-0.000 000 000 000 176 557 012 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 012(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
-0.000 000 000 000 176 557 012(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:

|-0.000 000 000 000 176 557 012| = 0.000 000 000 000 176 557 012


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: 0.000 000 000 000 176 557 012.

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) 0.000 000 000 000 176 557 012 × 2 = 0 + 0.000 000 000 000 353 114 024;
  • 2) 0.000 000 000 000 353 114 024 × 2 = 0 + 0.000 000 000 000 706 228 048;
  • 3) 0.000 000 000 000 706 228 048 × 2 = 0 + 0.000 000 000 001 412 456 096;
  • 4) 0.000 000 000 001 412 456 096 × 2 = 0 + 0.000 000 000 002 824 912 192;
  • 5) 0.000 000 000 002 824 912 192 × 2 = 0 + 0.000 000 000 005 649 824 384;
  • 6) 0.000 000 000 005 649 824 384 × 2 = 0 + 0.000 000 000 011 299 648 768;
  • 7) 0.000 000 000 011 299 648 768 × 2 = 0 + 0.000 000 000 022 599 297 536;
  • 8) 0.000 000 000 022 599 297 536 × 2 = 0 + 0.000 000 000 045 198 595 072;
  • 9) 0.000 000 000 045 198 595 072 × 2 = 0 + 0.000 000 000 090 397 190 144;
  • 10) 0.000 000 000 090 397 190 144 × 2 = 0 + 0.000 000 000 180 794 380 288;
  • 11) 0.000 000 000 180 794 380 288 × 2 = 0 + 0.000 000 000 361 588 760 576;
  • 12) 0.000 000 000 361 588 760 576 × 2 = 0 + 0.000 000 000 723 177 521 152;
  • 13) 0.000 000 000 723 177 521 152 × 2 = 0 + 0.000 000 001 446 355 042 304;
  • 14) 0.000 000 001 446 355 042 304 × 2 = 0 + 0.000 000 002 892 710 084 608;
  • 15) 0.000 000 002 892 710 084 608 × 2 = 0 + 0.000 000 005 785 420 169 216;
  • 16) 0.000 000 005 785 420 169 216 × 2 = 0 + 0.000 000 011 570 840 338 432;
  • 17) 0.000 000 011 570 840 338 432 × 2 = 0 + 0.000 000 023 141 680 676 864;
  • 18) 0.000 000 023 141 680 676 864 × 2 = 0 + 0.000 000 046 283 361 353 728;
  • 19) 0.000 000 046 283 361 353 728 × 2 = 0 + 0.000 000 092 566 722 707 456;
  • 20) 0.000 000 092 566 722 707 456 × 2 = 0 + 0.000 000 185 133 445 414 912;
  • 21) 0.000 000 185 133 445 414 912 × 2 = 0 + 0.000 000 370 266 890 829 824;
  • 22) 0.000 000 370 266 890 829 824 × 2 = 0 + 0.000 000 740 533 781 659 648;
  • 23) 0.000 000 740 533 781 659 648 × 2 = 0 + 0.000 001 481 067 563 319 296;
  • 24) 0.000 001 481 067 563 319 296 × 2 = 0 + 0.000 002 962 135 126 638 592;
  • 25) 0.000 002 962 135 126 638 592 × 2 = 0 + 0.000 005 924 270 253 277 184;
  • 26) 0.000 005 924 270 253 277 184 × 2 = 0 + 0.000 011 848 540 506 554 368;
  • 27) 0.000 011 848 540 506 554 368 × 2 = 0 + 0.000 023 697 081 013 108 736;
  • 28) 0.000 023 697 081 013 108 736 × 2 = 0 + 0.000 047 394 162 026 217 472;
  • 29) 0.000 047 394 162 026 217 472 × 2 = 0 + 0.000 094 788 324 052 434 944;
  • 30) 0.000 094 788 324 052 434 944 × 2 = 0 + 0.000 189 576 648 104 869 888;
  • 31) 0.000 189 576 648 104 869 888 × 2 = 0 + 0.000 379 153 296 209 739 776;
  • 32) 0.000 379 153 296 209 739 776 × 2 = 0 + 0.000 758 306 592 419 479 552;
  • 33) 0.000 758 306 592 419 479 552 × 2 = 0 + 0.001 516 613 184 838 959 104;
  • 34) 0.001 516 613 184 838 959 104 × 2 = 0 + 0.003 033 226 369 677 918 208;
  • 35) 0.003 033 226 369 677 918 208 × 2 = 0 + 0.006 066 452 739 355 836 416;
  • 36) 0.006 066 452 739 355 836 416 × 2 = 0 + 0.012 132 905 478 711 672 832;
  • 37) 0.012 132 905 478 711 672 832 × 2 = 0 + 0.024 265 810 957 423 345 664;
  • 38) 0.024 265 810 957 423 345 664 × 2 = 0 + 0.048 531 621 914 846 691 328;
  • 39) 0.048 531 621 914 846 691 328 × 2 = 0 + 0.097 063 243 829 693 382 656;
  • 40) 0.097 063 243 829 693 382 656 × 2 = 0 + 0.194 126 487 659 386 765 312;
  • 41) 0.194 126 487 659 386 765 312 × 2 = 0 + 0.388 252 975 318 773 530 624;
  • 42) 0.388 252 975 318 773 530 624 × 2 = 0 + 0.776 505 950 637 547 061 248;
  • 43) 0.776 505 950 637 547 061 248 × 2 = 1 + 0.553 011 901 275 094 122 496;
  • 44) 0.553 011 901 275 094 122 496 × 2 = 1 + 0.106 023 802 550 188 244 992;
  • 45) 0.106 023 802 550 188 244 992 × 2 = 0 + 0.212 047 605 100 376 489 984;
  • 46) 0.212 047 605 100 376 489 984 × 2 = 0 + 0.424 095 210 200 752 979 968;
  • 47) 0.424 095 210 200 752 979 968 × 2 = 0 + 0.848 190 420 401 505 959 936;
  • 48) 0.848 190 420 401 505 959 936 × 2 = 1 + 0.696 380 840 803 011 919 872;
  • 49) 0.696 380 840 803 011 919 872 × 2 = 1 + 0.392 761 681 606 023 839 744;
  • 50) 0.392 761 681 606 023 839 744 × 2 = 0 + 0.785 523 363 212 047 679 488;
  • 51) 0.785 523 363 212 047 679 488 × 2 = 1 + 0.571 046 726 424 095 358 976;
  • 52) 0.571 046 726 424 095 358 976 × 2 = 1 + 0.142 093 452 848 190 717 952;
  • 53) 0.142 093 452 848 190 717 952 × 2 = 0 + 0.284 186 905 696 381 435 904;
  • 54) 0.284 186 905 696 381 435 904 × 2 = 0 + 0.568 373 811 392 762 871 808;
  • 55) 0.568 373 811 392 762 871 808 × 2 = 1 + 0.136 747 622 785 525 743 616;
  • 56) 0.136 747 622 785 525 743 616 × 2 = 0 + 0.273 495 245 571 051 487 232;
  • 57) 0.273 495 245 571 051 487 232 × 2 = 0 + 0.546 990 491 142 102 974 464;
  • 58) 0.546 990 491 142 102 974 464 × 2 = 1 + 0.093 980 982 284 205 948 928;
  • 59) 0.093 980 982 284 205 948 928 × 2 = 0 + 0.187 961 964 568 411 897 856;
  • 60) 0.187 961 964 568 411 897 856 × 2 = 0 + 0.375 923 929 136 823 795 712;
  • 61) 0.375 923 929 136 823 795 712 × 2 = 0 + 0.751 847 858 273 647 591 424;
  • 62) 0.751 847 858 273 647 591 424 × 2 = 1 + 0.503 695 716 547 295 182 848;
  • 63) 0.503 695 716 547 295 182 848 × 2 = 1 + 0.007 391 433 094 590 365 696;
  • 64) 0.007 391 433 094 590 365 696 × 2 = 0 + 0.014 782 866 189 180 731 392;
  • 65) 0.014 782 866 189 180 731 392 × 2 = 0 + 0.029 565 732 378 361 462 784;
  • 66) 0.029 565 732 378 361 462 784 × 2 = 0 + 0.059 131 464 756 722 925 568;
  • 67) 0.059 131 464 756 722 925 568 × 2 = 0 + 0.118 262 929 513 445 851 136;
  • 68) 0.118 262 929 513 445 851 136 × 2 = 0 + 0.236 525 859 026 891 702 272;
  • 69) 0.236 525 859 026 891 702 272 × 2 = 0 + 0.473 051 718 053 783 404 544;
  • 70) 0.473 051 718 053 783 404 544 × 2 = 0 + 0.946 103 436 107 566 809 088;
  • 71) 0.946 103 436 107 566 809 088 × 2 = 1 + 0.892 206 872 215 133 618 176;
  • 72) 0.892 206 872 215 133 618 176 × 2 = 1 + 0.784 413 744 430 267 236 352;
  • 73) 0.784 413 744 430 267 236 352 × 2 = 1 + 0.568 827 488 860 534 472 704;
  • 74) 0.568 827 488 860 534 472 704 × 2 = 1 + 0.137 654 977 721 068 945 408;
  • 75) 0.137 654 977 721 068 945 408 × 2 = 0 + 0.275 309 955 442 137 890 816;
  • 76) 0.275 309 955 442 137 890 816 × 2 = 0 + 0.550 619 910 884 275 781 632;
  • 77) 0.550 619 910 884 275 781 632 × 2 = 1 + 0.101 239 821 768 551 563 264;
  • 78) 0.101 239 821 768 551 563 264 × 2 = 0 + 0.202 479 643 537 103 126 528;
  • 79) 0.202 479 643 537 103 126 528 × 2 = 0 + 0.404 959 287 074 206 253 056;
  • 80) 0.404 959 287 074 206 253 056 × 2 = 0 + 0.809 918 574 148 412 506 112;
  • 81) 0.809 918 574 148 412 506 112 × 2 = 1 + 0.619 837 148 296 825 012 224;
  • 82) 0.619 837 148 296 825 012 224 × 2 = 1 + 0.239 674 296 593 650 024 448;
  • 83) 0.239 674 296 593 650 024 448 × 2 = 0 + 0.479 348 593 187 300 048 896;
  • 84) 0.479 348 593 187 300 048 896 × 2 = 0 + 0.958 697 186 374 600 097 792;
  • 85) 0.958 697 186 374 600 097 792 × 2 = 1 + 0.917 394 372 749 200 195 584;
  • 86) 0.917 394 372 749 200 195 584 × 2 = 1 + 0.834 788 745 498 400 391 168;
  • 87) 0.834 788 745 498 400 391 168 × 2 = 1 + 0.669 577 490 996 800 782 336;
  • 88) 0.669 577 490 996 800 782 336 × 2 = 1 + 0.339 154 981 993 601 564 672;
  • 89) 0.339 154 981 993 601 564 672 × 2 = 0 + 0.678 309 963 987 203 129 344;
  • 90) 0.678 309 963 987 203 129 344 × 2 = 1 + 0.356 619 927 974 406 258 688;
  • 91) 0.356 619 927 974 406 258 688 × 2 = 0 + 0.713 239 855 948 812 517 376;
  • 92) 0.713 239 855 948 812 517 376 × 2 = 1 + 0.426 479 711 897 625 034 752;
  • 93) 0.426 479 711 897 625 034 752 × 2 = 0 + 0.852 959 423 795 250 069 504;
  • 94) 0.852 959 423 795 250 069 504 × 2 = 1 + 0.705 918 847 590 500 139 008;
  • 95) 0.705 918 847 590 500 139 008 × 2 = 1 + 0.411 837 695 181 000 278 016;

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:


0.000 000 000 000 176 557 012(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 0000 0011 1100 1000 1100 1111 0101 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 012(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 0000 0011 1100 1000 1100 1111 0101 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:


0.000 000 000 000 176 557 012(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 0000 0011 1100 1000 1100 1111 0101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 0000 0011 1100 1000 1100 1111 0101 011(2) × 20 =

1.1000 1101 1001 0010 0011 0000 0001 1110 0100 0110 0111 1010 1011(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 0011 0000 0001 1110 0100 0110 0111 1010 1011


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 0011 0000 0001 1110 0100 0110 0111 1010 1011 =

1000 1101 1001 0010 0011 0000 0001 1110 0100 0110 0111 1010 1011


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 0011 0000 0001 1110 0100 0110 0111 1010 1011


Decimal number -0.000 000 000 000 176 557 012 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0011 0000 0001 1110 0100 0110 0111 1010 1011

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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