-0.000 000 000 000 176 557 256 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 256(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
-0.000 000 000 000 176 557 256(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:

|-0.000 000 000 000 176 557 256| = 0.000 000 000 000 176 557 256


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: 0.000 000 000 000 176 557 256.

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) 0.000 000 000 000 176 557 256 × 2 = 0 + 0.000 000 000 000 353 114 512;
  • 2) 0.000 000 000 000 353 114 512 × 2 = 0 + 0.000 000 000 000 706 229 024;
  • 3) 0.000 000 000 000 706 229 024 × 2 = 0 + 0.000 000 000 001 412 458 048;
  • 4) 0.000 000 000 001 412 458 048 × 2 = 0 + 0.000 000 000 002 824 916 096;
  • 5) 0.000 000 000 002 824 916 096 × 2 = 0 + 0.000 000 000 005 649 832 192;
  • 6) 0.000 000 000 005 649 832 192 × 2 = 0 + 0.000 000 000 011 299 664 384;
  • 7) 0.000 000 000 011 299 664 384 × 2 = 0 + 0.000 000 000 022 599 328 768;
  • 8) 0.000 000 000 022 599 328 768 × 2 = 0 + 0.000 000 000 045 198 657 536;
  • 9) 0.000 000 000 045 198 657 536 × 2 = 0 + 0.000 000 000 090 397 315 072;
  • 10) 0.000 000 000 090 397 315 072 × 2 = 0 + 0.000 000 000 180 794 630 144;
  • 11) 0.000 000 000 180 794 630 144 × 2 = 0 + 0.000 000 000 361 589 260 288;
  • 12) 0.000 000 000 361 589 260 288 × 2 = 0 + 0.000 000 000 723 178 520 576;
  • 13) 0.000 000 000 723 178 520 576 × 2 = 0 + 0.000 000 001 446 357 041 152;
  • 14) 0.000 000 001 446 357 041 152 × 2 = 0 + 0.000 000 002 892 714 082 304;
  • 15) 0.000 000 002 892 714 082 304 × 2 = 0 + 0.000 000 005 785 428 164 608;
  • 16) 0.000 000 005 785 428 164 608 × 2 = 0 + 0.000 000 011 570 856 329 216;
  • 17) 0.000 000 011 570 856 329 216 × 2 = 0 + 0.000 000 023 141 712 658 432;
  • 18) 0.000 000 023 141 712 658 432 × 2 = 0 + 0.000 000 046 283 425 316 864;
  • 19) 0.000 000 046 283 425 316 864 × 2 = 0 + 0.000 000 092 566 850 633 728;
  • 20) 0.000 000 092 566 850 633 728 × 2 = 0 + 0.000 000 185 133 701 267 456;
  • 21) 0.000 000 185 133 701 267 456 × 2 = 0 + 0.000 000 370 267 402 534 912;
  • 22) 0.000 000 370 267 402 534 912 × 2 = 0 + 0.000 000 740 534 805 069 824;
  • 23) 0.000 000 740 534 805 069 824 × 2 = 0 + 0.000 001 481 069 610 139 648;
  • 24) 0.000 001 481 069 610 139 648 × 2 = 0 + 0.000 002 962 139 220 279 296;
  • 25) 0.000 002 962 139 220 279 296 × 2 = 0 + 0.000 005 924 278 440 558 592;
  • 26) 0.000 005 924 278 440 558 592 × 2 = 0 + 0.000 011 848 556 881 117 184;
  • 27) 0.000 011 848 556 881 117 184 × 2 = 0 + 0.000 023 697 113 762 234 368;
  • 28) 0.000 023 697 113 762 234 368 × 2 = 0 + 0.000 047 394 227 524 468 736;
  • 29) 0.000 047 394 227 524 468 736 × 2 = 0 + 0.000 094 788 455 048 937 472;
  • 30) 0.000 094 788 455 048 937 472 × 2 = 0 + 0.000 189 576 910 097 874 944;
  • 31) 0.000 189 576 910 097 874 944 × 2 = 0 + 0.000 379 153 820 195 749 888;
  • 32) 0.000 379 153 820 195 749 888 × 2 = 0 + 0.000 758 307 640 391 499 776;
  • 33) 0.000 758 307 640 391 499 776 × 2 = 0 + 0.001 516 615 280 782 999 552;
  • 34) 0.001 516 615 280 782 999 552 × 2 = 0 + 0.003 033 230 561 565 999 104;
  • 35) 0.003 033 230 561 565 999 104 × 2 = 0 + 0.006 066 461 123 131 998 208;
  • 36) 0.006 066 461 123 131 998 208 × 2 = 0 + 0.012 132 922 246 263 996 416;
  • 37) 0.012 132 922 246 263 996 416 × 2 = 0 + 0.024 265 844 492 527 992 832;
  • 38) 0.024 265 844 492 527 992 832 × 2 = 0 + 0.048 531 688 985 055 985 664;
  • 39) 0.048 531 688 985 055 985 664 × 2 = 0 + 0.097 063 377 970 111 971 328;
  • 40) 0.097 063 377 970 111 971 328 × 2 = 0 + 0.194 126 755 940 223 942 656;
  • 41) 0.194 126 755 940 223 942 656 × 2 = 0 + 0.388 253 511 880 447 885 312;
  • 42) 0.388 253 511 880 447 885 312 × 2 = 0 + 0.776 507 023 760 895 770 624;
  • 43) 0.776 507 023 760 895 770 624 × 2 = 1 + 0.553 014 047 521 791 541 248;
  • 44) 0.553 014 047 521 791 541 248 × 2 = 1 + 0.106 028 095 043 583 082 496;
  • 45) 0.106 028 095 043 583 082 496 × 2 = 0 + 0.212 056 190 087 166 164 992;
  • 46) 0.212 056 190 087 166 164 992 × 2 = 0 + 0.424 112 380 174 332 329 984;
  • 47) 0.424 112 380 174 332 329 984 × 2 = 0 + 0.848 224 760 348 664 659 968;
  • 48) 0.848 224 760 348 664 659 968 × 2 = 1 + 0.696 449 520 697 329 319 936;
  • 49) 0.696 449 520 697 329 319 936 × 2 = 1 + 0.392 899 041 394 658 639 872;
  • 50) 0.392 899 041 394 658 639 872 × 2 = 0 + 0.785 798 082 789 317 279 744;
  • 51) 0.785 798 082 789 317 279 744 × 2 = 1 + 0.571 596 165 578 634 559 488;
  • 52) 0.571 596 165 578 634 559 488 × 2 = 1 + 0.143 192 331 157 269 118 976;
  • 53) 0.143 192 331 157 269 118 976 × 2 = 0 + 0.286 384 662 314 538 237 952;
  • 54) 0.286 384 662 314 538 237 952 × 2 = 0 + 0.572 769 324 629 076 475 904;
  • 55) 0.572 769 324 629 076 475 904 × 2 = 1 + 0.145 538 649 258 152 951 808;
  • 56) 0.145 538 649 258 152 951 808 × 2 = 0 + 0.291 077 298 516 305 903 616;
  • 57) 0.291 077 298 516 305 903 616 × 2 = 0 + 0.582 154 597 032 611 807 232;
  • 58) 0.582 154 597 032 611 807 232 × 2 = 1 + 0.164 309 194 065 223 614 464;
  • 59) 0.164 309 194 065 223 614 464 × 2 = 0 + 0.328 618 388 130 447 228 928;
  • 60) 0.328 618 388 130 447 228 928 × 2 = 0 + 0.657 236 776 260 894 457 856;
  • 61) 0.657 236 776 260 894 457 856 × 2 = 1 + 0.314 473 552 521 788 915 712;
  • 62) 0.314 473 552 521 788 915 712 × 2 = 0 + 0.628 947 105 043 577 831 424;
  • 63) 0.628 947 105 043 577 831 424 × 2 = 1 + 0.257 894 210 087 155 662 848;
  • 64) 0.257 894 210 087 155 662 848 × 2 = 0 + 0.515 788 420 174 311 325 696;
  • 65) 0.515 788 420 174 311 325 696 × 2 = 1 + 0.031 576 840 348 622 651 392;
  • 66) 0.031 576 840 348 622 651 392 × 2 = 0 + 0.063 153 680 697 245 302 784;
  • 67) 0.063 153 680 697 245 302 784 × 2 = 0 + 0.126 307 361 394 490 605 568;
  • 68) 0.126 307 361 394 490 605 568 × 2 = 0 + 0.252 614 722 788 981 211 136;
  • 69) 0.252 614 722 788 981 211 136 × 2 = 0 + 0.505 229 445 577 962 422 272;
  • 70) 0.505 229 445 577 962 422 272 × 2 = 1 + 0.010 458 891 155 924 844 544;
  • 71) 0.010 458 891 155 924 844 544 × 2 = 0 + 0.020 917 782 311 849 689 088;
  • 72) 0.020 917 782 311 849 689 088 × 2 = 0 + 0.041 835 564 623 699 378 176;
  • 73) 0.041 835 564 623 699 378 176 × 2 = 0 + 0.083 671 129 247 398 756 352;
  • 74) 0.083 671 129 247 398 756 352 × 2 = 0 + 0.167 342 258 494 797 512 704;
  • 75) 0.167 342 258 494 797 512 704 × 2 = 0 + 0.334 684 516 989 595 025 408;
  • 76) 0.334 684 516 989 595 025 408 × 2 = 0 + 0.669 369 033 979 190 050 816;
  • 77) 0.669 369 033 979 190 050 816 × 2 = 1 + 0.338 738 067 958 380 101 632;
  • 78) 0.338 738 067 958 380 101 632 × 2 = 0 + 0.677 476 135 916 760 203 264;
  • 79) 0.677 476 135 916 760 203 264 × 2 = 1 + 0.354 952 271 833 520 406 528;
  • 80) 0.354 952 271 833 520 406 528 × 2 = 0 + 0.709 904 543 667 040 813 056;
  • 81) 0.709 904 543 667 040 813 056 × 2 = 1 + 0.419 809 087 334 081 626 112;
  • 82) 0.419 809 087 334 081 626 112 × 2 = 0 + 0.839 618 174 668 163 252 224;
  • 83) 0.839 618 174 668 163 252 224 × 2 = 1 + 0.679 236 349 336 326 504 448;
  • 84) 0.679 236 349 336 326 504 448 × 2 = 1 + 0.358 472 698 672 653 008 896;
  • 85) 0.358 472 698 672 653 008 896 × 2 = 0 + 0.716 945 397 345 306 017 792;
  • 86) 0.716 945 397 345 306 017 792 × 2 = 1 + 0.433 890 794 690 612 035 584;
  • 87) 0.433 890 794 690 612 035 584 × 2 = 0 + 0.867 781 589 381 224 071 168;
  • 88) 0.867 781 589 381 224 071 168 × 2 = 1 + 0.735 563 178 762 448 142 336;
  • 89) 0.735 563 178 762 448 142 336 × 2 = 1 + 0.471 126 357 524 896 284 672;
  • 90) 0.471 126 357 524 896 284 672 × 2 = 0 + 0.942 252 715 049 792 569 344;
  • 91) 0.942 252 715 049 792 569 344 × 2 = 1 + 0.884 505 430 099 585 138 688;
  • 92) 0.884 505 430 099 585 138 688 × 2 = 1 + 0.769 010 860 199 170 277 376;
  • 93) 0.769 010 860 199 170 277 376 × 2 = 1 + 0.538 021 720 398 340 554 752;
  • 94) 0.538 021 720 398 340 554 752 × 2 = 1 + 0.076 043 440 796 681 109 504;
  • 95) 0.076 043 440 796 681 109 504 × 2 = 0 + 0.152 086 881 593 362 219 008;

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:


0.000 000 000 000 176 557 256(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 1000 0100 0000 1010 1011 0101 1011 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 256(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 1000 0100 0000 1010 1011 0101 1011 110(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:


0.000 000 000 000 176 557 256(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 1000 0100 0000 1010 1011 0101 1011 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 1000 0100 0000 1010 1011 0101 1011 110(2) × 20 =

1.1000 1101 1001 0010 0101 0100 0010 0000 0101 0101 1010 1101 1110(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 0101 0100 0010 0000 0101 0101 1010 1101 1110


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 0101 0100 0010 0000 0101 0101 1010 1101 1110 =

1000 1101 1001 0010 0101 0100 0010 0000 0101 0101 1010 1101 1110


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 0101 0100 0010 0000 0101 0101 1010 1101 1110


Decimal number -0.000 000 000 000 176 557 256 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 0100 0010 0000 0101 0101 1010 1101 1110

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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