-24 821.204 798 741 659 033 112 227 916 717 529 289 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -24 821.204 798 741 659 033 112 227 916 717 529 289 49(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
-24 821.204 798 741 659 033 112 227 916 717 529 289 49(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:

|-24 821.204 798 741 659 033 112 227 916 717 529 289 49| = 24 821.204 798 741 659 033 112 227 916 717 529 289 49


2. First, convert to binary (in base 2) the integer part: 24 821.
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;
  • 24 821 ÷ 2 = 12 410 + 1;
  • 12 410 ÷ 2 = 6 205 + 0;
  • 6 205 ÷ 2 = 3 102 + 1;
  • 3 102 ÷ 2 = 1 551 + 0;
  • 1 551 ÷ 2 = 775 + 1;
  • 775 ÷ 2 = 387 + 1;
  • 387 ÷ 2 = 193 + 1;
  • 193 ÷ 2 = 96 + 1;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

24 821(10) =

110 0000 1111 0101(2)


4. Convert to binary (base 2) the fractional part: 0.204 798 741 659 033 112 227 916 717 529 289 49.

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) 0.204 798 741 659 033 112 227 916 717 529 289 49 × 2 = 0 + 0.409 597 483 318 066 224 455 833 435 058 578 98;
  • 2) 0.409 597 483 318 066 224 455 833 435 058 578 98 × 2 = 0 + 0.819 194 966 636 132 448 911 666 870 117 157 96;
  • 3) 0.819 194 966 636 132 448 911 666 870 117 157 96 × 2 = 1 + 0.638 389 933 272 264 897 823 333 740 234 315 92;
  • 4) 0.638 389 933 272 264 897 823 333 740 234 315 92 × 2 = 1 + 0.276 779 866 544 529 795 646 667 480 468 631 84;
  • 5) 0.276 779 866 544 529 795 646 667 480 468 631 84 × 2 = 0 + 0.553 559 733 089 059 591 293 334 960 937 263 68;
  • 6) 0.553 559 733 089 059 591 293 334 960 937 263 68 × 2 = 1 + 0.107 119 466 178 119 182 586 669 921 874 527 36;
  • 7) 0.107 119 466 178 119 182 586 669 921 874 527 36 × 2 = 0 + 0.214 238 932 356 238 365 173 339 843 749 054 72;
  • 8) 0.214 238 932 356 238 365 173 339 843 749 054 72 × 2 = 0 + 0.428 477 864 712 476 730 346 679 687 498 109 44;
  • 9) 0.428 477 864 712 476 730 346 679 687 498 109 44 × 2 = 0 + 0.856 955 729 424 953 460 693 359 374 996 218 88;
  • 10) 0.856 955 729 424 953 460 693 359 374 996 218 88 × 2 = 1 + 0.713 911 458 849 906 921 386 718 749 992 437 76;
  • 11) 0.713 911 458 849 906 921 386 718 749 992 437 76 × 2 = 1 + 0.427 822 917 699 813 842 773 437 499 984 875 52;
  • 12) 0.427 822 917 699 813 842 773 437 499 984 875 52 × 2 = 0 + 0.855 645 835 399 627 685 546 874 999 969 751 04;
  • 13) 0.855 645 835 399 627 685 546 874 999 969 751 04 × 2 = 1 + 0.711 291 670 799 255 371 093 749 999 939 502 08;
  • 14) 0.711 291 670 799 255 371 093 749 999 939 502 08 × 2 = 1 + 0.422 583 341 598 510 742 187 499 999 879 004 16;
  • 15) 0.422 583 341 598 510 742 187 499 999 879 004 16 × 2 = 0 + 0.845 166 683 197 021 484 374 999 999 758 008 32;
  • 16) 0.845 166 683 197 021 484 374 999 999 758 008 32 × 2 = 1 + 0.690 333 366 394 042 968 749 999 999 516 016 64;
  • 17) 0.690 333 366 394 042 968 749 999 999 516 016 64 × 2 = 1 + 0.380 666 732 788 085 937 499 999 999 032 033 28;
  • 18) 0.380 666 732 788 085 937 499 999 999 032 033 28 × 2 = 0 + 0.761 333 465 576 171 874 999 999 998 064 066 56;
  • 19) 0.761 333 465 576 171 874 999 999 998 064 066 56 × 2 = 1 + 0.522 666 931 152 343 749 999 999 996 128 133 12;
  • 20) 0.522 666 931 152 343 749 999 999 996 128 133 12 × 2 = 1 + 0.045 333 862 304 687 499 999 999 992 256 266 24;
  • 21) 0.045 333 862 304 687 499 999 999 992 256 266 24 × 2 = 0 + 0.090 667 724 609 374 999 999 999 984 512 532 48;
  • 22) 0.090 667 724 609 374 999 999 999 984 512 532 48 × 2 = 0 + 0.181 335 449 218 749 999 999 999 969 025 064 96;
  • 23) 0.181 335 449 218 749 999 999 999 969 025 064 96 × 2 = 0 + 0.362 670 898 437 499 999 999 999 938 050 129 92;
  • 24) 0.362 670 898 437 499 999 999 999 938 050 129 92 × 2 = 0 + 0.725 341 796 874 999 999 999 999 876 100 259 84;
  • 25) 0.725 341 796 874 999 999 999 999 876 100 259 84 × 2 = 1 + 0.450 683 593 749 999 999 999 999 752 200 519 68;
  • 26) 0.450 683 593 749 999 999 999 999 752 200 519 68 × 2 = 0 + 0.901 367 187 499 999 999 999 999 504 401 039 36;
  • 27) 0.901 367 187 499 999 999 999 999 504 401 039 36 × 2 = 1 + 0.802 734 374 999 999 999 999 999 008 802 078 72;
  • 28) 0.802 734 374 999 999 999 999 999 008 802 078 72 × 2 = 1 + 0.605 468 749 999 999 999 999 998 017 604 157 44;
  • 29) 0.605 468 749 999 999 999 999 998 017 604 157 44 × 2 = 1 + 0.210 937 499 999 999 999 999 996 035 208 314 88;
  • 30) 0.210 937 499 999 999 999 999 996 035 208 314 88 × 2 = 0 + 0.421 874 999 999 999 999 999 992 070 416 629 76;
  • 31) 0.421 874 999 999 999 999 999 992 070 416 629 76 × 2 = 0 + 0.843 749 999 999 999 999 999 984 140 833 259 52;
  • 32) 0.843 749 999 999 999 999 999 984 140 833 259 52 × 2 = 1 + 0.687 499 999 999 999 999 999 968 281 666 519 04;
  • 33) 0.687 499 999 999 999 999 999 968 281 666 519 04 × 2 = 1 + 0.374 999 999 999 999 999 999 936 563 333 038 08;
  • 34) 0.374 999 999 999 999 999 999 936 563 333 038 08 × 2 = 0 + 0.749 999 999 999 999 999 999 873 126 666 076 16;
  • 35) 0.749 999 999 999 999 999 999 873 126 666 076 16 × 2 = 1 + 0.499 999 999 999 999 999 999 746 253 332 152 32;
  • 36) 0.499 999 999 999 999 999 999 746 253 332 152 32 × 2 = 0 + 0.999 999 999 999 999 999 999 492 506 664 304 64;
  • 37) 0.999 999 999 999 999 999 999 492 506 664 304 64 × 2 = 1 + 0.999 999 999 999 999 999 998 985 013 328 609 28;
  • 38) 0.999 999 999 999 999 999 998 985 013 328 609 28 × 2 = 1 + 0.999 999 999 999 999 999 997 970 026 657 218 56;
  • 39) 0.999 999 999 999 999 999 997 970 026 657 218 56 × 2 = 1 + 0.999 999 999 999 999 999 995 940 053 314 437 12;
  • 40) 0.999 999 999 999 999 999 995 940 053 314 437 12 × 2 = 1 + 0.999 999 999 999 999 999 991 880 106 628 874 24;
  • 41) 0.999 999 999 999 999 999 991 880 106 628 874 24 × 2 = 1 + 0.999 999 999 999 999 999 983 760 213 257 748 48;
  • 42) 0.999 999 999 999 999 999 983 760 213 257 748 48 × 2 = 1 + 0.999 999 999 999 999 999 967 520 426 515 496 96;
  • 43) 0.999 999 999 999 999 999 967 520 426 515 496 96 × 2 = 1 + 0.999 999 999 999 999 999 935 040 853 030 993 92;
  • 44) 0.999 999 999 999 999 999 935 040 853 030 993 92 × 2 = 1 + 0.999 999 999 999 999 999 870 081 706 061 987 84;
  • 45) 0.999 999 999 999 999 999 870 081 706 061 987 84 × 2 = 1 + 0.999 999 999 999 999 999 740 163 412 123 975 68;
  • 46) 0.999 999 999 999 999 999 740 163 412 123 975 68 × 2 = 1 + 0.999 999 999 999 999 999 480 326 824 247 951 36;
  • 47) 0.999 999 999 999 999 999 480 326 824 247 951 36 × 2 = 1 + 0.999 999 999 999 999 998 960 653 648 495 902 72;
  • 48) 0.999 999 999 999 999 998 960 653 648 495 902 72 × 2 = 1 + 0.999 999 999 999 999 997 921 307 296 991 805 44;
  • 49) 0.999 999 999 999 999 997 921 307 296 991 805 44 × 2 = 1 + 0.999 999 999 999 999 995 842 614 593 983 610 88;
  • 50) 0.999 999 999 999 999 995 842 614 593 983 610 88 × 2 = 1 + 0.999 999 999 999 999 991 685 229 187 967 221 76;
  • 51) 0.999 999 999 999 999 991 685 229 187 967 221 76 × 2 = 1 + 0.999 999 999 999 999 983 370 458 375 934 443 52;
  • 52) 0.999 999 999 999 999 983 370 458 375 934 443 52 × 2 = 1 + 0.999 999 999 999 999 966 740 916 751 868 887 04;
  • 53) 0.999 999 999 999 999 966 740 916 751 868 887 04 × 2 = 1 + 0.999 999 999 999 999 933 481 833 503 737 774 08;

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:


0.204 798 741 659 033 112 227 916 717 529 289 49(10) =

0.0011 0100 0110 1101 1011 0000 1011 1001 1010 1111 1111 1111 1111 1(2)

6. Positive number before normalization:

24 821.204 798 741 659 033 112 227 916 717 529 289 49(10) =

110 0000 1111 0101.0011 0100 0110 1101 1011 0000 1011 1001 1010 1111 1111 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the left, so that only one non zero digit remains to the left of it:


24 821.204 798 741 659 033 112 227 916 717 529 289 49(10) =

110 0000 1111 0101.0011 0100 0110 1101 1011 0000 1011 1001 1010 1111 1111 1111 1111 1(2) =

110 0000 1111 0101.0011 0100 0110 1101 1011 0000 1011 1001 1010 1111 1111 1111 1111 1(2) × 20 =

1.1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011 1111 1111 1111 111(2) × 214


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): 14

Mantissa (not normalized):
1.1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011 1111 1111 1111 111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

14 + 2(11-1) - 1 =

(14 + 1 023)(10) =

1 037(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;
  • 1 037 ÷ 2 = 518 + 1;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1037(10) =

100 0000 1101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011 111 1111 1111 1111 =

1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011


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) =
100 0000 1101

Mantissa (52 bits) =
1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011


Decimal number -24 821.204 798 741 659 033 112 227 916 717 529 289 49 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 1101 - 1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 0110 1011

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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