-0.000 000 000 000 176 557 426 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 426(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
-0.000 000 000 000 176 557 426(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:

|-0.000 000 000 000 176 557 426| = 0.000 000 000 000 176 557 426


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: 0.000 000 000 000 176 557 426.

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) 0.000 000 000 000 176 557 426 × 2 = 0 + 0.000 000 000 000 353 114 852;
  • 2) 0.000 000 000 000 353 114 852 × 2 = 0 + 0.000 000 000 000 706 229 704;
  • 3) 0.000 000 000 000 706 229 704 × 2 = 0 + 0.000 000 000 001 412 459 408;
  • 4) 0.000 000 000 001 412 459 408 × 2 = 0 + 0.000 000 000 002 824 918 816;
  • 5) 0.000 000 000 002 824 918 816 × 2 = 0 + 0.000 000 000 005 649 837 632;
  • 6) 0.000 000 000 005 649 837 632 × 2 = 0 + 0.000 000 000 011 299 675 264;
  • 7) 0.000 000 000 011 299 675 264 × 2 = 0 + 0.000 000 000 022 599 350 528;
  • 8) 0.000 000 000 022 599 350 528 × 2 = 0 + 0.000 000 000 045 198 701 056;
  • 9) 0.000 000 000 045 198 701 056 × 2 = 0 + 0.000 000 000 090 397 402 112;
  • 10) 0.000 000 000 090 397 402 112 × 2 = 0 + 0.000 000 000 180 794 804 224;
  • 11) 0.000 000 000 180 794 804 224 × 2 = 0 + 0.000 000 000 361 589 608 448;
  • 12) 0.000 000 000 361 589 608 448 × 2 = 0 + 0.000 000 000 723 179 216 896;
  • 13) 0.000 000 000 723 179 216 896 × 2 = 0 + 0.000 000 001 446 358 433 792;
  • 14) 0.000 000 001 446 358 433 792 × 2 = 0 + 0.000 000 002 892 716 867 584;
  • 15) 0.000 000 002 892 716 867 584 × 2 = 0 + 0.000 000 005 785 433 735 168;
  • 16) 0.000 000 005 785 433 735 168 × 2 = 0 + 0.000 000 011 570 867 470 336;
  • 17) 0.000 000 011 570 867 470 336 × 2 = 0 + 0.000 000 023 141 734 940 672;
  • 18) 0.000 000 023 141 734 940 672 × 2 = 0 + 0.000 000 046 283 469 881 344;
  • 19) 0.000 000 046 283 469 881 344 × 2 = 0 + 0.000 000 092 566 939 762 688;
  • 20) 0.000 000 092 566 939 762 688 × 2 = 0 + 0.000 000 185 133 879 525 376;
  • 21) 0.000 000 185 133 879 525 376 × 2 = 0 + 0.000 000 370 267 759 050 752;
  • 22) 0.000 000 370 267 759 050 752 × 2 = 0 + 0.000 000 740 535 518 101 504;
  • 23) 0.000 000 740 535 518 101 504 × 2 = 0 + 0.000 001 481 071 036 203 008;
  • 24) 0.000 001 481 071 036 203 008 × 2 = 0 + 0.000 002 962 142 072 406 016;
  • 25) 0.000 002 962 142 072 406 016 × 2 = 0 + 0.000 005 924 284 144 812 032;
  • 26) 0.000 005 924 284 144 812 032 × 2 = 0 + 0.000 011 848 568 289 624 064;
  • 27) 0.000 011 848 568 289 624 064 × 2 = 0 + 0.000 023 697 136 579 248 128;
  • 28) 0.000 023 697 136 579 248 128 × 2 = 0 + 0.000 047 394 273 158 496 256;
  • 29) 0.000 047 394 273 158 496 256 × 2 = 0 + 0.000 094 788 546 316 992 512;
  • 30) 0.000 094 788 546 316 992 512 × 2 = 0 + 0.000 189 577 092 633 985 024;
  • 31) 0.000 189 577 092 633 985 024 × 2 = 0 + 0.000 379 154 185 267 970 048;
  • 32) 0.000 379 154 185 267 970 048 × 2 = 0 + 0.000 758 308 370 535 940 096;
  • 33) 0.000 758 308 370 535 940 096 × 2 = 0 + 0.001 516 616 741 071 880 192;
  • 34) 0.001 516 616 741 071 880 192 × 2 = 0 + 0.003 033 233 482 143 760 384;
  • 35) 0.003 033 233 482 143 760 384 × 2 = 0 + 0.006 066 466 964 287 520 768;
  • 36) 0.006 066 466 964 287 520 768 × 2 = 0 + 0.012 132 933 928 575 041 536;
  • 37) 0.012 132 933 928 575 041 536 × 2 = 0 + 0.024 265 867 857 150 083 072;
  • 38) 0.024 265 867 857 150 083 072 × 2 = 0 + 0.048 531 735 714 300 166 144;
  • 39) 0.048 531 735 714 300 166 144 × 2 = 0 + 0.097 063 471 428 600 332 288;
  • 40) 0.097 063 471 428 600 332 288 × 2 = 0 + 0.194 126 942 857 200 664 576;
  • 41) 0.194 126 942 857 200 664 576 × 2 = 0 + 0.388 253 885 714 401 329 152;
  • 42) 0.388 253 885 714 401 329 152 × 2 = 0 + 0.776 507 771 428 802 658 304;
  • 43) 0.776 507 771 428 802 658 304 × 2 = 1 + 0.553 015 542 857 605 316 608;
  • 44) 0.553 015 542 857 605 316 608 × 2 = 1 + 0.106 031 085 715 210 633 216;
  • 45) 0.106 031 085 715 210 633 216 × 2 = 0 + 0.212 062 171 430 421 266 432;
  • 46) 0.212 062 171 430 421 266 432 × 2 = 0 + 0.424 124 342 860 842 532 864;
  • 47) 0.424 124 342 860 842 532 864 × 2 = 0 + 0.848 248 685 721 685 065 728;
  • 48) 0.848 248 685 721 685 065 728 × 2 = 1 + 0.696 497 371 443 370 131 456;
  • 49) 0.696 497 371 443 370 131 456 × 2 = 1 + 0.392 994 742 886 740 262 912;
  • 50) 0.392 994 742 886 740 262 912 × 2 = 0 + 0.785 989 485 773 480 525 824;
  • 51) 0.785 989 485 773 480 525 824 × 2 = 1 + 0.571 978 971 546 961 051 648;
  • 52) 0.571 978 971 546 961 051 648 × 2 = 1 + 0.143 957 943 093 922 103 296;
  • 53) 0.143 957 943 093 922 103 296 × 2 = 0 + 0.287 915 886 187 844 206 592;
  • 54) 0.287 915 886 187 844 206 592 × 2 = 0 + 0.575 831 772 375 688 413 184;
  • 55) 0.575 831 772 375 688 413 184 × 2 = 1 + 0.151 663 544 751 376 826 368;
  • 56) 0.151 663 544 751 376 826 368 × 2 = 0 + 0.303 327 089 502 753 652 736;
  • 57) 0.303 327 089 502 753 652 736 × 2 = 0 + 0.606 654 179 005 507 305 472;
  • 58) 0.606 654 179 005 507 305 472 × 2 = 1 + 0.213 308 358 011 014 610 944;
  • 59) 0.213 308 358 011 014 610 944 × 2 = 0 + 0.426 616 716 022 029 221 888;
  • 60) 0.426 616 716 022 029 221 888 × 2 = 0 + 0.853 233 432 044 058 443 776;
  • 61) 0.853 233 432 044 058 443 776 × 2 = 1 + 0.706 466 864 088 116 887 552;
  • 62) 0.706 466 864 088 116 887 552 × 2 = 1 + 0.412 933 728 176 233 775 104;
  • 63) 0.412 933 728 176 233 775 104 × 2 = 0 + 0.825 867 456 352 467 550 208;
  • 64) 0.825 867 456 352 467 550 208 × 2 = 1 + 0.651 734 912 704 935 100 416;
  • 65) 0.651 734 912 704 935 100 416 × 2 = 1 + 0.303 469 825 409 870 200 832;
  • 66) 0.303 469 825 409 870 200 832 × 2 = 0 + 0.606 939 650 819 740 401 664;
  • 67) 0.606 939 650 819 740 401 664 × 2 = 1 + 0.213 879 301 639 480 803 328;
  • 68) 0.213 879 301 639 480 803 328 × 2 = 0 + 0.427 758 603 278 961 606 656;
  • 69) 0.427 758 603 278 961 606 656 × 2 = 0 + 0.855 517 206 557 923 213 312;
  • 70) 0.855 517 206 557 923 213 312 × 2 = 1 + 0.711 034 413 115 846 426 624;
  • 71) 0.711 034 413 115 846 426 624 × 2 = 1 + 0.422 068 826 231 692 853 248;
  • 72) 0.422 068 826 231 692 853 248 × 2 = 0 + 0.844 137 652 463 385 706 496;
  • 73) 0.844 137 652 463 385 706 496 × 2 = 1 + 0.688 275 304 926 771 412 992;
  • 74) 0.688 275 304 926 771 412 992 × 2 = 1 + 0.376 550 609 853 542 825 984;
  • 75) 0.376 550 609 853 542 825 984 × 2 = 0 + 0.753 101 219 707 085 651 968;
  • 76) 0.753 101 219 707 085 651 968 × 2 = 1 + 0.506 202 439 414 171 303 936;
  • 77) 0.506 202 439 414 171 303 936 × 2 = 1 + 0.012 404 878 828 342 607 872;
  • 78) 0.012 404 878 828 342 607 872 × 2 = 0 + 0.024 809 757 656 685 215 744;
  • 79) 0.024 809 757 656 685 215 744 × 2 = 0 + 0.049 619 515 313 370 431 488;
  • 80) 0.049 619 515 313 370 431 488 × 2 = 0 + 0.099 239 030 626 740 862 976;
  • 81) 0.099 239 030 626 740 862 976 × 2 = 0 + 0.198 478 061 253 481 725 952;
  • 82) 0.198 478 061 253 481 725 952 × 2 = 0 + 0.396 956 122 506 963 451 904;
  • 83) 0.396 956 122 506 963 451 904 × 2 = 0 + 0.793 912 245 013 926 903 808;
  • 84) 0.793 912 245 013 926 903 808 × 2 = 1 + 0.587 824 490 027 853 807 616;
  • 85) 0.587 824 490 027 853 807 616 × 2 = 1 + 0.175 648 980 055 707 615 232;
  • 86) 0.175 648 980 055 707 615 232 × 2 = 0 + 0.351 297 960 111 415 230 464;
  • 87) 0.351 297 960 111 415 230 464 × 2 = 0 + 0.702 595 920 222 830 460 928;
  • 88) 0.702 595 920 222 830 460 928 × 2 = 1 + 0.405 191 840 445 660 921 856;
  • 89) 0.405 191 840 445 660 921 856 × 2 = 0 + 0.810 383 680 891 321 843 712;
  • 90) 0.810 383 680 891 321 843 712 × 2 = 1 + 0.620 767 361 782 643 687 424;
  • 91) 0.620 767 361 782 643 687 424 × 2 = 1 + 0.241 534 723 565 287 374 848;
  • 92) 0.241 534 723 565 287 374 848 × 2 = 0 + 0.483 069 447 130 574 749 696;
  • 93) 0.483 069 447 130 574 749 696 × 2 = 0 + 0.966 138 894 261 149 499 392;
  • 94) 0.966 138 894 261 149 499 392 × 2 = 1 + 0.932 277 788 522 298 998 784;
  • 95) 0.932 277 788 522 298 998 784 × 2 = 1 + 0.864 555 577 044 597 997 568;

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:


0.000 000 000 000 176 557 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1010 0110 1101 1000 0001 1001 0110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1010 0110 1101 1000 0001 1001 0110 011(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:


0.000 000 000 000 176 557 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1010 0110 1101 1000 0001 1001 0110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1010 0110 1101 1000 0001 1001 0110 011(2) × 20 =

1.1000 1101 1001 0010 0110 1101 0011 0110 1100 0000 1100 1011 0011(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 0110 1101 0011 0110 1100 0000 1100 1011 0011


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 0110 1101 0011 0110 1100 0000 1100 1011 0011 =

1000 1101 1001 0010 0110 1101 0011 0110 1100 0000 1100 1011 0011


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 0110 1101 0011 0110 1100 0000 1100 1011 0011


Decimal number -0.000 000 000 000 176 557 426 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1101 0011 0110 1100 0000 1100 1011 0011

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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