0.620 928 906 036 742 024 296 838 734 382 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.620 928 906 036 742 024 296 838 734 382 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
0.620 928 906 036 742 024 296 838 734 382 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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.620 928 906 036 742 024 296 838 734 382 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) 0.620 928 906 036 742 024 296 838 734 382 8 × 2 = 1 + 0.241 857 812 073 484 048 593 677 468 765 6;
  • 2) 0.241 857 812 073 484 048 593 677 468 765 6 × 2 = 0 + 0.483 715 624 146 968 097 187 354 937 531 2;
  • 3) 0.483 715 624 146 968 097 187 354 937 531 2 × 2 = 0 + 0.967 431 248 293 936 194 374 709 875 062 4;
  • 4) 0.967 431 248 293 936 194 374 709 875 062 4 × 2 = 1 + 0.934 862 496 587 872 388 749 419 750 124 8;
  • 5) 0.934 862 496 587 872 388 749 419 750 124 8 × 2 = 1 + 0.869 724 993 175 744 777 498 839 500 249 6;
  • 6) 0.869 724 993 175 744 777 498 839 500 249 6 × 2 = 1 + 0.739 449 986 351 489 554 997 679 000 499 2;
  • 7) 0.739 449 986 351 489 554 997 679 000 499 2 × 2 = 1 + 0.478 899 972 702 979 109 995 358 000 998 4;
  • 8) 0.478 899 972 702 979 109 995 358 000 998 4 × 2 = 0 + 0.957 799 945 405 958 219 990 716 001 996 8;
  • 9) 0.957 799 945 405 958 219 990 716 001 996 8 × 2 = 1 + 0.915 599 890 811 916 439 981 432 003 993 6;
  • 10) 0.915 599 890 811 916 439 981 432 003 993 6 × 2 = 1 + 0.831 199 781 623 832 879 962 864 007 987 2;
  • 11) 0.831 199 781 623 832 879 962 864 007 987 2 × 2 = 1 + 0.662 399 563 247 665 759 925 728 015 974 4;
  • 12) 0.662 399 563 247 665 759 925 728 015 974 4 × 2 = 1 + 0.324 799 126 495 331 519 851 456 031 948 8;
  • 13) 0.324 799 126 495 331 519 851 456 031 948 8 × 2 = 0 + 0.649 598 252 990 663 039 702 912 063 897 6;
  • 14) 0.649 598 252 990 663 039 702 912 063 897 6 × 2 = 1 + 0.299 196 505 981 326 079 405 824 127 795 2;
  • 15) 0.299 196 505 981 326 079 405 824 127 795 2 × 2 = 0 + 0.598 393 011 962 652 158 811 648 255 590 4;
  • 16) 0.598 393 011 962 652 158 811 648 255 590 4 × 2 = 1 + 0.196 786 023 925 304 317 623 296 511 180 8;
  • 17) 0.196 786 023 925 304 317 623 296 511 180 8 × 2 = 0 + 0.393 572 047 850 608 635 246 593 022 361 6;
  • 18) 0.393 572 047 850 608 635 246 593 022 361 6 × 2 = 0 + 0.787 144 095 701 217 270 493 186 044 723 2;
  • 19) 0.787 144 095 701 217 270 493 186 044 723 2 × 2 = 1 + 0.574 288 191 402 434 540 986 372 089 446 4;
  • 20) 0.574 288 191 402 434 540 986 372 089 446 4 × 2 = 1 + 0.148 576 382 804 869 081 972 744 178 892 8;
  • 21) 0.148 576 382 804 869 081 972 744 178 892 8 × 2 = 0 + 0.297 152 765 609 738 163 945 488 357 785 6;
  • 22) 0.297 152 765 609 738 163 945 488 357 785 6 × 2 = 0 + 0.594 305 531 219 476 327 890 976 715 571 2;
  • 23) 0.594 305 531 219 476 327 890 976 715 571 2 × 2 = 1 + 0.188 611 062 438 952 655 781 953 431 142 4;
  • 24) 0.188 611 062 438 952 655 781 953 431 142 4 × 2 = 0 + 0.377 222 124 877 905 311 563 906 862 284 8;
  • 25) 0.377 222 124 877 905 311 563 906 862 284 8 × 2 = 0 + 0.754 444 249 755 810 623 127 813 724 569 6;
  • 26) 0.754 444 249 755 810 623 127 813 724 569 6 × 2 = 1 + 0.508 888 499 511 621 246 255 627 449 139 2;
  • 27) 0.508 888 499 511 621 246 255 627 449 139 2 × 2 = 1 + 0.017 776 999 023 242 492 511 254 898 278 4;
  • 28) 0.017 776 999 023 242 492 511 254 898 278 4 × 2 = 0 + 0.035 553 998 046 484 985 022 509 796 556 8;
  • 29) 0.035 553 998 046 484 985 022 509 796 556 8 × 2 = 0 + 0.071 107 996 092 969 970 045 019 593 113 6;
  • 30) 0.071 107 996 092 969 970 045 019 593 113 6 × 2 = 0 + 0.142 215 992 185 939 940 090 039 186 227 2;
  • 31) 0.142 215 992 185 939 940 090 039 186 227 2 × 2 = 0 + 0.284 431 984 371 879 880 180 078 372 454 4;
  • 32) 0.284 431 984 371 879 880 180 078 372 454 4 × 2 = 0 + 0.568 863 968 743 759 760 360 156 744 908 8;
  • 33) 0.568 863 968 743 759 760 360 156 744 908 8 × 2 = 1 + 0.137 727 937 487 519 520 720 313 489 817 6;
  • 34) 0.137 727 937 487 519 520 720 313 489 817 6 × 2 = 0 + 0.275 455 874 975 039 041 440 626 979 635 2;
  • 35) 0.275 455 874 975 039 041 440 626 979 635 2 × 2 = 0 + 0.550 911 749 950 078 082 881 253 959 270 4;
  • 36) 0.550 911 749 950 078 082 881 253 959 270 4 × 2 = 1 + 0.101 823 499 900 156 165 762 507 918 540 8;
  • 37) 0.101 823 499 900 156 165 762 507 918 540 8 × 2 = 0 + 0.203 646 999 800 312 331 525 015 837 081 6;
  • 38) 0.203 646 999 800 312 331 525 015 837 081 6 × 2 = 0 + 0.407 293 999 600 624 663 050 031 674 163 2;
  • 39) 0.407 293 999 600 624 663 050 031 674 163 2 × 2 = 0 + 0.814 587 999 201 249 326 100 063 348 326 4;
  • 40) 0.814 587 999 201 249 326 100 063 348 326 4 × 2 = 1 + 0.629 175 998 402 498 652 200 126 696 652 8;
  • 41) 0.629 175 998 402 498 652 200 126 696 652 8 × 2 = 1 + 0.258 351 996 804 997 304 400 253 393 305 6;
  • 42) 0.258 351 996 804 997 304 400 253 393 305 6 × 2 = 0 + 0.516 703 993 609 994 608 800 506 786 611 2;
  • 43) 0.516 703 993 609 994 608 800 506 786 611 2 × 2 = 1 + 0.033 407 987 219 989 217 601 013 573 222 4;
  • 44) 0.033 407 987 219 989 217 601 013 573 222 4 × 2 = 0 + 0.066 815 974 439 978 435 202 027 146 444 8;
  • 45) 0.066 815 974 439 978 435 202 027 146 444 8 × 2 = 0 + 0.133 631 948 879 956 870 404 054 292 889 6;
  • 46) 0.133 631 948 879 956 870 404 054 292 889 6 × 2 = 0 + 0.267 263 897 759 913 740 808 108 585 779 2;
  • 47) 0.267 263 897 759 913 740 808 108 585 779 2 × 2 = 0 + 0.534 527 795 519 827 481 616 217 171 558 4;
  • 48) 0.534 527 795 519 827 481 616 217 171 558 4 × 2 = 1 + 0.069 055 591 039 654 963 232 434 343 116 8;
  • 49) 0.069 055 591 039 654 963 232 434 343 116 8 × 2 = 0 + 0.138 111 182 079 309 926 464 868 686 233 6;
  • 50) 0.138 111 182 079 309 926 464 868 686 233 6 × 2 = 0 + 0.276 222 364 158 619 852 929 737 372 467 2;
  • 51) 0.276 222 364 158 619 852 929 737 372 467 2 × 2 = 0 + 0.552 444 728 317 239 705 859 474 744 934 4;
  • 52) 0.552 444 728 317 239 705 859 474 744 934 4 × 2 = 1 + 0.104 889 456 634 479 411 718 949 489 868 8;
  • 53) 0.104 889 456 634 479 411 718 949 489 868 8 × 2 = 0 + 0.209 778 913 268 958 823 437 898 979 737 6;

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).


4. 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:


0.620 928 906 036 742 024 296 838 734 382 8(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2)

5. Positive number before normalization:

0.620 928 906 036 742 024 296 838 734 382 8(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.620 928 906 036 742 024 296 838 734 382 8(10) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2) =

0.1001 1110 1111 0101 0011 0010 0110 0000 1001 0001 1010 0001 0001 0(2) × 20 =

1.0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010(2) × 2-1


7. 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 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(10)


9. 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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. 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. 0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010 =

0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010


Decimal number 0.620 928 906 036 742 024 296 838 734 382 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0011 1101 1110 1010 0110 0100 1100 0001 0010 0011 0100 0010 0010

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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