-0.000 000 000 000 014 372 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 372 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
-0.000 000 000 000 014 372 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:

|-0.000 000 000 000 014 372 3| = 0.000 000 000 000 014 372 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: 0.000 000 000 000 014 372 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) 0.000 000 000 000 014 372 3 × 2 = 0 + 0.000 000 000 000 028 744 6;
  • 2) 0.000 000 000 000 028 744 6 × 2 = 0 + 0.000 000 000 000 057 489 2;
  • 3) 0.000 000 000 000 057 489 2 × 2 = 0 + 0.000 000 000 000 114 978 4;
  • 4) 0.000 000 000 000 114 978 4 × 2 = 0 + 0.000 000 000 000 229 956 8;
  • 5) 0.000 000 000 000 229 956 8 × 2 = 0 + 0.000 000 000 000 459 913 6;
  • 6) 0.000 000 000 000 459 913 6 × 2 = 0 + 0.000 000 000 000 919 827 2;
  • 7) 0.000 000 000 000 919 827 2 × 2 = 0 + 0.000 000 000 001 839 654 4;
  • 8) 0.000 000 000 001 839 654 4 × 2 = 0 + 0.000 000 000 003 679 308 8;
  • 9) 0.000 000 000 003 679 308 8 × 2 = 0 + 0.000 000 000 007 358 617 6;
  • 10) 0.000 000 000 007 358 617 6 × 2 = 0 + 0.000 000 000 014 717 235 2;
  • 11) 0.000 000 000 014 717 235 2 × 2 = 0 + 0.000 000 000 029 434 470 4;
  • 12) 0.000 000 000 029 434 470 4 × 2 = 0 + 0.000 000 000 058 868 940 8;
  • 13) 0.000 000 000 058 868 940 8 × 2 = 0 + 0.000 000 000 117 737 881 6;
  • 14) 0.000 000 000 117 737 881 6 × 2 = 0 + 0.000 000 000 235 475 763 2;
  • 15) 0.000 000 000 235 475 763 2 × 2 = 0 + 0.000 000 000 470 951 526 4;
  • 16) 0.000 000 000 470 951 526 4 × 2 = 0 + 0.000 000 000 941 903 052 8;
  • 17) 0.000 000 000 941 903 052 8 × 2 = 0 + 0.000 000 001 883 806 105 6;
  • 18) 0.000 000 001 883 806 105 6 × 2 = 0 + 0.000 000 003 767 612 211 2;
  • 19) 0.000 000 003 767 612 211 2 × 2 = 0 + 0.000 000 007 535 224 422 4;
  • 20) 0.000 000 007 535 224 422 4 × 2 = 0 + 0.000 000 015 070 448 844 8;
  • 21) 0.000 000 015 070 448 844 8 × 2 = 0 + 0.000 000 030 140 897 689 6;
  • 22) 0.000 000 030 140 897 689 6 × 2 = 0 + 0.000 000 060 281 795 379 2;
  • 23) 0.000 000 060 281 795 379 2 × 2 = 0 + 0.000 000 120 563 590 758 4;
  • 24) 0.000 000 120 563 590 758 4 × 2 = 0 + 0.000 000 241 127 181 516 8;
  • 25) 0.000 000 241 127 181 516 8 × 2 = 0 + 0.000 000 482 254 363 033 6;
  • 26) 0.000 000 482 254 363 033 6 × 2 = 0 + 0.000 000 964 508 726 067 2;
  • 27) 0.000 000 964 508 726 067 2 × 2 = 0 + 0.000 001 929 017 452 134 4;
  • 28) 0.000 001 929 017 452 134 4 × 2 = 0 + 0.000 003 858 034 904 268 8;
  • 29) 0.000 003 858 034 904 268 8 × 2 = 0 + 0.000 007 716 069 808 537 6;
  • 30) 0.000 007 716 069 808 537 6 × 2 = 0 + 0.000 015 432 139 617 075 2;
  • 31) 0.000 015 432 139 617 075 2 × 2 = 0 + 0.000 030 864 279 234 150 4;
  • 32) 0.000 030 864 279 234 150 4 × 2 = 0 + 0.000 061 728 558 468 300 8;
  • 33) 0.000 061 728 558 468 300 8 × 2 = 0 + 0.000 123 457 116 936 601 6;
  • 34) 0.000 123 457 116 936 601 6 × 2 = 0 + 0.000 246 914 233 873 203 2;
  • 35) 0.000 246 914 233 873 203 2 × 2 = 0 + 0.000 493 828 467 746 406 4;
  • 36) 0.000 493 828 467 746 406 4 × 2 = 0 + 0.000 987 656 935 492 812 8;
  • 37) 0.000 987 656 935 492 812 8 × 2 = 0 + 0.001 975 313 870 985 625 6;
  • 38) 0.001 975 313 870 985 625 6 × 2 = 0 + 0.003 950 627 741 971 251 2;
  • 39) 0.003 950 627 741 971 251 2 × 2 = 0 + 0.007 901 255 483 942 502 4;
  • 40) 0.007 901 255 483 942 502 4 × 2 = 0 + 0.015 802 510 967 885 004 8;
  • 41) 0.015 802 510 967 885 004 8 × 2 = 0 + 0.031 605 021 935 770 009 6;
  • 42) 0.031 605 021 935 770 009 6 × 2 = 0 + 0.063 210 043 871 540 019 2;
  • 43) 0.063 210 043 871 540 019 2 × 2 = 0 + 0.126 420 087 743 080 038 4;
  • 44) 0.126 420 087 743 080 038 4 × 2 = 0 + 0.252 840 175 486 160 076 8;
  • 45) 0.252 840 175 486 160 076 8 × 2 = 0 + 0.505 680 350 972 320 153 6;
  • 46) 0.505 680 350 972 320 153 6 × 2 = 1 + 0.011 360 701 944 640 307 2;
  • 47) 0.011 360 701 944 640 307 2 × 2 = 0 + 0.022 721 403 889 280 614 4;
  • 48) 0.022 721 403 889 280 614 4 × 2 = 0 + 0.045 442 807 778 561 228 8;
  • 49) 0.045 442 807 778 561 228 8 × 2 = 0 + 0.090 885 615 557 122 457 6;
  • 50) 0.090 885 615 557 122 457 6 × 2 = 0 + 0.181 771 231 114 244 915 2;
  • 51) 0.181 771 231 114 244 915 2 × 2 = 0 + 0.363 542 462 228 489 830 4;
  • 52) 0.363 542 462 228 489 830 4 × 2 = 0 + 0.727 084 924 456 979 660 8;
  • 53) 0.727 084 924 456 979 660 8 × 2 = 1 + 0.454 169 848 913 959 321 6;
  • 54) 0.454 169 848 913 959 321 6 × 2 = 0 + 0.908 339 697 827 918 643 2;
  • 55) 0.908 339 697 827 918 643 2 × 2 = 1 + 0.816 679 395 655 837 286 4;
  • 56) 0.816 679 395 655 837 286 4 × 2 = 1 + 0.633 358 791 311 674 572 8;
  • 57) 0.633 358 791 311 674 572 8 × 2 = 1 + 0.266 717 582 623 349 145 6;
  • 58) 0.266 717 582 623 349 145 6 × 2 = 0 + 0.533 435 165 246 698 291 2;
  • 59) 0.533 435 165 246 698 291 2 × 2 = 1 + 0.066 870 330 493 396 582 4;
  • 60) 0.066 870 330 493 396 582 4 × 2 = 0 + 0.133 740 660 986 793 164 8;
  • 61) 0.133 740 660 986 793 164 8 × 2 = 0 + 0.267 481 321 973 586 329 6;
  • 62) 0.267 481 321 973 586 329 6 × 2 = 0 + 0.534 962 643 947 172 659 2;
  • 63) 0.534 962 643 947 172 659 2 × 2 = 1 + 0.069 925 287 894 345 318 4;
  • 64) 0.069 925 287 894 345 318 4 × 2 = 0 + 0.139 850 575 788 690 636 8;
  • 65) 0.139 850 575 788 690 636 8 × 2 = 0 + 0.279 701 151 577 381 273 6;
  • 66) 0.279 701 151 577 381 273 6 × 2 = 0 + 0.559 402 303 154 762 547 2;
  • 67) 0.559 402 303 154 762 547 2 × 2 = 1 + 0.118 804 606 309 525 094 4;
  • 68) 0.118 804 606 309 525 094 4 × 2 = 0 + 0.237 609 212 619 050 188 8;
  • 69) 0.237 609 212 619 050 188 8 × 2 = 0 + 0.475 218 425 238 100 377 6;
  • 70) 0.475 218 425 238 100 377 6 × 2 = 0 + 0.950 436 850 476 200 755 2;
  • 71) 0.950 436 850 476 200 755 2 × 2 = 1 + 0.900 873 700 952 401 510 4;
  • 72) 0.900 873 700 952 401 510 4 × 2 = 1 + 0.801 747 401 904 803 020 8;
  • 73) 0.801 747 401 904 803 020 8 × 2 = 1 + 0.603 494 803 809 606 041 6;
  • 74) 0.603 494 803 809 606 041 6 × 2 = 1 + 0.206 989 607 619 212 083 2;
  • 75) 0.206 989 607 619 212 083 2 × 2 = 0 + 0.413 979 215 238 424 166 4;
  • 76) 0.413 979 215 238 424 166 4 × 2 = 0 + 0.827 958 430 476 848 332 8;
  • 77) 0.827 958 430 476 848 332 8 × 2 = 1 + 0.655 916 860 953 696 665 6;
  • 78) 0.655 916 860 953 696 665 6 × 2 = 1 + 0.311 833 721 907 393 331 2;
  • 79) 0.311 833 721 907 393 331 2 × 2 = 0 + 0.623 667 443 814 786 662 4;
  • 80) 0.623 667 443 814 786 662 4 × 2 = 1 + 0.247 334 887 629 573 324 8;
  • 81) 0.247 334 887 629 573 324 8 × 2 = 0 + 0.494 669 775 259 146 649 6;
  • 82) 0.494 669 775 259 146 649 6 × 2 = 0 + 0.989 339 550 518 293 299 2;
  • 83) 0.989 339 550 518 293 299 2 × 2 = 1 + 0.978 679 101 036 586 598 4;
  • 84) 0.978 679 101 036 586 598 4 × 2 = 1 + 0.957 358 202 073 173 196 8;
  • 85) 0.957 358 202 073 173 196 8 × 2 = 1 + 0.914 716 404 146 346 393 6;
  • 86) 0.914 716 404 146 346 393 6 × 2 = 1 + 0.829 432 808 292 692 787 2;
  • 87) 0.829 432 808 292 692 787 2 × 2 = 1 + 0.658 865 616 585 385 574 4;
  • 88) 0.658 865 616 585 385 574 4 × 2 = 1 + 0.317 731 233 170 771 148 8;
  • 89) 0.317 731 233 170 771 148 8 × 2 = 0 + 0.635 462 466 341 542 297 6;
  • 90) 0.635 462 466 341 542 297 6 × 2 = 1 + 0.270 924 932 683 084 595 2;
  • 91) 0.270 924 932 683 084 595 2 × 2 = 0 + 0.541 849 865 366 169 190 4;
  • 92) 0.541 849 865 366 169 190 4 × 2 = 1 + 0.083 699 730 732 338 380 8;
  • 93) 0.083 699 730 732 338 380 8 × 2 = 0 + 0.167 399 461 464 676 761 6;
  • 94) 0.167 399 461 464 676 761 6 × 2 = 0 + 0.334 798 922 929 353 523 2;
  • 95) 0.334 798 922 929 353 523 2 × 2 = 0 + 0.669 597 845 858 707 046 4;
  • 96) 0.669 597 845 858 707 046 4 × 2 = 1 + 0.339 195 691 717 414 092 8;
  • 97) 0.339 195 691 717 414 092 8 × 2 = 0 + 0.678 391 383 434 828 185 6;
  • 98) 0.678 391 383 434 828 185 6 × 2 = 1 + 0.356 782 766 869 656 371 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:


0.000 000 000 000 014 372 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1010 0010 0010 0011 1100 1101 0011 1111 0101 0001 01(2)

6. Positive number before normalization:

0.000 000 000 000 014 372 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1010 0010 0010 0011 1100 1101 0011 1111 0101 0001 01(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:


0.000 000 000 000 014 372 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1010 0010 0010 0011 1100 1101 0011 1111 0101 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1010 0010 0010 0011 1100 1101 0011 1111 0101 0001 01(2) × 20 =

1.0000 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101(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 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101


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 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101 =

0000 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101


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 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101


Decimal number -0.000 000 000 000 014 372 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1110 1000 1000 1000 1111 0011 0100 1111 1101 0100 0101

» Convert decimal -0.000 000 000 000 014 370 9 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