-0.000 000 000 000 176 557 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 06(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
-0.000 000 000 000 176 557 06(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:

|-0.000 000 000 000 176 557 06| = 0.000 000 000 000 176 557 06


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: 0.000 000 000 000 176 557 06.

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) 0.000 000 000 000 176 557 06 × 2 = 0 + 0.000 000 000 000 353 114 12;
  • 2) 0.000 000 000 000 353 114 12 × 2 = 0 + 0.000 000 000 000 706 228 24;
  • 3) 0.000 000 000 000 706 228 24 × 2 = 0 + 0.000 000 000 001 412 456 48;
  • 4) 0.000 000 000 001 412 456 48 × 2 = 0 + 0.000 000 000 002 824 912 96;
  • 5) 0.000 000 000 002 824 912 96 × 2 = 0 + 0.000 000 000 005 649 825 92;
  • 6) 0.000 000 000 005 649 825 92 × 2 = 0 + 0.000 000 000 011 299 651 84;
  • 7) 0.000 000 000 011 299 651 84 × 2 = 0 + 0.000 000 000 022 599 303 68;
  • 8) 0.000 000 000 022 599 303 68 × 2 = 0 + 0.000 000 000 045 198 607 36;
  • 9) 0.000 000 000 045 198 607 36 × 2 = 0 + 0.000 000 000 090 397 214 72;
  • 10) 0.000 000 000 090 397 214 72 × 2 = 0 + 0.000 000 000 180 794 429 44;
  • 11) 0.000 000 000 180 794 429 44 × 2 = 0 + 0.000 000 000 361 588 858 88;
  • 12) 0.000 000 000 361 588 858 88 × 2 = 0 + 0.000 000 000 723 177 717 76;
  • 13) 0.000 000 000 723 177 717 76 × 2 = 0 + 0.000 000 001 446 355 435 52;
  • 14) 0.000 000 001 446 355 435 52 × 2 = 0 + 0.000 000 002 892 710 871 04;
  • 15) 0.000 000 002 892 710 871 04 × 2 = 0 + 0.000 000 005 785 421 742 08;
  • 16) 0.000 000 005 785 421 742 08 × 2 = 0 + 0.000 000 011 570 843 484 16;
  • 17) 0.000 000 011 570 843 484 16 × 2 = 0 + 0.000 000 023 141 686 968 32;
  • 18) 0.000 000 023 141 686 968 32 × 2 = 0 + 0.000 000 046 283 373 936 64;
  • 19) 0.000 000 046 283 373 936 64 × 2 = 0 + 0.000 000 092 566 747 873 28;
  • 20) 0.000 000 092 566 747 873 28 × 2 = 0 + 0.000 000 185 133 495 746 56;
  • 21) 0.000 000 185 133 495 746 56 × 2 = 0 + 0.000 000 370 266 991 493 12;
  • 22) 0.000 000 370 266 991 493 12 × 2 = 0 + 0.000 000 740 533 982 986 24;
  • 23) 0.000 000 740 533 982 986 24 × 2 = 0 + 0.000 001 481 067 965 972 48;
  • 24) 0.000 001 481 067 965 972 48 × 2 = 0 + 0.000 002 962 135 931 944 96;
  • 25) 0.000 002 962 135 931 944 96 × 2 = 0 + 0.000 005 924 271 863 889 92;
  • 26) 0.000 005 924 271 863 889 92 × 2 = 0 + 0.000 011 848 543 727 779 84;
  • 27) 0.000 011 848 543 727 779 84 × 2 = 0 + 0.000 023 697 087 455 559 68;
  • 28) 0.000 023 697 087 455 559 68 × 2 = 0 + 0.000 047 394 174 911 119 36;
  • 29) 0.000 047 394 174 911 119 36 × 2 = 0 + 0.000 094 788 349 822 238 72;
  • 30) 0.000 094 788 349 822 238 72 × 2 = 0 + 0.000 189 576 699 644 477 44;
  • 31) 0.000 189 576 699 644 477 44 × 2 = 0 + 0.000 379 153 399 288 954 88;
  • 32) 0.000 379 153 399 288 954 88 × 2 = 0 + 0.000 758 306 798 577 909 76;
  • 33) 0.000 758 306 798 577 909 76 × 2 = 0 + 0.001 516 613 597 155 819 52;
  • 34) 0.001 516 613 597 155 819 52 × 2 = 0 + 0.003 033 227 194 311 639 04;
  • 35) 0.003 033 227 194 311 639 04 × 2 = 0 + 0.006 066 454 388 623 278 08;
  • 36) 0.006 066 454 388 623 278 08 × 2 = 0 + 0.012 132 908 777 246 556 16;
  • 37) 0.012 132 908 777 246 556 16 × 2 = 0 + 0.024 265 817 554 493 112 32;
  • 38) 0.024 265 817 554 493 112 32 × 2 = 0 + 0.048 531 635 108 986 224 64;
  • 39) 0.048 531 635 108 986 224 64 × 2 = 0 + 0.097 063 270 217 972 449 28;
  • 40) 0.097 063 270 217 972 449 28 × 2 = 0 + 0.194 126 540 435 944 898 56;
  • 41) 0.194 126 540 435 944 898 56 × 2 = 0 + 0.388 253 080 871 889 797 12;
  • 42) 0.388 253 080 871 889 797 12 × 2 = 0 + 0.776 506 161 743 779 594 24;
  • 43) 0.776 506 161 743 779 594 24 × 2 = 1 + 0.553 012 323 487 559 188 48;
  • 44) 0.553 012 323 487 559 188 48 × 2 = 1 + 0.106 024 646 975 118 376 96;
  • 45) 0.106 024 646 975 118 376 96 × 2 = 0 + 0.212 049 293 950 236 753 92;
  • 46) 0.212 049 293 950 236 753 92 × 2 = 0 + 0.424 098 587 900 473 507 84;
  • 47) 0.424 098 587 900 473 507 84 × 2 = 0 + 0.848 197 175 800 947 015 68;
  • 48) 0.848 197 175 800 947 015 68 × 2 = 1 + 0.696 394 351 601 894 031 36;
  • 49) 0.696 394 351 601 894 031 36 × 2 = 1 + 0.392 788 703 203 788 062 72;
  • 50) 0.392 788 703 203 788 062 72 × 2 = 0 + 0.785 577 406 407 576 125 44;
  • 51) 0.785 577 406 407 576 125 44 × 2 = 1 + 0.571 154 812 815 152 250 88;
  • 52) 0.571 154 812 815 152 250 88 × 2 = 1 + 0.142 309 625 630 304 501 76;
  • 53) 0.142 309 625 630 304 501 76 × 2 = 0 + 0.284 619 251 260 609 003 52;
  • 54) 0.284 619 251 260 609 003 52 × 2 = 0 + 0.569 238 502 521 218 007 04;
  • 55) 0.569 238 502 521 218 007 04 × 2 = 1 + 0.138 477 005 042 436 014 08;
  • 56) 0.138 477 005 042 436 014 08 × 2 = 0 + 0.276 954 010 084 872 028 16;
  • 57) 0.276 954 010 084 872 028 16 × 2 = 0 + 0.553 908 020 169 744 056 32;
  • 58) 0.553 908 020 169 744 056 32 × 2 = 1 + 0.107 816 040 339 488 112 64;
  • 59) 0.107 816 040 339 488 112 64 × 2 = 0 + 0.215 632 080 678 976 225 28;
  • 60) 0.215 632 080 678 976 225 28 × 2 = 0 + 0.431 264 161 357 952 450 56;
  • 61) 0.431 264 161 357 952 450 56 × 2 = 0 + 0.862 528 322 715 904 901 12;
  • 62) 0.862 528 322 715 904 901 12 × 2 = 1 + 0.725 056 645 431 809 802 24;
  • 63) 0.725 056 645 431 809 802 24 × 2 = 1 + 0.450 113 290 863 619 604 48;
  • 64) 0.450 113 290 863 619 604 48 × 2 = 0 + 0.900 226 581 727 239 208 96;
  • 65) 0.900 226 581 727 239 208 96 × 2 = 1 + 0.800 453 163 454 478 417 92;
  • 66) 0.800 453 163 454 478 417 92 × 2 = 1 + 0.600 906 326 908 956 835 84;
  • 67) 0.600 906 326 908 956 835 84 × 2 = 1 + 0.201 812 653 817 913 671 68;
  • 68) 0.201 812 653 817 913 671 68 × 2 = 0 + 0.403 625 307 635 827 343 36;
  • 69) 0.403 625 307 635 827 343 36 × 2 = 0 + 0.807 250 615 271 654 686 72;
  • 70) 0.807 250 615 271 654 686 72 × 2 = 1 + 0.614 501 230 543 309 373 44;
  • 71) 0.614 501 230 543 309 373 44 × 2 = 1 + 0.229 002 461 086 618 746 88;
  • 72) 0.229 002 461 086 618 746 88 × 2 = 0 + 0.458 004 922 173 237 493 76;
  • 73) 0.458 004 922 173 237 493 76 × 2 = 0 + 0.916 009 844 346 474 987 52;
  • 74) 0.916 009 844 346 474 987 52 × 2 = 1 + 0.832 019 688 692 949 975 04;
  • 75) 0.832 019 688 692 949 975 04 × 2 = 1 + 0.664 039 377 385 899 950 08;
  • 76) 0.664 039 377 385 899 950 08 × 2 = 1 + 0.328 078 754 771 799 900 16;
  • 77) 0.328 078 754 771 799 900 16 × 2 = 0 + 0.656 157 509 543 599 800 32;
  • 78) 0.656 157 509 543 599 800 32 × 2 = 1 + 0.312 315 019 087 199 600 64;
  • 79) 0.312 315 019 087 199 600 64 × 2 = 0 + 0.624 630 038 174 399 201 28;
  • 80) 0.624 630 038 174 399 201 28 × 2 = 1 + 0.249 260 076 348 798 402 56;
  • 81) 0.249 260 076 348 798 402 56 × 2 = 0 + 0.498 520 152 697 596 805 12;
  • 82) 0.498 520 152 697 596 805 12 × 2 = 0 + 0.997 040 305 395 193 610 24;
  • 83) 0.997 040 305 395 193 610 24 × 2 = 1 + 0.994 080 610 790 387 220 48;
  • 84) 0.994 080 610 790 387 220 48 × 2 = 1 + 0.988 161 221 580 774 440 96;
  • 85) 0.988 161 221 580 774 440 96 × 2 = 1 + 0.976 322 443 161 548 881 92;
  • 86) 0.976 322 443 161 548 881 92 × 2 = 1 + 0.952 644 886 323 097 763 84;
  • 87) 0.952 644 886 323 097 763 84 × 2 = 1 + 0.905 289 772 646 195 527 68;
  • 88) 0.905 289 772 646 195 527 68 × 2 = 1 + 0.810 579 545 292 391 055 36;
  • 89) 0.810 579 545 292 391 055 36 × 2 = 1 + 0.621 159 090 584 782 110 72;
  • 90) 0.621 159 090 584 782 110 72 × 2 = 1 + 0.242 318 181 169 564 221 44;
  • 91) 0.242 318 181 169 564 221 44 × 2 = 0 + 0.484 636 362 339 128 442 88;
  • 92) 0.484 636 362 339 128 442 88 × 2 = 0 + 0.969 272 724 678 256 885 76;
  • 93) 0.969 272 724 678 256 885 76 × 2 = 1 + 0.938 545 449 356 513 771 52;
  • 94) 0.938 545 449 356 513 771 52 × 2 = 1 + 0.877 090 898 713 027 543 04;
  • 95) 0.877 090 898 713 027 543 04 × 2 = 1 + 0.754 181 797 426 055 086 08;

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:


0.000 000 000 000 176 557 06(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 1110 0110 0111 0101 0011 1111 1100 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 06(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 1110 0110 0111 0101 0011 1111 1100 111(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:


0.000 000 000 000 176 557 06(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 1110 0110 0111 0101 0011 1111 1100 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0110 1110 0110 0111 0101 0011 1111 1100 111(2) × 20 =

1.1000 1101 1001 0010 0011 0111 0011 0011 1010 1001 1111 1110 0111(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 0011 0111 0011 0011 1010 1001 1111 1110 0111


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 0011 0111 0011 0011 1010 1001 1111 1110 0111 =

1000 1101 1001 0010 0011 0111 0011 0011 1010 1001 1111 1110 0111


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 0011 0111 0011 0011 1010 1001 1111 1110 0111


Decimal number -0.000 000 000 000 176 557 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0011 0111 0011 0011 1010 1001 1111 1110 0111

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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