-0.000 000 000 000 293 892 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 892 1(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
-0.000 000 000 000 293 892 1(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:

|-0.000 000 000 000 293 892 1| = 0.000 000 000 000 293 892 1


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: 0.000 000 000 000 293 892 1.

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) 0.000 000 000 000 293 892 1 × 2 = 0 + 0.000 000 000 000 587 784 2;
  • 2) 0.000 000 000 000 587 784 2 × 2 = 0 + 0.000 000 000 001 175 568 4;
  • 3) 0.000 000 000 001 175 568 4 × 2 = 0 + 0.000 000 000 002 351 136 8;
  • 4) 0.000 000 000 002 351 136 8 × 2 = 0 + 0.000 000 000 004 702 273 6;
  • 5) 0.000 000 000 004 702 273 6 × 2 = 0 + 0.000 000 000 009 404 547 2;
  • 6) 0.000 000 000 009 404 547 2 × 2 = 0 + 0.000 000 000 018 809 094 4;
  • 7) 0.000 000 000 018 809 094 4 × 2 = 0 + 0.000 000 000 037 618 188 8;
  • 8) 0.000 000 000 037 618 188 8 × 2 = 0 + 0.000 000 000 075 236 377 6;
  • 9) 0.000 000 000 075 236 377 6 × 2 = 0 + 0.000 000 000 150 472 755 2;
  • 10) 0.000 000 000 150 472 755 2 × 2 = 0 + 0.000 000 000 300 945 510 4;
  • 11) 0.000 000 000 300 945 510 4 × 2 = 0 + 0.000 000 000 601 891 020 8;
  • 12) 0.000 000 000 601 891 020 8 × 2 = 0 + 0.000 000 001 203 782 041 6;
  • 13) 0.000 000 001 203 782 041 6 × 2 = 0 + 0.000 000 002 407 564 083 2;
  • 14) 0.000 000 002 407 564 083 2 × 2 = 0 + 0.000 000 004 815 128 166 4;
  • 15) 0.000 000 004 815 128 166 4 × 2 = 0 + 0.000 000 009 630 256 332 8;
  • 16) 0.000 000 009 630 256 332 8 × 2 = 0 + 0.000 000 019 260 512 665 6;
  • 17) 0.000 000 019 260 512 665 6 × 2 = 0 + 0.000 000 038 521 025 331 2;
  • 18) 0.000 000 038 521 025 331 2 × 2 = 0 + 0.000 000 077 042 050 662 4;
  • 19) 0.000 000 077 042 050 662 4 × 2 = 0 + 0.000 000 154 084 101 324 8;
  • 20) 0.000 000 154 084 101 324 8 × 2 = 0 + 0.000 000 308 168 202 649 6;
  • 21) 0.000 000 308 168 202 649 6 × 2 = 0 + 0.000 000 616 336 405 299 2;
  • 22) 0.000 000 616 336 405 299 2 × 2 = 0 + 0.000 001 232 672 810 598 4;
  • 23) 0.000 001 232 672 810 598 4 × 2 = 0 + 0.000 002 465 345 621 196 8;
  • 24) 0.000 002 465 345 621 196 8 × 2 = 0 + 0.000 004 930 691 242 393 6;
  • 25) 0.000 004 930 691 242 393 6 × 2 = 0 + 0.000 009 861 382 484 787 2;
  • 26) 0.000 009 861 382 484 787 2 × 2 = 0 + 0.000 019 722 764 969 574 4;
  • 27) 0.000 019 722 764 969 574 4 × 2 = 0 + 0.000 039 445 529 939 148 8;
  • 28) 0.000 039 445 529 939 148 8 × 2 = 0 + 0.000 078 891 059 878 297 6;
  • 29) 0.000 078 891 059 878 297 6 × 2 = 0 + 0.000 157 782 119 756 595 2;
  • 30) 0.000 157 782 119 756 595 2 × 2 = 0 + 0.000 315 564 239 513 190 4;
  • 31) 0.000 315 564 239 513 190 4 × 2 = 0 + 0.000 631 128 479 026 380 8;
  • 32) 0.000 631 128 479 026 380 8 × 2 = 0 + 0.001 262 256 958 052 761 6;
  • 33) 0.001 262 256 958 052 761 6 × 2 = 0 + 0.002 524 513 916 105 523 2;
  • 34) 0.002 524 513 916 105 523 2 × 2 = 0 + 0.005 049 027 832 211 046 4;
  • 35) 0.005 049 027 832 211 046 4 × 2 = 0 + 0.010 098 055 664 422 092 8;
  • 36) 0.010 098 055 664 422 092 8 × 2 = 0 + 0.020 196 111 328 844 185 6;
  • 37) 0.020 196 111 328 844 185 6 × 2 = 0 + 0.040 392 222 657 688 371 2;
  • 38) 0.040 392 222 657 688 371 2 × 2 = 0 + 0.080 784 445 315 376 742 4;
  • 39) 0.080 784 445 315 376 742 4 × 2 = 0 + 0.161 568 890 630 753 484 8;
  • 40) 0.161 568 890 630 753 484 8 × 2 = 0 + 0.323 137 781 261 506 969 6;
  • 41) 0.323 137 781 261 506 969 6 × 2 = 0 + 0.646 275 562 523 013 939 2;
  • 42) 0.646 275 562 523 013 939 2 × 2 = 1 + 0.292 551 125 046 027 878 4;
  • 43) 0.292 551 125 046 027 878 4 × 2 = 0 + 0.585 102 250 092 055 756 8;
  • 44) 0.585 102 250 092 055 756 8 × 2 = 1 + 0.170 204 500 184 111 513 6;
  • 45) 0.170 204 500 184 111 513 6 × 2 = 0 + 0.340 409 000 368 223 027 2;
  • 46) 0.340 409 000 368 223 027 2 × 2 = 0 + 0.680 818 000 736 446 054 4;
  • 47) 0.680 818 000 736 446 054 4 × 2 = 1 + 0.361 636 001 472 892 108 8;
  • 48) 0.361 636 001 472 892 108 8 × 2 = 0 + 0.723 272 002 945 784 217 6;
  • 49) 0.723 272 002 945 784 217 6 × 2 = 1 + 0.446 544 005 891 568 435 2;
  • 50) 0.446 544 005 891 568 435 2 × 2 = 0 + 0.893 088 011 783 136 870 4;
  • 51) 0.893 088 011 783 136 870 4 × 2 = 1 + 0.786 176 023 566 273 740 8;
  • 52) 0.786 176 023 566 273 740 8 × 2 = 1 + 0.572 352 047 132 547 481 6;
  • 53) 0.572 352 047 132 547 481 6 × 2 = 1 + 0.144 704 094 265 094 963 2;
  • 54) 0.144 704 094 265 094 963 2 × 2 = 0 + 0.289 408 188 530 189 926 4;
  • 55) 0.289 408 188 530 189 926 4 × 2 = 0 + 0.578 816 377 060 379 852 8;
  • 56) 0.578 816 377 060 379 852 8 × 2 = 1 + 0.157 632 754 120 759 705 6;
  • 57) 0.157 632 754 120 759 705 6 × 2 = 0 + 0.315 265 508 241 519 411 2;
  • 58) 0.315 265 508 241 519 411 2 × 2 = 0 + 0.630 531 016 483 038 822 4;
  • 59) 0.630 531 016 483 038 822 4 × 2 = 1 + 0.261 062 032 966 077 644 8;
  • 60) 0.261 062 032 966 077 644 8 × 2 = 0 + 0.522 124 065 932 155 289 6;
  • 61) 0.522 124 065 932 155 289 6 × 2 = 1 + 0.044 248 131 864 310 579 2;
  • 62) 0.044 248 131 864 310 579 2 × 2 = 0 + 0.088 496 263 728 621 158 4;
  • 63) 0.088 496 263 728 621 158 4 × 2 = 0 + 0.176 992 527 457 242 316 8;
  • 64) 0.176 992 527 457 242 316 8 × 2 = 0 + 0.353 985 054 914 484 633 6;
  • 65) 0.353 985 054 914 484 633 6 × 2 = 0 + 0.707 970 109 828 969 267 2;
  • 66) 0.707 970 109 828 969 267 2 × 2 = 1 + 0.415 940 219 657 938 534 4;
  • 67) 0.415 940 219 657 938 534 4 × 2 = 0 + 0.831 880 439 315 877 068 8;
  • 68) 0.831 880 439 315 877 068 8 × 2 = 1 + 0.663 760 878 631 754 137 6;
  • 69) 0.663 760 878 631 754 137 6 × 2 = 1 + 0.327 521 757 263 508 275 2;
  • 70) 0.327 521 757 263 508 275 2 × 2 = 0 + 0.655 043 514 527 016 550 4;
  • 71) 0.655 043 514 527 016 550 4 × 2 = 1 + 0.310 087 029 054 033 100 8;
  • 72) 0.310 087 029 054 033 100 8 × 2 = 0 + 0.620 174 058 108 066 201 6;
  • 73) 0.620 174 058 108 066 201 6 × 2 = 1 + 0.240 348 116 216 132 403 2;
  • 74) 0.240 348 116 216 132 403 2 × 2 = 0 + 0.480 696 232 432 264 806 4;
  • 75) 0.480 696 232 432 264 806 4 × 2 = 0 + 0.961 392 464 864 529 612 8;
  • 76) 0.961 392 464 864 529 612 8 × 2 = 1 + 0.922 784 929 729 059 225 6;
  • 77) 0.922 784 929 729 059 225 6 × 2 = 1 + 0.845 569 859 458 118 451 2;
  • 78) 0.845 569 859 458 118 451 2 × 2 = 1 + 0.691 139 718 916 236 902 4;
  • 79) 0.691 139 718 916 236 902 4 × 2 = 1 + 0.382 279 437 832 473 804 8;
  • 80) 0.382 279 437 832 473 804 8 × 2 = 0 + 0.764 558 875 664 947 609 6;
  • 81) 0.764 558 875 664 947 609 6 × 2 = 1 + 0.529 117 751 329 895 219 2;
  • 82) 0.529 117 751 329 895 219 2 × 2 = 1 + 0.058 235 502 659 790 438 4;
  • 83) 0.058 235 502 659 790 438 4 × 2 = 0 + 0.116 471 005 319 580 876 8;
  • 84) 0.116 471 005 319 580 876 8 × 2 = 0 + 0.232 942 010 639 161 753 6;
  • 85) 0.232 942 010 639 161 753 6 × 2 = 0 + 0.465 884 021 278 323 507 2;
  • 86) 0.465 884 021 278 323 507 2 × 2 = 0 + 0.931 768 042 556 647 014 4;
  • 87) 0.931 768 042 556 647 014 4 × 2 = 1 + 0.863 536 085 113 294 028 8;
  • 88) 0.863 536 085 113 294 028 8 × 2 = 1 + 0.727 072 170 226 588 057 6;
  • 89) 0.727 072 170 226 588 057 6 × 2 = 1 + 0.454 144 340 453 176 115 2;
  • 90) 0.454 144 340 453 176 115 2 × 2 = 0 + 0.908 288 680 906 352 230 4;
  • 91) 0.908 288 680 906 352 230 4 × 2 = 1 + 0.816 577 361 812 704 460 8;
  • 92) 0.816 577 361 812 704 460 8 × 2 = 1 + 0.633 154 723 625 408 921 6;
  • 93) 0.633 154 723 625 408 921 6 × 2 = 1 + 0.266 309 447 250 817 843 2;
  • 94) 0.266 309 447 250 817 843 2 × 2 = 0 + 0.532 618 894 501 635 686 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:


0.000 000 000 000 293 892 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0010 1000 0101 1010 1001 1110 1100 0011 1011 10(2)

6. Positive number before normalization:

0.000 000 000 000 293 892 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0010 1000 0101 1010 1001 1110 1100 0011 1011 10(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:


0.000 000 000 000 293 892 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0010 1000 0101 1010 1001 1110 1100 0011 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0010 1000 0101 1010 1001 1110 1100 0011 1011 10(2) × 20 =

1.0100 1010 1110 0100 1010 0001 0110 1010 0111 1011 0000 1110 1110(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 0100 1010 0001 0110 1010 0111 1011 0000 1110 1110


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 0100 1010 0001 0110 1010 0111 1011 0000 1110 1110 =

0100 1010 1110 0100 1010 0001 0110 1010 0111 1011 0000 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 0101

Mantissa (52 bits) =
0100 1010 1110 0100 1010 0001 0110 1010 0111 1011 0000 1110 1110


Decimal number -0.000 000 000 000 293 892 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 0100 1010 0001 0110 1010 0111 1011 0000 1110 1110

» Convert decimal -0.000 000 000 000 293 882 5 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