-0.000 000 000 000 176 557 113 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 113(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
-0.000 000 000 000 176 557 113(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:

|-0.000 000 000 000 176 557 113| = 0.000 000 000 000 176 557 113


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: 0.000 000 000 000 176 557 113.

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) 0.000 000 000 000 176 557 113 × 2 = 0 + 0.000 000 000 000 353 114 226;
  • 2) 0.000 000 000 000 353 114 226 × 2 = 0 + 0.000 000 000 000 706 228 452;
  • 3) 0.000 000 000 000 706 228 452 × 2 = 0 + 0.000 000 000 001 412 456 904;
  • 4) 0.000 000 000 001 412 456 904 × 2 = 0 + 0.000 000 000 002 824 913 808;
  • 5) 0.000 000 000 002 824 913 808 × 2 = 0 + 0.000 000 000 005 649 827 616;
  • 6) 0.000 000 000 005 649 827 616 × 2 = 0 + 0.000 000 000 011 299 655 232;
  • 7) 0.000 000 000 011 299 655 232 × 2 = 0 + 0.000 000 000 022 599 310 464;
  • 8) 0.000 000 000 022 599 310 464 × 2 = 0 + 0.000 000 000 045 198 620 928;
  • 9) 0.000 000 000 045 198 620 928 × 2 = 0 + 0.000 000 000 090 397 241 856;
  • 10) 0.000 000 000 090 397 241 856 × 2 = 0 + 0.000 000 000 180 794 483 712;
  • 11) 0.000 000 000 180 794 483 712 × 2 = 0 + 0.000 000 000 361 588 967 424;
  • 12) 0.000 000 000 361 588 967 424 × 2 = 0 + 0.000 000 000 723 177 934 848;
  • 13) 0.000 000 000 723 177 934 848 × 2 = 0 + 0.000 000 001 446 355 869 696;
  • 14) 0.000 000 001 446 355 869 696 × 2 = 0 + 0.000 000 002 892 711 739 392;
  • 15) 0.000 000 002 892 711 739 392 × 2 = 0 + 0.000 000 005 785 423 478 784;
  • 16) 0.000 000 005 785 423 478 784 × 2 = 0 + 0.000 000 011 570 846 957 568;
  • 17) 0.000 000 011 570 846 957 568 × 2 = 0 + 0.000 000 023 141 693 915 136;
  • 18) 0.000 000 023 141 693 915 136 × 2 = 0 + 0.000 000 046 283 387 830 272;
  • 19) 0.000 000 046 283 387 830 272 × 2 = 0 + 0.000 000 092 566 775 660 544;
  • 20) 0.000 000 092 566 775 660 544 × 2 = 0 + 0.000 000 185 133 551 321 088;
  • 21) 0.000 000 185 133 551 321 088 × 2 = 0 + 0.000 000 370 267 102 642 176;
  • 22) 0.000 000 370 267 102 642 176 × 2 = 0 + 0.000 000 740 534 205 284 352;
  • 23) 0.000 000 740 534 205 284 352 × 2 = 0 + 0.000 001 481 068 410 568 704;
  • 24) 0.000 001 481 068 410 568 704 × 2 = 0 + 0.000 002 962 136 821 137 408;
  • 25) 0.000 002 962 136 821 137 408 × 2 = 0 + 0.000 005 924 273 642 274 816;
  • 26) 0.000 005 924 273 642 274 816 × 2 = 0 + 0.000 011 848 547 284 549 632;
  • 27) 0.000 011 848 547 284 549 632 × 2 = 0 + 0.000 023 697 094 569 099 264;
  • 28) 0.000 023 697 094 569 099 264 × 2 = 0 + 0.000 047 394 189 138 198 528;
  • 29) 0.000 047 394 189 138 198 528 × 2 = 0 + 0.000 094 788 378 276 397 056;
  • 30) 0.000 094 788 378 276 397 056 × 2 = 0 + 0.000 189 576 756 552 794 112;
  • 31) 0.000 189 576 756 552 794 112 × 2 = 0 + 0.000 379 153 513 105 588 224;
  • 32) 0.000 379 153 513 105 588 224 × 2 = 0 + 0.000 758 307 026 211 176 448;
  • 33) 0.000 758 307 026 211 176 448 × 2 = 0 + 0.001 516 614 052 422 352 896;
  • 34) 0.001 516 614 052 422 352 896 × 2 = 0 + 0.003 033 228 104 844 705 792;
  • 35) 0.003 033 228 104 844 705 792 × 2 = 0 + 0.006 066 456 209 689 411 584;
  • 36) 0.006 066 456 209 689 411 584 × 2 = 0 + 0.012 132 912 419 378 823 168;
  • 37) 0.012 132 912 419 378 823 168 × 2 = 0 + 0.024 265 824 838 757 646 336;
  • 38) 0.024 265 824 838 757 646 336 × 2 = 0 + 0.048 531 649 677 515 292 672;
  • 39) 0.048 531 649 677 515 292 672 × 2 = 0 + 0.097 063 299 355 030 585 344;
  • 40) 0.097 063 299 355 030 585 344 × 2 = 0 + 0.194 126 598 710 061 170 688;
  • 41) 0.194 126 598 710 061 170 688 × 2 = 0 + 0.388 253 197 420 122 341 376;
  • 42) 0.388 253 197 420 122 341 376 × 2 = 0 + 0.776 506 394 840 244 682 752;
  • 43) 0.776 506 394 840 244 682 752 × 2 = 1 + 0.553 012 789 680 489 365 504;
  • 44) 0.553 012 789 680 489 365 504 × 2 = 1 + 0.106 025 579 360 978 731 008;
  • 45) 0.106 025 579 360 978 731 008 × 2 = 0 + 0.212 051 158 721 957 462 016;
  • 46) 0.212 051 158 721 957 462 016 × 2 = 0 + 0.424 102 317 443 914 924 032;
  • 47) 0.424 102 317 443 914 924 032 × 2 = 0 + 0.848 204 634 887 829 848 064;
  • 48) 0.848 204 634 887 829 848 064 × 2 = 1 + 0.696 409 269 775 659 696 128;
  • 49) 0.696 409 269 775 659 696 128 × 2 = 1 + 0.392 818 539 551 319 392 256;
  • 50) 0.392 818 539 551 319 392 256 × 2 = 0 + 0.785 637 079 102 638 784 512;
  • 51) 0.785 637 079 102 638 784 512 × 2 = 1 + 0.571 274 158 205 277 569 024;
  • 52) 0.571 274 158 205 277 569 024 × 2 = 1 + 0.142 548 316 410 555 138 048;
  • 53) 0.142 548 316 410 555 138 048 × 2 = 0 + 0.285 096 632 821 110 276 096;
  • 54) 0.285 096 632 821 110 276 096 × 2 = 0 + 0.570 193 265 642 220 552 192;
  • 55) 0.570 193 265 642 220 552 192 × 2 = 1 + 0.140 386 531 284 441 104 384;
  • 56) 0.140 386 531 284 441 104 384 × 2 = 0 + 0.280 773 062 568 882 208 768;
  • 57) 0.280 773 062 568 882 208 768 × 2 = 0 + 0.561 546 125 137 764 417 536;
  • 58) 0.561 546 125 137 764 417 536 × 2 = 1 + 0.123 092 250 275 528 835 072;
  • 59) 0.123 092 250 275 528 835 072 × 2 = 0 + 0.246 184 500 551 057 670 144;
  • 60) 0.246 184 500 551 057 670 144 × 2 = 0 + 0.492 369 001 102 115 340 288;
  • 61) 0.492 369 001 102 115 340 288 × 2 = 0 + 0.984 738 002 204 230 680 576;
  • 62) 0.984 738 002 204 230 680 576 × 2 = 1 + 0.969 476 004 408 461 361 152;
  • 63) 0.969 476 004 408 461 361 152 × 2 = 1 + 0.938 952 008 816 922 722 304;
  • 64) 0.938 952 008 816 922 722 304 × 2 = 1 + 0.877 904 017 633 845 444 608;
  • 65) 0.877 904 017 633 845 444 608 × 2 = 1 + 0.755 808 035 267 690 889 216;
  • 66) 0.755 808 035 267 690 889 216 × 2 = 1 + 0.511 616 070 535 381 778 432;
  • 67) 0.511 616 070 535 381 778 432 × 2 = 1 + 0.023 232 141 070 763 556 864;
  • 68) 0.023 232 141 070 763 556 864 × 2 = 0 + 0.046 464 282 141 527 113 728;
  • 69) 0.046 464 282 141 527 113 728 × 2 = 0 + 0.092 928 564 283 054 227 456;
  • 70) 0.092 928 564 283 054 227 456 × 2 = 0 + 0.185 857 128 566 108 454 912;
  • 71) 0.185 857 128 566 108 454 912 × 2 = 0 + 0.371 714 257 132 216 909 824;
  • 72) 0.371 714 257 132 216 909 824 × 2 = 0 + 0.743 428 514 264 433 819 648;
  • 73) 0.743 428 514 264 433 819 648 × 2 = 1 + 0.486 857 028 528 867 639 296;
  • 74) 0.486 857 028 528 867 639 296 × 2 = 0 + 0.973 714 057 057 735 278 592;
  • 75) 0.973 714 057 057 735 278 592 × 2 = 1 + 0.947 428 114 115 470 557 184;
  • 76) 0.947 428 114 115 470 557 184 × 2 = 1 + 0.894 856 228 230 941 114 368;
  • 77) 0.894 856 228 230 941 114 368 × 2 = 1 + 0.789 712 456 461 882 228 736;
  • 78) 0.789 712 456 461 882 228 736 × 2 = 1 + 0.579 424 912 923 764 457 472;
  • 79) 0.579 424 912 923 764 457 472 × 2 = 1 + 0.158 849 825 847 528 914 944;
  • 80) 0.158 849 825 847 528 914 944 × 2 = 0 + 0.317 699 651 695 057 829 888;
  • 81) 0.317 699 651 695 057 829 888 × 2 = 0 + 0.635 399 303 390 115 659 776;
  • 82) 0.635 399 303 390 115 659 776 × 2 = 1 + 0.270 798 606 780 231 319 552;
  • 83) 0.270 798 606 780 231 319 552 × 2 = 0 + 0.541 597 213 560 462 639 104;
  • 84) 0.541 597 213 560 462 639 104 × 2 = 1 + 0.083 194 427 120 925 278 208;
  • 85) 0.083 194 427 120 925 278 208 × 2 = 0 + 0.166 388 854 241 850 556 416;
  • 86) 0.166 388 854 241 850 556 416 × 2 = 0 + 0.332 777 708 483 701 112 832;
  • 87) 0.332 777 708 483 701 112 832 × 2 = 0 + 0.665 555 416 967 402 225 664;
  • 88) 0.665 555 416 967 402 225 664 × 2 = 1 + 0.331 110 833 934 804 451 328;
  • 89) 0.331 110 833 934 804 451 328 × 2 = 0 + 0.662 221 667 869 608 902 656;
  • 90) 0.662 221 667 869 608 902 656 × 2 = 1 + 0.324 443 335 739 217 805 312;
  • 91) 0.324 443 335 739 217 805 312 × 2 = 0 + 0.648 886 671 478 435 610 624;
  • 92) 0.648 886 671 478 435 610 624 × 2 = 1 + 0.297 773 342 956 871 221 248;
  • 93) 0.297 773 342 956 871 221 248 × 2 = 0 + 0.595 546 685 913 742 442 496;
  • 94) 0.595 546 685 913 742 442 496 × 2 = 1 + 0.191 093 371 827 484 884 992;
  • 95) 0.191 093 371 827 484 884 992 × 2 = 0 + 0.382 186 743 654 969 769 984;

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:


0.000 000 000 000 176 557 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1110 0000 1011 1110 0101 0001 0101 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1110 0000 1011 1110 0101 0001 0101 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:


0.000 000 000 000 176 557 113(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1110 0000 1011 1110 0101 0001 0101 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1110 0000 1011 1110 0101 0001 0101 010(2) × 20 =

1.1000 1101 1001 0010 0011 1111 0000 0101 1111 0010 1000 1010 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 0011 1111 0000 0101 1111 0010 1000 1010 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 0011 1111 0000 0101 1111 0010 1000 1010 1010 =

1000 1101 1001 0010 0011 1111 0000 0101 1111 0010 1000 1010 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 0011 1111 0000 0101 1111 0010 1000 1010 1010


Decimal number -0.000 000 000 000 176 557 113 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0011 1111 0000 0101 1111 0010 1000 1010 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