-0.000 000 000 000 176 544 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 544 2(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
-0.000 000 000 000 176 544 2(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:

|-0.000 000 000 000 176 544 2| = 0.000 000 000 000 176 544 2


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: 0.000 000 000 000 176 544 2.

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) 0.000 000 000 000 176 544 2 × 2 = 0 + 0.000 000 000 000 353 088 4;
  • 2) 0.000 000 000 000 353 088 4 × 2 = 0 + 0.000 000 000 000 706 176 8;
  • 3) 0.000 000 000 000 706 176 8 × 2 = 0 + 0.000 000 000 001 412 353 6;
  • 4) 0.000 000 000 001 412 353 6 × 2 = 0 + 0.000 000 000 002 824 707 2;
  • 5) 0.000 000 000 002 824 707 2 × 2 = 0 + 0.000 000 000 005 649 414 4;
  • 6) 0.000 000 000 005 649 414 4 × 2 = 0 + 0.000 000 000 011 298 828 8;
  • 7) 0.000 000 000 011 298 828 8 × 2 = 0 + 0.000 000 000 022 597 657 6;
  • 8) 0.000 000 000 022 597 657 6 × 2 = 0 + 0.000 000 000 045 195 315 2;
  • 9) 0.000 000 000 045 195 315 2 × 2 = 0 + 0.000 000 000 090 390 630 4;
  • 10) 0.000 000 000 090 390 630 4 × 2 = 0 + 0.000 000 000 180 781 260 8;
  • 11) 0.000 000 000 180 781 260 8 × 2 = 0 + 0.000 000 000 361 562 521 6;
  • 12) 0.000 000 000 361 562 521 6 × 2 = 0 + 0.000 000 000 723 125 043 2;
  • 13) 0.000 000 000 723 125 043 2 × 2 = 0 + 0.000 000 001 446 250 086 4;
  • 14) 0.000 000 001 446 250 086 4 × 2 = 0 + 0.000 000 002 892 500 172 8;
  • 15) 0.000 000 002 892 500 172 8 × 2 = 0 + 0.000 000 005 785 000 345 6;
  • 16) 0.000 000 005 785 000 345 6 × 2 = 0 + 0.000 000 011 570 000 691 2;
  • 17) 0.000 000 011 570 000 691 2 × 2 = 0 + 0.000 000 023 140 001 382 4;
  • 18) 0.000 000 023 140 001 382 4 × 2 = 0 + 0.000 000 046 280 002 764 8;
  • 19) 0.000 000 046 280 002 764 8 × 2 = 0 + 0.000 000 092 560 005 529 6;
  • 20) 0.000 000 092 560 005 529 6 × 2 = 0 + 0.000 000 185 120 011 059 2;
  • 21) 0.000 000 185 120 011 059 2 × 2 = 0 + 0.000 000 370 240 022 118 4;
  • 22) 0.000 000 370 240 022 118 4 × 2 = 0 + 0.000 000 740 480 044 236 8;
  • 23) 0.000 000 740 480 044 236 8 × 2 = 0 + 0.000 001 480 960 088 473 6;
  • 24) 0.000 001 480 960 088 473 6 × 2 = 0 + 0.000 002 961 920 176 947 2;
  • 25) 0.000 002 961 920 176 947 2 × 2 = 0 + 0.000 005 923 840 353 894 4;
  • 26) 0.000 005 923 840 353 894 4 × 2 = 0 + 0.000 011 847 680 707 788 8;
  • 27) 0.000 011 847 680 707 788 8 × 2 = 0 + 0.000 023 695 361 415 577 6;
  • 28) 0.000 023 695 361 415 577 6 × 2 = 0 + 0.000 047 390 722 831 155 2;
  • 29) 0.000 047 390 722 831 155 2 × 2 = 0 + 0.000 094 781 445 662 310 4;
  • 30) 0.000 094 781 445 662 310 4 × 2 = 0 + 0.000 189 562 891 324 620 8;
  • 31) 0.000 189 562 891 324 620 8 × 2 = 0 + 0.000 379 125 782 649 241 6;
  • 32) 0.000 379 125 782 649 241 6 × 2 = 0 + 0.000 758 251 565 298 483 2;
  • 33) 0.000 758 251 565 298 483 2 × 2 = 0 + 0.001 516 503 130 596 966 4;
  • 34) 0.001 516 503 130 596 966 4 × 2 = 0 + 0.003 033 006 261 193 932 8;
  • 35) 0.003 033 006 261 193 932 8 × 2 = 0 + 0.006 066 012 522 387 865 6;
  • 36) 0.006 066 012 522 387 865 6 × 2 = 0 + 0.012 132 025 044 775 731 2;
  • 37) 0.012 132 025 044 775 731 2 × 2 = 0 + 0.024 264 050 089 551 462 4;
  • 38) 0.024 264 050 089 551 462 4 × 2 = 0 + 0.048 528 100 179 102 924 8;
  • 39) 0.048 528 100 179 102 924 8 × 2 = 0 + 0.097 056 200 358 205 849 6;
  • 40) 0.097 056 200 358 205 849 6 × 2 = 0 + 0.194 112 400 716 411 699 2;
  • 41) 0.194 112 400 716 411 699 2 × 2 = 0 + 0.388 224 801 432 823 398 4;
  • 42) 0.388 224 801 432 823 398 4 × 2 = 0 + 0.776 449 602 865 646 796 8;
  • 43) 0.776 449 602 865 646 796 8 × 2 = 1 + 0.552 899 205 731 293 593 6;
  • 44) 0.552 899 205 731 293 593 6 × 2 = 1 + 0.105 798 411 462 587 187 2;
  • 45) 0.105 798 411 462 587 187 2 × 2 = 0 + 0.211 596 822 925 174 374 4;
  • 46) 0.211 596 822 925 174 374 4 × 2 = 0 + 0.423 193 645 850 348 748 8;
  • 47) 0.423 193 645 850 348 748 8 × 2 = 0 + 0.846 387 291 700 697 497 6;
  • 48) 0.846 387 291 700 697 497 6 × 2 = 1 + 0.692 774 583 401 394 995 2;
  • 49) 0.692 774 583 401 394 995 2 × 2 = 1 + 0.385 549 166 802 789 990 4;
  • 50) 0.385 549 166 802 789 990 4 × 2 = 0 + 0.771 098 333 605 579 980 8;
  • 51) 0.771 098 333 605 579 980 8 × 2 = 1 + 0.542 196 667 211 159 961 6;
  • 52) 0.542 196 667 211 159 961 6 × 2 = 1 + 0.084 393 334 422 319 923 2;
  • 53) 0.084 393 334 422 319 923 2 × 2 = 0 + 0.168 786 668 844 639 846 4;
  • 54) 0.168 786 668 844 639 846 4 × 2 = 0 + 0.337 573 337 689 279 692 8;
  • 55) 0.337 573 337 689 279 692 8 × 2 = 0 + 0.675 146 675 378 559 385 6;
  • 56) 0.675 146 675 378 559 385 6 × 2 = 1 + 0.350 293 350 757 118 771 2;
  • 57) 0.350 293 350 757 118 771 2 × 2 = 0 + 0.700 586 701 514 237 542 4;
  • 58) 0.700 586 701 514 237 542 4 × 2 = 1 + 0.401 173 403 028 475 084 8;
  • 59) 0.401 173 403 028 475 084 8 × 2 = 0 + 0.802 346 806 056 950 169 6;
  • 60) 0.802 346 806 056 950 169 6 × 2 = 1 + 0.604 693 612 113 900 339 2;
  • 61) 0.604 693 612 113 900 339 2 × 2 = 1 + 0.209 387 224 227 800 678 4;
  • 62) 0.209 387 224 227 800 678 4 × 2 = 0 + 0.418 774 448 455 601 356 8;
  • 63) 0.418 774 448 455 601 356 8 × 2 = 0 + 0.837 548 896 911 202 713 6;
  • 64) 0.837 548 896 911 202 713 6 × 2 = 1 + 0.675 097 793 822 405 427 2;
  • 65) 0.675 097 793 822 405 427 2 × 2 = 1 + 0.350 195 587 644 810 854 4;
  • 66) 0.350 195 587 644 810 854 4 × 2 = 0 + 0.700 391 175 289 621 708 8;
  • 67) 0.700 391 175 289 621 708 8 × 2 = 1 + 0.400 782 350 579 243 417 6;
  • 68) 0.400 782 350 579 243 417 6 × 2 = 0 + 0.801 564 701 158 486 835 2;
  • 69) 0.801 564 701 158 486 835 2 × 2 = 1 + 0.603 129 402 316 973 670 4;
  • 70) 0.603 129 402 316 973 670 4 × 2 = 1 + 0.206 258 804 633 947 340 8;
  • 71) 0.206 258 804 633 947 340 8 × 2 = 0 + 0.412 517 609 267 894 681 6;
  • 72) 0.412 517 609 267 894 681 6 × 2 = 0 + 0.825 035 218 535 789 363 2;
  • 73) 0.825 035 218 535 789 363 2 × 2 = 1 + 0.650 070 437 071 578 726 4;
  • 74) 0.650 070 437 071 578 726 4 × 2 = 1 + 0.300 140 874 143 157 452 8;
  • 75) 0.300 140 874 143 157 452 8 × 2 = 0 + 0.600 281 748 286 314 905 6;
  • 76) 0.600 281 748 286 314 905 6 × 2 = 1 + 0.200 563 496 572 629 811 2;
  • 77) 0.200 563 496 572 629 811 2 × 2 = 0 + 0.401 126 993 145 259 622 4;
  • 78) 0.401 126 993 145 259 622 4 × 2 = 0 + 0.802 253 986 290 519 244 8;
  • 79) 0.802 253 986 290 519 244 8 × 2 = 1 + 0.604 507 972 581 038 489 6;
  • 80) 0.604 507 972 581 038 489 6 × 2 = 1 + 0.209 015 945 162 076 979 2;
  • 81) 0.209 015 945 162 076 979 2 × 2 = 0 + 0.418 031 890 324 153 958 4;
  • 82) 0.418 031 890 324 153 958 4 × 2 = 0 + 0.836 063 780 648 307 916 8;
  • 83) 0.836 063 780 648 307 916 8 × 2 = 1 + 0.672 127 561 296 615 833 6;
  • 84) 0.672 127 561 296 615 833 6 × 2 = 1 + 0.344 255 122 593 231 667 2;
  • 85) 0.344 255 122 593 231 667 2 × 2 = 0 + 0.688 510 245 186 463 334 4;
  • 86) 0.688 510 245 186 463 334 4 × 2 = 1 + 0.377 020 490 372 926 668 8;
  • 87) 0.377 020 490 372 926 668 8 × 2 = 0 + 0.754 040 980 745 853 337 6;
  • 88) 0.754 040 980 745 853 337 6 × 2 = 1 + 0.508 081 961 491 706 675 2;
  • 89) 0.508 081 961 491 706 675 2 × 2 = 1 + 0.016 163 922 983 413 350 4;
  • 90) 0.016 163 922 983 413 350 4 × 2 = 0 + 0.032 327 845 966 826 700 8;
  • 91) 0.032 327 845 966 826 700 8 × 2 = 0 + 0.064 655 691 933 653 401 6;
  • 92) 0.064 655 691 933 653 401 6 × 2 = 0 + 0.129 311 383 867 306 803 2;
  • 93) 0.129 311 383 867 306 803 2 × 2 = 0 + 0.258 622 767 734 613 606 4;
  • 94) 0.258 622 767 734 613 606 4 × 2 = 0 + 0.517 245 535 469 227 212 8;
  • 95) 0.517 245 535 469 227 212 8 × 2 = 1 + 0.034 491 070 938 454 425 6;

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:


0.000 000 000 000 176 544 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0101 1001 1010 1100 1101 0011 0011 0101 1000 001(2)

6. Positive number before normalization:

0.000 000 000 000 176 544 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0101 1001 1010 1100 1101 0011 0011 0101 1000 001(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:


0.000 000 000 000 176 544 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0101 1001 1010 1100 1101 0011 0011 0101 1000 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0101 1001 1010 1100 1101 0011 0011 0101 1000 001(2) × 20 =

1.1000 1101 1000 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001(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 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001


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 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001 =

1000 1101 1000 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001


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 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001


Decimal number -0.000 000 000 000 176 544 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1010 1100 1101 0110 0110 1001 1001 1010 1100 0001

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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