-0.000 000 000 000 014 386 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 386 8(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 8(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 8| = 0.000 000 000 000 014 386 8


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 8.

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 8 × 2 = 0 + 0.000 000 000 000 028 773 6;
  • 2) 0.000 000 000 000 028 773 6 × 2 = 0 + 0.000 000 000 000 057 547 2;
  • 3) 0.000 000 000 000 057 547 2 × 2 = 0 + 0.000 000 000 000 115 094 4;
  • 4) 0.000 000 000 000 115 094 4 × 2 = 0 + 0.000 000 000 000 230 188 8;
  • 5) 0.000 000 000 000 230 188 8 × 2 = 0 + 0.000 000 000 000 460 377 6;
  • 6) 0.000 000 000 000 460 377 6 × 2 = 0 + 0.000 000 000 000 920 755 2;
  • 7) 0.000 000 000 000 920 755 2 × 2 = 0 + 0.000 000 000 001 841 510 4;
  • 8) 0.000 000 000 001 841 510 4 × 2 = 0 + 0.000 000 000 003 683 020 8;
  • 9) 0.000 000 000 003 683 020 8 × 2 = 0 + 0.000 000 000 007 366 041 6;
  • 10) 0.000 000 000 007 366 041 6 × 2 = 0 + 0.000 000 000 014 732 083 2;
  • 11) 0.000 000 000 014 732 083 2 × 2 = 0 + 0.000 000 000 029 464 166 4;
  • 12) 0.000 000 000 029 464 166 4 × 2 = 0 + 0.000 000 000 058 928 332 8;
  • 13) 0.000 000 000 058 928 332 8 × 2 = 0 + 0.000 000 000 117 856 665 6;
  • 14) 0.000 000 000 117 856 665 6 × 2 = 0 + 0.000 000 000 235 713 331 2;
  • 15) 0.000 000 000 235 713 331 2 × 2 = 0 + 0.000 000 000 471 426 662 4;
  • 16) 0.000 000 000 471 426 662 4 × 2 = 0 + 0.000 000 000 942 853 324 8;
  • 17) 0.000 000 000 942 853 324 8 × 2 = 0 + 0.000 000 001 885 706 649 6;
  • 18) 0.000 000 001 885 706 649 6 × 2 = 0 + 0.000 000 003 771 413 299 2;
  • 19) 0.000 000 003 771 413 299 2 × 2 = 0 + 0.000 000 007 542 826 598 4;
  • 20) 0.000 000 007 542 826 598 4 × 2 = 0 + 0.000 000 015 085 653 196 8;
  • 21) 0.000 000 015 085 653 196 8 × 2 = 0 + 0.000 000 030 171 306 393 6;
  • 22) 0.000 000 030 171 306 393 6 × 2 = 0 + 0.000 000 060 342 612 787 2;
  • 23) 0.000 000 060 342 612 787 2 × 2 = 0 + 0.000 000 120 685 225 574 4;
  • 24) 0.000 000 120 685 225 574 4 × 2 = 0 + 0.000 000 241 370 451 148 8;
  • 25) 0.000 000 241 370 451 148 8 × 2 = 0 + 0.000 000 482 740 902 297 6;
  • 26) 0.000 000 482 740 902 297 6 × 2 = 0 + 0.000 000 965 481 804 595 2;
  • 27) 0.000 000 965 481 804 595 2 × 2 = 0 + 0.000 001 930 963 609 190 4;
  • 28) 0.000 001 930 963 609 190 4 × 2 = 0 + 0.000 003 861 927 218 380 8;
  • 29) 0.000 003 861 927 218 380 8 × 2 = 0 + 0.000 007 723 854 436 761 6;
  • 30) 0.000 007 723 854 436 761 6 × 2 = 0 + 0.000 015 447 708 873 523 2;
  • 31) 0.000 015 447 708 873 523 2 × 2 = 0 + 0.000 030 895 417 747 046 4;
  • 32) 0.000 030 895 417 747 046 4 × 2 = 0 + 0.000 061 790 835 494 092 8;
  • 33) 0.000 061 790 835 494 092 8 × 2 = 0 + 0.000 123 581 670 988 185 6;
  • 34) 0.000 123 581 670 988 185 6 × 2 = 0 + 0.000 247 163 341 976 371 2;
  • 35) 0.000 247 163 341 976 371 2 × 2 = 0 + 0.000 494 326 683 952 742 4;
  • 36) 0.000 494 326 683 952 742 4 × 2 = 0 + 0.000 988 653 367 905 484 8;
  • 37) 0.000 988 653 367 905 484 8 × 2 = 0 + 0.001 977 306 735 810 969 6;
  • 38) 0.001 977 306 735 810 969 6 × 2 = 0 + 0.003 954 613 471 621 939 2;
  • 39) 0.003 954 613 471 621 939 2 × 2 = 0 + 0.007 909 226 943 243 878 4;
  • 40) 0.007 909 226 943 243 878 4 × 2 = 0 + 0.015 818 453 886 487 756 8;
  • 41) 0.015 818 453 886 487 756 8 × 2 = 0 + 0.031 636 907 772 975 513 6;
  • 42) 0.031 636 907 772 975 513 6 × 2 = 0 + 0.063 273 815 545 951 027 2;
  • 43) 0.063 273 815 545 951 027 2 × 2 = 0 + 0.126 547 631 091 902 054 4;
  • 44) 0.126 547 631 091 902 054 4 × 2 = 0 + 0.253 095 262 183 804 108 8;
  • 45) 0.253 095 262 183 804 108 8 × 2 = 0 + 0.506 190 524 367 608 217 6;
  • 46) 0.506 190 524 367 608 217 6 × 2 = 1 + 0.012 381 048 735 216 435 2;
  • 47) 0.012 381 048 735 216 435 2 × 2 = 0 + 0.024 762 097 470 432 870 4;
  • 48) 0.024 762 097 470 432 870 4 × 2 = 0 + 0.049 524 194 940 865 740 8;
  • 49) 0.049 524 194 940 865 740 8 × 2 = 0 + 0.099 048 389 881 731 481 6;
  • 50) 0.099 048 389 881 731 481 6 × 2 = 0 + 0.198 096 779 763 462 963 2;
  • 51) 0.198 096 779 763 462 963 2 × 2 = 0 + 0.396 193 559 526 925 926 4;
  • 52) 0.396 193 559 526 925 926 4 × 2 = 0 + 0.792 387 119 053 851 852 8;
  • 53) 0.792 387 119 053 851 852 8 × 2 = 1 + 0.584 774 238 107 703 705 6;
  • 54) 0.584 774 238 107 703 705 6 × 2 = 1 + 0.169 548 476 215 407 411 2;
  • 55) 0.169 548 476 215 407 411 2 × 2 = 0 + 0.339 096 952 430 814 822 4;
  • 56) 0.339 096 952 430 814 822 4 × 2 = 0 + 0.678 193 904 861 629 644 8;
  • 57) 0.678 193 904 861 629 644 8 × 2 = 1 + 0.356 387 809 723 259 289 6;
  • 58) 0.356 387 809 723 259 289 6 × 2 = 0 + 0.712 775 619 446 518 579 2;
  • 59) 0.712 775 619 446 518 579 2 × 2 = 1 + 0.425 551 238 893 037 158 4;
  • 60) 0.425 551 238 893 037 158 4 × 2 = 0 + 0.851 102 477 786 074 316 8;
  • 61) 0.851 102 477 786 074 316 8 × 2 = 1 + 0.702 204 955 572 148 633 6;
  • 62) 0.702 204 955 572 148 633 6 × 2 = 1 + 0.404 409 911 144 297 267 2;
  • 63) 0.404 409 911 144 297 267 2 × 2 = 0 + 0.808 819 822 288 594 534 4;
  • 64) 0.808 819 822 288 594 534 4 × 2 = 1 + 0.617 639 644 577 189 068 8;
  • 65) 0.617 639 644 577 189 068 8 × 2 = 1 + 0.235 279 289 154 378 137 6;
  • 66) 0.235 279 289 154 378 137 6 × 2 = 0 + 0.470 558 578 308 756 275 2;
  • 67) 0.470 558 578 308 756 275 2 × 2 = 0 + 0.941 117 156 617 512 550 4;
  • 68) 0.941 117 156 617 512 550 4 × 2 = 1 + 0.882 234 313 235 025 100 8;
  • 69) 0.882 234 313 235 025 100 8 × 2 = 1 + 0.764 468 626 470 050 201 6;
  • 70) 0.764 468 626 470 050 201 6 × 2 = 1 + 0.528 937 252 940 100 403 2;
  • 71) 0.528 937 252 940 100 403 2 × 2 = 1 + 0.057 874 505 880 200 806 4;
  • 72) 0.057 874 505 880 200 806 4 × 2 = 0 + 0.115 749 011 760 401 612 8;
  • 73) 0.115 749 011 760 401 612 8 × 2 = 0 + 0.231 498 023 520 803 225 6;
  • 74) 0.231 498 023 520 803 225 6 × 2 = 0 + 0.462 996 047 041 606 451 2;
  • 75) 0.462 996 047 041 606 451 2 × 2 = 0 + 0.925 992 094 083 212 902 4;
  • 76) 0.925 992 094 083 212 902 4 × 2 = 1 + 0.851 984 188 166 425 804 8;
  • 77) 0.851 984 188 166 425 804 8 × 2 = 1 + 0.703 968 376 332 851 609 6;
  • 78) 0.703 968 376 332 851 609 6 × 2 = 1 + 0.407 936 752 665 703 219 2;
  • 79) 0.407 936 752 665 703 219 2 × 2 = 0 + 0.815 873 505 331 406 438 4;
  • 80) 0.815 873 505 331 406 438 4 × 2 = 1 + 0.631 747 010 662 812 876 8;
  • 81) 0.631 747 010 662 812 876 8 × 2 = 1 + 0.263 494 021 325 625 753 6;
  • 82) 0.263 494 021 325 625 753 6 × 2 = 0 + 0.526 988 042 651 251 507 2;
  • 83) 0.526 988 042 651 251 507 2 × 2 = 1 + 0.053 976 085 302 503 014 4;
  • 84) 0.053 976 085 302 503 014 4 × 2 = 0 + 0.107 952 170 605 006 028 8;
  • 85) 0.107 952 170 605 006 028 8 × 2 = 0 + 0.215 904 341 210 012 057 6;
  • 86) 0.215 904 341 210 012 057 6 × 2 = 0 + 0.431 808 682 420 024 115 2;
  • 87) 0.431 808 682 420 024 115 2 × 2 = 0 + 0.863 617 364 840 048 230 4;
  • 88) 0.863 617 364 840 048 230 4 × 2 = 1 + 0.727 234 729 680 096 460 8;
  • 89) 0.727 234 729 680 096 460 8 × 2 = 1 + 0.454 469 459 360 192 921 6;
  • 90) 0.454 469 459 360 192 921 6 × 2 = 0 + 0.908 938 918 720 385 843 2;
  • 91) 0.908 938 918 720 385 843 2 × 2 = 1 + 0.817 877 837 440 771 686 4;
  • 92) 0.817 877 837 440 771 686 4 × 2 = 1 + 0.635 755 674 881 543 372 8;
  • 93) 0.635 755 674 881 543 372 8 × 2 = 1 + 0.271 511 349 763 086 745 6;
  • 94) 0.271 511 349 763 086 745 6 × 2 = 0 + 0.543 022 699 526 173 491 2;
  • 95) 0.543 022 699 526 173 491 2 × 2 = 1 + 0.086 045 399 052 346 982 4;
  • 96) 0.086 045 399 052 346 982 4 × 2 = 0 + 0.172 090 798 104 693 964 8;
  • 97) 0.172 090 798 104 693 964 8 × 2 = 0 + 0.344 181 596 209 387 929 6;
  • 98) 0.344 181 596 209 387 929 6 × 2 = 0 + 0.688 363 192 418 775 859 2;

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 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 1101 1001 1110 0001 1101 1010 0001 1011 1010 00(2)

6. Positive number before normalization:

0.000 000 000 000 014 386 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 1101 1001 1110 0001 1101 1010 0001 1011 1010 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:


0.000 000 000 000 014 386 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 1101 1001 1110 0001 1101 1010 0001 1011 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1010 1101 1001 1110 0001 1101 1010 0001 1011 1010 00(2) × 20 =

1.0000 0011 0010 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000(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 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000


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 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000 =

0000 0011 0010 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000


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 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000


Decimal number -0.000 000 000 000 014 386 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0010 1011 0110 0111 1000 0111 0110 1000 0110 1110 1000

» Convert decimal -0.000 000 000 000 014 385 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