39 643.788 742 023 025 406 524 538 993 835 449 218 02 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 39 643.788 742 023 025 406 524 538 993 835 449 218 02(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
39 643.788 742 023 025 406 524 538 993 835 449 218 02(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: 39 643.
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;
  • 39 643 ÷ 2 = 19 821 + 1;
  • 19 821 ÷ 2 = 9 910 + 1;
  • 9 910 ÷ 2 = 4 955 + 0;
  • 4 955 ÷ 2 = 2 477 + 1;
  • 2 477 ÷ 2 = 1 238 + 1;
  • 1 238 ÷ 2 = 619 + 0;
  • 619 ÷ 2 = 309 + 1;
  • 309 ÷ 2 = 154 + 1;
  • 154 ÷ 2 = 77 + 0;
  • 77 ÷ 2 = 38 + 1;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

39 643(10) =

1001 1010 1101 1011(2)


3. Convert to binary (base 2) the fractional part: 0.788 742 023 025 406 524 538 993 835 449 218 02.

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) 0.788 742 023 025 406 524 538 993 835 449 218 02 × 2 = 1 + 0.577 484 046 050 813 049 077 987 670 898 436 04;
  • 2) 0.577 484 046 050 813 049 077 987 670 898 436 04 × 2 = 1 + 0.154 968 092 101 626 098 155 975 341 796 872 08;
  • 3) 0.154 968 092 101 626 098 155 975 341 796 872 08 × 2 = 0 + 0.309 936 184 203 252 196 311 950 683 593 744 16;
  • 4) 0.309 936 184 203 252 196 311 950 683 593 744 16 × 2 = 0 + 0.619 872 368 406 504 392 623 901 367 187 488 32;
  • 5) 0.619 872 368 406 504 392 623 901 367 187 488 32 × 2 = 1 + 0.239 744 736 813 008 785 247 802 734 374 976 64;
  • 6) 0.239 744 736 813 008 785 247 802 734 374 976 64 × 2 = 0 + 0.479 489 473 626 017 570 495 605 468 749 953 28;
  • 7) 0.479 489 473 626 017 570 495 605 468 749 953 28 × 2 = 0 + 0.958 978 947 252 035 140 991 210 937 499 906 56;
  • 8) 0.958 978 947 252 035 140 991 210 937 499 906 56 × 2 = 1 + 0.917 957 894 504 070 281 982 421 874 999 813 12;
  • 9) 0.917 957 894 504 070 281 982 421 874 999 813 12 × 2 = 1 + 0.835 915 789 008 140 563 964 843 749 999 626 24;
  • 10) 0.835 915 789 008 140 563 964 843 749 999 626 24 × 2 = 1 + 0.671 831 578 016 281 127 929 687 499 999 252 48;
  • 11) 0.671 831 578 016 281 127 929 687 499 999 252 48 × 2 = 1 + 0.343 663 156 032 562 255 859 374 999 998 504 96;
  • 12) 0.343 663 156 032 562 255 859 374 999 998 504 96 × 2 = 0 + 0.687 326 312 065 124 511 718 749 999 997 009 92;
  • 13) 0.687 326 312 065 124 511 718 749 999 997 009 92 × 2 = 1 + 0.374 652 624 130 249 023 437 499 999 994 019 84;
  • 14) 0.374 652 624 130 249 023 437 499 999 994 019 84 × 2 = 0 + 0.749 305 248 260 498 046 874 999 999 988 039 68;
  • 15) 0.749 305 248 260 498 046 874 999 999 988 039 68 × 2 = 1 + 0.498 610 496 520 996 093 749 999 999 976 079 36;
  • 16) 0.498 610 496 520 996 093 749 999 999 976 079 36 × 2 = 0 + 0.997 220 993 041 992 187 499 999 999 952 158 72;
  • 17) 0.997 220 993 041 992 187 499 999 999 952 158 72 × 2 = 1 + 0.994 441 986 083 984 374 999 999 999 904 317 44;
  • 18) 0.994 441 986 083 984 374 999 999 999 904 317 44 × 2 = 1 + 0.988 883 972 167 968 749 999 999 999 808 634 88;
  • 19) 0.988 883 972 167 968 749 999 999 999 808 634 88 × 2 = 1 + 0.977 767 944 335 937 499 999 999 999 617 269 76;
  • 20) 0.977 767 944 335 937 499 999 999 999 617 269 76 × 2 = 1 + 0.955 535 888 671 874 999 999 999 999 234 539 52;
  • 21) 0.955 535 888 671 874 999 999 999 999 234 539 52 × 2 = 1 + 0.911 071 777 343 749 999 999 999 998 469 079 04;
  • 22) 0.911 071 777 343 749 999 999 999 998 469 079 04 × 2 = 1 + 0.822 143 554 687 499 999 999 999 996 938 158 08;
  • 23) 0.822 143 554 687 499 999 999 999 996 938 158 08 × 2 = 1 + 0.644 287 109 374 999 999 999 999 993 876 316 16;
  • 24) 0.644 287 109 374 999 999 999 999 993 876 316 16 × 2 = 1 + 0.288 574 218 749 999 999 999 999 987 752 632 32;
  • 25) 0.288 574 218 749 999 999 999 999 987 752 632 32 × 2 = 0 + 0.577 148 437 499 999 999 999 999 975 505 264 64;
  • 26) 0.577 148 437 499 999 999 999 999 975 505 264 64 × 2 = 1 + 0.154 296 874 999 999 999 999 999 951 010 529 28;
  • 27) 0.154 296 874 999 999 999 999 999 951 010 529 28 × 2 = 0 + 0.308 593 749 999 999 999 999 999 902 021 058 56;
  • 28) 0.308 593 749 999 999 999 999 999 902 021 058 56 × 2 = 0 + 0.617 187 499 999 999 999 999 999 804 042 117 12;
  • 29) 0.617 187 499 999 999 999 999 999 804 042 117 12 × 2 = 1 + 0.234 374 999 999 999 999 999 999 608 084 234 24;
  • 30) 0.234 374 999 999 999 999 999 999 608 084 234 24 × 2 = 0 + 0.468 749 999 999 999 999 999 999 216 168 468 48;
  • 31) 0.468 749 999 999 999 999 999 999 216 168 468 48 × 2 = 0 + 0.937 499 999 999 999 999 999 998 432 336 936 96;
  • 32) 0.937 499 999 999 999 999 999 998 432 336 936 96 × 2 = 1 + 0.874 999 999 999 999 999 999 996 864 673 873 92;
  • 33) 0.874 999 999 999 999 999 999 996 864 673 873 92 × 2 = 1 + 0.749 999 999 999 999 999 999 993 729 347 747 84;
  • 34) 0.749 999 999 999 999 999 999 993 729 347 747 84 × 2 = 1 + 0.499 999 999 999 999 999 999 987 458 695 495 68;
  • 35) 0.499 999 999 999 999 999 999 987 458 695 495 68 × 2 = 0 + 0.999 999 999 999 999 999 999 974 917 390 991 36;
  • 36) 0.999 999 999 999 999 999 999 974 917 390 991 36 × 2 = 1 + 0.999 999 999 999 999 999 999 949 834 781 982 72;
  • 37) 0.999 999 999 999 999 999 999 949 834 781 982 72 × 2 = 1 + 0.999 999 999 999 999 999 999 899 669 563 965 44;
  • 38) 0.999 999 999 999 999 999 999 899 669 563 965 44 × 2 = 1 + 0.999 999 999 999 999 999 999 799 339 127 930 88;
  • 39) 0.999 999 999 999 999 999 999 799 339 127 930 88 × 2 = 1 + 0.999 999 999 999 999 999 999 598 678 255 861 76;
  • 40) 0.999 999 999 999 999 999 999 598 678 255 861 76 × 2 = 1 + 0.999 999 999 999 999 999 999 197 356 511 723 52;
  • 41) 0.999 999 999 999 999 999 999 197 356 511 723 52 × 2 = 1 + 0.999 999 999 999 999 999 998 394 713 023 447 04;
  • 42) 0.999 999 999 999 999 999 998 394 713 023 447 04 × 2 = 1 + 0.999 999 999 999 999 999 996 789 426 046 894 08;
  • 43) 0.999 999 999 999 999 999 996 789 426 046 894 08 × 2 = 1 + 0.999 999 999 999 999 999 993 578 852 093 788 16;
  • 44) 0.999 999 999 999 999 999 993 578 852 093 788 16 × 2 = 1 + 0.999 999 999 999 999 999 987 157 704 187 576 32;
  • 45) 0.999 999 999 999 999 999 987 157 704 187 576 32 × 2 = 1 + 0.999 999 999 999 999 999 974 315 408 375 152 64;
  • 46) 0.999 999 999 999 999 999 974 315 408 375 152 64 × 2 = 1 + 0.999 999 999 999 999 999 948 630 816 750 305 28;
  • 47) 0.999 999 999 999 999 999 948 630 816 750 305 28 × 2 = 1 + 0.999 999 999 999 999 999 897 261 633 500 610 56;
  • 48) 0.999 999 999 999 999 999 897 261 633 500 610 56 × 2 = 1 + 0.999 999 999 999 999 999 794 523 267 001 221 12;
  • 49) 0.999 999 999 999 999 999 794 523 267 001 221 12 × 2 = 1 + 0.999 999 999 999 999 999 589 046 534 002 442 24;
  • 50) 0.999 999 999 999 999 999 589 046 534 002 442 24 × 2 = 1 + 0.999 999 999 999 999 999 178 093 068 004 884 48;
  • 51) 0.999 999 999 999 999 999 178 093 068 004 884 48 × 2 = 1 + 0.999 999 999 999 999 998 356 186 136 009 768 96;
  • 52) 0.999 999 999 999 999 998 356 186 136 009 768 96 × 2 = 1 + 0.999 999 999 999 999 996 712 372 272 019 537 92;
  • 53) 0.999 999 999 999 999 996 712 372 272 019 537 92 × 2 = 1 + 0.999 999 999 999 999 993 424 744 544 039 075 84;

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:


0.788 742 023 025 406 524 538 993 835 449 218 02(10) =

0.1100 1001 1110 1010 1111 1111 0100 1001 1101 1111 1111 1111 1111 1(2)

5. Positive number before normalization:

39 643.788 742 023 025 406 524 538 993 835 449 218 02(10) =

1001 1010 1101 1011.1100 1001 1110 1010 1111 1111 0100 1001 1101 1111 1111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


39 643.788 742 023 025 406 524 538 993 835 449 218 02(10) =

1001 1010 1101 1011.1100 1001 1110 1010 1111 1111 0100 1001 1101 1111 1111 1111 1111 1(2) =

1001 1010 1101 1011.1100 1001 1110 1010 1111 1111 0100 1001 1101 1111 1111 1111 1111 1(2) × 20 =

1.0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 1011 1111 1111 1111 1111(2) × 215


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): 15

Mantissa (not normalized):
1.0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 1011 1111 1111 1111 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

15 + 2(11-1) - 1 =

(15 + 1 023)(10) =

1 038(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;
  • 1 038 ÷ 2 = 519 + 0;
  • 519 ÷ 2 = 259 + 1;
  • 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;

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) =

1038(10) =

100 0000 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, 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. 0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 1011 1111 1111 1111 1111 =

0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 1011


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) =
100 0000 1110

Mantissa (52 bits) =
0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 1011


Decimal number 39 643.788 742 023 025 406 524 538 993 835 449 218 02 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1110 - 0011 0101 1011 0111 1001 0011 1101 0101 1111 1110 1001 0011 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