-0.000 000 000 000 176 557 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 74(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 74(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 74| = 0.000 000 000 000 176 557 74


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 74.

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 74 × 2 = 0 + 0.000 000 000 000 353 115 48;
  • 2) 0.000 000 000 000 353 115 48 × 2 = 0 + 0.000 000 000 000 706 230 96;
  • 3) 0.000 000 000 000 706 230 96 × 2 = 0 + 0.000 000 000 001 412 461 92;
  • 4) 0.000 000 000 001 412 461 92 × 2 = 0 + 0.000 000 000 002 824 923 84;
  • 5) 0.000 000 000 002 824 923 84 × 2 = 0 + 0.000 000 000 005 649 847 68;
  • 6) 0.000 000 000 005 649 847 68 × 2 = 0 + 0.000 000 000 011 299 695 36;
  • 7) 0.000 000 000 011 299 695 36 × 2 = 0 + 0.000 000 000 022 599 390 72;
  • 8) 0.000 000 000 022 599 390 72 × 2 = 0 + 0.000 000 000 045 198 781 44;
  • 9) 0.000 000 000 045 198 781 44 × 2 = 0 + 0.000 000 000 090 397 562 88;
  • 10) 0.000 000 000 090 397 562 88 × 2 = 0 + 0.000 000 000 180 795 125 76;
  • 11) 0.000 000 000 180 795 125 76 × 2 = 0 + 0.000 000 000 361 590 251 52;
  • 12) 0.000 000 000 361 590 251 52 × 2 = 0 + 0.000 000 000 723 180 503 04;
  • 13) 0.000 000 000 723 180 503 04 × 2 = 0 + 0.000 000 001 446 361 006 08;
  • 14) 0.000 000 001 446 361 006 08 × 2 = 0 + 0.000 000 002 892 722 012 16;
  • 15) 0.000 000 002 892 722 012 16 × 2 = 0 + 0.000 000 005 785 444 024 32;
  • 16) 0.000 000 005 785 444 024 32 × 2 = 0 + 0.000 000 011 570 888 048 64;
  • 17) 0.000 000 011 570 888 048 64 × 2 = 0 + 0.000 000 023 141 776 097 28;
  • 18) 0.000 000 023 141 776 097 28 × 2 = 0 + 0.000 000 046 283 552 194 56;
  • 19) 0.000 000 046 283 552 194 56 × 2 = 0 + 0.000 000 092 567 104 389 12;
  • 20) 0.000 000 092 567 104 389 12 × 2 = 0 + 0.000 000 185 134 208 778 24;
  • 21) 0.000 000 185 134 208 778 24 × 2 = 0 + 0.000 000 370 268 417 556 48;
  • 22) 0.000 000 370 268 417 556 48 × 2 = 0 + 0.000 000 740 536 835 112 96;
  • 23) 0.000 000 740 536 835 112 96 × 2 = 0 + 0.000 001 481 073 670 225 92;
  • 24) 0.000 001 481 073 670 225 92 × 2 = 0 + 0.000 002 962 147 340 451 84;
  • 25) 0.000 002 962 147 340 451 84 × 2 = 0 + 0.000 005 924 294 680 903 68;
  • 26) 0.000 005 924 294 680 903 68 × 2 = 0 + 0.000 011 848 589 361 807 36;
  • 27) 0.000 011 848 589 361 807 36 × 2 = 0 + 0.000 023 697 178 723 614 72;
  • 28) 0.000 023 697 178 723 614 72 × 2 = 0 + 0.000 047 394 357 447 229 44;
  • 29) 0.000 047 394 357 447 229 44 × 2 = 0 + 0.000 094 788 714 894 458 88;
  • 30) 0.000 094 788 714 894 458 88 × 2 = 0 + 0.000 189 577 429 788 917 76;
  • 31) 0.000 189 577 429 788 917 76 × 2 = 0 + 0.000 379 154 859 577 835 52;
  • 32) 0.000 379 154 859 577 835 52 × 2 = 0 + 0.000 758 309 719 155 671 04;
  • 33) 0.000 758 309 719 155 671 04 × 2 = 0 + 0.001 516 619 438 311 342 08;
  • 34) 0.001 516 619 438 311 342 08 × 2 = 0 + 0.003 033 238 876 622 684 16;
  • 35) 0.003 033 238 876 622 684 16 × 2 = 0 + 0.006 066 477 753 245 368 32;
  • 36) 0.006 066 477 753 245 368 32 × 2 = 0 + 0.012 132 955 506 490 736 64;
  • 37) 0.012 132 955 506 490 736 64 × 2 = 0 + 0.024 265 911 012 981 473 28;
  • 38) 0.024 265 911 012 981 473 28 × 2 = 0 + 0.048 531 822 025 962 946 56;
  • 39) 0.048 531 822 025 962 946 56 × 2 = 0 + 0.097 063 644 051 925 893 12;
  • 40) 0.097 063 644 051 925 893 12 × 2 = 0 + 0.194 127 288 103 851 786 24;
  • 41) 0.194 127 288 103 851 786 24 × 2 = 0 + 0.388 254 576 207 703 572 48;
  • 42) 0.388 254 576 207 703 572 48 × 2 = 0 + 0.776 509 152 415 407 144 96;
  • 43) 0.776 509 152 415 407 144 96 × 2 = 1 + 0.553 018 304 830 814 289 92;
  • 44) 0.553 018 304 830 814 289 92 × 2 = 1 + 0.106 036 609 661 628 579 84;
  • 45) 0.106 036 609 661 628 579 84 × 2 = 0 + 0.212 073 219 323 257 159 68;
  • 46) 0.212 073 219 323 257 159 68 × 2 = 0 + 0.424 146 438 646 514 319 36;
  • 47) 0.424 146 438 646 514 319 36 × 2 = 0 + 0.848 292 877 293 028 638 72;
  • 48) 0.848 292 877 293 028 638 72 × 2 = 1 + 0.696 585 754 586 057 277 44;
  • 49) 0.696 585 754 586 057 277 44 × 2 = 1 + 0.393 171 509 172 114 554 88;
  • 50) 0.393 171 509 172 114 554 88 × 2 = 0 + 0.786 343 018 344 229 109 76;
  • 51) 0.786 343 018 344 229 109 76 × 2 = 1 + 0.572 686 036 688 458 219 52;
  • 52) 0.572 686 036 688 458 219 52 × 2 = 1 + 0.145 372 073 376 916 439 04;
  • 53) 0.145 372 073 376 916 439 04 × 2 = 0 + 0.290 744 146 753 832 878 08;
  • 54) 0.290 744 146 753 832 878 08 × 2 = 0 + 0.581 488 293 507 665 756 16;
  • 55) 0.581 488 293 507 665 756 16 × 2 = 1 + 0.162 976 587 015 331 512 32;
  • 56) 0.162 976 587 015 331 512 32 × 2 = 0 + 0.325 953 174 030 663 024 64;
  • 57) 0.325 953 174 030 663 024 64 × 2 = 0 + 0.651 906 348 061 326 049 28;
  • 58) 0.651 906 348 061 326 049 28 × 2 = 1 + 0.303 812 696 122 652 098 56;
  • 59) 0.303 812 696 122 652 098 56 × 2 = 0 + 0.607 625 392 245 304 197 12;
  • 60) 0.607 625 392 245 304 197 12 × 2 = 1 + 0.215 250 784 490 608 394 24;
  • 61) 0.215 250 784 490 608 394 24 × 2 = 0 + 0.430 501 568 981 216 788 48;
  • 62) 0.430 501 568 981 216 788 48 × 2 = 0 + 0.861 003 137 962 433 576 96;
  • 63) 0.861 003 137 962 433 576 96 × 2 = 1 + 0.722 006 275 924 867 153 92;
  • 64) 0.722 006 275 924 867 153 92 × 2 = 1 + 0.444 012 551 849 734 307 84;
  • 65) 0.444 012 551 849 734 307 84 × 2 = 0 + 0.888 025 103 699 468 615 68;
  • 66) 0.888 025 103 699 468 615 68 × 2 = 1 + 0.776 050 207 398 937 231 36;
  • 67) 0.776 050 207 398 937 231 36 × 2 = 1 + 0.552 100 414 797 874 462 72;
  • 68) 0.552 100 414 797 874 462 72 × 2 = 1 + 0.104 200 829 595 748 925 44;
  • 69) 0.104 200 829 595 748 925 44 × 2 = 0 + 0.208 401 659 191 497 850 88;
  • 70) 0.208 401 659 191 497 850 88 × 2 = 0 + 0.416 803 318 382 995 701 76;
  • 71) 0.416 803 318 382 995 701 76 × 2 = 0 + 0.833 606 636 765 991 403 52;
  • 72) 0.833 606 636 765 991 403 52 × 2 = 1 + 0.667 213 273 531 982 807 04;
  • 73) 0.667 213 273 531 982 807 04 × 2 = 1 + 0.334 426 547 063 965 614 08;
  • 74) 0.334 426 547 063 965 614 08 × 2 = 0 + 0.668 853 094 127 931 228 16;
  • 75) 0.668 853 094 127 931 228 16 × 2 = 1 + 0.337 706 188 255 862 456 32;
  • 76) 0.337 706 188 255 862 456 32 × 2 = 0 + 0.675 412 376 511 724 912 64;
  • 77) 0.675 412 376 511 724 912 64 × 2 = 1 + 0.350 824 753 023 449 825 28;
  • 78) 0.350 824 753 023 449 825 28 × 2 = 0 + 0.701 649 506 046 899 650 56;
  • 79) 0.701 649 506 046 899 650 56 × 2 = 1 + 0.403 299 012 093 799 301 12;
  • 80) 0.403 299 012 093 799 301 12 × 2 = 0 + 0.806 598 024 187 598 602 24;
  • 81) 0.806 598 024 187 598 602 24 × 2 = 1 + 0.613 196 048 375 197 204 48;
  • 82) 0.613 196 048 375 197 204 48 × 2 = 1 + 0.226 392 096 750 394 408 96;
  • 83) 0.226 392 096 750 394 408 96 × 2 = 0 + 0.452 784 193 500 788 817 92;
  • 84) 0.452 784 193 500 788 817 92 × 2 = 0 + 0.905 568 387 001 577 635 84;
  • 85) 0.905 568 387 001 577 635 84 × 2 = 1 + 0.811 136 774 003 155 271 68;
  • 86) 0.811 136 774 003 155 271 68 × 2 = 1 + 0.622 273 548 006 310 543 36;
  • 87) 0.622 273 548 006 310 543 36 × 2 = 1 + 0.244 547 096 012 621 086 72;
  • 88) 0.244 547 096 012 621 086 72 × 2 = 0 + 0.489 094 192 025 242 173 44;
  • 89) 0.489 094 192 025 242 173 44 × 2 = 0 + 0.978 188 384 050 484 346 88;
  • 90) 0.978 188 384 050 484 346 88 × 2 = 1 + 0.956 376 768 100 968 693 76;
  • 91) 0.956 376 768 100 968 693 76 × 2 = 1 + 0.912 753 536 201 937 387 52;
  • 92) 0.912 753 536 201 937 387 52 × 2 = 1 + 0.825 507 072 403 874 775 04;
  • 93) 0.825 507 072 403 874 775 04 × 2 = 1 + 0.651 014 144 807 749 550 08;
  • 94) 0.651 014 144 807 749 550 08 × 2 = 1 + 0.302 028 289 615 499 100 16;
  • 95) 0.302 028 289 615 499 100 16 × 2 = 0 + 0.604 056 579 230 998 200 32;

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 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0011 0111 0001 1010 1010 1100 1110 0111 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0011 0111 0001 1010 1010 1100 1110 0111 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:


0.000 000 000 000 176 557 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0011 0111 0001 1010 1010 1100 1110 0111 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0011 0111 0001 1010 1010 1100 1110 0111 110(2) × 20 =

1.1000 1101 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 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 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 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 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 1110 =

1000 1101 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 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 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 1110


Decimal number -0.000 000 000 000 176 557 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1001 1011 1000 1101 0101 0110 0111 0011 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