-0.000 000 000 000 176 557 582 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 582(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
-0.000 000 000 000 176 557 582(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:

|-0.000 000 000 000 176 557 582| = 0.000 000 000 000 176 557 582


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: 0.000 000 000 000 176 557 582.

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) 0.000 000 000 000 176 557 582 × 2 = 0 + 0.000 000 000 000 353 115 164;
  • 2) 0.000 000 000 000 353 115 164 × 2 = 0 + 0.000 000 000 000 706 230 328;
  • 3) 0.000 000 000 000 706 230 328 × 2 = 0 + 0.000 000 000 001 412 460 656;
  • 4) 0.000 000 000 001 412 460 656 × 2 = 0 + 0.000 000 000 002 824 921 312;
  • 5) 0.000 000 000 002 824 921 312 × 2 = 0 + 0.000 000 000 005 649 842 624;
  • 6) 0.000 000 000 005 649 842 624 × 2 = 0 + 0.000 000 000 011 299 685 248;
  • 7) 0.000 000 000 011 299 685 248 × 2 = 0 + 0.000 000 000 022 599 370 496;
  • 8) 0.000 000 000 022 599 370 496 × 2 = 0 + 0.000 000 000 045 198 740 992;
  • 9) 0.000 000 000 045 198 740 992 × 2 = 0 + 0.000 000 000 090 397 481 984;
  • 10) 0.000 000 000 090 397 481 984 × 2 = 0 + 0.000 000 000 180 794 963 968;
  • 11) 0.000 000 000 180 794 963 968 × 2 = 0 + 0.000 000 000 361 589 927 936;
  • 12) 0.000 000 000 361 589 927 936 × 2 = 0 + 0.000 000 000 723 179 855 872;
  • 13) 0.000 000 000 723 179 855 872 × 2 = 0 + 0.000 000 001 446 359 711 744;
  • 14) 0.000 000 001 446 359 711 744 × 2 = 0 + 0.000 000 002 892 719 423 488;
  • 15) 0.000 000 002 892 719 423 488 × 2 = 0 + 0.000 000 005 785 438 846 976;
  • 16) 0.000 000 005 785 438 846 976 × 2 = 0 + 0.000 000 011 570 877 693 952;
  • 17) 0.000 000 011 570 877 693 952 × 2 = 0 + 0.000 000 023 141 755 387 904;
  • 18) 0.000 000 023 141 755 387 904 × 2 = 0 + 0.000 000 046 283 510 775 808;
  • 19) 0.000 000 046 283 510 775 808 × 2 = 0 + 0.000 000 092 567 021 551 616;
  • 20) 0.000 000 092 567 021 551 616 × 2 = 0 + 0.000 000 185 134 043 103 232;
  • 21) 0.000 000 185 134 043 103 232 × 2 = 0 + 0.000 000 370 268 086 206 464;
  • 22) 0.000 000 370 268 086 206 464 × 2 = 0 + 0.000 000 740 536 172 412 928;
  • 23) 0.000 000 740 536 172 412 928 × 2 = 0 + 0.000 001 481 072 344 825 856;
  • 24) 0.000 001 481 072 344 825 856 × 2 = 0 + 0.000 002 962 144 689 651 712;
  • 25) 0.000 002 962 144 689 651 712 × 2 = 0 + 0.000 005 924 289 379 303 424;
  • 26) 0.000 005 924 289 379 303 424 × 2 = 0 + 0.000 011 848 578 758 606 848;
  • 27) 0.000 011 848 578 758 606 848 × 2 = 0 + 0.000 023 697 157 517 213 696;
  • 28) 0.000 023 697 157 517 213 696 × 2 = 0 + 0.000 047 394 315 034 427 392;
  • 29) 0.000 047 394 315 034 427 392 × 2 = 0 + 0.000 094 788 630 068 854 784;
  • 30) 0.000 094 788 630 068 854 784 × 2 = 0 + 0.000 189 577 260 137 709 568;
  • 31) 0.000 189 577 260 137 709 568 × 2 = 0 + 0.000 379 154 520 275 419 136;
  • 32) 0.000 379 154 520 275 419 136 × 2 = 0 + 0.000 758 309 040 550 838 272;
  • 33) 0.000 758 309 040 550 838 272 × 2 = 0 + 0.001 516 618 081 101 676 544;
  • 34) 0.001 516 618 081 101 676 544 × 2 = 0 + 0.003 033 236 162 203 353 088;
  • 35) 0.003 033 236 162 203 353 088 × 2 = 0 + 0.006 066 472 324 406 706 176;
  • 36) 0.006 066 472 324 406 706 176 × 2 = 0 + 0.012 132 944 648 813 412 352;
  • 37) 0.012 132 944 648 813 412 352 × 2 = 0 + 0.024 265 889 297 626 824 704;
  • 38) 0.024 265 889 297 626 824 704 × 2 = 0 + 0.048 531 778 595 253 649 408;
  • 39) 0.048 531 778 595 253 649 408 × 2 = 0 + 0.097 063 557 190 507 298 816;
  • 40) 0.097 063 557 190 507 298 816 × 2 = 0 + 0.194 127 114 381 014 597 632;
  • 41) 0.194 127 114 381 014 597 632 × 2 = 0 + 0.388 254 228 762 029 195 264;
  • 42) 0.388 254 228 762 029 195 264 × 2 = 0 + 0.776 508 457 524 058 390 528;
  • 43) 0.776 508 457 524 058 390 528 × 2 = 1 + 0.553 016 915 048 116 781 056;
  • 44) 0.553 016 915 048 116 781 056 × 2 = 1 + 0.106 033 830 096 233 562 112;
  • 45) 0.106 033 830 096 233 562 112 × 2 = 0 + 0.212 067 660 192 467 124 224;
  • 46) 0.212 067 660 192 467 124 224 × 2 = 0 + 0.424 135 320 384 934 248 448;
  • 47) 0.424 135 320 384 934 248 448 × 2 = 0 + 0.848 270 640 769 868 496 896;
  • 48) 0.848 270 640 769 868 496 896 × 2 = 1 + 0.696 541 281 539 736 993 792;
  • 49) 0.696 541 281 539 736 993 792 × 2 = 1 + 0.393 082 563 079 473 987 584;
  • 50) 0.393 082 563 079 473 987 584 × 2 = 0 + 0.786 165 126 158 947 975 168;
  • 51) 0.786 165 126 158 947 975 168 × 2 = 1 + 0.572 330 252 317 895 950 336;
  • 52) 0.572 330 252 317 895 950 336 × 2 = 1 + 0.144 660 504 635 791 900 672;
  • 53) 0.144 660 504 635 791 900 672 × 2 = 0 + 0.289 321 009 271 583 801 344;
  • 54) 0.289 321 009 271 583 801 344 × 2 = 0 + 0.578 642 018 543 167 602 688;
  • 55) 0.578 642 018 543 167 602 688 × 2 = 1 + 0.157 284 037 086 335 205 376;
  • 56) 0.157 284 037 086 335 205 376 × 2 = 0 + 0.314 568 074 172 670 410 752;
  • 57) 0.314 568 074 172 670 410 752 × 2 = 0 + 0.629 136 148 345 340 821 504;
  • 58) 0.629 136 148 345 340 821 504 × 2 = 1 + 0.258 272 296 690 681 643 008;
  • 59) 0.258 272 296 690 681 643 008 × 2 = 0 + 0.516 544 593 381 363 286 016;
  • 60) 0.516 544 593 381 363 286 016 × 2 = 1 + 0.033 089 186 762 726 572 032;
  • 61) 0.033 089 186 762 726 572 032 × 2 = 0 + 0.066 178 373 525 453 144 064;
  • 62) 0.066 178 373 525 453 144 064 × 2 = 0 + 0.132 356 747 050 906 288 128;
  • 63) 0.132 356 747 050 906 288 128 × 2 = 0 + 0.264 713 494 101 812 576 256;
  • 64) 0.264 713 494 101 812 576 256 × 2 = 0 + 0.529 426 988 203 625 152 512;
  • 65) 0.529 426 988 203 625 152 512 × 2 = 1 + 0.058 853 976 407 250 305 024;
  • 66) 0.058 853 976 407 250 305 024 × 2 = 0 + 0.117 707 952 814 500 610 048;
  • 67) 0.117 707 952 814 500 610 048 × 2 = 0 + 0.235 415 905 629 001 220 096;
  • 68) 0.235 415 905 629 001 220 096 × 2 = 0 + 0.470 831 811 258 002 440 192;
  • 69) 0.470 831 811 258 002 440 192 × 2 = 0 + 0.941 663 622 516 004 880 384;
  • 70) 0.941 663 622 516 004 880 384 × 2 = 1 + 0.883 327 245 032 009 760 768;
  • 71) 0.883 327 245 032 009 760 768 × 2 = 1 + 0.766 654 490 064 019 521 536;
  • 72) 0.766 654 490 064 019 521 536 × 2 = 1 + 0.533 308 980 128 039 043 072;
  • 73) 0.533 308 980 128 039 043 072 × 2 = 1 + 0.066 617 960 256 078 086 144;
  • 74) 0.066 617 960 256 078 086 144 × 2 = 0 + 0.133 235 920 512 156 172 288;
  • 75) 0.133 235 920 512 156 172 288 × 2 = 0 + 0.266 471 841 024 312 344 576;
  • 76) 0.266 471 841 024 312 344 576 × 2 = 0 + 0.532 943 682 048 624 689 152;
  • 77) 0.532 943 682 048 624 689 152 × 2 = 1 + 0.065 887 364 097 249 378 304;
  • 78) 0.065 887 364 097 249 378 304 × 2 = 0 + 0.131 774 728 194 498 756 608;
  • 79) 0.131 774 728 194 498 756 608 × 2 = 0 + 0.263 549 456 388 997 513 216;
  • 80) 0.263 549 456 388 997 513 216 × 2 = 0 + 0.527 098 912 777 995 026 432;
  • 81) 0.527 098 912 777 995 026 432 × 2 = 1 + 0.054 197 825 555 990 052 864;
  • 82) 0.054 197 825 555 990 052 864 × 2 = 0 + 0.108 395 651 111 980 105 728;
  • 83) 0.108 395 651 111 980 105 728 × 2 = 0 + 0.216 791 302 223 960 211 456;
  • 84) 0.216 791 302 223 960 211 456 × 2 = 0 + 0.433 582 604 447 920 422 912;
  • 85) 0.433 582 604 447 920 422 912 × 2 = 0 + 0.867 165 208 895 840 845 824;
  • 86) 0.867 165 208 895 840 845 824 × 2 = 1 + 0.734 330 417 791 681 691 648;
  • 87) 0.734 330 417 791 681 691 648 × 2 = 1 + 0.468 660 835 583 363 383 296;
  • 88) 0.468 660 835 583 363 383 296 × 2 = 0 + 0.937 321 671 166 726 766 592;
  • 89) 0.937 321 671 166 726 766 592 × 2 = 1 + 0.874 643 342 333 453 533 184;
  • 90) 0.874 643 342 333 453 533 184 × 2 = 1 + 0.749 286 684 666 907 066 368;
  • 91) 0.749 286 684 666 907 066 368 × 2 = 1 + 0.498 573 369 333 814 132 736;
  • 92) 0.498 573 369 333 814 132 736 × 2 = 0 + 0.997 146 738 667 628 265 472;
  • 93) 0.997 146 738 667 628 265 472 × 2 = 1 + 0.994 293 477 335 256 530 944;
  • 94) 0.994 293 477 335 256 530 944 × 2 = 1 + 0.988 586 954 670 513 061 888;
  • 95) 0.988 586 954 670 513 061 888 × 2 = 1 + 0.977 173 909 341 026 123 776;

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:


0.000 000 000 000 176 557 582(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 1000 0111 1000 1000 1000 0110 1110 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 582(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 1000 0111 1000 1000 1000 0110 1110 111(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:


0.000 000 000 000 176 557 582(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 1000 0111 1000 1000 1000 0110 1110 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 1000 0111 1000 1000 1000 0110 1110 111(2) × 20 =

1.1000 1101 1001 0010 1000 0100 0011 1100 0100 0100 0011 0111 0111(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 1000 0100 0011 1100 0100 0100 0011 0111 0111


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 1000 0100 0011 1100 0100 0100 0011 0111 0111 =

1000 1101 1001 0010 1000 0100 0011 1100 0100 0100 0011 0111 0111


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 1000 0100 0011 1100 0100 0100 0011 0111 0111


Decimal number -0.000 000 000 000 176 557 582 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1000 0100 0011 1100 0100 0100 0011 0111 0111

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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