-0.000 000 000 000 176 557 395 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 395(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
-0.000 000 000 000 176 557 395(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:

|-0.000 000 000 000 176 557 395| = 0.000 000 000 000 176 557 395


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: 0.000 000 000 000 176 557 395.

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) 0.000 000 000 000 176 557 395 × 2 = 0 + 0.000 000 000 000 353 114 79;
  • 2) 0.000 000 000 000 353 114 79 × 2 = 0 + 0.000 000 000 000 706 229 58;
  • 3) 0.000 000 000 000 706 229 58 × 2 = 0 + 0.000 000 000 001 412 459 16;
  • 4) 0.000 000 000 001 412 459 16 × 2 = 0 + 0.000 000 000 002 824 918 32;
  • 5) 0.000 000 000 002 824 918 32 × 2 = 0 + 0.000 000 000 005 649 836 64;
  • 6) 0.000 000 000 005 649 836 64 × 2 = 0 + 0.000 000 000 011 299 673 28;
  • 7) 0.000 000 000 011 299 673 28 × 2 = 0 + 0.000 000 000 022 599 346 56;
  • 8) 0.000 000 000 022 599 346 56 × 2 = 0 + 0.000 000 000 045 198 693 12;
  • 9) 0.000 000 000 045 198 693 12 × 2 = 0 + 0.000 000 000 090 397 386 24;
  • 10) 0.000 000 000 090 397 386 24 × 2 = 0 + 0.000 000 000 180 794 772 48;
  • 11) 0.000 000 000 180 794 772 48 × 2 = 0 + 0.000 000 000 361 589 544 96;
  • 12) 0.000 000 000 361 589 544 96 × 2 = 0 + 0.000 000 000 723 179 089 92;
  • 13) 0.000 000 000 723 179 089 92 × 2 = 0 + 0.000 000 001 446 358 179 84;
  • 14) 0.000 000 001 446 358 179 84 × 2 = 0 + 0.000 000 002 892 716 359 68;
  • 15) 0.000 000 002 892 716 359 68 × 2 = 0 + 0.000 000 005 785 432 719 36;
  • 16) 0.000 000 005 785 432 719 36 × 2 = 0 + 0.000 000 011 570 865 438 72;
  • 17) 0.000 000 011 570 865 438 72 × 2 = 0 + 0.000 000 023 141 730 877 44;
  • 18) 0.000 000 023 141 730 877 44 × 2 = 0 + 0.000 000 046 283 461 754 88;
  • 19) 0.000 000 046 283 461 754 88 × 2 = 0 + 0.000 000 092 566 923 509 76;
  • 20) 0.000 000 092 566 923 509 76 × 2 = 0 + 0.000 000 185 133 847 019 52;
  • 21) 0.000 000 185 133 847 019 52 × 2 = 0 + 0.000 000 370 267 694 039 04;
  • 22) 0.000 000 370 267 694 039 04 × 2 = 0 + 0.000 000 740 535 388 078 08;
  • 23) 0.000 000 740 535 388 078 08 × 2 = 0 + 0.000 001 481 070 776 156 16;
  • 24) 0.000 001 481 070 776 156 16 × 2 = 0 + 0.000 002 962 141 552 312 32;
  • 25) 0.000 002 962 141 552 312 32 × 2 = 0 + 0.000 005 924 283 104 624 64;
  • 26) 0.000 005 924 283 104 624 64 × 2 = 0 + 0.000 011 848 566 209 249 28;
  • 27) 0.000 011 848 566 209 249 28 × 2 = 0 + 0.000 023 697 132 418 498 56;
  • 28) 0.000 023 697 132 418 498 56 × 2 = 0 + 0.000 047 394 264 836 997 12;
  • 29) 0.000 047 394 264 836 997 12 × 2 = 0 + 0.000 094 788 529 673 994 24;
  • 30) 0.000 094 788 529 673 994 24 × 2 = 0 + 0.000 189 577 059 347 988 48;
  • 31) 0.000 189 577 059 347 988 48 × 2 = 0 + 0.000 379 154 118 695 976 96;
  • 32) 0.000 379 154 118 695 976 96 × 2 = 0 + 0.000 758 308 237 391 953 92;
  • 33) 0.000 758 308 237 391 953 92 × 2 = 0 + 0.001 516 616 474 783 907 84;
  • 34) 0.001 516 616 474 783 907 84 × 2 = 0 + 0.003 033 232 949 567 815 68;
  • 35) 0.003 033 232 949 567 815 68 × 2 = 0 + 0.006 066 465 899 135 631 36;
  • 36) 0.006 066 465 899 135 631 36 × 2 = 0 + 0.012 132 931 798 271 262 72;
  • 37) 0.012 132 931 798 271 262 72 × 2 = 0 + 0.024 265 863 596 542 525 44;
  • 38) 0.024 265 863 596 542 525 44 × 2 = 0 + 0.048 531 727 193 085 050 88;
  • 39) 0.048 531 727 193 085 050 88 × 2 = 0 + 0.097 063 454 386 170 101 76;
  • 40) 0.097 063 454 386 170 101 76 × 2 = 0 + 0.194 126 908 772 340 203 52;
  • 41) 0.194 126 908 772 340 203 52 × 2 = 0 + 0.388 253 817 544 680 407 04;
  • 42) 0.388 253 817 544 680 407 04 × 2 = 0 + 0.776 507 635 089 360 814 08;
  • 43) 0.776 507 635 089 360 814 08 × 2 = 1 + 0.553 015 270 178 721 628 16;
  • 44) 0.553 015 270 178 721 628 16 × 2 = 1 + 0.106 030 540 357 443 256 32;
  • 45) 0.106 030 540 357 443 256 32 × 2 = 0 + 0.212 061 080 714 886 512 64;
  • 46) 0.212 061 080 714 886 512 64 × 2 = 0 + 0.424 122 161 429 773 025 28;
  • 47) 0.424 122 161 429 773 025 28 × 2 = 0 + 0.848 244 322 859 546 050 56;
  • 48) 0.848 244 322 859 546 050 56 × 2 = 1 + 0.696 488 645 719 092 101 12;
  • 49) 0.696 488 645 719 092 101 12 × 2 = 1 + 0.392 977 291 438 184 202 24;
  • 50) 0.392 977 291 438 184 202 24 × 2 = 0 + 0.785 954 582 876 368 404 48;
  • 51) 0.785 954 582 876 368 404 48 × 2 = 1 + 0.571 909 165 752 736 808 96;
  • 52) 0.571 909 165 752 736 808 96 × 2 = 1 + 0.143 818 331 505 473 617 92;
  • 53) 0.143 818 331 505 473 617 92 × 2 = 0 + 0.287 636 663 010 947 235 84;
  • 54) 0.287 636 663 010 947 235 84 × 2 = 0 + 0.575 273 326 021 894 471 68;
  • 55) 0.575 273 326 021 894 471 68 × 2 = 1 + 0.150 546 652 043 788 943 36;
  • 56) 0.150 546 652 043 788 943 36 × 2 = 0 + 0.301 093 304 087 577 886 72;
  • 57) 0.301 093 304 087 577 886 72 × 2 = 0 + 0.602 186 608 175 155 773 44;
  • 58) 0.602 186 608 175 155 773 44 × 2 = 1 + 0.204 373 216 350 311 546 88;
  • 59) 0.204 373 216 350 311 546 88 × 2 = 0 + 0.408 746 432 700 623 093 76;
  • 60) 0.408 746 432 700 623 093 76 × 2 = 0 + 0.817 492 865 401 246 187 52;
  • 61) 0.817 492 865 401 246 187 52 × 2 = 1 + 0.634 985 730 802 492 375 04;
  • 62) 0.634 985 730 802 492 375 04 × 2 = 1 + 0.269 971 461 604 984 750 08;
  • 63) 0.269 971 461 604 984 750 08 × 2 = 0 + 0.539 942 923 209 969 500 16;
  • 64) 0.539 942 923 209 969 500 16 × 2 = 1 + 0.079 885 846 419 939 000 32;
  • 65) 0.079 885 846 419 939 000 32 × 2 = 0 + 0.159 771 692 839 878 000 64;
  • 66) 0.159 771 692 839 878 000 64 × 2 = 0 + 0.319 543 385 679 756 001 28;
  • 67) 0.319 543 385 679 756 001 28 × 2 = 0 + 0.639 086 771 359 512 002 56;
  • 68) 0.639 086 771 359 512 002 56 × 2 = 1 + 0.278 173 542 719 024 005 12;
  • 69) 0.278 173 542 719 024 005 12 × 2 = 0 + 0.556 347 085 438 048 010 24;
  • 70) 0.556 347 085 438 048 010 24 × 2 = 1 + 0.112 694 170 876 096 020 48;
  • 71) 0.112 694 170 876 096 020 48 × 2 = 0 + 0.225 388 341 752 192 040 96;
  • 72) 0.225 388 341 752 192 040 96 × 2 = 0 + 0.450 776 683 504 384 081 92;
  • 73) 0.450 776 683 504 384 081 92 × 2 = 0 + 0.901 553 367 008 768 163 84;
  • 74) 0.901 553 367 008 768 163 84 × 2 = 1 + 0.803 106 734 017 536 327 68;
  • 75) 0.803 106 734 017 536 327 68 × 2 = 1 + 0.606 213 468 035 072 655 36;
  • 76) 0.606 213 468 035 072 655 36 × 2 = 1 + 0.212 426 936 070 145 310 72;
  • 77) 0.212 426 936 070 145 310 72 × 2 = 0 + 0.424 853 872 140 290 621 44;
  • 78) 0.424 853 872 140 290 621 44 × 2 = 0 + 0.849 707 744 280 581 242 88;
  • 79) 0.849 707 744 280 581 242 88 × 2 = 1 + 0.699 415 488 561 162 485 76;
  • 80) 0.699 415 488 561 162 485 76 × 2 = 1 + 0.398 830 977 122 324 971 52;
  • 81) 0.398 830 977 122 324 971 52 × 2 = 0 + 0.797 661 954 244 649 943 04;
  • 82) 0.797 661 954 244 649 943 04 × 2 = 1 + 0.595 323 908 489 299 886 08;
  • 83) 0.595 323 908 489 299 886 08 × 2 = 1 + 0.190 647 816 978 599 772 16;
  • 84) 0.190 647 816 978 599 772 16 × 2 = 0 + 0.381 295 633 957 199 544 32;
  • 85) 0.381 295 633 957 199 544 32 × 2 = 0 + 0.762 591 267 914 399 088 64;
  • 86) 0.762 591 267 914 399 088 64 × 2 = 1 + 0.525 182 535 828 798 177 28;
  • 87) 0.525 182 535 828 798 177 28 × 2 = 1 + 0.050 365 071 657 596 354 56;
  • 88) 0.050 365 071 657 596 354 56 × 2 = 0 + 0.100 730 143 315 192 709 12;
  • 89) 0.100 730 143 315 192 709 12 × 2 = 0 + 0.201 460 286 630 385 418 24;
  • 90) 0.201 460 286 630 385 418 24 × 2 = 0 + 0.402 920 573 260 770 836 48;
  • 91) 0.402 920 573 260 770 836 48 × 2 = 0 + 0.805 841 146 521 541 672 96;
  • 92) 0.805 841 146 521 541 672 96 × 2 = 1 + 0.611 682 293 043 083 345 92;
  • 93) 0.611 682 293 043 083 345 92 × 2 = 1 + 0.223 364 586 086 166 691 84;
  • 94) 0.223 364 586 086 166 691 84 × 2 = 0 + 0.446 729 172 172 333 383 68;
  • 95) 0.446 729 172 172 333 383 68 × 2 = 0 + 0.893 458 344 344 666 767 36;

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:


0.000 000 000 000 176 557 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0001 0100 0111 0011 0110 0110 0001 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0001 0100 0111 0011 0110 0110 0001 100(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:


0.000 000 000 000 176 557 395(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0001 0100 0111 0011 0110 0110 0001 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0001 0100 0111 0011 0110 0110 0001 100(2) × 20 =

1.1000 1101 1001 0010 0110 1000 1010 0011 1001 1011 0011 0000 1100(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 0110 1000 1010 0011 1001 1011 0011 0000 1100


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 0110 1000 1010 0011 1001 1011 0011 0000 1100 =

1000 1101 1001 0010 0110 1000 1010 0011 1001 1011 0011 0000 1100


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 0110 1000 1010 0011 1001 1011 0011 0000 1100


Decimal number -0.000 000 000 000 176 557 395 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1000 1010 0011 1001 1011 0011 0000 1100

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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