-0.000 000 000 000 014 384 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 384 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
-0.000 000 000 000 014 384 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:

|-0.000 000 000 000 014 384 2| = 0.000 000 000 000 014 384 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: 0.000 000 000 000 014 384 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) 0.000 000 000 000 014 384 2 × 2 = 0 + 0.000 000 000 000 028 768 4;
  • 2) 0.000 000 000 000 028 768 4 × 2 = 0 + 0.000 000 000 000 057 536 8;
  • 3) 0.000 000 000 000 057 536 8 × 2 = 0 + 0.000 000 000 000 115 073 6;
  • 4) 0.000 000 000 000 115 073 6 × 2 = 0 + 0.000 000 000 000 230 147 2;
  • 5) 0.000 000 000 000 230 147 2 × 2 = 0 + 0.000 000 000 000 460 294 4;
  • 6) 0.000 000 000 000 460 294 4 × 2 = 0 + 0.000 000 000 000 920 588 8;
  • 7) 0.000 000 000 000 920 588 8 × 2 = 0 + 0.000 000 000 001 841 177 6;
  • 8) 0.000 000 000 001 841 177 6 × 2 = 0 + 0.000 000 000 003 682 355 2;
  • 9) 0.000 000 000 003 682 355 2 × 2 = 0 + 0.000 000 000 007 364 710 4;
  • 10) 0.000 000 000 007 364 710 4 × 2 = 0 + 0.000 000 000 014 729 420 8;
  • 11) 0.000 000 000 014 729 420 8 × 2 = 0 + 0.000 000 000 029 458 841 6;
  • 12) 0.000 000 000 029 458 841 6 × 2 = 0 + 0.000 000 000 058 917 683 2;
  • 13) 0.000 000 000 058 917 683 2 × 2 = 0 + 0.000 000 000 117 835 366 4;
  • 14) 0.000 000 000 117 835 366 4 × 2 = 0 + 0.000 000 000 235 670 732 8;
  • 15) 0.000 000 000 235 670 732 8 × 2 = 0 + 0.000 000 000 471 341 465 6;
  • 16) 0.000 000 000 471 341 465 6 × 2 = 0 + 0.000 000 000 942 682 931 2;
  • 17) 0.000 000 000 942 682 931 2 × 2 = 0 + 0.000 000 001 885 365 862 4;
  • 18) 0.000 000 001 885 365 862 4 × 2 = 0 + 0.000 000 003 770 731 724 8;
  • 19) 0.000 000 003 770 731 724 8 × 2 = 0 + 0.000 000 007 541 463 449 6;
  • 20) 0.000 000 007 541 463 449 6 × 2 = 0 + 0.000 000 015 082 926 899 2;
  • 21) 0.000 000 015 082 926 899 2 × 2 = 0 + 0.000 000 030 165 853 798 4;
  • 22) 0.000 000 030 165 853 798 4 × 2 = 0 + 0.000 000 060 331 707 596 8;
  • 23) 0.000 000 060 331 707 596 8 × 2 = 0 + 0.000 000 120 663 415 193 6;
  • 24) 0.000 000 120 663 415 193 6 × 2 = 0 + 0.000 000 241 326 830 387 2;
  • 25) 0.000 000 241 326 830 387 2 × 2 = 0 + 0.000 000 482 653 660 774 4;
  • 26) 0.000 000 482 653 660 774 4 × 2 = 0 + 0.000 000 965 307 321 548 8;
  • 27) 0.000 000 965 307 321 548 8 × 2 = 0 + 0.000 001 930 614 643 097 6;
  • 28) 0.000 001 930 614 643 097 6 × 2 = 0 + 0.000 003 861 229 286 195 2;
  • 29) 0.000 003 861 229 286 195 2 × 2 = 0 + 0.000 007 722 458 572 390 4;
  • 30) 0.000 007 722 458 572 390 4 × 2 = 0 + 0.000 015 444 917 144 780 8;
  • 31) 0.000 015 444 917 144 780 8 × 2 = 0 + 0.000 030 889 834 289 561 6;
  • 32) 0.000 030 889 834 289 561 6 × 2 = 0 + 0.000 061 779 668 579 123 2;
  • 33) 0.000 061 779 668 579 123 2 × 2 = 0 + 0.000 123 559 337 158 246 4;
  • 34) 0.000 123 559 337 158 246 4 × 2 = 0 + 0.000 247 118 674 316 492 8;
  • 35) 0.000 247 118 674 316 492 8 × 2 = 0 + 0.000 494 237 348 632 985 6;
  • 36) 0.000 494 237 348 632 985 6 × 2 = 0 + 0.000 988 474 697 265 971 2;
  • 37) 0.000 988 474 697 265 971 2 × 2 = 0 + 0.001 976 949 394 531 942 4;
  • 38) 0.001 976 949 394 531 942 4 × 2 = 0 + 0.003 953 898 789 063 884 8;
  • 39) 0.003 953 898 789 063 884 8 × 2 = 0 + 0.007 907 797 578 127 769 6;
  • 40) 0.007 907 797 578 127 769 6 × 2 = 0 + 0.015 815 595 156 255 539 2;
  • 41) 0.015 815 595 156 255 539 2 × 2 = 0 + 0.031 631 190 312 511 078 4;
  • 42) 0.031 631 190 312 511 078 4 × 2 = 0 + 0.063 262 380 625 022 156 8;
  • 43) 0.063 262 380 625 022 156 8 × 2 = 0 + 0.126 524 761 250 044 313 6;
  • 44) 0.126 524 761 250 044 313 6 × 2 = 0 + 0.253 049 522 500 088 627 2;
  • 45) 0.253 049 522 500 088 627 2 × 2 = 0 + 0.506 099 045 000 177 254 4;
  • 46) 0.506 099 045 000 177 254 4 × 2 = 1 + 0.012 198 090 000 354 508 8;
  • 47) 0.012 198 090 000 354 508 8 × 2 = 0 + 0.024 396 180 000 709 017 6;
  • 48) 0.024 396 180 000 709 017 6 × 2 = 0 + 0.048 792 360 001 418 035 2;
  • 49) 0.048 792 360 001 418 035 2 × 2 = 0 + 0.097 584 720 002 836 070 4;
  • 50) 0.097 584 720 002 836 070 4 × 2 = 0 + 0.195 169 440 005 672 140 8;
  • 51) 0.195 169 440 005 672 140 8 × 2 = 0 + 0.390 338 880 011 344 281 6;
  • 52) 0.390 338 880 011 344 281 6 × 2 = 0 + 0.780 677 760 022 688 563 2;
  • 53) 0.780 677 760 022 688 563 2 × 2 = 1 + 0.561 355 520 045 377 126 4;
  • 54) 0.561 355 520 045 377 126 4 × 2 = 1 + 0.122 711 040 090 754 252 8;
  • 55) 0.122 711 040 090 754 252 8 × 2 = 0 + 0.245 422 080 181 508 505 6;
  • 56) 0.245 422 080 181 508 505 6 × 2 = 0 + 0.490 844 160 363 017 011 2;
  • 57) 0.490 844 160 363 017 011 2 × 2 = 0 + 0.981 688 320 726 034 022 4;
  • 58) 0.981 688 320 726 034 022 4 × 2 = 1 + 0.963 376 641 452 068 044 8;
  • 59) 0.963 376 641 452 068 044 8 × 2 = 1 + 0.926 753 282 904 136 089 6;
  • 60) 0.926 753 282 904 136 089 6 × 2 = 1 + 0.853 506 565 808 272 179 2;
  • 61) 0.853 506 565 808 272 179 2 × 2 = 1 + 0.707 013 131 616 544 358 4;
  • 62) 0.707 013 131 616 544 358 4 × 2 = 1 + 0.414 026 263 233 088 716 8;
  • 63) 0.414 026 263 233 088 716 8 × 2 = 0 + 0.828 052 526 466 177 433 6;
  • 64) 0.828 052 526 466 177 433 6 × 2 = 1 + 0.656 105 052 932 354 867 2;
  • 65) 0.656 105 052 932 354 867 2 × 2 = 1 + 0.312 210 105 864 709 734 4;
  • 66) 0.312 210 105 864 709 734 4 × 2 = 0 + 0.624 420 211 729 419 468 8;
  • 67) 0.624 420 211 729 419 468 8 × 2 = 1 + 0.248 840 423 458 838 937 6;
  • 68) 0.248 840 423 458 838 937 6 × 2 = 0 + 0.497 680 846 917 677 875 2;
  • 69) 0.497 680 846 917 677 875 2 × 2 = 0 + 0.995 361 693 835 355 750 4;
  • 70) 0.995 361 693 835 355 750 4 × 2 = 1 + 0.990 723 387 670 711 500 8;
  • 71) 0.990 723 387 670 711 500 8 × 2 = 1 + 0.981 446 775 341 423 001 6;
  • 72) 0.981 446 775 341 423 001 6 × 2 = 1 + 0.962 893 550 682 846 003 2;
  • 73) 0.962 893 550 682 846 003 2 × 2 = 1 + 0.925 787 101 365 692 006 4;
  • 74) 0.925 787 101 365 692 006 4 × 2 = 1 + 0.851 574 202 731 384 012 8;
  • 75) 0.851 574 202 731 384 012 8 × 2 = 1 + 0.703 148 405 462 768 025 6;
  • 76) 0.703 148 405 462 768 025 6 × 2 = 1 + 0.406 296 810 925 536 051 2;
  • 77) 0.406 296 810 925 536 051 2 × 2 = 0 + 0.812 593 621 851 072 102 4;
  • 78) 0.812 593 621 851 072 102 4 × 2 = 1 + 0.625 187 243 702 144 204 8;
  • 79) 0.625 187 243 702 144 204 8 × 2 = 1 + 0.250 374 487 404 288 409 6;
  • 80) 0.250 374 487 404 288 409 6 × 2 = 0 + 0.500 748 974 808 576 819 2;
  • 81) 0.500 748 974 808 576 819 2 × 2 = 1 + 0.001 497 949 617 153 638 4;
  • 82) 0.001 497 949 617 153 638 4 × 2 = 0 + 0.002 995 899 234 307 276 8;
  • 83) 0.002 995 899 234 307 276 8 × 2 = 0 + 0.005 991 798 468 614 553 6;
  • 84) 0.005 991 798 468 614 553 6 × 2 = 0 + 0.011 983 596 937 229 107 2;
  • 85) 0.011 983 596 937 229 107 2 × 2 = 0 + 0.023 967 193 874 458 214 4;
  • 86) 0.023 967 193 874 458 214 4 × 2 = 0 + 0.047 934 387 748 916 428 8;
  • 87) 0.047 934 387 748 916 428 8 × 2 = 0 + 0.095 868 775 497 832 857 6;
  • 88) 0.095 868 775 497 832 857 6 × 2 = 0 + 0.191 737 550 995 665 715 2;
  • 89) 0.191 737 550 995 665 715 2 × 2 = 0 + 0.383 475 101 991 331 430 4;
  • 90) 0.383 475 101 991 331 430 4 × 2 = 0 + 0.766 950 203 982 662 860 8;
  • 91) 0.766 950 203 982 662 860 8 × 2 = 1 + 0.533 900 407 965 325 721 6;
  • 92) 0.533 900 407 965 325 721 6 × 2 = 1 + 0.067 800 815 930 651 443 2;
  • 93) 0.067 800 815 930 651 443 2 × 2 = 0 + 0.135 601 631 861 302 886 4;
  • 94) 0.135 601 631 861 302 886 4 × 2 = 0 + 0.271 203 263 722 605 772 8;
  • 95) 0.271 203 263 722 605 772 8 × 2 = 0 + 0.542 406 527 445 211 545 6;
  • 96) 0.542 406 527 445 211 545 6 × 2 = 1 + 0.084 813 054 890 423 091 2;
  • 97) 0.084 813 054 890 423 091 2 × 2 = 0 + 0.169 626 109 780 846 182 4;
  • 98) 0.169 626 109 780 846 182 4 × 2 = 0 + 0.339 252 219 561 692 364 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:


0.000 000 000 000 014 384 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0111 1101 1010 0111 1111 0110 1000 0000 0011 0001 00(2)

6. Positive number before normalization:

0.000 000 000 000 014 384 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0111 1101 1010 0111 1111 0110 1000 0000 0011 0001 00(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:


0.000 000 000 000 014 384 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0111 1101 1010 0111 1111 0110 1000 0000 0011 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0111 1101 1010 0111 1111 0110 1000 0000 0011 0001 00(2) × 20 =

1.0000 0011 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100(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 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100


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 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100 =

0000 0011 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100


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 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100


Decimal number -0.000 000 000 000 014 384 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0001 1111 0110 1001 1111 1101 1010 0000 0000 1100 0100

» Convert decimal -0.000 000 000 000 014 387 6 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