803 138 890 667 602 739 2;
  • 83) 0.803 138 890 667 602 739 2 × 2 = 1 + 0.606 277 781 335 205 478 4;
  • 84) 0.606 277 781 335 205 478 4 × 2 = 1 + 0.212 555 562 670 410 956 8;
  • 85) 0.212 555 562 670 410 956 8 × 2 = 0 + 0.425 111 125 340 821 913 6;
  • 86) 0.425 111 125 340 821 913 6 × 2 = 0 + 0.850 222 250 681 643 827 2;
  • 87) 0.850 222 250 681 643 827 2 × 2 = 1 + 0.700 444 501 363 287 654 4;
  • 88) 0.700 444 501 363 287 654 4 × 2 = 1 + 0.400 889 002 726 575 308 8;
  • 89) 0.400 889 002 726 575 308 8 × 2 = 0 + 0.801 778 005 453 150 617 6;
  • 90) 0.801 778 005 453 150 617 6 × 2 = 1 + 0.603 556 010 906 301 235 2;
  • 91) 0.603 556 010 906 301 235 2 × 2 = 1 + 0.207 112 021 812 602 470 4;
  • 92) 0.207 112 021 812 602 470 4 × 2 = 0 + 0.414 224 043 625 204 940 8;
  • 93) 0.414 224 043 625 204 940 8 × 2 = 0 + 0.828 448 087 250 409 881 6;
  • 94) 0.828 448 087 250 409 881 6 × 2 = 1 + 0.656 896 174 500 819 763 2;
  • 95) 0.656 896 174 500 819 763 2 × 2 = 1 + 0.313 792 349 001 639 526 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:


0.000 000 000 000 176 544 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0110 0100 1011 1110 0011 1110 1011 0011 0110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 544 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0110 0100 1011 1110 0011 1110 1011 0011 0110 011(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:


0.000 000 000 000 176 544 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0110 0100 1011 1110 0011 1110 1011 0011 0110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0110 0100 1011 1110 0011 1110 1011 0011 0110 011(2) × 20 =

1.1000 1101 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 0011(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 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 0011


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 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 0011 =

1000 1101 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 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 0100

Mantissa (52 bits) =
1000 1101 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 0011


Decimal number -0.000 000 000 000 176 544 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1011 0010 0101 1111 0001 1111 0101 1001 1011 0011

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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