-0.000 000 000 000 176 557 438 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 438(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
-0.000 000 000 000 176 557 438(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:

|-0.000 000 000 000 176 557 438| = 0.000 000 000 000 176 557 438


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: 0.000 000 000 000 176 557 438.

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) 0.000 000 000 000 176 557 438 × 2 = 0 + 0.000 000 000 000 353 114 876;
  • 2) 0.000 000 000 000 353 114 876 × 2 = 0 + 0.000 000 000 000 706 229 752;
  • 3) 0.000 000 000 000 706 229 752 × 2 = 0 + 0.000 000 000 001 412 459 504;
  • 4) 0.000 000 000 001 412 459 504 × 2 = 0 + 0.000 000 000 002 824 919 008;
  • 5) 0.000 000 000 002 824 919 008 × 2 = 0 + 0.000 000 000 005 649 838 016;
  • 6) 0.000 000 000 005 649 838 016 × 2 = 0 + 0.000 000 000 011 299 676 032;
  • 7) 0.000 000 000 011 299 676 032 × 2 = 0 + 0.000 000 000 022 599 352 064;
  • 8) 0.000 000 000 022 599 352 064 × 2 = 0 + 0.000 000 000 045 198 704 128;
  • 9) 0.000 000 000 045 198 704 128 × 2 = 0 + 0.000 000 000 090 397 408 256;
  • 10) 0.000 000 000 090 397 408 256 × 2 = 0 + 0.000 000 000 180 794 816 512;
  • 11) 0.000 000 000 180 794 816 512 × 2 = 0 + 0.000 000 000 361 589 633 024;
  • 12) 0.000 000 000 361 589 633 024 × 2 = 0 + 0.000 000 000 723 179 266 048;
  • 13) 0.000 000 000 723 179 266 048 × 2 = 0 + 0.000 000 001 446 358 532 096;
  • 14) 0.000 000 001 446 358 532 096 × 2 = 0 + 0.000 000 002 892 717 064 192;
  • 15) 0.000 000 002 892 717 064 192 × 2 = 0 + 0.000 000 005 785 434 128 384;
  • 16) 0.000 000 005 785 434 128 384 × 2 = 0 + 0.000 000 011 570 868 256 768;
  • 17) 0.000 000 011 570 868 256 768 × 2 = 0 + 0.000 000 023 141 736 513 536;
  • 18) 0.000 000 023 141 736 513 536 × 2 = 0 + 0.000 000 046 283 473 027 072;
  • 19) 0.000 000 046 283 473 027 072 × 2 = 0 + 0.000 000 092 566 946 054 144;
  • 20) 0.000 000 092 566 946 054 144 × 2 = 0 + 0.000 000 185 133 892 108 288;
  • 21) 0.000 000 185 133 892 108 288 × 2 = 0 + 0.000 000 370 267 784 216 576;
  • 22) 0.000 000 370 267 784 216 576 × 2 = 0 + 0.000 000 740 535 568 433 152;
  • 23) 0.000 000 740 535 568 433 152 × 2 = 0 + 0.000 001 481 071 136 866 304;
  • 24) 0.000 001 481 071 136 866 304 × 2 = 0 + 0.000 002 962 142 273 732 608;
  • 25) 0.000 002 962 142 273 732 608 × 2 = 0 + 0.000 005 924 284 547 465 216;
  • 26) 0.000 005 924 284 547 465 216 × 2 = 0 + 0.000 011 848 569 094 930 432;
  • 27) 0.000 011 848 569 094 930 432 × 2 = 0 + 0.000 023 697 138 189 860 864;
  • 28) 0.000 023 697 138 189 860 864 × 2 = 0 + 0.000 047 394 276 379 721 728;
  • 29) 0.000 047 394 276 379 721 728 × 2 = 0 + 0.000 094 788 552 759 443 456;
  • 30) 0.000 094 788 552 759 443 456 × 2 = 0 + 0.000 189 577 105 518 886 912;
  • 31) 0.000 189 577 105 518 886 912 × 2 = 0 + 0.000 379 154 211 037 773 824;
  • 32) 0.000 379 154 211 037 773 824 × 2 = 0 + 0.000 758 308 422 075 547 648;
  • 33) 0.000 758 308 422 075 547 648 × 2 = 0 + 0.001 516 616 844 151 095 296;
  • 34) 0.001 516 616 844 151 095 296 × 2 = 0 + 0.003 033 233 688 302 190 592;
  • 35) 0.003 033 233 688 302 190 592 × 2 = 0 + 0.006 066 467 376 604 381 184;
  • 36) 0.006 066 467 376 604 381 184 × 2 = 0 + 0.012 132 934 753 208 762 368;
  • 37) 0.012 132 934 753 208 762 368 × 2 = 0 + 0.024 265 869 506 417 524 736;
  • 38) 0.024 265 869 506 417 524 736 × 2 = 0 + 0.048 531 739 012 835 049 472;
  • 39) 0.048 531 739 012 835 049 472 × 2 = 0 + 0.097 063 478 025 670 098 944;
  • 40) 0.097 063 478 025 670 098 944 × 2 = 0 + 0.194 126 956 051 340 197 888;
  • 41) 0.194 126 956 051 340 197 888 × 2 = 0 + 0.388 253 912 102 680 395 776;
  • 42) 0.388 253 912 102 680 395 776 × 2 = 0 + 0.776 507 824 205 360 791 552;
  • 43) 0.776 507 824 205 360 791 552 × 2 = 1 + 0.553 015 648 410 721 583 104;
  • 44) 0.553 015 648 410 721 583 104 × 2 = 1 + 0.106 031 296 821 443 166 208;
  • 45) 0.106 031 296 821 443 166 208 × 2 = 0 + 0.212 062 593 642 886 332 416;
  • 46) 0.212 062 593 642 886 332 416 × 2 = 0 + 0.424 125 187 285 772 664 832;
  • 47) 0.424 125 187 285 772 664 832 × 2 = 0 + 0.848 250 374 571 545 329 664;
  • 48) 0.848 250 374 571 545 329 664 × 2 = 1 + 0.696 500 749 143 090 659 328;
  • 49) 0.696 500 749 143 090 659 328 × 2 = 1 + 0.393 001 498 286 181 318 656;
  • 50) 0.393 001 498 286 181 318 656 × 2 = 0 + 0.786 002 996 572 362 637 312;
  • 51) 0.786 002 996 572 362 637 312 × 2 = 1 + 0.572 005 993 144 725 274 624;
  • 52) 0.572 005 993 144 725 274 624 × 2 = 1 + 0.144 011 986 289 450 549 248;
  • 53) 0.144 011 986 289 450 549 248 × 2 = 0 + 0.288 023 972 578 901 098 496;
  • 54) 0.288 023 972 578 901 098 496 × 2 = 0 + 0.576 047 945 157 802 196 992;
  • 55) 0.576 047 945 157 802 196 992 × 2 = 1 + 0.152 095 890 315 604 393 984;
  • 56) 0.152 095 890 315 604 393 984 × 2 = 0 + 0.304 191 780 631 208 787 968;
  • 57) 0.304 191 780 631 208 787 968 × 2 = 0 + 0.608 383 561 262 417 575 936;
  • 58) 0.608 383 561 262 417 575 936 × 2 = 1 + 0.216 767 122 524 835 151 872;
  • 59) 0.216 767 122 524 835 151 872 × 2 = 0 + 0.433 534 245 049 670 303 744;
  • 60) 0.433 534 245 049 670 303 744 × 2 = 0 + 0.867 068 490 099 340 607 488;
  • 61) 0.867 068 490 099 340 607 488 × 2 = 1 + 0.734 136 980 198 681 214 976;
  • 62) 0.734 136 980 198 681 214 976 × 2 = 1 + 0.468 273 960 397 362 429 952;
  • 63) 0.468 273 960 397 362 429 952 × 2 = 0 + 0.936 547 920 794 724 859 904;
  • 64) 0.936 547 920 794 724 859 904 × 2 = 1 + 0.873 095 841 589 449 719 808;
  • 65) 0.873 095 841 589 449 719 808 × 2 = 1 + 0.746 191 683 178 899 439 616;
  • 66) 0.746 191 683 178 899 439 616 × 2 = 1 + 0.492 383 366 357 798 879 232;
  • 67) 0.492 383 366 357 798 879 232 × 2 = 0 + 0.984 766 732 715 597 758 464;
  • 68) 0.984 766 732 715 597 758 464 × 2 = 1 + 0.969 533 465 431 195 516 928;
  • 69) 0.969 533 465 431 195 516 928 × 2 = 1 + 0.939 066 930 862 391 033 856;
  • 70) 0.939 066 930 862 391 033 856 × 2 = 1 + 0.878 133 861 724 782 067 712;
  • 71) 0.878 133 861 724 782 067 712 × 2 = 1 + 0.756 267 723 449 564 135 424;
  • 72) 0.756 267 723 449 564 135 424 × 2 = 1 + 0.512 535 446 899 128 270 848;
  • 73) 0.512 535 446 899 128 270 848 × 2 = 1 + 0.025 070 893 798 256 541 696;
  • 74) 0.025 070 893 798 256 541 696 × 2 = 0 + 0.050 141 787 596 513 083 392;
  • 75) 0.050 141 787 596 513 083 392 × 2 = 0 + 0.100 283 575 193 026 166 784;
  • 76) 0.100 283 575 193 026 166 784 × 2 = 0 + 0.200 567 150 386 052 333 568;
  • 77) 0.200 567 150 386 052 333 568 × 2 = 0 + 0.401 134 300 772 104 667 136;
  • 78) 0.401 134 300 772 104 667 136 × 2 = 0 + 0.802 268 601 544 209 334 272;
  • 79) 0.802 268 601 544 209 334 272 × 2 = 1 + 0.604 537 203 088 418 668 544;
  • 80) 0.604 537 203 088 418 668 544 × 2 = 1 + 0.209 074 406 176 837 337 088;
  • 81) 0.209 074 406 176 837 337 088 × 2 = 0 + 0.418 148 812 353 674 674 176;
  • 82) 0.418 148 812 353 674 674 176 × 2 = 0 + 0.836 297 624 707 349 348 352;
  • 83) 0.836 297 624 707 349 348 352 × 2 = 1 + 0.672 595 249 414 698 696 704;
  • 84) 0.672 595 249 414 698 696 704 × 2 = 1 + 0.345 190 498 829 397 393 408;
  • 85) 0.345 190 498 829 397 393 408 × 2 = 0 + 0.690 380 997 658 794 786 816;
  • 86) 0.690 380 997 658 794 786 816 × 2 = 1 + 0.380 761 995 317 589 573 632;
  • 87) 0.380 761 995 317 589 573 632 × 2 = 0 + 0.761 523 990 635 179 147 264;
  • 88) 0.761 523 990 635 179 147 264 × 2 = 1 + 0.523 047 981 270 358 294 528;
  • 89) 0.523 047 981 270 358 294 528 × 2 = 1 + 0.046 095 962 540 716 589 056;
  • 90) 0.046 095 962 540 716 589 056 × 2 = 0 + 0.092 191 925 081 433 178 112;
  • 91) 0.092 191 925 081 433 178 112 × 2 = 0 + 0.184 383 850 162 866 356 224;
  • 92) 0.184 383 850 162 866 356 224 × 2 = 0 + 0.368 767 700 325 732 712 448;
  • 93) 0.368 767 700 325 732 712 448 × 2 = 0 + 0.737 535 400 651 465 424 896;
  • 94) 0.737 535 400 651 465 424 896 × 2 = 1 + 0.475 070 801 302 930 849 792;
  • 95) 0.475 070 801 302 930 849 792 × 2 = 0 + 0.950 141 602 605 861 699 584;

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:


0.000 000 000 000 176 557 438(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1101 1111 1000 0011 0011 0101 1000 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 438(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1101 1111 1000 0011 0011 0101 1000 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:


0.000 000 000 000 176 557 438(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1101 1111 1000 0011 0011 0101 1000 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1101 1111 1000 0011 0011 0101 1000 010(2) × 20 =

1.1000 1101 1001 0010 0110 1110 1111 1100 0001 1001 1010 1100 0010(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 1110 1111 1100 0001 1001 1010 1100 0010


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 1110 1111 1100 0001 1001 1010 1100 0010 =

1000 1101 1001 0010 0110 1110 1111 1100 0001 1001 1010 1100 0010


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 1110 1111 1100 0001 1001 1010 1100 0010


Decimal number -0.000 000 000 000 176 557 438 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1110 1111 1100 0001 1001 1010 1100 0010

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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