-0.000 000 000 000 176 557 524 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 524(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
-0.000 000 000 000 176 557 524(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:

|-0.000 000 000 000 176 557 524| = 0.000 000 000 000 176 557 524


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: 0.000 000 000 000 176 557 524.

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) 0.000 000 000 000 176 557 524 × 2 = 0 + 0.000 000 000 000 353 115 048;
  • 2) 0.000 000 000 000 353 115 048 × 2 = 0 + 0.000 000 000 000 706 230 096;
  • 3) 0.000 000 000 000 706 230 096 × 2 = 0 + 0.000 000 000 001 412 460 192;
  • 4) 0.000 000 000 001 412 460 192 × 2 = 0 + 0.000 000 000 002 824 920 384;
  • 5) 0.000 000 000 002 824 920 384 × 2 = 0 + 0.000 000 000 005 649 840 768;
  • 6) 0.000 000 000 005 649 840 768 × 2 = 0 + 0.000 000 000 011 299 681 536;
  • 7) 0.000 000 000 011 299 681 536 × 2 = 0 + 0.000 000 000 022 599 363 072;
  • 8) 0.000 000 000 022 599 363 072 × 2 = 0 + 0.000 000 000 045 198 726 144;
  • 9) 0.000 000 000 045 198 726 144 × 2 = 0 + 0.000 000 000 090 397 452 288;
  • 10) 0.000 000 000 090 397 452 288 × 2 = 0 + 0.000 000 000 180 794 904 576;
  • 11) 0.000 000 000 180 794 904 576 × 2 = 0 + 0.000 000 000 361 589 809 152;
  • 12) 0.000 000 000 361 589 809 152 × 2 = 0 + 0.000 000 000 723 179 618 304;
  • 13) 0.000 000 000 723 179 618 304 × 2 = 0 + 0.000 000 001 446 359 236 608;
  • 14) 0.000 000 001 446 359 236 608 × 2 = 0 + 0.000 000 002 892 718 473 216;
  • 15) 0.000 000 002 892 718 473 216 × 2 = 0 + 0.000 000 005 785 436 946 432;
  • 16) 0.000 000 005 785 436 946 432 × 2 = 0 + 0.000 000 011 570 873 892 864;
  • 17) 0.000 000 011 570 873 892 864 × 2 = 0 + 0.000 000 023 141 747 785 728;
  • 18) 0.000 000 023 141 747 785 728 × 2 = 0 + 0.000 000 046 283 495 571 456;
  • 19) 0.000 000 046 283 495 571 456 × 2 = 0 + 0.000 000 092 566 991 142 912;
  • 20) 0.000 000 092 566 991 142 912 × 2 = 0 + 0.000 000 185 133 982 285 824;
  • 21) 0.000 000 185 133 982 285 824 × 2 = 0 + 0.000 000 370 267 964 571 648;
  • 22) 0.000 000 370 267 964 571 648 × 2 = 0 + 0.000 000 740 535 929 143 296;
  • 23) 0.000 000 740 535 929 143 296 × 2 = 0 + 0.000 001 481 071 858 286 592;
  • 24) 0.000 001 481 071 858 286 592 × 2 = 0 + 0.000 002 962 143 716 573 184;
  • 25) 0.000 002 962 143 716 573 184 × 2 = 0 + 0.000 005 924 287 433 146 368;
  • 26) 0.000 005 924 287 433 146 368 × 2 = 0 + 0.000 011 848 574 866 292 736;
  • 27) 0.000 011 848 574 866 292 736 × 2 = 0 + 0.000 023 697 149 732 585 472;
  • 28) 0.000 023 697 149 732 585 472 × 2 = 0 + 0.000 047 394 299 465 170 944;
  • 29) 0.000 047 394 299 465 170 944 × 2 = 0 + 0.000 094 788 598 930 341 888;
  • 30) 0.000 094 788 598 930 341 888 × 2 = 0 + 0.000 189 577 197 860 683 776;
  • 31) 0.000 189 577 197 860 683 776 × 2 = 0 + 0.000 379 154 395 721 367 552;
  • 32) 0.000 379 154 395 721 367 552 × 2 = 0 + 0.000 758 308 791 442 735 104;
  • 33) 0.000 758 308 791 442 735 104 × 2 = 0 + 0.001 516 617 582 885 470 208;
  • 34) 0.001 516 617 582 885 470 208 × 2 = 0 + 0.003 033 235 165 770 940 416;
  • 35) 0.003 033 235 165 770 940 416 × 2 = 0 + 0.006 066 470 331 541 880 832;
  • 36) 0.006 066 470 331 541 880 832 × 2 = 0 + 0.012 132 940 663 083 761 664;
  • 37) 0.012 132 940 663 083 761 664 × 2 = 0 + 0.024 265 881 326 167 523 328;
  • 38) 0.024 265 881 326 167 523 328 × 2 = 0 + 0.048 531 762 652 335 046 656;
  • 39) 0.048 531 762 652 335 046 656 × 2 = 0 + 0.097 063 525 304 670 093 312;
  • 40) 0.097 063 525 304 670 093 312 × 2 = 0 + 0.194 127 050 609 340 186 624;
  • 41) 0.194 127 050 609 340 186 624 × 2 = 0 + 0.388 254 101 218 680 373 248;
  • 42) 0.388 254 101 218 680 373 248 × 2 = 0 + 0.776 508 202 437 360 746 496;
  • 43) 0.776 508 202 437 360 746 496 × 2 = 1 + 0.553 016 404 874 721 492 992;
  • 44) 0.553 016 404 874 721 492 992 × 2 = 1 + 0.106 032 809 749 442 985 984;
  • 45) 0.106 032 809 749 442 985 984 × 2 = 0 + 0.212 065 619 498 885 971 968;
  • 46) 0.212 065 619 498 885 971 968 × 2 = 0 + 0.424 131 238 997 771 943 936;
  • 47) 0.424 131 238 997 771 943 936 × 2 = 0 + 0.848 262 477 995 543 887 872;
  • 48) 0.848 262 477 995 543 887 872 × 2 = 1 + 0.696 524 955 991 087 775 744;
  • 49) 0.696 524 955 991 087 775 744 × 2 = 1 + 0.393 049 911 982 175 551 488;
  • 50) 0.393 049 911 982 175 551 488 × 2 = 0 + 0.786 099 823 964 351 102 976;
  • 51) 0.786 099 823 964 351 102 976 × 2 = 1 + 0.572 199 647 928 702 205 952;
  • 52) 0.572 199 647 928 702 205 952 × 2 = 1 + 0.144 399 295 857 404 411 904;
  • 53) 0.144 399 295 857 404 411 904 × 2 = 0 + 0.288 798 591 714 808 823 808;
  • 54) 0.288 798 591 714 808 823 808 × 2 = 0 + 0.577 597 183 429 617 647 616;
  • 55) 0.577 597 183 429 617 647 616 × 2 = 1 + 0.155 194 366 859 235 295 232;
  • 56) 0.155 194 366 859 235 295 232 × 2 = 0 + 0.310 388 733 718 470 590 464;
  • 57) 0.310 388 733 718 470 590 464 × 2 = 0 + 0.620 777 467 436 941 180 928;
  • 58) 0.620 777 467 436 941 180 928 × 2 = 1 + 0.241 554 934 873 882 361 856;
  • 59) 0.241 554 934 873 882 361 856 × 2 = 0 + 0.483 109 869 747 764 723 712;
  • 60) 0.483 109 869 747 764 723 712 × 2 = 0 + 0.966 219 739 495 529 447 424;
  • 61) 0.966 219 739 495 529 447 424 × 2 = 1 + 0.932 439 478 991 058 894 848;
  • 62) 0.932 439 478 991 058 894 848 × 2 = 1 + 0.864 878 957 982 117 789 696;
  • 63) 0.864 878 957 982 117 789 696 × 2 = 1 + 0.729 757 915 964 235 579 392;
  • 64) 0.729 757 915 964 235 579 392 × 2 = 1 + 0.459 515 831 928 471 158 784;
  • 65) 0.459 515 831 928 471 158 784 × 2 = 0 + 0.919 031 663 856 942 317 568;
  • 66) 0.919 031 663 856 942 317 568 × 2 = 1 + 0.838 063 327 713 884 635 136;
  • 67) 0.838 063 327 713 884 635 136 × 2 = 1 + 0.676 126 655 427 769 270 272;
  • 68) 0.676 126 655 427 769 270 272 × 2 = 1 + 0.352 253 310 855 538 540 544;
  • 69) 0.352 253 310 855 538 540 544 × 2 = 0 + 0.704 506 621 711 077 081 088;
  • 70) 0.704 506 621 711 077 081 088 × 2 = 1 + 0.409 013 243 422 154 162 176;
  • 71) 0.409 013 243 422 154 162 176 × 2 = 0 + 0.818 026 486 844 308 324 352;
  • 72) 0.818 026 486 844 308 324 352 × 2 = 1 + 0.636 052 973 688 616 648 704;
  • 73) 0.636 052 973 688 616 648 704 × 2 = 1 + 0.272 105 947 377 233 297 408;
  • 74) 0.272 105 947 377 233 297 408 × 2 = 0 + 0.544 211 894 754 466 594 816;
  • 75) 0.544 211 894 754 466 594 816 × 2 = 1 + 0.088 423 789 508 933 189 632;
  • 76) 0.088 423 789 508 933 189 632 × 2 = 0 + 0.176 847 579 017 866 379 264;
  • 77) 0.176 847 579 017 866 379 264 × 2 = 0 + 0.353 695 158 035 732 758 528;
  • 78) 0.353 695 158 035 732 758 528 × 2 = 0 + 0.707 390 316 071 465 517 056;
  • 79) 0.707 390 316 071 465 517 056 × 2 = 1 + 0.414 780 632 142 931 034 112;
  • 80) 0.414 780 632 142 931 034 112 × 2 = 0 + 0.829 561 264 285 862 068 224;
  • 81) 0.829 561 264 285 862 068 224 × 2 = 1 + 0.659 122 528 571 724 136 448;
  • 82) 0.659 122 528 571 724 136 448 × 2 = 1 + 0.318 245 057 143 448 272 896;
  • 83) 0.318 245 057 143 448 272 896 × 2 = 0 + 0.636 490 114 286 896 545 792;
  • 84) 0.636 490 114 286 896 545 792 × 2 = 1 + 0.272 980 228 573 793 091 584;
  • 85) 0.272 980 228 573 793 091 584 × 2 = 0 + 0.545 960 457 147 586 183 168;
  • 86) 0.545 960 457 147 586 183 168 × 2 = 1 + 0.091 920 914 295 172 366 336;
  • 87) 0.091 920 914 295 172 366 336 × 2 = 0 + 0.183 841 828 590 344 732 672;
  • 88) 0.183 841 828 590 344 732 672 × 2 = 0 + 0.367 683 657 180 689 465 344;
  • 89) 0.367 683 657 180 689 465 344 × 2 = 0 + 0.735 367 314 361 378 930 688;
  • 90) 0.735 367 314 361 378 930 688 × 2 = 1 + 0.470 734 628 722 757 861 376;
  • 91) 0.470 734 628 722 757 861 376 × 2 = 0 + 0.941 469 257 445 515 722 752;
  • 92) 0.941 469 257 445 515 722 752 × 2 = 1 + 0.882 938 514 891 031 445 504;
  • 93) 0.882 938 514 891 031 445 504 × 2 = 1 + 0.765 877 029 782 062 891 008;
  • 94) 0.765 877 029 782 062 891 008 × 2 = 1 + 0.531 754 059 564 125 782 016;
  • 95) 0.531 754 059 564 125 782 016 × 2 = 1 + 0.063 508 119 128 251 564 032;

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:


0.000 000 000 000 176 557 524(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0111 0101 1010 0010 1101 0100 0101 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 524(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0111 0101 1010 0010 1101 0100 0101 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:


0.000 000 000 000 176 557 524(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0111 0101 1010 0010 1101 0100 0101 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0111 0101 1010 0010 1101 0100 0101 111(2) × 20 =

1.1000 1101 1001 0010 0111 1011 1010 1101 0001 0110 1010 0010 1111(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 1011 1010 1101 0001 0110 1010 0010 1111


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 1011 1010 1101 0001 0110 1010 0010 1111 =

1000 1101 1001 0010 0111 1011 1010 1101 0001 0110 1010 0010 1111


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 1011 1010 1101 0001 0110 1010 0010 1111


Decimal number -0.000 000 000 000 176 557 524 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 1011 1010 1101 0001 0110 1010 0010 1111

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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