-0.000 000 000 000 176 557 501 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 501(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
-0.000 000 000 000 176 557 501(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:

|-0.000 000 000 000 176 557 501| = 0.000 000 000 000 176 557 501


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: 0.000 000 000 000 176 557 501.

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) 0.000 000 000 000 176 557 501 × 2 = 0 + 0.000 000 000 000 353 115 002;
  • 2) 0.000 000 000 000 353 115 002 × 2 = 0 + 0.000 000 000 000 706 230 004;
  • 3) 0.000 000 000 000 706 230 004 × 2 = 0 + 0.000 000 000 001 412 460 008;
  • 4) 0.000 000 000 001 412 460 008 × 2 = 0 + 0.000 000 000 002 824 920 016;
  • 5) 0.000 000 000 002 824 920 016 × 2 = 0 + 0.000 000 000 005 649 840 032;
  • 6) 0.000 000 000 005 649 840 032 × 2 = 0 + 0.000 000 000 011 299 680 064;
  • 7) 0.000 000 000 011 299 680 064 × 2 = 0 + 0.000 000 000 022 599 360 128;
  • 8) 0.000 000 000 022 599 360 128 × 2 = 0 + 0.000 000 000 045 198 720 256;
  • 9) 0.000 000 000 045 198 720 256 × 2 = 0 + 0.000 000 000 090 397 440 512;
  • 10) 0.000 000 000 090 397 440 512 × 2 = 0 + 0.000 000 000 180 794 881 024;
  • 11) 0.000 000 000 180 794 881 024 × 2 = 0 + 0.000 000 000 361 589 762 048;
  • 12) 0.000 000 000 361 589 762 048 × 2 = 0 + 0.000 000 000 723 179 524 096;
  • 13) 0.000 000 000 723 179 524 096 × 2 = 0 + 0.000 000 001 446 359 048 192;
  • 14) 0.000 000 001 446 359 048 192 × 2 = 0 + 0.000 000 002 892 718 096 384;
  • 15) 0.000 000 002 892 718 096 384 × 2 = 0 + 0.000 000 005 785 436 192 768;
  • 16) 0.000 000 005 785 436 192 768 × 2 = 0 + 0.000 000 011 570 872 385 536;
  • 17) 0.000 000 011 570 872 385 536 × 2 = 0 + 0.000 000 023 141 744 771 072;
  • 18) 0.000 000 023 141 744 771 072 × 2 = 0 + 0.000 000 046 283 489 542 144;
  • 19) 0.000 000 046 283 489 542 144 × 2 = 0 + 0.000 000 092 566 979 084 288;
  • 20) 0.000 000 092 566 979 084 288 × 2 = 0 + 0.000 000 185 133 958 168 576;
  • 21) 0.000 000 185 133 958 168 576 × 2 = 0 + 0.000 000 370 267 916 337 152;
  • 22) 0.000 000 370 267 916 337 152 × 2 = 0 + 0.000 000 740 535 832 674 304;
  • 23) 0.000 000 740 535 832 674 304 × 2 = 0 + 0.000 001 481 071 665 348 608;
  • 24) 0.000 001 481 071 665 348 608 × 2 = 0 + 0.000 002 962 143 330 697 216;
  • 25) 0.000 002 962 143 330 697 216 × 2 = 0 + 0.000 005 924 286 661 394 432;
  • 26) 0.000 005 924 286 661 394 432 × 2 = 0 + 0.000 011 848 573 322 788 864;
  • 27) 0.000 011 848 573 322 788 864 × 2 = 0 + 0.000 023 697 146 645 577 728;
  • 28) 0.000 023 697 146 645 577 728 × 2 = 0 + 0.000 047 394 293 291 155 456;
  • 29) 0.000 047 394 293 291 155 456 × 2 = 0 + 0.000 094 788 586 582 310 912;
  • 30) 0.000 094 788 586 582 310 912 × 2 = 0 + 0.000 189 577 173 164 621 824;
  • 31) 0.000 189 577 173 164 621 824 × 2 = 0 + 0.000 379 154 346 329 243 648;
  • 32) 0.000 379 154 346 329 243 648 × 2 = 0 + 0.000 758 308 692 658 487 296;
  • 33) 0.000 758 308 692 658 487 296 × 2 = 0 + 0.001 516 617 385 316 974 592;
  • 34) 0.001 516 617 385 316 974 592 × 2 = 0 + 0.003 033 234 770 633 949 184;
  • 35) 0.003 033 234 770 633 949 184 × 2 = 0 + 0.006 066 469 541 267 898 368;
  • 36) 0.006 066 469 541 267 898 368 × 2 = 0 + 0.012 132 939 082 535 796 736;
  • 37) 0.012 132 939 082 535 796 736 × 2 = 0 + 0.024 265 878 165 071 593 472;
  • 38) 0.024 265 878 165 071 593 472 × 2 = 0 + 0.048 531 756 330 143 186 944;
  • 39) 0.048 531 756 330 143 186 944 × 2 = 0 + 0.097 063 512 660 286 373 888;
  • 40) 0.097 063 512 660 286 373 888 × 2 = 0 + 0.194 127 025 320 572 747 776;
  • 41) 0.194 127 025 320 572 747 776 × 2 = 0 + 0.388 254 050 641 145 495 552;
  • 42) 0.388 254 050 641 145 495 552 × 2 = 0 + 0.776 508 101 282 290 991 104;
  • 43) 0.776 508 101 282 290 991 104 × 2 = 1 + 0.553 016 202 564 581 982 208;
  • 44) 0.553 016 202 564 581 982 208 × 2 = 1 + 0.106 032 405 129 163 964 416;
  • 45) 0.106 032 405 129 163 964 416 × 2 = 0 + 0.212 064 810 258 327 928 832;
  • 46) 0.212 064 810 258 327 928 832 × 2 = 0 + 0.424 129 620 516 655 857 664;
  • 47) 0.424 129 620 516 655 857 664 × 2 = 0 + 0.848 259 241 033 311 715 328;
  • 48) 0.848 259 241 033 311 715 328 × 2 = 1 + 0.696 518 482 066 623 430 656;
  • 49) 0.696 518 482 066 623 430 656 × 2 = 1 + 0.393 036 964 133 246 861 312;
  • 50) 0.393 036 964 133 246 861 312 × 2 = 0 + 0.786 073 928 266 493 722 624;
  • 51) 0.786 073 928 266 493 722 624 × 2 = 1 + 0.572 147 856 532 987 445 248;
  • 52) 0.572 147 856 532 987 445 248 × 2 = 1 + 0.144 295 713 065 974 890 496;
  • 53) 0.144 295 713 065 974 890 496 × 2 = 0 + 0.288 591 426 131 949 780 992;
  • 54) 0.288 591 426 131 949 780 992 × 2 = 0 + 0.577 182 852 263 899 561 984;
  • 55) 0.577 182 852 263 899 561 984 × 2 = 1 + 0.154 365 704 527 799 123 968;
  • 56) 0.154 365 704 527 799 123 968 × 2 = 0 + 0.308 731 409 055 598 247 936;
  • 57) 0.308 731 409 055 598 247 936 × 2 = 0 + 0.617 462 818 111 196 495 872;
  • 58) 0.617 462 818 111 196 495 872 × 2 = 1 + 0.234 925 636 222 392 991 744;
  • 59) 0.234 925 636 222 392 991 744 × 2 = 0 + 0.469 851 272 444 785 983 488;
  • 60) 0.469 851 272 444 785 983 488 × 2 = 0 + 0.939 702 544 889 571 966 976;
  • 61) 0.939 702 544 889 571 966 976 × 2 = 1 + 0.879 405 089 779 143 933 952;
  • 62) 0.879 405 089 779 143 933 952 × 2 = 1 + 0.758 810 179 558 287 867 904;
  • 63) 0.758 810 179 558 287 867 904 × 2 = 1 + 0.517 620 359 116 575 735 808;
  • 64) 0.517 620 359 116 575 735 808 × 2 = 1 + 0.035 240 718 233 151 471 616;
  • 65) 0.035 240 718 233 151 471 616 × 2 = 0 + 0.070 481 436 466 302 943 232;
  • 66) 0.070 481 436 466 302 943 232 × 2 = 0 + 0.140 962 872 932 605 886 464;
  • 67) 0.140 962 872 932 605 886 464 × 2 = 0 + 0.281 925 745 865 211 772 928;
  • 68) 0.281 925 745 865 211 772 928 × 2 = 0 + 0.563 851 491 730 423 545 856;
  • 69) 0.563 851 491 730 423 545 856 × 2 = 1 + 0.127 702 983 460 847 091 712;
  • 70) 0.127 702 983 460 847 091 712 × 2 = 0 + 0.255 405 966 921 694 183 424;
  • 71) 0.255 405 966 921 694 183 424 × 2 = 0 + 0.510 811 933 843 388 366 848;
  • 72) 0.510 811 933 843 388 366 848 × 2 = 1 + 0.021 623 867 686 776 733 696;
  • 73) 0.021 623 867 686 776 733 696 × 2 = 0 + 0.043 247 735 373 553 467 392;
  • 74) 0.043 247 735 373 553 467 392 × 2 = 0 + 0.086 495 470 747 106 934 784;
  • 75) 0.086 495 470 747 106 934 784 × 2 = 0 + 0.172 990 941 494 213 869 568;
  • 76) 0.172 990 941 494 213 869 568 × 2 = 0 + 0.345 981 882 988 427 739 136;
  • 77) 0.345 981 882 988 427 739 136 × 2 = 0 + 0.691 963 765 976 855 478 272;
  • 78) 0.691 963 765 976 855 478 272 × 2 = 1 + 0.383 927 531 953 710 956 544;
  • 79) 0.383 927 531 953 710 956 544 × 2 = 0 + 0.767 855 063 907 421 913 088;
  • 80) 0.767 855 063 907 421 913 088 × 2 = 1 + 0.535 710 127 814 843 826 176;
  • 81) 0.535 710 127 814 843 826 176 × 2 = 1 + 0.071 420 255 629 687 652 352;
  • 82) 0.071 420 255 629 687 652 352 × 2 = 0 + 0.142 840 511 259 375 304 704;
  • 83) 0.142 840 511 259 375 304 704 × 2 = 0 + 0.285 681 022 518 750 609 408;
  • 84) 0.285 681 022 518 750 609 408 × 2 = 0 + 0.571 362 045 037 501 218 816;
  • 85) 0.571 362 045 037 501 218 816 × 2 = 1 + 0.142 724 090 075 002 437 632;
  • 86) 0.142 724 090 075 002 437 632 × 2 = 0 + 0.285 448 180 150 004 875 264;
  • 87) 0.285 448 180 150 004 875 264 × 2 = 0 + 0.570 896 360 300 009 750 528;
  • 88) 0.570 896 360 300 009 750 528 × 2 = 1 + 0.141 792 720 600 019 501 056;
  • 89) 0.141 792 720 600 019 501 056 × 2 = 0 + 0.283 585 441 200 039 002 112;
  • 90) 0.283 585 441 200 039 002 112 × 2 = 0 + 0.567 170 882 400 078 004 224;
  • 91) 0.567 170 882 400 078 004 224 × 2 = 1 + 0.134 341 764 800 156 008 448;
  • 92) 0.134 341 764 800 156 008 448 × 2 = 0 + 0.268 683 529 600 312 016 896;
  • 93) 0.268 683 529 600 312 016 896 × 2 = 0 + 0.537 367 059 200 624 033 792;
  • 94) 0.537 367 059 200 624 033 792 × 2 = 1 + 0.074 734 118 401 248 067 584;
  • 95) 0.074 734 118 401 248 067 584 × 2 = 0 + 0.149 468 236 802 496 135 168;

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:


0.000 000 000 000 176 557 501(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0000 1001 0000 0101 1000 1001 0010 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 501(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0000 1001 0000 0101 1000 1001 0010 010(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:


0.000 000 000 000 176 557 501(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0000 1001 0000 0101 1000 1001 0010 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0000 1001 0000 0101 1000 1001 0010 010(2) × 20 =

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


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 1000 0100 1000 0010 1100 0100 1001 0010 =

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


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 1000 0100 1000 0010 1100 0100 1001 0010


Decimal number -0.000 000 000 000 176 557 501 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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