-0.000 000 000 000 176 557 497 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 497(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
-0.000 000 000 000 176 557 497(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:

|-0.000 000 000 000 176 557 497| = 0.000 000 000 000 176 557 497


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: 0.000 000 000 000 176 557 497.

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) 0.000 000 000 000 176 557 497 × 2 = 0 + 0.000 000 000 000 353 114 994;
  • 2) 0.000 000 000 000 353 114 994 × 2 = 0 + 0.000 000 000 000 706 229 988;
  • 3) 0.000 000 000 000 706 229 988 × 2 = 0 + 0.000 000 000 001 412 459 976;
  • 4) 0.000 000 000 001 412 459 976 × 2 = 0 + 0.000 000 000 002 824 919 952;
  • 5) 0.000 000 000 002 824 919 952 × 2 = 0 + 0.000 000 000 005 649 839 904;
  • 6) 0.000 000 000 005 649 839 904 × 2 = 0 + 0.000 000 000 011 299 679 808;
  • 7) 0.000 000 000 011 299 679 808 × 2 = 0 + 0.000 000 000 022 599 359 616;
  • 8) 0.000 000 000 022 599 359 616 × 2 = 0 + 0.000 000 000 045 198 719 232;
  • 9) 0.000 000 000 045 198 719 232 × 2 = 0 + 0.000 000 000 090 397 438 464;
  • 10) 0.000 000 000 090 397 438 464 × 2 = 0 + 0.000 000 000 180 794 876 928;
  • 11) 0.000 000 000 180 794 876 928 × 2 = 0 + 0.000 000 000 361 589 753 856;
  • 12) 0.000 000 000 361 589 753 856 × 2 = 0 + 0.000 000 000 723 179 507 712;
  • 13) 0.000 000 000 723 179 507 712 × 2 = 0 + 0.000 000 001 446 359 015 424;
  • 14) 0.000 000 001 446 359 015 424 × 2 = 0 + 0.000 000 002 892 718 030 848;
  • 15) 0.000 000 002 892 718 030 848 × 2 = 0 + 0.000 000 005 785 436 061 696;
  • 16) 0.000 000 005 785 436 061 696 × 2 = 0 + 0.000 000 011 570 872 123 392;
  • 17) 0.000 000 011 570 872 123 392 × 2 = 0 + 0.000 000 023 141 744 246 784;
  • 18) 0.000 000 023 141 744 246 784 × 2 = 0 + 0.000 000 046 283 488 493 568;
  • 19) 0.000 000 046 283 488 493 568 × 2 = 0 + 0.000 000 092 566 976 987 136;
  • 20) 0.000 000 092 566 976 987 136 × 2 = 0 + 0.000 000 185 133 953 974 272;
  • 21) 0.000 000 185 133 953 974 272 × 2 = 0 + 0.000 000 370 267 907 948 544;
  • 22) 0.000 000 370 267 907 948 544 × 2 = 0 + 0.000 000 740 535 815 897 088;
  • 23) 0.000 000 740 535 815 897 088 × 2 = 0 + 0.000 001 481 071 631 794 176;
  • 24) 0.000 001 481 071 631 794 176 × 2 = 0 + 0.000 002 962 143 263 588 352;
  • 25) 0.000 002 962 143 263 588 352 × 2 = 0 + 0.000 005 924 286 527 176 704;
  • 26) 0.000 005 924 286 527 176 704 × 2 = 0 + 0.000 011 848 573 054 353 408;
  • 27) 0.000 011 848 573 054 353 408 × 2 = 0 + 0.000 023 697 146 108 706 816;
  • 28) 0.000 023 697 146 108 706 816 × 2 = 0 + 0.000 047 394 292 217 413 632;
  • 29) 0.000 047 394 292 217 413 632 × 2 = 0 + 0.000 094 788 584 434 827 264;
  • 30) 0.000 094 788 584 434 827 264 × 2 = 0 + 0.000 189 577 168 869 654 528;
  • 31) 0.000 189 577 168 869 654 528 × 2 = 0 + 0.000 379 154 337 739 309 056;
  • 32) 0.000 379 154 337 739 309 056 × 2 = 0 + 0.000 758 308 675 478 618 112;
  • 33) 0.000 758 308 675 478 618 112 × 2 = 0 + 0.001 516 617 350 957 236 224;
  • 34) 0.001 516 617 350 957 236 224 × 2 = 0 + 0.003 033 234 701 914 472 448;
  • 35) 0.003 033 234 701 914 472 448 × 2 = 0 + 0.006 066 469 403 828 944 896;
  • 36) 0.006 066 469 403 828 944 896 × 2 = 0 + 0.012 132 938 807 657 889 792;
  • 37) 0.012 132 938 807 657 889 792 × 2 = 0 + 0.024 265 877 615 315 779 584;
  • 38) 0.024 265 877 615 315 779 584 × 2 = 0 + 0.048 531 755 230 631 559 168;
  • 39) 0.048 531 755 230 631 559 168 × 2 = 0 + 0.097 063 510 461 263 118 336;
  • 40) 0.097 063 510 461 263 118 336 × 2 = 0 + 0.194 127 020 922 526 236 672;
  • 41) 0.194 127 020 922 526 236 672 × 2 = 0 + 0.388 254 041 845 052 473 344;
  • 42) 0.388 254 041 845 052 473 344 × 2 = 0 + 0.776 508 083 690 104 946 688;
  • 43) 0.776 508 083 690 104 946 688 × 2 = 1 + 0.553 016 167 380 209 893 376;
  • 44) 0.553 016 167 380 209 893 376 × 2 = 1 + 0.106 032 334 760 419 786 752;
  • 45) 0.106 032 334 760 419 786 752 × 2 = 0 + 0.212 064 669 520 839 573 504;
  • 46) 0.212 064 669 520 839 573 504 × 2 = 0 + 0.424 129 339 041 679 147 008;
  • 47) 0.424 129 339 041 679 147 008 × 2 = 0 + 0.848 258 678 083 358 294 016;
  • 48) 0.848 258 678 083 358 294 016 × 2 = 1 + 0.696 517 356 166 716 588 032;
  • 49) 0.696 517 356 166 716 588 032 × 2 = 1 + 0.393 034 712 333 433 176 064;
  • 50) 0.393 034 712 333 433 176 064 × 2 = 0 + 0.786 069 424 666 866 352 128;
  • 51) 0.786 069 424 666 866 352 128 × 2 = 1 + 0.572 138 849 333 732 704 256;
  • 52) 0.572 138 849 333 732 704 256 × 2 = 1 + 0.144 277 698 667 465 408 512;
  • 53) 0.144 277 698 667 465 408 512 × 2 = 0 + 0.288 555 397 334 930 817 024;
  • 54) 0.288 555 397 334 930 817 024 × 2 = 0 + 0.577 110 794 669 861 634 048;
  • 55) 0.577 110 794 669 861 634 048 × 2 = 1 + 0.154 221 589 339 723 268 096;
  • 56) 0.154 221 589 339 723 268 096 × 2 = 0 + 0.308 443 178 679 446 536 192;
  • 57) 0.308 443 178 679 446 536 192 × 2 = 0 + 0.616 886 357 358 893 072 384;
  • 58) 0.616 886 357 358 893 072 384 × 2 = 1 + 0.233 772 714 717 786 144 768;
  • 59) 0.233 772 714 717 786 144 768 × 2 = 0 + 0.467 545 429 435 572 289 536;
  • 60) 0.467 545 429 435 572 289 536 × 2 = 0 + 0.935 090 858 871 144 579 072;
  • 61) 0.935 090 858 871 144 579 072 × 2 = 1 + 0.870 181 717 742 289 158 144;
  • 62) 0.870 181 717 742 289 158 144 × 2 = 1 + 0.740 363 435 484 578 316 288;
  • 63) 0.740 363 435 484 578 316 288 × 2 = 1 + 0.480 726 870 969 156 632 576;
  • 64) 0.480 726 870 969 156 632 576 × 2 = 0 + 0.961 453 741 938 313 265 152;
  • 65) 0.961 453 741 938 313 265 152 × 2 = 1 + 0.922 907 483 876 626 530 304;
  • 66) 0.922 907 483 876 626 530 304 × 2 = 1 + 0.845 814 967 753 253 060 608;
  • 67) 0.845 814 967 753 253 060 608 × 2 = 1 + 0.691 629 935 506 506 121 216;
  • 68) 0.691 629 935 506 506 121 216 × 2 = 1 + 0.383 259 871 013 012 242 432;
  • 69) 0.383 259 871 013 012 242 432 × 2 = 0 + 0.766 519 742 026 024 484 864;
  • 70) 0.766 519 742 026 024 484 864 × 2 = 1 + 0.533 039 484 052 048 969 728;
  • 71) 0.533 039 484 052 048 969 728 × 2 = 1 + 0.066 078 968 104 097 939 456;
  • 72) 0.066 078 968 104 097 939 456 × 2 = 0 + 0.132 157 936 208 195 878 912;
  • 73) 0.132 157 936 208 195 878 912 × 2 = 0 + 0.264 315 872 416 391 757 824;
  • 74) 0.264 315 872 416 391 757 824 × 2 = 0 + 0.528 631 744 832 783 515 648;
  • 75) 0.528 631 744 832 783 515 648 × 2 = 1 + 0.057 263 489 665 567 031 296;
  • 76) 0.057 263 489 665 567 031 296 × 2 = 0 + 0.114 526 979 331 134 062 592;
  • 77) 0.114 526 979 331 134 062 592 × 2 = 0 + 0.229 053 958 662 268 125 184;
  • 78) 0.229 053 958 662 268 125 184 × 2 = 0 + 0.458 107 917 324 536 250 368;
  • 79) 0.458 107 917 324 536 250 368 × 2 = 0 + 0.916 215 834 649 072 500 736;
  • 80) 0.916 215 834 649 072 500 736 × 2 = 1 + 0.832 431 669 298 145 001 472;
  • 81) 0.832 431 669 298 145 001 472 × 2 = 1 + 0.664 863 338 596 290 002 944;
  • 82) 0.664 863 338 596 290 002 944 × 2 = 1 + 0.329 726 677 192 580 005 888;
  • 83) 0.329 726 677 192 580 005 888 × 2 = 0 + 0.659 453 354 385 160 011 776;
  • 84) 0.659 453 354 385 160 011 776 × 2 = 1 + 0.318 906 708 770 320 023 552;
  • 85) 0.318 906 708 770 320 023 552 × 2 = 0 + 0.637 813 417 540 640 047 104;
  • 86) 0.637 813 417 540 640 047 104 × 2 = 1 + 0.275 626 835 081 280 094 208;
  • 87) 0.275 626 835 081 280 094 208 × 2 = 0 + 0.551 253 670 162 560 188 416;
  • 88) 0.551 253 670 162 560 188 416 × 2 = 1 + 0.102 507 340 325 120 376 832;
  • 89) 0.102 507 340 325 120 376 832 × 2 = 0 + 0.205 014 680 650 240 753 664;
  • 90) 0.205 014 680 650 240 753 664 × 2 = 0 + 0.410 029 361 300 481 507 328;
  • 91) 0.410 029 361 300 481 507 328 × 2 = 0 + 0.820 058 722 600 963 014 656;
  • 92) 0.820 058 722 600 963 014 656 × 2 = 1 + 0.640 117 445 201 926 029 312;
  • 93) 0.640 117 445 201 926 029 312 × 2 = 1 + 0.280 234 890 403 852 058 624;
  • 94) 0.280 234 890 403 852 058 624 × 2 = 0 + 0.560 469 780 807 704 117 248;
  • 95) 0.560 469 780 807 704 117 248 × 2 = 1 + 0.120 939 561 615 408 234 496;

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:


0.000 000 000 000 176 557 497(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0110 0010 0001 1101 0101 0001 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 497(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0110 0010 0001 1101 0101 0001 101(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:


0.000 000 000 000 176 557 497(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0110 0010 0001 1101 0101 0001 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0110 0010 0001 1101 0101 0001 101(2) × 20 =

1.1000 1101 1001 0010 0111 0111 1011 0001 0000 1110 1010 1000 1101(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 1011 0001 0000 1110 1010 1000 1101


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 1011 0001 0000 1110 1010 1000 1101 =

1000 1101 1001 0010 0111 0111 1011 0001 0000 1110 1010 1000 1101


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 1011 0001 0000 1110 1010 1000 1101


Decimal number -0.000 000 000 000 176 557 497 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0111 1011 0001 0000 1110 1010 1000 1101

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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