-0.000 000 000 000 176 564 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 564 3(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
-0.000 000 000 000 176 564 3(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:

|-0.000 000 000 000 176 564 3| = 0.000 000 000 000 176 564 3


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: 0.000 000 000 000 176 564 3.

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) 0.000 000 000 000 176 564 3 × 2 = 0 + 0.000 000 000 000 353 128 6;
  • 2) 0.000 000 000 000 353 128 6 × 2 = 0 + 0.000 000 000 000 706 257 2;
  • 3) 0.000 000 000 000 706 257 2 × 2 = 0 + 0.000 000 000 001 412 514 4;
  • 4) 0.000 000 000 001 412 514 4 × 2 = 0 + 0.000 000 000 002 825 028 8;
  • 5) 0.000 000 000 002 825 028 8 × 2 = 0 + 0.000 000 000 005 650 057 6;
  • 6) 0.000 000 000 005 650 057 6 × 2 = 0 + 0.000 000 000 011 300 115 2;
  • 7) 0.000 000 000 011 300 115 2 × 2 = 0 + 0.000 000 000 022 600 230 4;
  • 8) 0.000 000 000 022 600 230 4 × 2 = 0 + 0.000 000 000 045 200 460 8;
  • 9) 0.000 000 000 045 200 460 8 × 2 = 0 + 0.000 000 000 090 400 921 6;
  • 10) 0.000 000 000 090 400 921 6 × 2 = 0 + 0.000 000 000 180 801 843 2;
  • 11) 0.000 000 000 180 801 843 2 × 2 = 0 + 0.000 000 000 361 603 686 4;
  • 12) 0.000 000 000 361 603 686 4 × 2 = 0 + 0.000 000 000 723 207 372 8;
  • 13) 0.000 000 000 723 207 372 8 × 2 = 0 + 0.000 000 001 446 414 745 6;
  • 14) 0.000 000 001 446 414 745 6 × 2 = 0 + 0.000 000 002 892 829 491 2;
  • 15) 0.000 000 002 892 829 491 2 × 2 = 0 + 0.000 000 005 785 658 982 4;
  • 16) 0.000 000 005 785 658 982 4 × 2 = 0 + 0.000 000 011 571 317 964 8;
  • 17) 0.000 000 011 571 317 964 8 × 2 = 0 + 0.000 000 023 142 635 929 6;
  • 18) 0.000 000 023 142 635 929 6 × 2 = 0 + 0.000 000 046 285 271 859 2;
  • 19) 0.000 000 046 285 271 859 2 × 2 = 0 + 0.000 000 092 570 543 718 4;
  • 20) 0.000 000 092 570 543 718 4 × 2 = 0 + 0.000 000 185 141 087 436 8;
  • 21) 0.000 000 185 141 087 436 8 × 2 = 0 + 0.000 000 370 282 174 873 6;
  • 22) 0.000 000 370 282 174 873 6 × 2 = 0 + 0.000 000 740 564 349 747 2;
  • 23) 0.000 000 740 564 349 747 2 × 2 = 0 + 0.000 001 481 128 699 494 4;
  • 24) 0.000 001 481 128 699 494 4 × 2 = 0 + 0.000 002 962 257 398 988 8;
  • 25) 0.000 002 962 257 398 988 8 × 2 = 0 + 0.000 005 924 514 797 977 6;
  • 26) 0.000 005 924 514 797 977 6 × 2 = 0 + 0.000 011 849 029 595 955 2;
  • 27) 0.000 011 849 029 595 955 2 × 2 = 0 + 0.000 023 698 059 191 910 4;
  • 28) 0.000 023 698 059 191 910 4 × 2 = 0 + 0.000 047 396 118 383 820 8;
  • 29) 0.000 047 396 118 383 820 8 × 2 = 0 + 0.000 094 792 236 767 641 6;
  • 30) 0.000 094 792 236 767 641 6 × 2 = 0 + 0.000 189 584 473 535 283 2;
  • 31) 0.000 189 584 473 535 283 2 × 2 = 0 + 0.000 379 168 947 070 566 4;
  • 32) 0.000 379 168 947 070 566 4 × 2 = 0 + 0.000 758 337 894 141 132 8;
  • 33) 0.000 758 337 894 141 132 8 × 2 = 0 + 0.001 516 675 788 282 265 6;
  • 34) 0.001 516 675 788 282 265 6 × 2 = 0 + 0.003 033 351 576 564 531 2;
  • 35) 0.003 033 351 576 564 531 2 × 2 = 0 + 0.006 066 703 153 129 062 4;
  • 36) 0.006 066 703 153 129 062 4 × 2 = 0 + 0.012 133 406 306 258 124 8;
  • 37) 0.012 133 406 306 258 124 8 × 2 = 0 + 0.024 266 812 612 516 249 6;
  • 38) 0.024 266 812 612 516 249 6 × 2 = 0 + 0.048 533 625 225 032 499 2;
  • 39) 0.048 533 625 225 032 499 2 × 2 = 0 + 0.097 067 250 450 064 998 4;
  • 40) 0.097 067 250 450 064 998 4 × 2 = 0 + 0.194 134 500 900 129 996 8;
  • 41) 0.194 134 500 900 129 996 8 × 2 = 0 + 0.388 269 001 800 259 993 6;
  • 42) 0.388 269 001 800 259 993 6 × 2 = 0 + 0.776 538 003 600 519 987 2;
  • 43) 0.776 538 003 600 519 987 2 × 2 = 1 + 0.553 076 007 201 039 974 4;
  • 44) 0.553 076 007 201 039 974 4 × 2 = 1 + 0.106 152 014 402 079 948 8;
  • 45) 0.106 152 014 402 079 948 8 × 2 = 0 + 0.212 304 028 804 159 897 6;
  • 46) 0.212 304 028 804 159 897 6 × 2 = 0 + 0.424 608 057 608 319 795 2;
  • 47) 0.424 608 057 608 319 795 2 × 2 = 0 + 0.849 216 115 216 639 590 4;
  • 48) 0.849 216 115 216 639 590 4 × 2 = 1 + 0.698 432 230 433 279 180 8;
  • 49) 0.698 432 230 433 279 180 8 × 2 = 1 + 0.396 864 460 866 558 361 6;
  • 50) 0.396 864 460 866 558 361 6 × 2 = 0 + 0.793 728 921 733 116 723 2;
  • 51) 0.793 728 921 733 116 723 2 × 2 = 1 + 0.587 457 843 466 233 446 4;
  • 52) 0.587 457 843 466 233 446 4 × 2 = 1 + 0.174 915 686 932 466 892 8;
  • 53) 0.174 915 686 932 466 892 8 × 2 = 0 + 0.349 831 373 864 933 785 6;
  • 54) 0.349 831 373 864 933 785 6 × 2 = 0 + 0.699 662 747 729 867 571 2;
  • 55) 0.699 662 747 729 867 571 2 × 2 = 1 + 0.399 325 495 459 735 142 4;
  • 56) 0.399 325 495 459 735 142 4 × 2 = 0 + 0.798 650 990 919 470 284 8;
  • 57) 0.798 650 990 919 470 284 8 × 2 = 1 + 0.597 301 981 838 940 569 6;
  • 58) 0.597 301 981 838 940 569 6 × 2 = 1 + 0.194 603 963 677 881 139 2;
  • 59) 0.194 603 963 677 881 139 2 × 2 = 0 + 0.389 207 927 355 762 278 4;
  • 60) 0.389 207 927 355 762 278 4 × 2 = 0 + 0.778 415 854 711 524 556 8;
  • 61) 0.778 415 854 711 524 556 8 × 2 = 1 + 0.556 831 709 423 049 113 6;
  • 62) 0.556 831 709 423 049 113 6 × 2 = 1 + 0.113 663 418 846 098 227 2;
  • 63) 0.113 663 418 846 098 227 2 × 2 = 0 + 0.227 326 837 692 196 454 4;
  • 64) 0.227 326 837 692 196 454 4 × 2 = 0 + 0.454 653 675 384 392 908 8;
  • 65) 0.454 653 675 384 392 908 8 × 2 = 0 + 0.909 307 350 768 785 817 6;
  • 66) 0.909 307 350 768 785 817 6 × 2 = 1 + 0.818 614 701 537 571 635 2;
  • 67) 0.818 614 701 537 571 635 2 × 2 = 1 + 0.637 229 403 075 143 270 4;
  • 68) 0.637 229 403 075 143 270 4 × 2 = 1 + 0.274 458 806 150 286 540 8;
  • 69) 0.274 458 806 150 286 540 8 × 2 = 0 + 0.548 917 612 300 573 081 6;
  • 70) 0.548 917 612 300 573 081 6 × 2 = 1 + 0.097 835 224 601 146 163 2;
  • 71) 0.097 835 224 601 146 163 2 × 2 = 0 + 0.195 670 449 202 292 326 4;
  • 72) 0.195 670 449 202 292 326 4 × 2 = 0 + 0.391 340 898 404 584 652 8;
  • 73) 0.391 340 898 404 584 652 8 × 2 = 0 + 0.782 681 796 809 169 305 6;
  • 74) 0.782 681 796 809 169 305 6 × 2 = 1 + 0.565 363 593 618 338 611 2;
  • 75) 0.565 363 593 618 338 611 2 × 2 = 1 + 0.130 727 187 236 677 222 4;
  • 76) 0.130 727 187 236 677 222 4 × 2 = 0 + 0.261 454 374 473 354 444 8;
  • 77) 0.261 454 374 473 354 444 8 × 2 = 0 + 0.522 908 748 946 708 889 6;
  • 78) 0.522 908 748 946 708 889 6 × 2 = 1 + 0.045 817 497 893 417 779 2;
  • 79) 0.045 817 497 893 417 779 2 × 2 = 0 + 0.091 634 995 786 835 558 4;
  • 80) 0.091 634 995 786 835 558 4 × 2 = 0 + 0.183 269 991 573 671 116 8;
  • 81) 0.183 269 991 573 671 116 8 × 2 = 0 + 0.366 539 983 147 342 233 6;
  • 82) 0.366 539 983 147 342 233 6 × 2 = 0 + 0.733 079 966 294 684 467 2;
  • 83) 0.733 079 966 294 684 467 2 × 2 = 1 + 0.466 159 932 589 368 934 4;
  • 84) 0.466 159 932 589 368 934 4 × 2 = 0 + 0.932 319 865 178 737 868 8;
  • 85) 0.932 319 865 178 737 868 8 × 2 = 1 + 0.864 639 730 357 475 737 6;
  • 86) 0.864 639 730 357 475 737 6 × 2 = 1 + 0.729 279 460 714 951 475 2;
  • 87) 0.729 279 460 714 951 475 2 × 2 = 1 + 0.458 558 921 429 902 950 4;
  • 88) 0.458 558 921 429 902 950 4 × 2 = 0 + 0.917 117 842 859 805 900 8;
  • 89) 0.917 117 842 859 805 900 8 × 2 = 1 + 0.834 235 685 719 611 801 6;
  • 90) 0.834 235 685 719 611 801 6 × 2 = 1 + 0.668 471 371 439 223 603 2;
  • 91) 0.668 471 371 439 223 603 2 × 2 = 1 + 0.336 942 742 878 447 206 4;
  • 92) 0.336 942 742 878 447 206 4 × 2 = 0 + 0.673 885 485 756 894 412 8;
  • 93) 0.673 885 485 756 894 412 8 × 2 = 1 + 0.347 770 971 513 788 825 6;
  • 94) 0.347 770 971 513 788 825 6 × 2 = 0 + 0.695 541 943 027 577 651 2;
  • 95) 0.695 541 943 027 577 651 2 × 2 = 1 + 0.391 083 886 055 155 302 4;

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:


0.000 000 000 000 176 564 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1100 1100 0111 0100 0110 0100 0010 1110 1110 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 564 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1100 1100 0111 0100 0110 0100 0010 1110 1110 101(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:


0.000 000 000 000 176 564 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1100 1100 0111 0100 0110 0100 0010 1110 1110 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1100 1100 0111 0100 0110 0100 0010 1110 1110 101(2) × 20 =

1.1000 1101 1001 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101(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 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101


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 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101 =

1000 1101 1001 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101


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 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101


Decimal number -0.000 000 000 000 176 564 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0110 0110 0011 1010 0011 0010 0001 0111 0111 0101

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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