-0.000 000 000 000 176 556 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 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
-0.000 000 000 000 176 556 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:

|-0.000 000 000 000 176 556 3| = 0.000 000 000 000 176 556 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: 0.000 000 000 000 176 556 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) 0.000 000 000 000 176 556 3 × 2 = 0 + 0.000 000 000 000 353 112 6;
  • 2) 0.000 000 000 000 353 112 6 × 2 = 0 + 0.000 000 000 000 706 225 2;
  • 3) 0.000 000 000 000 706 225 2 × 2 = 0 + 0.000 000 000 001 412 450 4;
  • 4) 0.000 000 000 001 412 450 4 × 2 = 0 + 0.000 000 000 002 824 900 8;
  • 5) 0.000 000 000 002 824 900 8 × 2 = 0 + 0.000 000 000 005 649 801 6;
  • 6) 0.000 000 000 005 649 801 6 × 2 = 0 + 0.000 000 000 011 299 603 2;
  • 7) 0.000 000 000 011 299 603 2 × 2 = 0 + 0.000 000 000 022 599 206 4;
  • 8) 0.000 000 000 022 599 206 4 × 2 = 0 + 0.000 000 000 045 198 412 8;
  • 9) 0.000 000 000 045 198 412 8 × 2 = 0 + 0.000 000 000 090 396 825 6;
  • 10) 0.000 000 000 090 396 825 6 × 2 = 0 + 0.000 000 000 180 793 651 2;
  • 11) 0.000 000 000 180 793 651 2 × 2 = 0 + 0.000 000 000 361 587 302 4;
  • 12) 0.000 000 000 361 587 302 4 × 2 = 0 + 0.000 000 000 723 174 604 8;
  • 13) 0.000 000 000 723 174 604 8 × 2 = 0 + 0.000 000 001 446 349 209 6;
  • 14) 0.000 000 001 446 349 209 6 × 2 = 0 + 0.000 000 002 892 698 419 2;
  • 15) 0.000 000 002 892 698 419 2 × 2 = 0 + 0.000 000 005 785 396 838 4;
  • 16) 0.000 000 005 785 396 838 4 × 2 = 0 + 0.000 000 011 570 793 676 8;
  • 17) 0.000 000 011 570 793 676 8 × 2 = 0 + 0.000 000 023 141 587 353 6;
  • 18) 0.000 000 023 141 587 353 6 × 2 = 0 + 0.000 000 046 283 174 707 2;
  • 19) 0.000 000 046 283 174 707 2 × 2 = 0 + 0.000 000 092 566 349 414 4;
  • 20) 0.000 000 092 566 349 414 4 × 2 = 0 + 0.000 000 185 132 698 828 8;
  • 21) 0.000 000 185 132 698 828 8 × 2 = 0 + 0.000 000 370 265 397 657 6;
  • 22) 0.000 000 370 265 397 657 6 × 2 = 0 + 0.000 000 740 530 795 315 2;
  • 23) 0.000 000 740 530 795 315 2 × 2 = 0 + 0.000 001 481 061 590 630 4;
  • 24) 0.000 001 481 061 590 630 4 × 2 = 0 + 0.000 002 962 123 181 260 8;
  • 25) 0.000 002 962 123 181 260 8 × 2 = 0 + 0.000 005 924 246 362 521 6;
  • 26) 0.000 005 924 246 362 521 6 × 2 = 0 + 0.000 011 848 492 725 043 2;
  • 27) 0.000 011 848 492 725 043 2 × 2 = 0 + 0.000 023 696 985 450 086 4;
  • 28) 0.000 023 696 985 450 086 4 × 2 = 0 + 0.000 047 393 970 900 172 8;
  • 29) 0.000 047 393 970 900 172 8 × 2 = 0 + 0.000 094 787 941 800 345 6;
  • 30) 0.000 094 787 941 800 345 6 × 2 = 0 + 0.000 189 575 883 600 691 2;
  • 31) 0.000 189 575 883 600 691 2 × 2 = 0 + 0.000 379 151 767 201 382 4;
  • 32) 0.000 379 151 767 201 382 4 × 2 = 0 + 0.000 758 303 534 402 764 8;
  • 33) 0.000 758 303 534 402 764 8 × 2 = 0 + 0.001 516 607 068 805 529 6;
  • 34) 0.001 516 607 068 805 529 6 × 2 = 0 + 0.003 033 214 137 611 059 2;
  • 35) 0.003 033 214 137 611 059 2 × 2 = 0 + 0.006 066 428 275 222 118 4;
  • 36) 0.006 066 428 275 222 118 4 × 2 = 0 + 0.012 132 856 550 444 236 8;
  • 37) 0.012 132 856 550 444 236 8 × 2 = 0 + 0.024 265 713 100 888 473 6;
  • 38) 0.024 265 713 100 888 473 6 × 2 = 0 + 0.048 531 426 201 776 947 2;
  • 39) 0.048 531 426 201 776 947 2 × 2 = 0 + 0.097 062 852 403 553 894 4;
  • 40) 0.097 062 852 403 553 894 4 × 2 = 0 + 0.194 125 704 807 107 788 8;
  • 41) 0.194 125 704 807 107 788 8 × 2 = 0 + 0.388 251 409 614 215 577 6;
  • 42) 0.388 251 409 614 215 577 6 × 2 = 0 + 0.776 502 819 228 431 155 2;
  • 43) 0.776 502 819 228 431 155 2 × 2 = 1 + 0.553 005 638 456 862 310 4;
  • 44) 0.553 005 638 456 862 310 4 × 2 = 1 + 0.106 011 276 913 724 620 8;
  • 45) 0.106 011 276 913 724 620 8 × 2 = 0 + 0.212 022 553 827 449 241 6;
  • 46) 0.212 022 553 827 449 241 6 × 2 = 0 + 0.424 045 107 654 898 483 2;
  • 47) 0.424 045 107 654 898 483 2 × 2 = 0 + 0.848 090 215 309 796 966 4;
  • 48) 0.848 090 215 309 796 966 4 × 2 = 1 + 0.696 180 430 619 593 932 8;
  • 49) 0.696 180 430 619 593 932 8 × 2 = 1 + 0.392 360 861 239 187 865 6;
  • 50) 0.392 360 861 239 187 865 6 × 2 = 0 + 0.784 721 722 478 375 731 2;
  • 51) 0.784 721 722 478 375 731 2 × 2 = 1 + 0.569 443 444 956 751 462 4;
  • 52) 0.569 443 444 956 751 462 4 × 2 = 1 + 0.138 886 889 913 502 924 8;
  • 53) 0.138 886 889 913 502 924 8 × 2 = 0 + 0.277 773 779 827 005 849 6;
  • 54) 0.277 773 779 827 005 849 6 × 2 = 0 + 0.555 547 559 654 011 699 2;
  • 55) 0.555 547 559 654 011 699 2 × 2 = 1 + 0.111 095 119 308 023 398 4;
  • 56) 0.111 095 119 308 023 398 4 × 2 = 0 + 0.222 190 238 616 046 796 8;
  • 57) 0.222 190 238 616 046 796 8 × 2 = 0 + 0.444 380 477 232 093 593 6;
  • 58) 0.444 380 477 232 093 593 6 × 2 = 0 + 0.888 760 954 464 187 187 2;
  • 59) 0.888 760 954 464 187 187 2 × 2 = 1 + 0.777 521 908 928 374 374 4;
  • 60) 0.777 521 908 928 374 374 4 × 2 = 1 + 0.555 043 817 856 748 748 8;
  • 61) 0.555 043 817 856 748 748 8 × 2 = 1 + 0.110 087 635 713 497 497 6;
  • 62) 0.110 087 635 713 497 497 6 × 2 = 0 + 0.220 175 271 426 994 995 2;
  • 63) 0.220 175 271 426 994 995 2 × 2 = 0 + 0.440 350 542 853 989 990 4;
  • 64) 0.440 350 542 853 989 990 4 × 2 = 0 + 0.880 701 085 707 979 980 8;
  • 65) 0.880 701 085 707 979 980 8 × 2 = 1 + 0.761 402 171 415 959 961 6;
  • 66) 0.761 402 171 415 959 961 6 × 2 = 1 + 0.522 804 342 831 919 923 2;
  • 67) 0.522 804 342 831 919 923 2 × 2 = 1 + 0.045 608 685 663 839 846 4;
  • 68) 0.045 608 685 663 839 846 4 × 2 = 0 + 0.091 217 371 327 679 692 8;
  • 69) 0.091 217 371 327 679 692 8 × 2 = 0 + 0.182 434 742 655 359 385 6;
  • 70) 0.182 434 742 655 359 385 6 × 2 = 0 + 0.364 869 485 310 718 771 2;
  • 71) 0.364 869 485 310 718 771 2 × 2 = 0 + 0.729 738 970 621 437 542 4;
  • 72) 0.729 738 970 621 437 542 4 × 2 = 1 + 0.459 477 941 242 875 084 8;
  • 73) 0.459 477 941 242 875 084 8 × 2 = 0 + 0.918 955 882 485 750 169 6;
  • 74) 0.918 955 882 485 750 169 6 × 2 = 1 + 0.837 911 764 971 500 339 2;
  • 75) 0.837 911 764 971 500 339 2 × 2 = 1 + 0.675 823 529 943 000 678 4;
  • 76) 0.675 823 529 943 000 678 4 × 2 = 1 + 0.351 647 059 886 001 356 8;
  • 77) 0.351 647 059 886 001 356 8 × 2 = 0 + 0.703 294 119 772 002 713 6;
  • 78) 0.703 294 119 772 002 713 6 × 2 = 1 + 0.406 588 239 544 005 427 2;
  • 79) 0.406 588 239 544 005 427 2 × 2 = 0 + 0.813 176 479 088 010 854 4;
  • 80) 0.813 176 479 088 010 854 4 × 2 = 1 + 0.626 352 958 176 021 708 8;
  • 81) 0.626 352 958 176 021 708 8 × 2 = 1 + 0.252 705 916 352 043 417 6;
  • 82) 0.252 705 916 352 043 417 6 × 2 = 0 + 0.505 411 832 704 086 835 2;
  • 83) 0.505 411 832 704 086 835 2 × 2 = 1 + 0.010 823 665 408 173 670 4;
  • 84) 0.010 823 665 408 173 670 4 × 2 = 0 + 0.021 647 330 816 347 340 8;
  • 85) 0.021 647 330 816 347 340 8 × 2 = 0 + 0.043 294 661 632 694 681 6;
  • 86) 0.043 294 661 632 694 681 6 × 2 = 0 + 0.086 589 323 265 389 363 2;
  • 87) 0.086 589 323 265 389 363 2 × 2 = 0 + 0.173 178 646 530 778 726 4;
  • 88) 0.173 178 646 530 778 726 4 × 2 = 0 + 0.346 357 293 061 557 452 8;
  • 89) 0.346 357 293 061 557 452 8 × 2 = 0 + 0.692 714 586 123 114 905 6;
  • 90) 0.692 714 586 123 114 905 6 × 2 = 1 + 0.385 429 172 246 229 811 2;
  • 91) 0.385 429 172 246 229 811 2 × 2 = 0 + 0.770 858 344 492 459 622 4;
  • 92) 0.770 858 344 492 459 622 4 × 2 = 1 + 0.541 716 688 984 919 244 8;
  • 93) 0.541 716 688 984 919 244 8 × 2 = 1 + 0.083 433 377 969 838 489 6;
  • 94) 0.083 433 377 969 838 489 6 × 2 = 0 + 0.166 866 755 939 676 979 2;
  • 95) 0.166 866 755 939 676 979 2 × 2 = 0 + 0.333 733 511 879 353 958 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:


0.000 000 000 000 176 556 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1000 1110 0001 0111 0101 1010 0000 0101 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1000 1110 0001 0111 0101 1010 0000 0101 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:


0.000 000 000 000 176 556 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1000 1110 0001 0111 0101 1010 0000 0101 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1000 1110 0001 0111 0101 1010 0000 0101 100(2) × 20 =

1.1000 1101 1001 0001 1100 0111 0000 1011 1010 1101 0000 0010 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 0001 1100 0111 0000 1011 1010 1101 0000 0010 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 0001 1100 0111 0000 1011 1010 1101 0000 0010 1100 =

1000 1101 1001 0001 1100 0111 0000 1011 1010 1101 0000 0010 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 0001 1100 0111 0000 1011 1010 1101 0000 0010 1100


Decimal number -0.000 000 000 000 176 556 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1100 0111 0000 1011 1010 1101 0000 0010 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