-0.000 000 000 000 176 558 14 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 558 14(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
-0.000 000 000 000 176 558 14(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:

|-0.000 000 000 000 176 558 14| = 0.000 000 000 000 176 558 14


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: 0.000 000 000 000 176 558 14.

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) 0.000 000 000 000 176 558 14 × 2 = 0 + 0.000 000 000 000 353 116 28;
  • 2) 0.000 000 000 000 353 116 28 × 2 = 0 + 0.000 000 000 000 706 232 56;
  • 3) 0.000 000 000 000 706 232 56 × 2 = 0 + 0.000 000 000 001 412 465 12;
  • 4) 0.000 000 000 001 412 465 12 × 2 = 0 + 0.000 000 000 002 824 930 24;
  • 5) 0.000 000 000 002 824 930 24 × 2 = 0 + 0.000 000 000 005 649 860 48;
  • 6) 0.000 000 000 005 649 860 48 × 2 = 0 + 0.000 000 000 011 299 720 96;
  • 7) 0.000 000 000 011 299 720 96 × 2 = 0 + 0.000 000 000 022 599 441 92;
  • 8) 0.000 000 000 022 599 441 92 × 2 = 0 + 0.000 000 000 045 198 883 84;
  • 9) 0.000 000 000 045 198 883 84 × 2 = 0 + 0.000 000 000 090 397 767 68;
  • 10) 0.000 000 000 090 397 767 68 × 2 = 0 + 0.000 000 000 180 795 535 36;
  • 11) 0.000 000 000 180 795 535 36 × 2 = 0 + 0.000 000 000 361 591 070 72;
  • 12) 0.000 000 000 361 591 070 72 × 2 = 0 + 0.000 000 000 723 182 141 44;
  • 13) 0.000 000 000 723 182 141 44 × 2 = 0 + 0.000 000 001 446 364 282 88;
  • 14) 0.000 000 001 446 364 282 88 × 2 = 0 + 0.000 000 002 892 728 565 76;
  • 15) 0.000 000 002 892 728 565 76 × 2 = 0 + 0.000 000 005 785 457 131 52;
  • 16) 0.000 000 005 785 457 131 52 × 2 = 0 + 0.000 000 011 570 914 263 04;
  • 17) 0.000 000 011 570 914 263 04 × 2 = 0 + 0.000 000 023 141 828 526 08;
  • 18) 0.000 000 023 141 828 526 08 × 2 = 0 + 0.000 000 046 283 657 052 16;
  • 19) 0.000 000 046 283 657 052 16 × 2 = 0 + 0.000 000 092 567 314 104 32;
  • 20) 0.000 000 092 567 314 104 32 × 2 = 0 + 0.000 000 185 134 628 208 64;
  • 21) 0.000 000 185 134 628 208 64 × 2 = 0 + 0.000 000 370 269 256 417 28;
  • 22) 0.000 000 370 269 256 417 28 × 2 = 0 + 0.000 000 740 538 512 834 56;
  • 23) 0.000 000 740 538 512 834 56 × 2 = 0 + 0.000 001 481 077 025 669 12;
  • 24) 0.000 001 481 077 025 669 12 × 2 = 0 + 0.000 002 962 154 051 338 24;
  • 25) 0.000 002 962 154 051 338 24 × 2 = 0 + 0.000 005 924 308 102 676 48;
  • 26) 0.000 005 924 308 102 676 48 × 2 = 0 + 0.000 011 848 616 205 352 96;
  • 27) 0.000 011 848 616 205 352 96 × 2 = 0 + 0.000 023 697 232 410 705 92;
  • 28) 0.000 023 697 232 410 705 92 × 2 = 0 + 0.000 047 394 464 821 411 84;
  • 29) 0.000 047 394 464 821 411 84 × 2 = 0 + 0.000 094 788 929 642 823 68;
  • 30) 0.000 094 788 929 642 823 68 × 2 = 0 + 0.000 189 577 859 285 647 36;
  • 31) 0.000 189 577 859 285 647 36 × 2 = 0 + 0.000 379 155 718 571 294 72;
  • 32) 0.000 379 155 718 571 294 72 × 2 = 0 + 0.000 758 311 437 142 589 44;
  • 33) 0.000 758 311 437 142 589 44 × 2 = 0 + 0.001 516 622 874 285 178 88;
  • 34) 0.001 516 622 874 285 178 88 × 2 = 0 + 0.003 033 245 748 570 357 76;
  • 35) 0.003 033 245 748 570 357 76 × 2 = 0 + 0.006 066 491 497 140 715 52;
  • 36) 0.006 066 491 497 140 715 52 × 2 = 0 + 0.012 132 982 994 281 431 04;
  • 37) 0.012 132 982 994 281 431 04 × 2 = 0 + 0.024 265 965 988 562 862 08;
  • 38) 0.024 265 965 988 562 862 08 × 2 = 0 + 0.048 531 931 977 125 724 16;
  • 39) 0.048 531 931 977 125 724 16 × 2 = 0 + 0.097 063 863 954 251 448 32;
  • 40) 0.097 063 863 954 251 448 32 × 2 = 0 + 0.194 127 727 908 502 896 64;
  • 41) 0.194 127 727 908 502 896 64 × 2 = 0 + 0.388 255 455 817 005 793 28;
  • 42) 0.388 255 455 817 005 793 28 × 2 = 0 + 0.776 510 911 634 011 586 56;
  • 43) 0.776 510 911 634 011 586 56 × 2 = 1 + 0.553 021 823 268 023 173 12;
  • 44) 0.553 021 823 268 023 173 12 × 2 = 1 + 0.106 043 646 536 046 346 24;
  • 45) 0.106 043 646 536 046 346 24 × 2 = 0 + 0.212 087 293 072 092 692 48;
  • 46) 0.212 087 293 072 092 692 48 × 2 = 0 + 0.424 174 586 144 185 384 96;
  • 47) 0.424 174 586 144 185 384 96 × 2 = 0 + 0.848 349 172 288 370 769 92;
  • 48) 0.848 349 172 288 370 769 92 × 2 = 1 + 0.696 698 344 576 741 539 84;
  • 49) 0.696 698 344 576 741 539 84 × 2 = 1 + 0.393 396 689 153 483 079 68;
  • 50) 0.393 396 689 153 483 079 68 × 2 = 0 + 0.786 793 378 306 966 159 36;
  • 51) 0.786 793 378 306 966 159 36 × 2 = 1 + 0.573 586 756 613 932 318 72;
  • 52) 0.573 586 756 613 932 318 72 × 2 = 1 + 0.147 173 513 227 864 637 44;
  • 53) 0.147 173 513 227 864 637 44 × 2 = 0 + 0.294 347 026 455 729 274 88;
  • 54) 0.294 347 026 455 729 274 88 × 2 = 0 + 0.588 694 052 911 458 549 76;
  • 55) 0.588 694 052 911 458 549 76 × 2 = 1 + 0.177 388 105 822 917 099 52;
  • 56) 0.177 388 105 822 917 099 52 × 2 = 0 + 0.354 776 211 645 834 199 04;
  • 57) 0.354 776 211 645 834 199 04 × 2 = 0 + 0.709 552 423 291 668 398 08;
  • 58) 0.709 552 423 291 668 398 08 × 2 = 1 + 0.419 104 846 583 336 796 16;
  • 59) 0.419 104 846 583 336 796 16 × 2 = 0 + 0.838 209 693 166 673 592 32;
  • 60) 0.838 209 693 166 673 592 32 × 2 = 1 + 0.676 419 386 333 347 184 64;
  • 61) 0.676 419 386 333 347 184 64 × 2 = 1 + 0.352 838 772 666 694 369 28;
  • 62) 0.352 838 772 666 694 369 28 × 2 = 0 + 0.705 677 545 333 388 738 56;
  • 63) 0.705 677 545 333 388 738 56 × 2 = 1 + 0.411 355 090 666 777 477 12;
  • 64) 0.411 355 090 666 777 477 12 × 2 = 0 + 0.822 710 181 333 554 954 24;
  • 65) 0.822 710 181 333 554 954 24 × 2 = 1 + 0.645 420 362 667 109 908 48;
  • 66) 0.645 420 362 667 109 908 48 × 2 = 1 + 0.290 840 725 334 219 816 96;
  • 67) 0.290 840 725 334 219 816 96 × 2 = 0 + 0.581 681 450 668 439 633 92;
  • 68) 0.581 681 450 668 439 633 92 × 2 = 1 + 0.163 362 901 336 879 267 84;
  • 69) 0.163 362 901 336 879 267 84 × 2 = 0 + 0.326 725 802 673 758 535 68;
  • 70) 0.326 725 802 673 758 535 68 × 2 = 0 + 0.653 451 605 347 517 071 36;
  • 71) 0.653 451 605 347 517 071 36 × 2 = 1 + 0.306 903 210 695 034 142 72;
  • 72) 0.306 903 210 695 034 142 72 × 2 = 0 + 0.613 806 421 390 068 285 44;
  • 73) 0.613 806 421 390 068 285 44 × 2 = 1 + 0.227 612 842 780 136 570 88;
  • 74) 0.227 612 842 780 136 570 88 × 2 = 0 + 0.455 225 685 560 273 141 76;
  • 75) 0.455 225 685 560 273 141 76 × 2 = 0 + 0.910 451 371 120 546 283 52;
  • 76) 0.910 451 371 120 546 283 52 × 2 = 1 + 0.820 902 742 241 092 567 04;
  • 77) 0.820 902 742 241 092 567 04 × 2 = 1 + 0.641 805 484 482 185 134 08;
  • 78) 0.641 805 484 482 185 134 08 × 2 = 1 + 0.283 610 968 964 370 268 16;
  • 79) 0.283 610 968 964 370 268 16 × 2 = 0 + 0.567 221 937 928 740 536 32;
  • 80) 0.567 221 937 928 740 536 32 × 2 = 1 + 0.134 443 875 857 481 072 64;
  • 81) 0.134 443 875 857 481 072 64 × 2 = 0 + 0.268 887 751 714 962 145 28;
  • 82) 0.268 887 751 714 962 145 28 × 2 = 0 + 0.537 775 503 429 924 290 56;
  • 83) 0.537 775 503 429 924 290 56 × 2 = 1 + 0.075 551 006 859 848 581 12;
  • 84) 0.075 551 006 859 848 581 12 × 2 = 0 + 0.151 102 013 719 697 162 24;
  • 85) 0.151 102 013 719 697 162 24 × 2 = 0 + 0.302 204 027 439 394 324 48;
  • 86) 0.302 204 027 439 394 324 48 × 2 = 0 + 0.604 408 054 878 788 648 96;
  • 87) 0.604 408 054 878 788 648 96 × 2 = 1 + 0.208 816 109 757 577 297 92;
  • 88) 0.208 816 109 757 577 297 92 × 2 = 0 + 0.417 632 219 515 154 595 84;
  • 89) 0.417 632 219 515 154 595 84 × 2 = 0 + 0.835 264 439 030 309 191 68;
  • 90) 0.835 264 439 030 309 191 68 × 2 = 1 + 0.670 528 878 060 618 383 36;
  • 91) 0.670 528 878 060 618 383 36 × 2 = 1 + 0.341 057 756 121 236 766 72;
  • 92) 0.341 057 756 121 236 766 72 × 2 = 0 + 0.682 115 512 242 473 533 44;
  • 93) 0.682 115 512 242 473 533 44 × 2 = 1 + 0.364 231 024 484 947 066 88;
  • 94) 0.364 231 024 484 947 066 88 × 2 = 0 + 0.728 462 048 969 894 133 76;
  • 95) 0.728 462 048 969 894 133 76 × 2 = 1 + 0.456 924 097 939 788 267 52;

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:


0.000 000 000 000 176 558 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1010 1101 0010 1001 1101 0010 0010 0110 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 558 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1010 1101 0010 1001 1101 0010 0010 0110 101(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:


0.000 000 000 000 176 558 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1010 1101 0010 1001 1101 0010 0010 0110 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1010 1101 0010 1001 1101 0010 0010 0110 101(2) × 20 =

1.1000 1101 1001 0010 1101 0110 1001 0100 1110 1001 0001 0011 0101(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 1101 0110 1001 0100 1110 1001 0001 0011 0101


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 1101 0110 1001 0100 1110 1001 0001 0011 0101 =

1000 1101 1001 0010 1101 0110 1001 0100 1110 1001 0001 0011 0101


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 1101 0110 1001 0100 1110 1001 0001 0011 0101


Decimal number -0.000 000 000 000 176 558 14 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1101 0110 1001 0100 1110 1001 0001 0011 0101

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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