-0.000 000 000 000 293 888 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 888 4(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
-0.000 000 000 000 293 888 4(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:

|-0.000 000 000 000 293 888 4| = 0.000 000 000 000 293 888 4


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: 0.000 000 000 000 293 888 4.

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) 0.000 000 000 000 293 888 4 × 2 = 0 + 0.000 000 000 000 587 776 8;
  • 2) 0.000 000 000 000 587 776 8 × 2 = 0 + 0.000 000 000 001 175 553 6;
  • 3) 0.000 000 000 001 175 553 6 × 2 = 0 + 0.000 000 000 002 351 107 2;
  • 4) 0.000 000 000 002 351 107 2 × 2 = 0 + 0.000 000 000 004 702 214 4;
  • 5) 0.000 000 000 004 702 214 4 × 2 = 0 + 0.000 000 000 009 404 428 8;
  • 6) 0.000 000 000 009 404 428 8 × 2 = 0 + 0.000 000 000 018 808 857 6;
  • 7) 0.000 000 000 018 808 857 6 × 2 = 0 + 0.000 000 000 037 617 715 2;
  • 8) 0.000 000 000 037 617 715 2 × 2 = 0 + 0.000 000 000 075 235 430 4;
  • 9) 0.000 000 000 075 235 430 4 × 2 = 0 + 0.000 000 000 150 470 860 8;
  • 10) 0.000 000 000 150 470 860 8 × 2 = 0 + 0.000 000 000 300 941 721 6;
  • 11) 0.000 000 000 300 941 721 6 × 2 = 0 + 0.000 000 000 601 883 443 2;
  • 12) 0.000 000 000 601 883 443 2 × 2 = 0 + 0.000 000 001 203 766 886 4;
  • 13) 0.000 000 001 203 766 886 4 × 2 = 0 + 0.000 000 002 407 533 772 8;
  • 14) 0.000 000 002 407 533 772 8 × 2 = 0 + 0.000 000 004 815 067 545 6;
  • 15) 0.000 000 004 815 067 545 6 × 2 = 0 + 0.000 000 009 630 135 091 2;
  • 16) 0.000 000 009 630 135 091 2 × 2 = 0 + 0.000 000 019 260 270 182 4;
  • 17) 0.000 000 019 260 270 182 4 × 2 = 0 + 0.000 000 038 520 540 364 8;
  • 18) 0.000 000 038 520 540 364 8 × 2 = 0 + 0.000 000 077 041 080 729 6;
  • 19) 0.000 000 077 041 080 729 6 × 2 = 0 + 0.000 000 154 082 161 459 2;
  • 20) 0.000 000 154 082 161 459 2 × 2 = 0 + 0.000 000 308 164 322 918 4;
  • 21) 0.000 000 308 164 322 918 4 × 2 = 0 + 0.000 000 616 328 645 836 8;
  • 22) 0.000 000 616 328 645 836 8 × 2 = 0 + 0.000 001 232 657 291 673 6;
  • 23) 0.000 001 232 657 291 673 6 × 2 = 0 + 0.000 002 465 314 583 347 2;
  • 24) 0.000 002 465 314 583 347 2 × 2 = 0 + 0.000 004 930 629 166 694 4;
  • 25) 0.000 004 930 629 166 694 4 × 2 = 0 + 0.000 009 861 258 333 388 8;
  • 26) 0.000 009 861 258 333 388 8 × 2 = 0 + 0.000 019 722 516 666 777 6;
  • 27) 0.000 019 722 516 666 777 6 × 2 = 0 + 0.000 039 445 033 333 555 2;
  • 28) 0.000 039 445 033 333 555 2 × 2 = 0 + 0.000 078 890 066 667 110 4;
  • 29) 0.000 078 890 066 667 110 4 × 2 = 0 + 0.000 157 780 133 334 220 8;
  • 30) 0.000 157 780 133 334 220 8 × 2 = 0 + 0.000 315 560 266 668 441 6;
  • 31) 0.000 315 560 266 668 441 6 × 2 = 0 + 0.000 631 120 533 336 883 2;
  • 32) 0.000 631 120 533 336 883 2 × 2 = 0 + 0.001 262 241 066 673 766 4;
  • 33) 0.001 262 241 066 673 766 4 × 2 = 0 + 0.002 524 482 133 347 532 8;
  • 34) 0.002 524 482 133 347 532 8 × 2 = 0 + 0.005 048 964 266 695 065 6;
  • 35) 0.005 048 964 266 695 065 6 × 2 = 0 + 0.010 097 928 533 390 131 2;
  • 36) 0.010 097 928 533 390 131 2 × 2 = 0 + 0.020 195 857 066 780 262 4;
  • 37) 0.020 195 857 066 780 262 4 × 2 = 0 + 0.040 391 714 133 560 524 8;
  • 38) 0.040 391 714 133 560 524 8 × 2 = 0 + 0.080 783 428 267 121 049 6;
  • 39) 0.080 783 428 267 121 049 6 × 2 = 0 + 0.161 566 856 534 242 099 2;
  • 40) 0.161 566 856 534 242 099 2 × 2 = 0 + 0.323 133 713 068 484 198 4;
  • 41) 0.323 133 713 068 484 198 4 × 2 = 0 + 0.646 267 426 136 968 396 8;
  • 42) 0.646 267 426 136 968 396 8 × 2 = 1 + 0.292 534 852 273 936 793 6;
  • 43) 0.292 534 852 273 936 793 6 × 2 = 0 + 0.585 069 704 547 873 587 2;
  • 44) 0.585 069 704 547 873 587 2 × 2 = 1 + 0.170 139 409 095 747 174 4;
  • 45) 0.170 139 409 095 747 174 4 × 2 = 0 + 0.340 278 818 191 494 348 8;
  • 46) 0.340 278 818 191 494 348 8 × 2 = 0 + 0.680 557 636 382 988 697 6;
  • 47) 0.680 557 636 382 988 697 6 × 2 = 1 + 0.361 115 272 765 977 395 2;
  • 48) 0.361 115 272 765 977 395 2 × 2 = 0 + 0.722 230 545 531 954 790 4;
  • 49) 0.722 230 545 531 954 790 4 × 2 = 1 + 0.444 461 091 063 909 580 8;
  • 50) 0.444 461 091 063 909 580 8 × 2 = 0 + 0.888 922 182 127 819 161 6;
  • 51) 0.888 922 182 127 819 161 6 × 2 = 1 + 0.777 844 364 255 638 323 2;
  • 52) 0.777 844 364 255 638 323 2 × 2 = 1 + 0.555 688 728 511 276 646 4;
  • 53) 0.555 688 728 511 276 646 4 × 2 = 1 + 0.111 377 457 022 553 292 8;
  • 54) 0.111 377 457 022 553 292 8 × 2 = 0 + 0.222 754 914 045 106 585 6;
  • 55) 0.222 754 914 045 106 585 6 × 2 = 0 + 0.445 509 828 090 213 171 2;
  • 56) 0.445 509 828 090 213 171 2 × 2 = 0 + 0.891 019 656 180 426 342 4;
  • 57) 0.891 019 656 180 426 342 4 × 2 = 1 + 0.782 039 312 360 852 684 8;
  • 58) 0.782 039 312 360 852 684 8 × 2 = 1 + 0.564 078 624 721 705 369 6;
  • 59) 0.564 078 624 721 705 369 6 × 2 = 1 + 0.128 157 249 443 410 739 2;
  • 60) 0.128 157 249 443 410 739 2 × 2 = 0 + 0.256 314 498 886 821 478 4;
  • 61) 0.256 314 498 886 821 478 4 × 2 = 0 + 0.512 628 997 773 642 956 8;
  • 62) 0.512 628 997 773 642 956 8 × 2 = 1 + 0.025 257 995 547 285 913 6;
  • 63) 0.025 257 995 547 285 913 6 × 2 = 0 + 0.050 515 991 094 571 827 2;
  • 64) 0.050 515 991 094 571 827 2 × 2 = 0 + 0.101 031 982 189 143 654 4;
  • 65) 0.101 031 982 189 143 654 4 × 2 = 0 + 0.202 063 964 378 287 308 8;
  • 66) 0.202 063 964 378 287 308 8 × 2 = 0 + 0.404 127 928 756 574 617 6;
  • 67) 0.404 127 928 756 574 617 6 × 2 = 0 + 0.808 255 857 513 149 235 2;
  • 68) 0.808 255 857 513 149 235 2 × 2 = 1 + 0.616 511 715 026 298 470 4;
  • 69) 0.616 511 715 026 298 470 4 × 2 = 1 + 0.233 023 430 052 596 940 8;
  • 70) 0.233 023 430 052 596 940 8 × 2 = 0 + 0.466 046 860 105 193 881 6;
  • 71) 0.466 046 860 105 193 881 6 × 2 = 0 + 0.932 093 720 210 387 763 2;
  • 72) 0.932 093 720 210 387 763 2 × 2 = 1 + 0.864 187 440 420 775 526 4;
  • 73) 0.864 187 440 420 775 526 4 × 2 = 1 + 0.728 374 880 841 551 052 8;
  • 74) 0.728 374 880 841 551 052 8 × 2 = 1 + 0.456 749 761 683 102 105 6;
  • 75) 0.456 749 761 683 102 105 6 × 2 = 0 + 0.913 499 523 366 204 211 2;
  • 76) 0.913 499 523 366 204 211 2 × 2 = 1 + 0.826 999 046 732 408 422 4;
  • 77) 0.826 999 046 732 408 422 4 × 2 = 1 + 0.653 998 093 464 816 844 8;
  • 78) 0.653 998 093 464 816 844 8 × 2 = 1 + 0.307 996 186 929 633 689 6;
  • 79) 0.307 996 186 929 633 689 6 × 2 = 0 + 0.615 992 373 859 267 379 2;
  • 80) 0.615 992 373 859 267 379 2 × 2 = 1 + 0.231 984 747 718 534 758 4;
  • 81) 0.231 984 747 718 534 758 4 × 2 = 0 + 0.463 969 495 437 069 516 8;
  • 82) 0.463 969 495 437 069 516 8 × 2 = 0 + 0.927 938 990 874 139 033 6;
  • 83) 0.927 938 990 874 139 033 6 × 2 = 1 + 0.855 877 981 748 278 067 2;
  • 84) 0.855 877 981 748 278 067 2 × 2 = 1 + 0.711 755 963 496 556 134 4;
  • 85) 0.711 755 963 496 556 134 4 × 2 = 1 + 0.423 511 926 993 112 268 8;
  • 86) 0.423 511 926 993 112 268 8 × 2 = 0 + 0.847 023 853 986 224 537 6;
  • 87) 0.847 023 853 986 224 537 6 × 2 = 1 + 0.694 047 707 972 449 075 2;
  • 88) 0.694 047 707 972 449 075 2 × 2 = 1 + 0.388 095 415 944 898 150 4;
  • 89) 0.388 095 415 944 898 150 4 × 2 = 0 + 0.776 190 831 889 796 300 8;
  • 90) 0.776 190 831 889 796 300 8 × 2 = 1 + 0.552 381 663 779 592 601 6;
  • 91) 0.552 381 663 779 592 601 6 × 2 = 1 + 0.104 763 327 559 185 203 2;
  • 92) 0.104 763 327 559 185 203 2 × 2 = 0 + 0.209 526 655 118 370 406 4;
  • 93) 0.209 526 655 118 370 406 4 × 2 = 0 + 0.419 053 310 236 740 812 8;
  • 94) 0.419 053 310 236 740 812 8 × 2 = 0 + 0.838 106 620 473 481 625 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:


0.000 000 000 000 293 888 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1110 0100 0001 1001 1101 1101 0011 1011 0110 00(2)

6. Positive number before normalization:

0.000 000 000 000 293 888 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1110 0100 0001 1001 1101 1101 0011 1011 0110 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 293 888 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1110 0100 0001 1001 1101 1101 0011 1011 0110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1000 1110 0100 0001 1001 1101 1101 0011 1011 0110 00(2) × 20 =

1.0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000(2) × 2-42


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): -42

Mantissa (not normalized):
1.0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 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) =

981(10) =

011 1101 0101(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. 0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000 =

0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000


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 0101

Mantissa (52 bits) =
0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000


Decimal number -0.000 000 000 000 293 888 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 0011 1001 0000 0110 0111 0111 0100 1110 1101 1000

» Convert decimal -0.000 000 000 000 293 886 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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