-0.000 000 000 000 176 551 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 551 6(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
-0.000 000 000 000 176 551 6(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:

|-0.000 000 000 000 176 551 6| = 0.000 000 000 000 176 551 6


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: 0.000 000 000 000 176 551 6.

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) 0.000 000 000 000 176 551 6 × 2 = 0 + 0.000 000 000 000 353 103 2;
  • 2) 0.000 000 000 000 353 103 2 × 2 = 0 + 0.000 000 000 000 706 206 4;
  • 3) 0.000 000 000 000 706 206 4 × 2 = 0 + 0.000 000 000 001 412 412 8;
  • 4) 0.000 000 000 001 412 412 8 × 2 = 0 + 0.000 000 000 002 824 825 6;
  • 5) 0.000 000 000 002 824 825 6 × 2 = 0 + 0.000 000 000 005 649 651 2;
  • 6) 0.000 000 000 005 649 651 2 × 2 = 0 + 0.000 000 000 011 299 302 4;
  • 7) 0.000 000 000 011 299 302 4 × 2 = 0 + 0.000 000 000 022 598 604 8;
  • 8) 0.000 000 000 022 598 604 8 × 2 = 0 + 0.000 000 000 045 197 209 6;
  • 9) 0.000 000 000 045 197 209 6 × 2 = 0 + 0.000 000 000 090 394 419 2;
  • 10) 0.000 000 000 090 394 419 2 × 2 = 0 + 0.000 000 000 180 788 838 4;
  • 11) 0.000 000 000 180 788 838 4 × 2 = 0 + 0.000 000 000 361 577 676 8;
  • 12) 0.000 000 000 361 577 676 8 × 2 = 0 + 0.000 000 000 723 155 353 6;
  • 13) 0.000 000 000 723 155 353 6 × 2 = 0 + 0.000 000 001 446 310 707 2;
  • 14) 0.000 000 001 446 310 707 2 × 2 = 0 + 0.000 000 002 892 621 414 4;
  • 15) 0.000 000 002 892 621 414 4 × 2 = 0 + 0.000 000 005 785 242 828 8;
  • 16) 0.000 000 005 785 242 828 8 × 2 = 0 + 0.000 000 011 570 485 657 6;
  • 17) 0.000 000 011 570 485 657 6 × 2 = 0 + 0.000 000 023 140 971 315 2;
  • 18) 0.000 000 023 140 971 315 2 × 2 = 0 + 0.000 000 046 281 942 630 4;
  • 19) 0.000 000 046 281 942 630 4 × 2 = 0 + 0.000 000 092 563 885 260 8;
  • 20) 0.000 000 092 563 885 260 8 × 2 = 0 + 0.000 000 185 127 770 521 6;
  • 21) 0.000 000 185 127 770 521 6 × 2 = 0 + 0.000 000 370 255 541 043 2;
  • 22) 0.000 000 370 255 541 043 2 × 2 = 0 + 0.000 000 740 511 082 086 4;
  • 23) 0.000 000 740 511 082 086 4 × 2 = 0 + 0.000 001 481 022 164 172 8;
  • 24) 0.000 001 481 022 164 172 8 × 2 = 0 + 0.000 002 962 044 328 345 6;
  • 25) 0.000 002 962 044 328 345 6 × 2 = 0 + 0.000 005 924 088 656 691 2;
  • 26) 0.000 005 924 088 656 691 2 × 2 = 0 + 0.000 011 848 177 313 382 4;
  • 27) 0.000 011 848 177 313 382 4 × 2 = 0 + 0.000 023 696 354 626 764 8;
  • 28) 0.000 023 696 354 626 764 8 × 2 = 0 + 0.000 047 392 709 253 529 6;
  • 29) 0.000 047 392 709 253 529 6 × 2 = 0 + 0.000 094 785 418 507 059 2;
  • 30) 0.000 094 785 418 507 059 2 × 2 = 0 + 0.000 189 570 837 014 118 4;
  • 31) 0.000 189 570 837 014 118 4 × 2 = 0 + 0.000 379 141 674 028 236 8;
  • 32) 0.000 379 141 674 028 236 8 × 2 = 0 + 0.000 758 283 348 056 473 6;
  • 33) 0.000 758 283 348 056 473 6 × 2 = 0 + 0.001 516 566 696 112 947 2;
  • 34) 0.001 516 566 696 112 947 2 × 2 = 0 + 0.003 033 133 392 225 894 4;
  • 35) 0.003 033 133 392 225 894 4 × 2 = 0 + 0.006 066 266 784 451 788 8;
  • 36) 0.006 066 266 784 451 788 8 × 2 = 0 + 0.012 132 533 568 903 577 6;
  • 37) 0.012 132 533 568 903 577 6 × 2 = 0 + 0.024 265 067 137 807 155 2;
  • 38) 0.024 265 067 137 807 155 2 × 2 = 0 + 0.048 530 134 275 614 310 4;
  • 39) 0.048 530 134 275 614 310 4 × 2 = 0 + 0.097 060 268 551 228 620 8;
  • 40) 0.097 060 268 551 228 620 8 × 2 = 0 + 0.194 120 537 102 457 241 6;
  • 41) 0.194 120 537 102 457 241 6 × 2 = 0 + 0.388 241 074 204 914 483 2;
  • 42) 0.388 241 074 204 914 483 2 × 2 = 0 + 0.776 482 148 409 828 966 4;
  • 43) 0.776 482 148 409 828 966 4 × 2 = 1 + 0.552 964 296 819 657 932 8;
  • 44) 0.552 964 296 819 657 932 8 × 2 = 1 + 0.105 928 593 639 315 865 6;
  • 45) 0.105 928 593 639 315 865 6 × 2 = 0 + 0.211 857 187 278 631 731 2;
  • 46) 0.211 857 187 278 631 731 2 × 2 = 0 + 0.423 714 374 557 263 462 4;
  • 47) 0.423 714 374 557 263 462 4 × 2 = 0 + 0.847 428 749 114 526 924 8;
  • 48) 0.847 428 749 114 526 924 8 × 2 = 1 + 0.694 857 498 229 053 849 6;
  • 49) 0.694 857 498 229 053 849 6 × 2 = 1 + 0.389 714 996 458 107 699 2;
  • 50) 0.389 714 996 458 107 699 2 × 2 = 0 + 0.779 429 992 916 215 398 4;
  • 51) 0.779 429 992 916 215 398 4 × 2 = 1 + 0.558 859 985 832 430 796 8;
  • 52) 0.558 859 985 832 430 796 8 × 2 = 1 + 0.117 719 971 664 861 593 6;
  • 53) 0.117 719 971 664 861 593 6 × 2 = 0 + 0.235 439 943 329 723 187 2;
  • 54) 0.235 439 943 329 723 187 2 × 2 = 0 + 0.470 879 886 659 446 374 4;
  • 55) 0.470 879 886 659 446 374 4 × 2 = 0 + 0.941 759 773 318 892 748 8;
  • 56) 0.941 759 773 318 892 748 8 × 2 = 1 + 0.883 519 546 637 785 497 6;
  • 57) 0.883 519 546 637 785 497 6 × 2 = 1 + 0.767 039 093 275 570 995 2;
  • 58) 0.767 039 093 275 570 995 2 × 2 = 1 + 0.534 078 186 551 141 990 4;
  • 59) 0.534 078 186 551 141 990 4 × 2 = 1 + 0.068 156 373 102 283 980 8;
  • 60) 0.068 156 373 102 283 980 8 × 2 = 0 + 0.136 312 746 204 567 961 6;
  • 61) 0.136 312 746 204 567 961 6 × 2 = 0 + 0.272 625 492 409 135 923 2;
  • 62) 0.272 625 492 409 135 923 2 × 2 = 0 + 0.545 250 984 818 271 846 4;
  • 63) 0.545 250 984 818 271 846 4 × 2 = 1 + 0.090 501 969 636 543 692 8;
  • 64) 0.090 501 969 636 543 692 8 × 2 = 0 + 0.181 003 939 273 087 385 6;
  • 65) 0.181 003 939 273 087 385 6 × 2 = 0 + 0.362 007 878 546 174 771 2;
  • 66) 0.362 007 878 546 174 771 2 × 2 = 0 + 0.724 015 757 092 349 542 4;
  • 67) 0.724 015 757 092 349 542 4 × 2 = 1 + 0.448 031 514 184 699 084 8;
  • 68) 0.448 031 514 184 699 084 8 × 2 = 0 + 0.896 063 028 369 398 169 6;
  • 69) 0.896 063 028 369 398 169 6 × 2 = 1 + 0.792 126 056 738 796 339 2;
  • 70) 0.792 126 056 738 796 339 2 × 2 = 1 + 0.584 252 113 477 592 678 4;
  • 71) 0.584 252 113 477 592 678 4 × 2 = 1 + 0.168 504 226 955 185 356 8;
  • 72) 0.168 504 226 955 185 356 8 × 2 = 0 + 0.337 008 453 910 370 713 6;
  • 73) 0.337 008 453 910 370 713 6 × 2 = 0 + 0.674 016 907 820 741 427 2;
  • 74) 0.674 016 907 820 741 427 2 × 2 = 1 + 0.348 033 815 641 482 854 4;
  • 75) 0.348 033 815 641 482 854 4 × 2 = 0 + 0.696 067 631 282 965 708 8;
  • 76) 0.696 067 631 282 965 708 8 × 2 = 1 + 0.392 135 262 565 931 417 6;
  • 77) 0.392 135 262 565 931 417 6 × 2 = 0 + 0.784 270 525 131 862 835 2;
  • 78) 0.784 270 525 131 862 835 2 × 2 = 1 + 0.568 541 050 263 725 670 4;
  • 79) 0.568 541 050 263 725 670 4 × 2 = 1 + 0.137 082 100 527 451 340 8;
  • 80) 0.137 082 100 527 451 340 8 × 2 = 0 + 0.274 164 201 054 902 681 6;
  • 81) 0.274 164 201 054 902 681 6 × 2 = 0 + 0.548 328 402 109 805 363 2;
  • 82) 0.548 328 402 109 805 363 2 × 2 = 1 + 0.096 656 804 219 610 726 4;
  • 83) 0.096 656 804 219 610 726 4 × 2 = 0 + 0.193 313 608 439 221 452 8;
  • 84) 0.193 313 608 439 221 452 8 × 2 = 0 + 0.386 627 216 878 442 905 6;
  • 85) 0.386 627 216 878 442 905 6 × 2 = 0 + 0.773 254 433 756 885 811 2;
  • 86) 0.773 254 433 756 885 811 2 × 2 = 1 + 0.546 508 867 513 771 622 4;
  • 87) 0.546 508 867 513 771 622 4 × 2 = 1 + 0.093 017 735 027 543 244 8;
  • 88) 0.093 017 735 027 543 244 8 × 2 = 0 + 0.186 035 470 055 086 489 6;
  • 89) 0.186 035 470 055 086 489 6 × 2 = 0 + 0.372 070 940 110 172 979 2;
  • 90) 0.372 070 940 110 172 979 2 × 2 = 0 + 0.744 141 880 220 345 958 4;
  • 91) 0.744 141 880 220 345 958 4 × 2 = 1 + 0.488 283 760 440 691 916 8;
  • 92) 0.488 283 760 440 691 916 8 × 2 = 0 + 0.976 567 520 881 383 833 6;
  • 93) 0.976 567 520 881 383 833 6 × 2 = 1 + 0.953 135 041 762 767 667 2;
  • 94) 0.953 135 041 762 767 667 2 × 2 = 1 + 0.906 270 083 525 535 334 4;
  • 95) 0.906 270 083 525 535 334 4 × 2 = 1 + 0.812 540 167 051 070 668 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:


0.000 000 000 000 176 551 6(10) =

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

6. Positive number before normalization:

0.000 000 000 000 176 551 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1110 0010 0010 1110 0101 0110 0100 0110 0010 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:


0.000 000 000 000 176 551 6(10) =

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

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

1.1000 1101 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 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 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 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 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 0111 =

1000 1101 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 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 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 0111


Decimal number -0.000 000 000 000 176 551 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1111 0001 0001 0111 0010 1011 0010 0011 0001 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