-0.000 000 000 000 176 542 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 542 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
-0.000 000 000 000 176 542 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:

|-0.000 000 000 000 176 542 2| = 0.000 000 000 000 176 542 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: 0.000 000 000 000 176 542 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) 0.000 000 000 000 176 542 2 × 2 = 0 + 0.000 000 000 000 353 084 4;
  • 2) 0.000 000 000 000 353 084 4 × 2 = 0 + 0.000 000 000 000 706 168 8;
  • 3) 0.000 000 000 000 706 168 8 × 2 = 0 + 0.000 000 000 001 412 337 6;
  • 4) 0.000 000 000 001 412 337 6 × 2 = 0 + 0.000 000 000 002 824 675 2;
  • 5) 0.000 000 000 002 824 675 2 × 2 = 0 + 0.000 000 000 005 649 350 4;
  • 6) 0.000 000 000 005 649 350 4 × 2 = 0 + 0.000 000 000 011 298 700 8;
  • 7) 0.000 000 000 011 298 700 8 × 2 = 0 + 0.000 000 000 022 597 401 6;
  • 8) 0.000 000 000 022 597 401 6 × 2 = 0 + 0.000 000 000 045 194 803 2;
  • 9) 0.000 000 000 045 194 803 2 × 2 = 0 + 0.000 000 000 090 389 606 4;
  • 10) 0.000 000 000 090 389 606 4 × 2 = 0 + 0.000 000 000 180 779 212 8;
  • 11) 0.000 000 000 180 779 212 8 × 2 = 0 + 0.000 000 000 361 558 425 6;
  • 12) 0.000 000 000 361 558 425 6 × 2 = 0 + 0.000 000 000 723 116 851 2;
  • 13) 0.000 000 000 723 116 851 2 × 2 = 0 + 0.000 000 001 446 233 702 4;
  • 14) 0.000 000 001 446 233 702 4 × 2 = 0 + 0.000 000 002 892 467 404 8;
  • 15) 0.000 000 002 892 467 404 8 × 2 = 0 + 0.000 000 005 784 934 809 6;
  • 16) 0.000 000 005 784 934 809 6 × 2 = 0 + 0.000 000 011 569 869 619 2;
  • 17) 0.000 000 011 569 869 619 2 × 2 = 0 + 0.000 000 023 139 739 238 4;
  • 18) 0.000 000 023 139 739 238 4 × 2 = 0 + 0.000 000 046 279 478 476 8;
  • 19) 0.000 000 046 279 478 476 8 × 2 = 0 + 0.000 000 092 558 956 953 6;
  • 20) 0.000 000 092 558 956 953 6 × 2 = 0 + 0.000 000 185 117 913 907 2;
  • 21) 0.000 000 185 117 913 907 2 × 2 = 0 + 0.000 000 370 235 827 814 4;
  • 22) 0.000 000 370 235 827 814 4 × 2 = 0 + 0.000 000 740 471 655 628 8;
  • 23) 0.000 000 740 471 655 628 8 × 2 = 0 + 0.000 001 480 943 311 257 6;
  • 24) 0.000 001 480 943 311 257 6 × 2 = 0 + 0.000 002 961 886 622 515 2;
  • 25) 0.000 002 961 886 622 515 2 × 2 = 0 + 0.000 005 923 773 245 030 4;
  • 26) 0.000 005 923 773 245 030 4 × 2 = 0 + 0.000 011 847 546 490 060 8;
  • 27) 0.000 011 847 546 490 060 8 × 2 = 0 + 0.000 023 695 092 980 121 6;
  • 28) 0.000 023 695 092 980 121 6 × 2 = 0 + 0.000 047 390 185 960 243 2;
  • 29) 0.000 047 390 185 960 243 2 × 2 = 0 + 0.000 094 780 371 920 486 4;
  • 30) 0.000 094 780 371 920 486 4 × 2 = 0 + 0.000 189 560 743 840 972 8;
  • 31) 0.000 189 560 743 840 972 8 × 2 = 0 + 0.000 379 121 487 681 945 6;
  • 32) 0.000 379 121 487 681 945 6 × 2 = 0 + 0.000 758 242 975 363 891 2;
  • 33) 0.000 758 242 975 363 891 2 × 2 = 0 + 0.001 516 485 950 727 782 4;
  • 34) 0.001 516 485 950 727 782 4 × 2 = 0 + 0.003 032 971 901 455 564 8;
  • 35) 0.003 032 971 901 455 564 8 × 2 = 0 + 0.006 065 943 802 911 129 6;
  • 36) 0.006 065 943 802 911 129 6 × 2 = 0 + 0.012 131 887 605 822 259 2;
  • 37) 0.012 131 887 605 822 259 2 × 2 = 0 + 0.024 263 775 211 644 518 4;
  • 38) 0.024 263 775 211 644 518 4 × 2 = 0 + 0.048 527 550 423 289 036 8;
  • 39) 0.048 527 550 423 289 036 8 × 2 = 0 + 0.097 055 100 846 578 073 6;
  • 40) 0.097 055 100 846 578 073 6 × 2 = 0 + 0.194 110 201 693 156 147 2;
  • 41) 0.194 110 201 693 156 147 2 × 2 = 0 + 0.388 220 403 386 312 294 4;
  • 42) 0.388 220 403 386 312 294 4 × 2 = 0 + 0.776 440 806 772 624 588 8;
  • 43) 0.776 440 806 772 624 588 8 × 2 = 1 + 0.552 881 613 545 249 177 6;
  • 44) 0.552 881 613 545 249 177 6 × 2 = 1 + 0.105 763 227 090 498 355 2;
  • 45) 0.105 763 227 090 498 355 2 × 2 = 0 + 0.211 526 454 180 996 710 4;
  • 46) 0.211 526 454 180 996 710 4 × 2 = 0 + 0.423 052 908 361 993 420 8;
  • 47) 0.423 052 908 361 993 420 8 × 2 = 0 + 0.846 105 816 723 986 841 6;
  • 48) 0.846 105 816 723 986 841 6 × 2 = 1 + 0.692 211 633 447 973 683 2;
  • 49) 0.692 211 633 447 973 683 2 × 2 = 1 + 0.384 423 266 895 947 366 4;
  • 50) 0.384 423 266 895 947 366 4 × 2 = 0 + 0.768 846 533 791 894 732 8;
  • 51) 0.768 846 533 791 894 732 8 × 2 = 1 + 0.537 693 067 583 789 465 6;
  • 52) 0.537 693 067 583 789 465 6 × 2 = 1 + 0.075 386 135 167 578 931 2;
  • 53) 0.075 386 135 167 578 931 2 × 2 = 0 + 0.150 772 270 335 157 862 4;
  • 54) 0.150 772 270 335 157 862 4 × 2 = 0 + 0.301 544 540 670 315 724 8;
  • 55) 0.301 544 540 670 315 724 8 × 2 = 0 + 0.603 089 081 340 631 449 6;
  • 56) 0.603 089 081 340 631 449 6 × 2 = 1 + 0.206 178 162 681 262 899 2;
  • 57) 0.206 178 162 681 262 899 2 × 2 = 0 + 0.412 356 325 362 525 798 4;
  • 58) 0.412 356 325 362 525 798 4 × 2 = 0 + 0.824 712 650 725 051 596 8;
  • 59) 0.824 712 650 725 051 596 8 × 2 = 1 + 0.649 425 301 450 103 193 6;
  • 60) 0.649 425 301 450 103 193 6 × 2 = 1 + 0.298 850 602 900 206 387 2;
  • 61) 0.298 850 602 900 206 387 2 × 2 = 0 + 0.597 701 205 800 412 774 4;
  • 62) 0.597 701 205 800 412 774 4 × 2 = 1 + 0.195 402 411 600 825 548 8;
  • 63) 0.195 402 411 600 825 548 8 × 2 = 0 + 0.390 804 823 201 651 097 6;
  • 64) 0.390 804 823 201 651 097 6 × 2 = 0 + 0.781 609 646 403 302 195 2;
  • 65) 0.781 609 646 403 302 195 2 × 2 = 1 + 0.563 219 292 806 604 390 4;
  • 66) 0.563 219 292 806 604 390 4 × 2 = 1 + 0.126 438 585 613 208 780 8;
  • 67) 0.126 438 585 613 208 780 8 × 2 = 0 + 0.252 877 171 226 417 561 6;
  • 68) 0.252 877 171 226 417 561 6 × 2 = 0 + 0.505 754 342 452 835 123 2;
  • 69) 0.505 754 342 452 835 123 2 × 2 = 1 + 0.011 508 684 905 670 246 4;
  • 70) 0.011 508 684 905 670 246 4 × 2 = 0 + 0.023 017 369 811 340 492 8;
  • 71) 0.023 017 369 811 340 492 8 × 2 = 0 + 0.046 034 739 622 680 985 6;
  • 72) 0.046 034 739 622 680 985 6 × 2 = 0 + 0.092 069 479 245 361 971 2;
  • 73) 0.092 069 479 245 361 971 2 × 2 = 0 + 0.184 138 958 490 723 942 4;
  • 74) 0.184 138 958 490 723 942 4 × 2 = 0 + 0.368 277 916 981 447 884 8;
  • 75) 0.368 277 916 981 447 884 8 × 2 = 0 + 0.736 555 833 962 895 769 6;
  • 76) 0.736 555 833 962 895 769 6 × 2 = 1 + 0.473 111 667 925 791 539 2;
  • 77) 0.473 111 667 925 791 539 2 × 2 = 0 + 0.946 223 335 851 583 078 4;
  • 78) 0.946 223 335 851 583 078 4 × 2 = 1 + 0.892 446 671 703 166 156 8;
  • 79) 0.892 446 671 703 166 156 8 × 2 = 1 + 0.784 893 343 406 332 313 6;
  • 80) 0.784 893 343 406 332 313 6 × 2 = 1 + 0.569 786 686 812 664 627 2;
  • 81) 0.569 786 686 812 664 627 2 × 2 = 1 + 0.139 573 373 625 329 254 4;
  • 82) 0.139 573 373 625 329 254 4 × 2 = 0 + 0.279 146 747 250 658 508 8;
  • 83) 0.279 146 747 250 658 508 8 × 2 = 0 + 0.558 293 494 501 317 017 6;
  • 84) 0.558 293 494 501 317 017 6 × 2 = 1 + 0.116 586 989 002 634 035 2;
  • 85) 0.116 586 989 002 634 035 2 × 2 = 0 + 0.233 173 978 005 268 070 4;
  • 86) 0.233 173 978 005 268 070 4 × 2 = 0 + 0.466 347 956 010 536 140 8;
  • 87) 0.466 347 956 010 536 140 8 × 2 = 0 + 0.932 695 912 021 072 281 6;
  • 88) 0.932 695 912 021 072 281 6 × 2 = 1 + 0.865 391 824 042 144 563 2;
  • 89) 0.865 391 824 042 144 563 2 × 2 = 1 + 0.730 783 648 084 289 126 4;
  • 90) 0.730 783 648 084 289 126 4 × 2 = 1 + 0.461 567 296 168 578 252 8;
  • 91) 0.461 567 296 168 578 252 8 × 2 = 0 + 0.923 134 592 337 156 505 6;
  • 92) 0.923 134 592 337 156 505 6 × 2 = 1 + 0.846 269 184 674 313 011 2;
  • 93) 0.846 269 184 674 313 011 2 × 2 = 1 + 0.692 538 369 348 626 022 4;
  • 94) 0.692 538 369 348 626 022 4 × 2 = 1 + 0.385 076 738 697 252 044 8;
  • 95) 0.385 076 738 697 252 044 8 × 2 = 0 + 0.770 153 477 394 504 089 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:


0.000 000 000 000 176 542 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0011 0100 1100 1000 0001 0111 1001 0001 1101 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 542 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0011 0100 1100 1000 0001 0111 1001 0001 1101 110(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:


0.000 000 000 000 176 542 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0011 0100 1100 1000 0001 0111 1001 0001 1101 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0011 0100 1100 1000 0001 0111 1001 0001 1101 110(2) × 20 =

1.1000 1101 1000 1001 1010 0110 0100 0000 1011 1100 1000 1110 1110(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 1010 0110 0100 0000 1011 1100 1000 1110 1110


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 1010 0110 0100 0000 1011 1100 1000 1110 1110 =

1000 1101 1000 1001 1010 0110 0100 0000 1011 1100 1000 1110 1110


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 1010 0110 0100 0000 1011 1100 1000 1110 1110


Decimal number -0.000 000 000 000 176 542 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1001 1010 0110 0100 0000 1011 1100 1000 1110 1110

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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