-0.000 000 000 000 176 557 482 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 482(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
-0.000 000 000 000 176 557 482(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:

|-0.000 000 000 000 176 557 482| = 0.000 000 000 000 176 557 482


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: 0.000 000 000 000 176 557 482.

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) 0.000 000 000 000 176 557 482 × 2 = 0 + 0.000 000 000 000 353 114 964;
  • 2) 0.000 000 000 000 353 114 964 × 2 = 0 + 0.000 000 000 000 706 229 928;
  • 3) 0.000 000 000 000 706 229 928 × 2 = 0 + 0.000 000 000 001 412 459 856;
  • 4) 0.000 000 000 001 412 459 856 × 2 = 0 + 0.000 000 000 002 824 919 712;
  • 5) 0.000 000 000 002 824 919 712 × 2 = 0 + 0.000 000 000 005 649 839 424;
  • 6) 0.000 000 000 005 649 839 424 × 2 = 0 + 0.000 000 000 011 299 678 848;
  • 7) 0.000 000 000 011 299 678 848 × 2 = 0 + 0.000 000 000 022 599 357 696;
  • 8) 0.000 000 000 022 599 357 696 × 2 = 0 + 0.000 000 000 045 198 715 392;
  • 9) 0.000 000 000 045 198 715 392 × 2 = 0 + 0.000 000 000 090 397 430 784;
  • 10) 0.000 000 000 090 397 430 784 × 2 = 0 + 0.000 000 000 180 794 861 568;
  • 11) 0.000 000 000 180 794 861 568 × 2 = 0 + 0.000 000 000 361 589 723 136;
  • 12) 0.000 000 000 361 589 723 136 × 2 = 0 + 0.000 000 000 723 179 446 272;
  • 13) 0.000 000 000 723 179 446 272 × 2 = 0 + 0.000 000 001 446 358 892 544;
  • 14) 0.000 000 001 446 358 892 544 × 2 = 0 + 0.000 000 002 892 717 785 088;
  • 15) 0.000 000 002 892 717 785 088 × 2 = 0 + 0.000 000 005 785 435 570 176;
  • 16) 0.000 000 005 785 435 570 176 × 2 = 0 + 0.000 000 011 570 871 140 352;
  • 17) 0.000 000 011 570 871 140 352 × 2 = 0 + 0.000 000 023 141 742 280 704;
  • 18) 0.000 000 023 141 742 280 704 × 2 = 0 + 0.000 000 046 283 484 561 408;
  • 19) 0.000 000 046 283 484 561 408 × 2 = 0 + 0.000 000 092 566 969 122 816;
  • 20) 0.000 000 092 566 969 122 816 × 2 = 0 + 0.000 000 185 133 938 245 632;
  • 21) 0.000 000 185 133 938 245 632 × 2 = 0 + 0.000 000 370 267 876 491 264;
  • 22) 0.000 000 370 267 876 491 264 × 2 = 0 + 0.000 000 740 535 752 982 528;
  • 23) 0.000 000 740 535 752 982 528 × 2 = 0 + 0.000 001 481 071 505 965 056;
  • 24) 0.000 001 481 071 505 965 056 × 2 = 0 + 0.000 002 962 143 011 930 112;
  • 25) 0.000 002 962 143 011 930 112 × 2 = 0 + 0.000 005 924 286 023 860 224;
  • 26) 0.000 005 924 286 023 860 224 × 2 = 0 + 0.000 011 848 572 047 720 448;
  • 27) 0.000 011 848 572 047 720 448 × 2 = 0 + 0.000 023 697 144 095 440 896;
  • 28) 0.000 023 697 144 095 440 896 × 2 = 0 + 0.000 047 394 288 190 881 792;
  • 29) 0.000 047 394 288 190 881 792 × 2 = 0 + 0.000 094 788 576 381 763 584;
  • 30) 0.000 094 788 576 381 763 584 × 2 = 0 + 0.000 189 577 152 763 527 168;
  • 31) 0.000 189 577 152 763 527 168 × 2 = 0 + 0.000 379 154 305 527 054 336;
  • 32) 0.000 379 154 305 527 054 336 × 2 = 0 + 0.000 758 308 611 054 108 672;
  • 33) 0.000 758 308 611 054 108 672 × 2 = 0 + 0.001 516 617 222 108 217 344;
  • 34) 0.001 516 617 222 108 217 344 × 2 = 0 + 0.003 033 234 444 216 434 688;
  • 35) 0.003 033 234 444 216 434 688 × 2 = 0 + 0.006 066 468 888 432 869 376;
  • 36) 0.006 066 468 888 432 869 376 × 2 = 0 + 0.012 132 937 776 865 738 752;
  • 37) 0.012 132 937 776 865 738 752 × 2 = 0 + 0.024 265 875 553 731 477 504;
  • 38) 0.024 265 875 553 731 477 504 × 2 = 0 + 0.048 531 751 107 462 955 008;
  • 39) 0.048 531 751 107 462 955 008 × 2 = 0 + 0.097 063 502 214 925 910 016;
  • 40) 0.097 063 502 214 925 910 016 × 2 = 0 + 0.194 127 004 429 851 820 032;
  • 41) 0.194 127 004 429 851 820 032 × 2 = 0 + 0.388 254 008 859 703 640 064;
  • 42) 0.388 254 008 859 703 640 064 × 2 = 0 + 0.776 508 017 719 407 280 128;
  • 43) 0.776 508 017 719 407 280 128 × 2 = 1 + 0.553 016 035 438 814 560 256;
  • 44) 0.553 016 035 438 814 560 256 × 2 = 1 + 0.106 032 070 877 629 120 512;
  • 45) 0.106 032 070 877 629 120 512 × 2 = 0 + 0.212 064 141 755 258 241 024;
  • 46) 0.212 064 141 755 258 241 024 × 2 = 0 + 0.424 128 283 510 516 482 048;
  • 47) 0.424 128 283 510 516 482 048 × 2 = 0 + 0.848 256 567 021 032 964 096;
  • 48) 0.848 256 567 021 032 964 096 × 2 = 1 + 0.696 513 134 042 065 928 192;
  • 49) 0.696 513 134 042 065 928 192 × 2 = 1 + 0.393 026 268 084 131 856 384;
  • 50) 0.393 026 268 084 131 856 384 × 2 = 0 + 0.786 052 536 168 263 712 768;
  • 51) 0.786 052 536 168 263 712 768 × 2 = 1 + 0.572 105 072 336 527 425 536;
  • 52) 0.572 105 072 336 527 425 536 × 2 = 1 + 0.144 210 144 673 054 851 072;
  • 53) 0.144 210 144 673 054 851 072 × 2 = 0 + 0.288 420 289 346 109 702 144;
  • 54) 0.288 420 289 346 109 702 144 × 2 = 0 + 0.576 840 578 692 219 404 288;
  • 55) 0.576 840 578 692 219 404 288 × 2 = 1 + 0.153 681 157 384 438 808 576;
  • 56) 0.153 681 157 384 438 808 576 × 2 = 0 + 0.307 362 314 768 877 617 152;
  • 57) 0.307 362 314 768 877 617 152 × 2 = 0 + 0.614 724 629 537 755 234 304;
  • 58) 0.614 724 629 537 755 234 304 × 2 = 1 + 0.229 449 259 075 510 468 608;
  • 59) 0.229 449 259 075 510 468 608 × 2 = 0 + 0.458 898 518 151 020 937 216;
  • 60) 0.458 898 518 151 020 937 216 × 2 = 0 + 0.917 797 036 302 041 874 432;
  • 61) 0.917 797 036 302 041 874 432 × 2 = 1 + 0.835 594 072 604 083 748 864;
  • 62) 0.835 594 072 604 083 748 864 × 2 = 1 + 0.671 188 145 208 167 497 728;
  • 63) 0.671 188 145 208 167 497 728 × 2 = 1 + 0.342 376 290 416 334 995 456;
  • 64) 0.342 376 290 416 334 995 456 × 2 = 0 + 0.684 752 580 832 669 990 912;
  • 65) 0.684 752 580 832 669 990 912 × 2 = 1 + 0.369 505 161 665 339 981 824;
  • 66) 0.369 505 161 665 339 981 824 × 2 = 0 + 0.739 010 323 330 679 963 648;
  • 67) 0.739 010 323 330 679 963 648 × 2 = 1 + 0.478 020 646 661 359 927 296;
  • 68) 0.478 020 646 661 359 927 296 × 2 = 0 + 0.956 041 293 322 719 854 592;
  • 69) 0.956 041 293 322 719 854 592 × 2 = 1 + 0.912 082 586 645 439 709 184;
  • 70) 0.912 082 586 645 439 709 184 × 2 = 1 + 0.824 165 173 290 879 418 368;
  • 71) 0.824 165 173 290 879 418 368 × 2 = 1 + 0.648 330 346 581 758 836 736;
  • 72) 0.648 330 346 581 758 836 736 × 2 = 1 + 0.296 660 693 163 517 673 472;
  • 73) 0.296 660 693 163 517 673 472 × 2 = 0 + 0.593 321 386 327 035 346 944;
  • 74) 0.593 321 386 327 035 346 944 × 2 = 1 + 0.186 642 772 654 070 693 888;
  • 75) 0.186 642 772 654 070 693 888 × 2 = 0 + 0.373 285 545 308 141 387 776;
  • 76) 0.373 285 545 308 141 387 776 × 2 = 0 + 0.746 571 090 616 282 775 552;
  • 77) 0.746 571 090 616 282 775 552 × 2 = 1 + 0.493 142 181 232 565 551 104;
  • 78) 0.493 142 181 232 565 551 104 × 2 = 0 + 0.986 284 362 465 131 102 208;
  • 79) 0.986 284 362 465 131 102 208 × 2 = 1 + 0.972 568 724 930 262 204 416;
  • 80) 0.972 568 724 930 262 204 416 × 2 = 1 + 0.945 137 449 860 524 408 832;
  • 81) 0.945 137 449 860 524 408 832 × 2 = 1 + 0.890 274 899 721 048 817 664;
  • 82) 0.890 274 899 721 048 817 664 × 2 = 1 + 0.780 549 799 442 097 635 328;
  • 83) 0.780 549 799 442 097 635 328 × 2 = 1 + 0.561 099 598 884 195 270 656;
  • 84) 0.561 099 598 884 195 270 656 × 2 = 1 + 0.122 199 197 768 390 541 312;
  • 85) 0.122 199 197 768 390 541 312 × 2 = 0 + 0.244 398 395 536 781 082 624;
  • 86) 0.244 398 395 536 781 082 624 × 2 = 0 + 0.488 796 791 073 562 165 248;
  • 87) 0.488 796 791 073 562 165 248 × 2 = 0 + 0.977 593 582 147 124 330 496;
  • 88) 0.977 593 582 147 124 330 496 × 2 = 1 + 0.955 187 164 294 248 660 992;
  • 89) 0.955 187 164 294 248 660 992 × 2 = 1 + 0.910 374 328 588 497 321 984;
  • 90) 0.910 374 328 588 497 321 984 × 2 = 1 + 0.820 748 657 176 994 643 968;
  • 91) 0.820 748 657 176 994 643 968 × 2 = 1 + 0.641 497 314 353 989 287 936;
  • 92) 0.641 497 314 353 989 287 936 × 2 = 1 + 0.282 994 628 707 978 575 872;
  • 93) 0.282 994 628 707 978 575 872 × 2 = 0 + 0.565 989 257 415 957 151 744;
  • 94) 0.565 989 257 415 957 151 744 × 2 = 1 + 0.131 978 514 831 914 303 488;
  • 95) 0.131 978 514 831 914 303 488 × 2 = 0 + 0.263 957 029 663 828 606 976;

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:


0.000 000 000 000 176 557 482(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1010 1111 0100 1011 1111 0001 1111 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 482(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1010 1111 0100 1011 1111 0001 1111 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:


0.000 000 000 000 176 557 482(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1010 1111 0100 1011 1111 0001 1111 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1010 1111 0100 1011 1111 0001 1111 010(2) × 20 =

1.1000 1101 1001 0010 0111 0101 0111 1010 0101 1111 1000 1111 1010(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 0101 0111 1010 0101 1111 1000 1111 1010


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 0101 0111 1010 0101 1111 1000 1111 1010 =

1000 1101 1001 0010 0111 0101 0111 1010 0101 1111 1000 1111 1010


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 0101 0111 1010 0101 1111 1000 1111 1010


Decimal number -0.000 000 000 000 176 557 482 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0101 0111 1010 0101 1111 1000 1111 1010

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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