-0.000 000 000 000 014 388 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 388 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
-0.000 000 000 000 014 388 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:

|-0.000 000 000 000 014 388 3| = 0.000 000 000 000 014 388 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: 0.000 000 000 000 014 388 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) 0.000 000 000 000 014 388 3 × 2 = 0 + 0.000 000 000 000 028 776 6;
  • 2) 0.000 000 000 000 028 776 6 × 2 = 0 + 0.000 000 000 000 057 553 2;
  • 3) 0.000 000 000 000 057 553 2 × 2 = 0 + 0.000 000 000 000 115 106 4;
  • 4) 0.000 000 000 000 115 106 4 × 2 = 0 + 0.000 000 000 000 230 212 8;
  • 5) 0.000 000 000 000 230 212 8 × 2 = 0 + 0.000 000 000 000 460 425 6;
  • 6) 0.000 000 000 000 460 425 6 × 2 = 0 + 0.000 000 000 000 920 851 2;
  • 7) 0.000 000 000 000 920 851 2 × 2 = 0 + 0.000 000 000 001 841 702 4;
  • 8) 0.000 000 000 001 841 702 4 × 2 = 0 + 0.000 000 000 003 683 404 8;
  • 9) 0.000 000 000 003 683 404 8 × 2 = 0 + 0.000 000 000 007 366 809 6;
  • 10) 0.000 000 000 007 366 809 6 × 2 = 0 + 0.000 000 000 014 733 619 2;
  • 11) 0.000 000 000 014 733 619 2 × 2 = 0 + 0.000 000 000 029 467 238 4;
  • 12) 0.000 000 000 029 467 238 4 × 2 = 0 + 0.000 000 000 058 934 476 8;
  • 13) 0.000 000 000 058 934 476 8 × 2 = 0 + 0.000 000 000 117 868 953 6;
  • 14) 0.000 000 000 117 868 953 6 × 2 = 0 + 0.000 000 000 235 737 907 2;
  • 15) 0.000 000 000 235 737 907 2 × 2 = 0 + 0.000 000 000 471 475 814 4;
  • 16) 0.000 000 000 471 475 814 4 × 2 = 0 + 0.000 000 000 942 951 628 8;
  • 17) 0.000 000 000 942 951 628 8 × 2 = 0 + 0.000 000 001 885 903 257 6;
  • 18) 0.000 000 001 885 903 257 6 × 2 = 0 + 0.000 000 003 771 806 515 2;
  • 19) 0.000 000 003 771 806 515 2 × 2 = 0 + 0.000 000 007 543 613 030 4;
  • 20) 0.000 000 007 543 613 030 4 × 2 = 0 + 0.000 000 015 087 226 060 8;
  • 21) 0.000 000 015 087 226 060 8 × 2 = 0 + 0.000 000 030 174 452 121 6;
  • 22) 0.000 000 030 174 452 121 6 × 2 = 0 + 0.000 000 060 348 904 243 2;
  • 23) 0.000 000 060 348 904 243 2 × 2 = 0 + 0.000 000 120 697 808 486 4;
  • 24) 0.000 000 120 697 808 486 4 × 2 = 0 + 0.000 000 241 395 616 972 8;
  • 25) 0.000 000 241 395 616 972 8 × 2 = 0 + 0.000 000 482 791 233 945 6;
  • 26) 0.000 000 482 791 233 945 6 × 2 = 0 + 0.000 000 965 582 467 891 2;
  • 27) 0.000 000 965 582 467 891 2 × 2 = 0 + 0.000 001 931 164 935 782 4;
  • 28) 0.000 001 931 164 935 782 4 × 2 = 0 + 0.000 003 862 329 871 564 8;
  • 29) 0.000 003 862 329 871 564 8 × 2 = 0 + 0.000 007 724 659 743 129 6;
  • 30) 0.000 007 724 659 743 129 6 × 2 = 0 + 0.000 015 449 319 486 259 2;
  • 31) 0.000 015 449 319 486 259 2 × 2 = 0 + 0.000 030 898 638 972 518 4;
  • 32) 0.000 030 898 638 972 518 4 × 2 = 0 + 0.000 061 797 277 945 036 8;
  • 33) 0.000 061 797 277 945 036 8 × 2 = 0 + 0.000 123 594 555 890 073 6;
  • 34) 0.000 123 594 555 890 073 6 × 2 = 0 + 0.000 247 189 111 780 147 2;
  • 35) 0.000 247 189 111 780 147 2 × 2 = 0 + 0.000 494 378 223 560 294 4;
  • 36) 0.000 494 378 223 560 294 4 × 2 = 0 + 0.000 988 756 447 120 588 8;
  • 37) 0.000 988 756 447 120 588 8 × 2 = 0 + 0.001 977 512 894 241 177 6;
  • 38) 0.001 977 512 894 241 177 6 × 2 = 0 + 0.003 955 025 788 482 355 2;
  • 39) 0.003 955 025 788 482 355 2 × 2 = 0 + 0.007 910 051 576 964 710 4;
  • 40) 0.007 910 051 576 964 710 4 × 2 = 0 + 0.015 820 103 153 929 420 8;
  • 41) 0.015 820 103 153 929 420 8 × 2 = 0 + 0.031 640 206 307 858 841 6;
  • 42) 0.031 640 206 307 858 841 6 × 2 = 0 + 0.063 280 412 615 717 683 2;
  • 43) 0.063 280 412 615 717 683 2 × 2 = 0 + 0.126 560 825 231 435 366 4;
  • 44) 0.126 560 825 231 435 366 4 × 2 = 0 + 0.253 121 650 462 870 732 8;
  • 45) 0.253 121 650 462 870 732 8 × 2 = 0 + 0.506 243 300 925 741 465 6;
  • 46) 0.506 243 300 925 741 465 6 × 2 = 1 + 0.012 486 601 851 482 931 2;
  • 47) 0.012 486 601 851 482 931 2 × 2 = 0 + 0.024 973 203 702 965 862 4;
  • 48) 0.024 973 203 702 965 862 4 × 2 = 0 + 0.049 946 407 405 931 724 8;
  • 49) 0.049 946 407 405 931 724 8 × 2 = 0 + 0.099 892 814 811 863 449 6;
  • 50) 0.099 892 814 811 863 449 6 × 2 = 0 + 0.199 785 629 623 726 899 2;
  • 51) 0.199 785 629 623 726 899 2 × 2 = 0 + 0.399 571 259 247 453 798 4;
  • 52) 0.399 571 259 247 453 798 4 × 2 = 0 + 0.799 142 518 494 907 596 8;
  • 53) 0.799 142 518 494 907 596 8 × 2 = 1 + 0.598 285 036 989 815 193 6;
  • 54) 0.598 285 036 989 815 193 6 × 2 = 1 + 0.196 570 073 979 630 387 2;
  • 55) 0.196 570 073 979 630 387 2 × 2 = 0 + 0.393 140 147 959 260 774 4;
  • 56) 0.393 140 147 959 260 774 4 × 2 = 0 + 0.786 280 295 918 521 548 8;
  • 57) 0.786 280 295 918 521 548 8 × 2 = 1 + 0.572 560 591 837 043 097 6;
  • 58) 0.572 560 591 837 043 097 6 × 2 = 1 + 0.145 121 183 674 086 195 2;
  • 59) 0.145 121 183 674 086 195 2 × 2 = 0 + 0.290 242 367 348 172 390 4;
  • 60) 0.290 242 367 348 172 390 4 × 2 = 0 + 0.580 484 734 696 344 780 8;
  • 61) 0.580 484 734 696 344 780 8 × 2 = 1 + 0.160 969 469 392 689 561 6;
  • 62) 0.160 969 469 392 689 561 6 × 2 = 0 + 0.321 938 938 785 379 123 2;
  • 63) 0.321 938 938 785 379 123 2 × 2 = 0 + 0.643 877 877 570 758 246 4;
  • 64) 0.643 877 877 570 758 246 4 × 2 = 1 + 0.287 755 755 141 516 492 8;
  • 65) 0.287 755 755 141 516 492 8 × 2 = 0 + 0.575 511 510 283 032 985 6;
  • 66) 0.575 511 510 283 032 985 6 × 2 = 1 + 0.151 023 020 566 065 971 2;
  • 67) 0.151 023 020 566 065 971 2 × 2 = 0 + 0.302 046 041 132 131 942 4;
  • 68) 0.302 046 041 132 131 942 4 × 2 = 0 + 0.604 092 082 264 263 884 8;
  • 69) 0.604 092 082 264 263 884 8 × 2 = 1 + 0.208 184 164 528 527 769 6;
  • 70) 0.208 184 164 528 527 769 6 × 2 = 0 + 0.416 368 329 057 055 539 2;
  • 71) 0.416 368 329 057 055 539 2 × 2 = 0 + 0.832 736 658 114 111 078 4;
  • 72) 0.832 736 658 114 111 078 4 × 2 = 1 + 0.665 473 316 228 222 156 8;
  • 73) 0.665 473 316 228 222 156 8 × 2 = 1 + 0.330 946 632 456 444 313 6;
  • 74) 0.330 946 632 456 444 313 6 × 2 = 0 + 0.661 893 264 912 888 627 2;
  • 75) 0.661 893 264 912 888 627 2 × 2 = 1 + 0.323 786 529 825 777 254 4;
  • 76) 0.323 786 529 825 777 254 4 × 2 = 0 + 0.647 573 059 651 554 508 8;
  • 77) 0.647 573 059 651 554 508 8 × 2 = 1 + 0.295 146 119 303 109 017 6;
  • 78) 0.295 146 119 303 109 017 6 × 2 = 0 + 0.590 292 238 606 218 035 2;
  • 79) 0.590 292 238 606 218 035 2 × 2 = 1 + 0.180 584 477 212 436 070 4;
  • 80) 0.180 584 477 212 436 070 4 × 2 = 0 + 0.361 168 954 424 872 140 8;
  • 81) 0.361 168 954 424 872 140 8 × 2 = 0 + 0.722 337 908 849 744 281 6;
  • 82) 0.722 337 908 849 744 281 6 × 2 = 1 + 0.444 675 817 699 488 563 2;
  • 83) 0.444 675 817 699 488 563 2 × 2 = 0 + 0.889 351 635 398 977 126 4;
  • 84) 0.889 351 635 398 977 126 4 × 2 = 1 + 0.778 703 270 797 954 252 8;
  • 85) 0.778 703 270 797 954 252 8 × 2 = 1 + 0.557 406 541 595 908 505 6;
  • 86) 0.557 406 541 595 908 505 6 × 2 = 1 + 0.114 813 083 191 817 011 2;
  • 87) 0.114 813 083 191 817 011 2 × 2 = 0 + 0.229 626 166 383 634 022 4;
  • 88) 0.229 626 166 383 634 022 4 × 2 = 0 + 0.459 252 332 767 268 044 8;
  • 89) 0.459 252 332 767 268 044 8 × 2 = 0 + 0.918 504 665 534 536 089 6;
  • 90) 0.918 504 665 534 536 089 6 × 2 = 1 + 0.837 009 331 069 072 179 2;
  • 91) 0.837 009 331 069 072 179 2 × 2 = 1 + 0.674 018 662 138 144 358 4;
  • 92) 0.674 018 662 138 144 358 4 × 2 = 1 + 0.348 037 324 276 288 716 8;
  • 93) 0.348 037 324 276 288 716 8 × 2 = 0 + 0.696 074 648 552 577 433 6;
  • 94) 0.696 074 648 552 577 433 6 × 2 = 1 + 0.392 149 297 105 154 867 2;
  • 95) 0.392 149 297 105 154 867 2 × 2 = 0 + 0.784 298 594 210 309 734 4;
  • 96) 0.784 298 594 210 309 734 4 × 2 = 1 + 0.568 597 188 420 619 468 8;
  • 97) 0.568 597 188 420 619 468 8 × 2 = 1 + 0.137 194 376 841 238 937 6;
  • 98) 0.137 194 376 841 238 937 6 × 2 = 0 + 0.274 388 753 682 477 875 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:


0.000 000 000 000 014 388 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1100 1001 0100 1001 1010 1010 0101 1100 0111 0101 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 388 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1100 1001 0100 1001 1010 1010 0101 1100 0111 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:


0.000 000 000 000 014 388 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1100 1001 0100 1001 1010 1010 0101 1100 0111 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1100 1001 0100 1001 1010 1010 0101 1100 0111 0101 10(2) × 20 =

1.0000 0011 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 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 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 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 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 0110 =

0000 0011 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 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 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 0110


Decimal number -0.000 000 000 000 014 388 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0011 0010 0101 0010 0110 1010 1001 0111 0001 1101 0110

» Convert decimal -0.000 000 000 000 014 386 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