Decimal to 64 Bit IEEE 754 Binary: Convert Number 11.023 897 499 996 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 11.023 897 499 996(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 11.
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;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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.

11(10) =

1011(2)


3. Convert to binary (base 2) the fractional part: 0.023 897 499 996.

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) 0.023 897 499 996 × 2 = 0 + 0.047 794 999 992;
  • 2) 0.047 794 999 992 × 2 = 0 + 0.095 589 999 984;
  • 3) 0.095 589 999 984 × 2 = 0 + 0.191 179 999 968;
  • 4) 0.191 179 999 968 × 2 = 0 + 0.382 359 999 936;
  • 5) 0.382 359 999 936 × 2 = 0 + 0.764 719 999 872;
  • 6) 0.764 719 999 872 × 2 = 1 + 0.529 439 999 744;
  • 7) 0.529 439 999 744 × 2 = 1 + 0.058 879 999 488;
  • 8) 0.058 879 999 488 × 2 = 0 + 0.117 759 998 976;
  • 9) 0.117 759 998 976 × 2 = 0 + 0.235 519 997 952;
  • 10) 0.235 519 997 952 × 2 = 0 + 0.471 039 995 904;
  • 11) 0.471 039 995 904 × 2 = 0 + 0.942 079 991 808;
  • 12) 0.942 079 991 808 × 2 = 1 + 0.884 159 983 616;
  • 13) 0.884 159 983 616 × 2 = 1 + 0.768 319 967 232;
  • 14) 0.768 319 967 232 × 2 = 1 + 0.536 639 934 464;
  • 15) 0.536 639 934 464 × 2 = 1 + 0.073 279 868 928;
  • 16) 0.073 279 868 928 × 2 = 0 + 0.146 559 737 856;
  • 17) 0.146 559 737 856 × 2 = 0 + 0.293 119 475 712;
  • 18) 0.293 119 475 712 × 2 = 0 + 0.586 238 951 424;
  • 19) 0.586 238 951 424 × 2 = 1 + 0.172 477 902 848;
  • 20) 0.172 477 902 848 × 2 = 0 + 0.344 955 805 696;
  • 21) 0.344 955 805 696 × 2 = 0 + 0.689 911 611 392;
  • 22) 0.689 911 611 392 × 2 = 1 + 0.379 823 222 784;
  • 23) 0.379 823 222 784 × 2 = 0 + 0.759 646 445 568;
  • 24) 0.759 646 445 568 × 2 = 1 + 0.519 292 891 136;
  • 25) 0.519 292 891 136 × 2 = 1 + 0.038 585 782 272;
  • 26) 0.038 585 782 272 × 2 = 0 + 0.077 171 564 544;
  • 27) 0.077 171 564 544 × 2 = 0 + 0.154 343 129 088;
  • 28) 0.154 343 129 088 × 2 = 0 + 0.308 686 258 176;
  • 29) 0.308 686 258 176 × 2 = 0 + 0.617 372 516 352;
  • 30) 0.617 372 516 352 × 2 = 1 + 0.234 745 032 704;
  • 31) 0.234 745 032 704 × 2 = 0 + 0.469 490 065 408;
  • 32) 0.469 490 065 408 × 2 = 0 + 0.938 980 130 816;
  • 33) 0.938 980 130 816 × 2 = 1 + 0.877 960 261 632;
  • 34) 0.877 960 261 632 × 2 = 1 + 0.755 920 523 264;
  • 35) 0.755 920 523 264 × 2 = 1 + 0.511 841 046 528;
  • 36) 0.511 841 046 528 × 2 = 1 + 0.023 682 093 056;
  • 37) 0.023 682 093 056 × 2 = 0 + 0.047 364 186 112;
  • 38) 0.047 364 186 112 × 2 = 0 + 0.094 728 372 224;
  • 39) 0.094 728 372 224 × 2 = 0 + 0.189 456 744 448;
  • 40) 0.189 456 744 448 × 2 = 0 + 0.378 913 488 896;
  • 41) 0.378 913 488 896 × 2 = 0 + 0.757 826 977 792;
  • 42) 0.757 826 977 792 × 2 = 1 + 0.515 653 955 584;
  • 43) 0.515 653 955 584 × 2 = 1 + 0.031 307 911 168;
  • 44) 0.031 307 911 168 × 2 = 0 + 0.062 615 822 336;
  • 45) 0.062 615 822 336 × 2 = 0 + 0.125 231 644 672;
  • 46) 0.125 231 644 672 × 2 = 0 + 0.250 463 289 344;
  • 47) 0.250 463 289 344 × 2 = 0 + 0.500 926 578 688;
  • 48) 0.500 926 578 688 × 2 = 1 + 0.001 853 157 376;
  • 49) 0.001 853 157 376 × 2 = 0 + 0.003 706 314 752;
  • 50) 0.003 706 314 752 × 2 = 0 + 0.007 412 629 504;
  • 51) 0.007 412 629 504 × 2 = 0 + 0.014 825 259 008;
  • 52) 0.014 825 259 008 × 2 = 0 + 0.029 650 518 016;
  • 53) 0.029 650 518 016 × 2 = 0 + 0.059 301 036 032;

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:


0.023 897 499 996(10) =

0.0000 0110 0001 1110 0010 0101 1000 0100 1111 0000 0110 0001 0000 0(2)

5. Positive number before normalization:

11.023 897 499 996(10) =

1011.0000 0110 0001 1110 0010 0101 1000 0100 1111 0000 0110 0001 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


11.023 897 499 996(10) =

1011.0000 0110 0001 1110 0010 0101 1000 0100 1111 0000 0110 0001 0000 0(2) =

1011.0000 0110 0001 1110 0010 0101 1000 0100 1111 0000 0110 0001 0000 0(2) × 20 =

1.0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 0010 0000(2) × 23


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): 3

Mantissa (not normalized):
1.0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 0010 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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) =

1026(10) =

100 0000 0010(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. 0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 0010 0000 =

0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 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) =
100 0000 0010

Mantissa (52 bits) =
0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 0010


The base ten decimal number 11.023 897 499 996 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0010 - 0110 0000 1100 0011 1100 0100 1011 0000 1001 1110 0000 1100 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