-0.000 000 000 000 014 386 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 386 2(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
-0.000 000 000 000 014 386 2(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:

|-0.000 000 000 000 014 386 2| = 0.000 000 000 000 014 386 2


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: 0.000 000 000 000 014 386 2.

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) 0.000 000 000 000 014 386 2 × 2 = 0 + 0.000 000 000 000 028 772 4;
  • 2) 0.000 000 000 000 028 772 4 × 2 = 0 + 0.000 000 000 000 057 544 8;
  • 3) 0.000 000 000 000 057 544 8 × 2 = 0 + 0.000 000 000 000 115 089 6;
  • 4) 0.000 000 000 000 115 089 6 × 2 = 0 + 0.000 000 000 000 230 179 2;
  • 5) 0.000 000 000 000 230 179 2 × 2 = 0 + 0.000 000 000 000 460 358 4;
  • 6) 0.000 000 000 000 460 358 4 × 2 = 0 + 0.000 000 000 000 920 716 8;
  • 7) 0.000 000 000 000 920 716 8 × 2 = 0 + 0.000 000 000 001 841 433 6;
  • 8) 0.000 000 000 001 841 433 6 × 2 = 0 + 0.000 000 000 003 682 867 2;
  • 9) 0.000 000 000 003 682 867 2 × 2 = 0 + 0.000 000 000 007 365 734 4;
  • 10) 0.000 000 000 007 365 734 4 × 2 = 0 + 0.000 000 000 014 731 468 8;
  • 11) 0.000 000 000 014 731 468 8 × 2 = 0 + 0.000 000 000 029 462 937 6;
  • 12) 0.000 000 000 029 462 937 6 × 2 = 0 + 0.000 000 000 058 925 875 2;
  • 13) 0.000 000 000 058 925 875 2 × 2 = 0 + 0.000 000 000 117 851 750 4;
  • 14) 0.000 000 000 117 851 750 4 × 2 = 0 + 0.000 000 000 235 703 500 8;
  • 15) 0.000 000 000 235 703 500 8 × 2 = 0 + 0.000 000 000 471 407 001 6;
  • 16) 0.000 000 000 471 407 001 6 × 2 = 0 + 0.000 000 000 942 814 003 2;
  • 17) 0.000 000 000 942 814 003 2 × 2 = 0 + 0.000 000 001 885 628 006 4;
  • 18) 0.000 000 001 885 628 006 4 × 2 = 0 + 0.000 000 003 771 256 012 8;
  • 19) 0.000 000 003 771 256 012 8 × 2 = 0 + 0.000 000 007 542 512 025 6;
  • 20) 0.000 000 007 542 512 025 6 × 2 = 0 + 0.000 000 015 085 024 051 2;
  • 21) 0.000 000 015 085 024 051 2 × 2 = 0 + 0.000 000 030 170 048 102 4;
  • 22) 0.000 000 030 170 048 102 4 × 2 = 0 + 0.000 000 060 340 096 204 8;
  • 23) 0.000 000 060 340 096 204 8 × 2 = 0 + 0.000 000 120 680 192 409 6;
  • 24) 0.000 000 120 680 192 409 6 × 2 = 0 + 0.000 000 241 360 384 819 2;
  • 25) 0.000 000 241 360 384 819 2 × 2 = 0 + 0.000 000 482 720 769 638 4;
  • 26) 0.000 000 482 720 769 638 4 × 2 = 0 + 0.000 000 965 441 539 276 8;
  • 27) 0.000 000 965 441 539 276 8 × 2 = 0 + 0.000 001 930 883 078 553 6;
  • 28) 0.000 001 930 883 078 553 6 × 2 = 0 + 0.000 003 861 766 157 107 2;
  • 29) 0.000 003 861 766 157 107 2 × 2 = 0 + 0.000 007 723 532 314 214 4;
  • 30) 0.000 007 723 532 314 214 4 × 2 = 0 + 0.000 015 447 064 628 428 8;
  • 31) 0.000 015 447 064 628 428 8 × 2 = 0 + 0.000 030 894 129 256 857 6;
  • 32) 0.000 030 894 129 256 857 6 × 2 = 0 + 0.000 061 788 258 513 715 2;
  • 33) 0.000 061 788 258 513 715 2 × 2 = 0 + 0.000 123 576 517 027 430 4;
  • 34) 0.000 123 576 517 027 430 4 × 2 = 0 + 0.000 247 153 034 054 860 8;
  • 35) 0.000 247 153 034 054 860 8 × 2 = 0 + 0.000 494 306 068 109 721 6;
  • 36) 0.000 494 306 068 109 721 6 × 2 = 0 + 0.000 988 612 136 219 443 2;
  • 37) 0.000 988 612 136 219 443 2 × 2 = 0 + 0.001 977 224 272 438 886 4;
  • 38) 0.001 977 224 272 438 886 4 × 2 = 0 + 0.003 954 448 544 877 772 8;
  • 39) 0.003 954 448 544 877 772 8 × 2 = 0 + 0.007 908 897 089 755 545 6;
  • 40) 0.007 908 897 089 755 545 6 × 2 = 0 + 0.015 817 794 179 511 091 2;
  • 41) 0.015 817 794 179 511 091 2 × 2 = 0 + 0.031 635 588 359 022 182 4;
  • 42) 0.031 635 588 359 022 182 4 × 2 = 0 + 0.063 271 176 718 044 364 8;
  • 43) 0.063 271 176 718 044 364 8 × 2 = 0 + 0.126 542 353 436 088 729 6;
  • 44) 0.126 542 353 436 088 729 6 × 2 = 0 + 0.253 084 706 872 177 459 2;
  • 45) 0.253 084 706 872 177 459 2 × 2 = 0 + 0.506 169 413 744 354 918 4;
  • 46) 0.506 169 413 744 354 918 4 × 2 = 1 + 0.012 338 827 488 709 836 8;
  • 47) 0.012 338 827 488 709 836 8 × 2 = 0 + 0.024 677 654 977 419 673 6;
  • 48) 0.024 677 654 977 419 673 6 × 2 = 0 + 0.049 355 309 954 839 347 2;
  • 49) 0.049 355 309 954 839 347 2 × 2 = 0 + 0.098 710 619 909 678 694 4;
  • 50) 0.098 710 619 909 678 694 4 × 2 = 0 + 0.197 421 239 819 357 388 8;
  • 51) 0.197 421 239 819 357 388 8 × 2 = 0 + 0.394 842 479 638 714 777 6;
  • 52) 0.394 842 479 638 714 777 6 × 2 = 0 + 0.789 684 959 277 429 555 2;
  • 53) 0.789 684 959 277 429 555 2 × 2 = 1 + 0.579 369 918 554 859 110 4;
  • 54) 0.579 369 918 554 859 110 4 × 2 = 1 + 0.158 739 837 109 718 220 8;
  • 55) 0.158 739 837 109 718 220 8 × 2 = 0 + 0.317 479 674 219 436 441 6;
  • 56) 0.317 479 674 219 436 441 6 × 2 = 0 + 0.634 959 348 438 872 883 2;
  • 57) 0.634 959 348 438 872 883 2 × 2 = 1 + 0.269 918 696 877 745 766 4;
  • 58) 0.269 918 696 877 745 766 4 × 2 = 0 + 0.539 837 393 755 491 532 8;
  • 59) 0.539 837 393 755 491 532 8 × 2 = 1 + 0.079 674 787 510 983 065 6;
  • 60) 0.079 674 787 510 983 065 6 × 2 = 0 + 0.159 349 575 021 966 131 2;
  • 61) 0.159 349 575 021 966 131 2 × 2 = 0 + 0.318 699 150 043 932 262 4;
  • 62) 0.318 699 150 043 932 262 4 × 2 = 0 + 0.637 398 300 087 864 524 8;
  • 63) 0.637 398 300 087 864 524 8 × 2 = 1 + 0.274 796 600 175 729 049 6;
  • 64) 0.274 796 600 175 729 049 6 × 2 = 0 + 0.549 593 200 351 458 099 2;
  • 65) 0.549 593 200 351 458 099 2 × 2 = 1 + 0.099 186 400 702 916 198 4;
  • 66) 0.099 186 400 702 916 198 4 × 2 = 0 + 0.198 372 801 405 832 396 8;
  • 67) 0.198 372 801 405 832 396 8 × 2 = 0 + 0.396 745 602 811 664 793 6;
  • 68) 0.396 745 602 811 664 793 6 × 2 = 0 + 0.793 491 205 623 329 587 2;
  • 69) 0.793 491 205 623 329 587 2 × 2 = 1 + 0.586 982 411 246 659 174 4;
  • 70) 0.586 982 411 246 659 174 4 × 2 = 1 + 0.173 964 822 493 318 348 8;
  • 71) 0.173 964 822 493 318 348 8 × 2 = 0 + 0.347 929 644 986 636 697 6;
  • 72) 0.347 929 644 986 636 697 6 × 2 = 0 + 0.695 859 289 973 273 395 2;
  • 73) 0.695 859 289 973 273 395 2 × 2 = 1 + 0.391 718 579 946 546 790 4;
  • 74) 0.391 718 579 946 546 790 4 × 2 = 0 + 0.783 437 159 893 093 580 8;
  • 75) 0.783 437 159 893 093 580 8 × 2 = 1 + 0.566 874 319 786 187 161 6;
  • 76) 0.566 874 319 786 187 161 6 × 2 = 1 + 0.133 748 639 572 374 323 2;
  • 77) 0.133 748 639 572 374 323 2 × 2 = 0 + 0.267 497 279 144 748 646 4;
  • 78) 0.267 497 279 144 748 646 4 × 2 = 0 + 0.534 994 558 289 497 292 8;
  • 79) 0.534 994 558 289 497 292 8 × 2 = 1 + 0.069 989 116 578 994 585 6;
  • 80) 0.069 989 116 578 994 585 6 × 2 = 0 + 0.139 978 233 157 989 171 2;
  • 81) 0.139 978 233 157 989 171 2 × 2 = 0 + 0.279 956 466 315 978 342 4;
  • 82) 0.279 956 466 315 978 342 4 × 2 = 0 + 0.559 912 932 631 956 684 8;
  • 83) 0.559 912 932 631 956 684 8 × 2 = 1 + 0.119 825 865 263 913 369 6;
  • 84) 0.119 825 865 263 913 369 6 × 2 = 0 + 0.239 651 730 527 826 739 2;
  • 85) 0.239 651 730 527 826 739 2 × 2 = 0 + 0.479 303 461 055 653 478 4;
  • 86) 0.479 303 461 055 653 478 4 × 2 = 0 + 0.958 606 922 111 306 956 8;
  • 87) 0.958 606 922 111 306 956 8 × 2 = 1 + 0.917 213 844 222 613 913 6;
  • 88) 0.917 213 844 222 613 913 6 × 2 = 1 + 0.834 427 688 445 227 827 2;
  • 89) 0.834 427 688 445 227 827 2 × 2 = 1 + 0.668 855 376 890 455 654 4;
  • 90) 0.668 855 376 890 455 654 4 × 2 = 1 + 0.337 710 753 780 911 308 8;
  • 91) 0.337 710 753 780 911 308 8 × 2 = 0 + 0.675 421 507 561 822 617 6;
  • 92) 0.675 421 507 561 822 617 6 × 2 = 1 + 0.350 843 015 123 645 235 2;
  • 93) 0.350 843 015 123 645 235 2 × 2 = 0 + 0.701 686 030 247 290 470 4;
  • 94) 0.701 686 030 247 290 470 4 × 2 = 1 + 0.403 372 060 494 580 940 8;
  • 95) 0.403 372 060 494 580 940 8 × 2 = 0 + 0.806 744 120 989 161 881 6;
  • 96) 0.806 744 120 989 161 881 6 × 2 = 1 + 0.613 488 241 978 323 763 2;
  • 97) 0.613 488 241 978 323 763 2 × 2 = 1 + 0.226 976 483 956 647 526 4;
  • 98) 0.226 976 483 956 647 526 4 × 2 = 0 + 0.453 952 967 913 295 052 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:


0.000 000 000 000 014 386 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 0010 1000 1100 1011 0010 0010 0011 1101 0101 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 386 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 0010 1000 1100 1011 0010 0010 0011 1101 0101 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 386 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 0010 1000 1100 1011 0010 0010 0011 1101 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 0010 1000 1100 1011 0010 0010 0011 1101 0101 10(2) × 20 =

1.0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110 =

0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110


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 0001

Mantissa (52 bits) =
0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110


Decimal number -0.000 000 000 000 014 386 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0010 1000 1010 0011 0010 1100 1000 1000 1111 0101 0110

» Convert decimal -0.000 000 000 000 014 381 8 to 64 bit double precision IEEE 754 binary floating point representation standard

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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