-0.000 000 000 000 176 562 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 562 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
-0.000 000 000 000 176 562 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:

|-0.000 000 000 000 176 562 1| = 0.000 000 000 000 176 562 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: 0.000 000 000 000 176 562 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) 0.000 000 000 000 176 562 1 × 2 = 0 + 0.000 000 000 000 353 124 2;
  • 2) 0.000 000 000 000 353 124 2 × 2 = 0 + 0.000 000 000 000 706 248 4;
  • 3) 0.000 000 000 000 706 248 4 × 2 = 0 + 0.000 000 000 001 412 496 8;
  • 4) 0.000 000 000 001 412 496 8 × 2 = 0 + 0.000 000 000 002 824 993 6;
  • 5) 0.000 000 000 002 824 993 6 × 2 = 0 + 0.000 000 000 005 649 987 2;
  • 6) 0.000 000 000 005 649 987 2 × 2 = 0 + 0.000 000 000 011 299 974 4;
  • 7) 0.000 000 000 011 299 974 4 × 2 = 0 + 0.000 000 000 022 599 948 8;
  • 8) 0.000 000 000 022 599 948 8 × 2 = 0 + 0.000 000 000 045 199 897 6;
  • 9) 0.000 000 000 045 199 897 6 × 2 = 0 + 0.000 000 000 090 399 795 2;
  • 10) 0.000 000 000 090 399 795 2 × 2 = 0 + 0.000 000 000 180 799 590 4;
  • 11) 0.000 000 000 180 799 590 4 × 2 = 0 + 0.000 000 000 361 599 180 8;
  • 12) 0.000 000 000 361 599 180 8 × 2 = 0 + 0.000 000 000 723 198 361 6;
  • 13) 0.000 000 000 723 198 361 6 × 2 = 0 + 0.000 000 001 446 396 723 2;
  • 14) 0.000 000 001 446 396 723 2 × 2 = 0 + 0.000 000 002 892 793 446 4;
  • 15) 0.000 000 002 892 793 446 4 × 2 = 0 + 0.000 000 005 785 586 892 8;
  • 16) 0.000 000 005 785 586 892 8 × 2 = 0 + 0.000 000 011 571 173 785 6;
  • 17) 0.000 000 011 571 173 785 6 × 2 = 0 + 0.000 000 023 142 347 571 2;
  • 18) 0.000 000 023 142 347 571 2 × 2 = 0 + 0.000 000 046 284 695 142 4;
  • 19) 0.000 000 046 284 695 142 4 × 2 = 0 + 0.000 000 092 569 390 284 8;
  • 20) 0.000 000 092 569 390 284 8 × 2 = 0 + 0.000 000 185 138 780 569 6;
  • 21) 0.000 000 185 138 780 569 6 × 2 = 0 + 0.000 000 370 277 561 139 2;
  • 22) 0.000 000 370 277 561 139 2 × 2 = 0 + 0.000 000 740 555 122 278 4;
  • 23) 0.000 000 740 555 122 278 4 × 2 = 0 + 0.000 001 481 110 244 556 8;
  • 24) 0.000 001 481 110 244 556 8 × 2 = 0 + 0.000 002 962 220 489 113 6;
  • 25) 0.000 002 962 220 489 113 6 × 2 = 0 + 0.000 005 924 440 978 227 2;
  • 26) 0.000 005 924 440 978 227 2 × 2 = 0 + 0.000 011 848 881 956 454 4;
  • 27) 0.000 011 848 881 956 454 4 × 2 = 0 + 0.000 023 697 763 912 908 8;
  • 28) 0.000 023 697 763 912 908 8 × 2 = 0 + 0.000 047 395 527 825 817 6;
  • 29) 0.000 047 395 527 825 817 6 × 2 = 0 + 0.000 094 791 055 651 635 2;
  • 30) 0.000 094 791 055 651 635 2 × 2 = 0 + 0.000 189 582 111 303 270 4;
  • 31) 0.000 189 582 111 303 270 4 × 2 = 0 + 0.000 379 164 222 606 540 8;
  • 32) 0.000 379 164 222 606 540 8 × 2 = 0 + 0.000 758 328 445 213 081 6;
  • 33) 0.000 758 328 445 213 081 6 × 2 = 0 + 0.001 516 656 890 426 163 2;
  • 34) 0.001 516 656 890 426 163 2 × 2 = 0 + 0.003 033 313 780 852 326 4;
  • 35) 0.003 033 313 780 852 326 4 × 2 = 0 + 0.006 066 627 561 704 652 8;
  • 36) 0.006 066 627 561 704 652 8 × 2 = 0 + 0.012 133 255 123 409 305 6;
  • 37) 0.012 133 255 123 409 305 6 × 2 = 0 + 0.024 266 510 246 818 611 2;
  • 38) 0.024 266 510 246 818 611 2 × 2 = 0 + 0.048 533 020 493 637 222 4;
  • 39) 0.048 533 020 493 637 222 4 × 2 = 0 + 0.097 066 040 987 274 444 8;
  • 40) 0.097 066 040 987 274 444 8 × 2 = 0 + 0.194 132 081 974 548 889 6;
  • 41) 0.194 132 081 974 548 889 6 × 2 = 0 + 0.388 264 163 949 097 779 2;
  • 42) 0.388 264 163 949 097 779 2 × 2 = 0 + 0.776 528 327 898 195 558 4;
  • 43) 0.776 528 327 898 195 558 4 × 2 = 1 + 0.553 056 655 796 391 116 8;
  • 44) 0.553 056 655 796 391 116 8 × 2 = 1 + 0.106 113 311 592 782 233 6;
  • 45) 0.106 113 311 592 782 233 6 × 2 = 0 + 0.212 226 623 185 564 467 2;
  • 46) 0.212 226 623 185 564 467 2 × 2 = 0 + 0.424 453 246 371 128 934 4;
  • 47) 0.424 453 246 371 128 934 4 × 2 = 0 + 0.848 906 492 742 257 868 8;
  • 48) 0.848 906 492 742 257 868 8 × 2 = 1 + 0.697 812 985 484 515 737 6;
  • 49) 0.697 812 985 484 515 737 6 × 2 = 1 + 0.395 625 970 969 031 475 2;
  • 50) 0.395 625 970 969 031 475 2 × 2 = 0 + 0.791 251 941 938 062 950 4;
  • 51) 0.791 251 941 938 062 950 4 × 2 = 1 + 0.582 503 883 876 125 900 8;
  • 52) 0.582 503 883 876 125 900 8 × 2 = 1 + 0.165 007 767 752 251 801 6;
  • 53) 0.165 007 767 752 251 801 6 × 2 = 0 + 0.330 015 535 504 503 603 2;
  • 54) 0.330 015 535 504 503 603 2 × 2 = 0 + 0.660 031 071 009 007 206 4;
  • 55) 0.660 031 071 009 007 206 4 × 2 = 1 + 0.320 062 142 018 014 412 8;
  • 56) 0.320 062 142 018 014 412 8 × 2 = 0 + 0.640 124 284 036 028 825 6;
  • 57) 0.640 124 284 036 028 825 6 × 2 = 1 + 0.280 248 568 072 057 651 2;
  • 58) 0.280 248 568 072 057 651 2 × 2 = 0 + 0.560 497 136 144 115 302 4;
  • 59) 0.560 497 136 144 115 302 4 × 2 = 1 + 0.120 994 272 288 230 604 8;
  • 60) 0.120 994 272 288 230 604 8 × 2 = 0 + 0.241 988 544 576 461 209 6;
  • 61) 0.241 988 544 576 461 209 6 × 2 = 0 + 0.483 977 089 152 922 419 2;
  • 62) 0.483 977 089 152 922 419 2 × 2 = 0 + 0.967 954 178 305 844 838 4;
  • 63) 0.967 954 178 305 844 838 4 × 2 = 1 + 0.935 908 356 611 689 676 8;
  • 64) 0.935 908 356 611 689 676 8 × 2 = 1 + 0.871 816 713 223 379 353 6;
  • 65) 0.871 816 713 223 379 353 6 × 2 = 1 + 0.743 633 426 446 758 707 2;
  • 66) 0.743 633 426 446 758 707 2 × 2 = 1 + 0.487 266 852 893 517 414 4;
  • 67) 0.487 266 852 893 517 414 4 × 2 = 0 + 0.974 533 705 787 034 828 8;
  • 68) 0.974 533 705 787 034 828 8 × 2 = 1 + 0.949 067 411 574 069 657 6;
  • 69) 0.949 067 411 574 069 657 6 × 2 = 1 + 0.898 134 823 148 139 315 2;
  • 70) 0.898 134 823 148 139 315 2 × 2 = 1 + 0.796 269 646 296 278 630 4;
  • 71) 0.796 269 646 296 278 630 4 × 2 = 1 + 0.592 539 292 592 557 260 8;
  • 72) 0.592 539 292 592 557 260 8 × 2 = 1 + 0.185 078 585 185 114 521 6;
  • 73) 0.185 078 585 185 114 521 6 × 2 = 0 + 0.370 157 170 370 229 043 2;
  • 74) 0.370 157 170 370 229 043 2 × 2 = 0 + 0.740 314 340 740 458 086 4;
  • 75) 0.740 314 340 740 458 086 4 × 2 = 1 + 0.480 628 681 480 916 172 8;
  • 76) 0.480 628 681 480 916 172 8 × 2 = 0 + 0.961 257 362 961 832 345 6;
  • 77) 0.961 257 362 961 832 345 6 × 2 = 1 + 0.922 514 725 923 664 691 2;
  • 78) 0.922 514 725 923 664 691 2 × 2 = 1 + 0.845 029 451 847 329 382 4;
  • 79) 0.845 029 451 847 329 382 4 × 2 = 1 + 0.690 058 903 694 658 764 8;
  • 80) 0.690 058 903 694 658 764 8 × 2 = 1 + 0.380 117 807 389 317 529 6;
  • 81) 0.380 117 807 389 317 529 6 × 2 = 0 + 0.760 235 614 778 635 059 2;
  • 82) 0.760 235 614 778 635 059 2 × 2 = 1 + 0.520 471 229 557 270 118 4;
  • 83) 0.520 471 229 557 270 118 4 × 2 = 1 + 0.040 942 459 114 540 236 8;
  • 84) 0.040 942 459 114 540 236 8 × 2 = 0 + 0.081 884 918 229 080 473 6;
  • 85) 0.081 884 918 229 080 473 6 × 2 = 0 + 0.163 769 836 458 160 947 2;
  • 86) 0.163 769 836 458 160 947 2 × 2 = 0 + 0.327 539 672 916 321 894 4;
  • 87) 0.327 539 672 916 321 894 4 × 2 = 0 + 0.655 079 345 832 643 788 8;
  • 88) 0.655 079 345 832 643 788 8 × 2 = 1 + 0.310 158 691 665 287 577 6;
  • 89) 0.310 158 691 665 287 577 6 × 2 = 0 + 0.620 317 383 330 575 155 2;
  • 90) 0.620 317 383 330 575 155 2 × 2 = 1 + 0.240 634 766 661 150 310 4;
  • 91) 0.240 634 766 661 150 310 4 × 2 = 0 + 0.481 269 533 322 300 620 8;
  • 92) 0.481 269 533 322 300 620 8 × 2 = 0 + 0.962 539 066 644 601 241 6;
  • 93) 0.962 539 066 644 601 241 6 × 2 = 1 + 0.925 078 133 289 202 483 2;
  • 94) 0.925 078 133 289 202 483 2 × 2 = 1 + 0.850 156 266 578 404 966 4;
  • 95) 0.850 156 266 578 404 966 4 × 2 = 1 + 0.700 312 533 156 809 932 8;

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:


0.000 000 000 000 176 562 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 0011 1101 1111 0010 1111 0110 0001 0100 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 562 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 0011 1101 1111 0010 1111 0110 0001 0100 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:


0.000 000 000 000 176 562 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 0011 1101 1111 0010 1111 0110 0001 0100 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 0011 1101 1111 0010 1111 0110 0001 0100 111(2) × 20 =

1.1000 1101 1001 0101 0001 1110 1111 1001 0111 1011 0000 1010 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 0101 0001 1110 1111 1001 0111 1011 0000 1010 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 0101 0001 1110 1111 1001 0111 1011 0000 1010 0111 =

1000 1101 1001 0101 0001 1110 1111 1001 0111 1011 0000 1010 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 0101 0001 1110 1111 1001 0111 1011 0000 1010 0111


Decimal number -0.000 000 000 000 176 562 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0101 0001 1110 1111 1001 0111 1011 0000 1010 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