-0.000 000 000 000 176 541 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 541 3(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
-0.000 000 000 000 176 541 3(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:

|-0.000 000 000 000 176 541 3| = 0.000 000 000 000 176 541 3


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: 0.000 000 000 000 176 541 3.

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) 0.000 000 000 000 176 541 3 × 2 = 0 + 0.000 000 000 000 353 082 6;
  • 2) 0.000 000 000 000 353 082 6 × 2 = 0 + 0.000 000 000 000 706 165 2;
  • 3) 0.000 000 000 000 706 165 2 × 2 = 0 + 0.000 000 000 001 412 330 4;
  • 4) 0.000 000 000 001 412 330 4 × 2 = 0 + 0.000 000 000 002 824 660 8;
  • 5) 0.000 000 000 002 824 660 8 × 2 = 0 + 0.000 000 000 005 649 321 6;
  • 6) 0.000 000 000 005 649 321 6 × 2 = 0 + 0.000 000 000 011 298 643 2;
  • 7) 0.000 000 000 011 298 643 2 × 2 = 0 + 0.000 000 000 022 597 286 4;
  • 8) 0.000 000 000 022 597 286 4 × 2 = 0 + 0.000 000 000 045 194 572 8;
  • 9) 0.000 000 000 045 194 572 8 × 2 = 0 + 0.000 000 000 090 389 145 6;
  • 10) 0.000 000 000 090 389 145 6 × 2 = 0 + 0.000 000 000 180 778 291 2;
  • 11) 0.000 000 000 180 778 291 2 × 2 = 0 + 0.000 000 000 361 556 582 4;
  • 12) 0.000 000 000 361 556 582 4 × 2 = 0 + 0.000 000 000 723 113 164 8;
  • 13) 0.000 000 000 723 113 164 8 × 2 = 0 + 0.000 000 001 446 226 329 6;
  • 14) 0.000 000 001 446 226 329 6 × 2 = 0 + 0.000 000 002 892 452 659 2;
  • 15) 0.000 000 002 892 452 659 2 × 2 = 0 + 0.000 000 005 784 905 318 4;
  • 16) 0.000 000 005 784 905 318 4 × 2 = 0 + 0.000 000 011 569 810 636 8;
  • 17) 0.000 000 011 569 810 636 8 × 2 = 0 + 0.000 000 023 139 621 273 6;
  • 18) 0.000 000 023 139 621 273 6 × 2 = 0 + 0.000 000 046 279 242 547 2;
  • 19) 0.000 000 046 279 242 547 2 × 2 = 0 + 0.000 000 092 558 485 094 4;
  • 20) 0.000 000 092 558 485 094 4 × 2 = 0 + 0.000 000 185 116 970 188 8;
  • 21) 0.000 000 185 116 970 188 8 × 2 = 0 + 0.000 000 370 233 940 377 6;
  • 22) 0.000 000 370 233 940 377 6 × 2 = 0 + 0.000 000 740 467 880 755 2;
  • 23) 0.000 000 740 467 880 755 2 × 2 = 0 + 0.000 001 480 935 761 510 4;
  • 24) 0.000 001 480 935 761 510 4 × 2 = 0 + 0.000 002 961 871 523 020 8;
  • 25) 0.000 002 961 871 523 020 8 × 2 = 0 + 0.000 005 923 743 046 041 6;
  • 26) 0.000 005 923 743 046 041 6 × 2 = 0 + 0.000 011 847 486 092 083 2;
  • 27) 0.000 011 847 486 092 083 2 × 2 = 0 + 0.000 023 694 972 184 166 4;
  • 28) 0.000 023 694 972 184 166 4 × 2 = 0 + 0.000 047 389 944 368 332 8;
  • 29) 0.000 047 389 944 368 332 8 × 2 = 0 + 0.000 094 779 888 736 665 6;
  • 30) 0.000 094 779 888 736 665 6 × 2 = 0 + 0.000 189 559 777 473 331 2;
  • 31) 0.000 189 559 777 473 331 2 × 2 = 0 + 0.000 379 119 554 946 662 4;
  • 32) 0.000 379 119 554 946 662 4 × 2 = 0 + 0.000 758 239 109 893 324 8;
  • 33) 0.000 758 239 109 893 324 8 × 2 = 0 + 0.001 516 478 219 786 649 6;
  • 34) 0.001 516 478 219 786 649 6 × 2 = 0 + 0.003 032 956 439 573 299 2;
  • 35) 0.003 032 956 439 573 299 2 × 2 = 0 + 0.006 065 912 879 146 598 4;
  • 36) 0.006 065 912 879 146 598 4 × 2 = 0 + 0.012 131 825 758 293 196 8;
  • 37) 0.012 131 825 758 293 196 8 × 2 = 0 + 0.024 263 651 516 586 393 6;
  • 38) 0.024 263 651 516 586 393 6 × 2 = 0 + 0.048 527 303 033 172 787 2;
  • 39) 0.048 527 303 033 172 787 2 × 2 = 0 + 0.097 054 606 066 345 574 4;
  • 40) 0.097 054 606 066 345 574 4 × 2 = 0 + 0.194 109 212 132 691 148 8;
  • 41) 0.194 109 212 132 691 148 8 × 2 = 0 + 0.388 218 424 265 382 297 6;
  • 42) 0.388 218 424 265 382 297 6 × 2 = 0 + 0.776 436 848 530 764 595 2;
  • 43) 0.776 436 848 530 764 595 2 × 2 = 1 + 0.552 873 697 061 529 190 4;
  • 44) 0.552 873 697 061 529 190 4 × 2 = 1 + 0.105 747 394 123 058 380 8;
  • 45) 0.105 747 394 123 058 380 8 × 2 = 0 + 0.211 494 788 246 116 761 6;
  • 46) 0.211 494 788 246 116 761 6 × 2 = 0 + 0.422 989 576 492 233 523 2;
  • 47) 0.422 989 576 492 233 523 2 × 2 = 0 + 0.845 979 152 984 467 046 4;
  • 48) 0.845 979 152 984 467 046 4 × 2 = 1 + 0.691 958 305 968 934 092 8;
  • 49) 0.691 958 305 968 934 092 8 × 2 = 1 + 0.383 916 611 937 868 185 6;
  • 50) 0.383 916 611 937 868 185 6 × 2 = 0 + 0.767 833 223 875 736 371 2;
  • 51) 0.767 833 223 875 736 371 2 × 2 = 1 + 0.535 666 447 751 472 742 4;
  • 52) 0.535 666 447 751 472 742 4 × 2 = 1 + 0.071 332 895 502 945 484 8;
  • 53) 0.071 332 895 502 945 484 8 × 2 = 0 + 0.142 665 791 005 890 969 6;
  • 54) 0.142 665 791 005 890 969 6 × 2 = 0 + 0.285 331 582 011 781 939 2;
  • 55) 0.285 331 582 011 781 939 2 × 2 = 0 + 0.570 663 164 023 563 878 4;
  • 56) 0.570 663 164 023 563 878 4 × 2 = 1 + 0.141 326 328 047 127 756 8;
  • 57) 0.141 326 328 047 127 756 8 × 2 = 0 + 0.282 652 656 094 255 513 6;
  • 58) 0.282 652 656 094 255 513 6 × 2 = 0 + 0.565 305 312 188 511 027 2;
  • 59) 0.565 305 312 188 511 027 2 × 2 = 1 + 0.130 610 624 377 022 054 4;
  • 60) 0.130 610 624 377 022 054 4 × 2 = 0 + 0.261 221 248 754 044 108 8;
  • 61) 0.261 221 248 754 044 108 8 × 2 = 0 + 0.522 442 497 508 088 217 6;
  • 62) 0.522 442 497 508 088 217 6 × 2 = 1 + 0.044 884 995 016 176 435 2;
  • 63) 0.044 884 995 016 176 435 2 × 2 = 0 + 0.089 769 990 032 352 870 4;
  • 64) 0.089 769 990 032 352 870 4 × 2 = 0 + 0.179 539 980 064 705 740 8;
  • 65) 0.179 539 980 064 705 740 8 × 2 = 0 + 0.359 079 960 129 411 481 6;
  • 66) 0.359 079 960 129 411 481 6 × 2 = 0 + 0.718 159 920 258 822 963 2;
  • 67) 0.718 159 920 258 822 963 2 × 2 = 1 + 0.436 319 840 517 645 926 4;
  • 68) 0.436 319 840 517 645 926 4 × 2 = 0 + 0.872 639 681 035 291 852 8;
  • 69) 0.872 639 681 035 291 852 8 × 2 = 1 + 0.745 279 362 070 583 705 6;
  • 70) 0.745 279 362 070 583 705 6 × 2 = 1 + 0.490 558 724 141 167 411 2;
  • 71) 0.490 558 724 141 167 411 2 × 2 = 0 + 0.981 117 448 282 334 822 4;
  • 72) 0.981 117 448 282 334 822 4 × 2 = 1 + 0.962 234 896 564 669 644 8;
  • 73) 0.962 234 896 564 669 644 8 × 2 = 1 + 0.924 469 793 129 339 289 6;
  • 74) 0.924 469 793 129 339 289 6 × 2 = 1 + 0.848 939 586 258 678 579 2;
  • 75) 0.848 939 586 258 678 579 2 × 2 = 1 + 0.697 879 172 517 357 158 4;
  • 76) 0.697 879 172 517 357 158 4 × 2 = 1 + 0.395 758 345 034 714 316 8;
  • 77) 0.395 758 345 034 714 316 8 × 2 = 0 + 0.791 516 690 069 428 633 6;
  • 78) 0.791 516 690 069 428 633 6 × 2 = 1 + 0.583 033 380 138 857 267 2;
  • 79) 0.583 033 380 138 857 267 2 × 2 = 1 + 0.166 066 760 277 714 534 4;
  • 80) 0.166 066 760 277 714 534 4 × 2 = 0 + 0.332 133 520 555 429 068 8;
  • 81) 0.332 133 520 555 429 068 8 × 2 = 0 + 0.664 267 041 110 858 137 6;
  • 82) 0.664 267 041 110 858 137 6 × 2 = 1 + 0.328 534 082 221 716 275 2;
  • 83) 0.328 534 082 221 716 275 2 × 2 = 0 + 0.657 068 164 443 432 550 4;
  • 84) 0.657 068 164 443 432 550 4 × 2 = 1 + 0.314 136 328 886 865 100 8;
  • 85) 0.314 136 328 886 865 100 8 × 2 = 0 + 0.628 272 657 773 730 201 6;
  • 86) 0.628 272 657 773 730 201 6 × 2 = 1 + 0.256 545 315 547 460 403 2;
  • 87) 0.256 545 315 547 460 403 2 × 2 = 0 + 0.513 090 631 094 920 806 4;
  • 88) 0.513 090 631 094 920 806 4 × 2 = 1 + 0.026 181 262 189 841 612 8;
  • 89) 0.026 181 262 189 841 612 8 × 2 = 0 + 0.052 362 524 379 683 225 6;
  • 90) 0.052 362 524 379 683 225 6 × 2 = 0 + 0.104 725 048 759 366 451 2;
  • 91) 0.104 725 048 759 366 451 2 × 2 = 0 + 0.209 450 097 518 732 902 4;
  • 92) 0.209 450 097 518 732 902 4 × 2 = 0 + 0.418 900 195 037 465 804 8;
  • 93) 0.418 900 195 037 465 804 8 × 2 = 0 + 0.837 800 390 074 931 609 6;
  • 94) 0.837 800 390 074 931 609 6 × 2 = 1 + 0.675 600 780 149 863 219 2;
  • 95) 0.675 600 780 149 863 219 2 × 2 = 1 + 0.351 201 560 299 726 438 4;

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:


0.000 000 000 000 176 541 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0010 0100 0010 1101 1111 0110 0101 0101 0000 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 541 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0010 0100 0010 1101 1111 0110 0101 0101 0000 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:


0.000 000 000 000 176 541 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0010 0100 0010 1101 1111 0110 0101 0101 0000 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0010 0100 0010 1101 1111 0110 0101 0101 0000 011(2) × 20 =

1.1000 1101 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 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 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 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 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 0011 =

1000 1101 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 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 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 0011


Decimal number -0.000 000 000 000 176 541 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1001 0010 0001 0110 1111 1011 0010 1010 1000 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