1.004 999 998 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.004 999 998 99(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
1.004 999 998 99(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.004 999 998 99.

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) 0.004 999 998 99 × 2 = 0 + 0.009 999 997 98;
  • 2) 0.009 999 997 98 × 2 = 0 + 0.019 999 995 96;
  • 3) 0.019 999 995 96 × 2 = 0 + 0.039 999 991 92;
  • 4) 0.039 999 991 92 × 2 = 0 + 0.079 999 983 84;
  • 5) 0.079 999 983 84 × 2 = 0 + 0.159 999 967 68;
  • 6) 0.159 999 967 68 × 2 = 0 + 0.319 999 935 36;
  • 7) 0.319 999 935 36 × 2 = 0 + 0.639 999 870 72;
  • 8) 0.639 999 870 72 × 2 = 1 + 0.279 999 741 44;
  • 9) 0.279 999 741 44 × 2 = 0 + 0.559 999 482 88;
  • 10) 0.559 999 482 88 × 2 = 1 + 0.119 998 965 76;
  • 11) 0.119 998 965 76 × 2 = 0 + 0.239 997 931 52;
  • 12) 0.239 997 931 52 × 2 = 0 + 0.479 995 863 04;
  • 13) 0.479 995 863 04 × 2 = 0 + 0.959 991 726 08;
  • 14) 0.959 991 726 08 × 2 = 1 + 0.919 983 452 16;
  • 15) 0.919 983 452 16 × 2 = 1 + 0.839 966 904 32;
  • 16) 0.839 966 904 32 × 2 = 1 + 0.679 933 808 64;
  • 17) 0.679 933 808 64 × 2 = 1 + 0.359 867 617 28;
  • 18) 0.359 867 617 28 × 2 = 0 + 0.719 735 234 56;
  • 19) 0.719 735 234 56 × 2 = 1 + 0.439 470 469 12;
  • 20) 0.439 470 469 12 × 2 = 0 + 0.878 940 938 24;
  • 21) 0.878 940 938 24 × 2 = 1 + 0.757 881 876 48;
  • 22) 0.757 881 876 48 × 2 = 1 + 0.515 763 752 96;
  • 23) 0.515 763 752 96 × 2 = 1 + 0.031 527 505 92;
  • 24) 0.031 527 505 92 × 2 = 0 + 0.063 055 011 84;
  • 25) 0.063 055 011 84 × 2 = 0 + 0.126 110 023 68;
  • 26) 0.126 110 023 68 × 2 = 0 + 0.252 220 047 36;
  • 27) 0.252 220 047 36 × 2 = 0 + 0.504 440 094 72;
  • 28) 0.504 440 094 72 × 2 = 1 + 0.008 880 189 44;
  • 29) 0.008 880 189 44 × 2 = 0 + 0.017 760 378 88;
  • 30) 0.017 760 378 88 × 2 = 0 + 0.035 520 757 76;
  • 31) 0.035 520 757 76 × 2 = 0 + 0.071 041 515 52;
  • 32) 0.071 041 515 52 × 2 = 0 + 0.142 083 031 04;
  • 33) 0.142 083 031 04 × 2 = 0 + 0.284 166 062 08;
  • 34) 0.284 166 062 08 × 2 = 0 + 0.568 332 124 16;
  • 35) 0.568 332 124 16 × 2 = 1 + 0.136 664 248 32;
  • 36) 0.136 664 248 32 × 2 = 0 + 0.273 328 496 64;
  • 37) 0.273 328 496 64 × 2 = 0 + 0.546 656 993 28;
  • 38) 0.546 656 993 28 × 2 = 1 + 0.093 313 986 56;
  • 39) 0.093 313 986 56 × 2 = 0 + 0.186 627 973 12;
  • 40) 0.186 627 973 12 × 2 = 0 + 0.373 255 946 24;
  • 41) 0.373 255 946 24 × 2 = 0 + 0.746 511 892 48;
  • 42) 0.746 511 892 48 × 2 = 1 + 0.493 023 784 96;
  • 43) 0.493 023 784 96 × 2 = 0 + 0.986 047 569 92;
  • 44) 0.986 047 569 92 × 2 = 1 + 0.972 095 139 84;
  • 45) 0.972 095 139 84 × 2 = 1 + 0.944 190 279 68;
  • 46) 0.944 190 279 68 × 2 = 1 + 0.888 380 559 36;
  • 47) 0.888 380 559 36 × 2 = 1 + 0.776 761 118 72;
  • 48) 0.776 761 118 72 × 2 = 1 + 0.553 522 237 44;
  • 49) 0.553 522 237 44 × 2 = 1 + 0.107 044 474 88;
  • 50) 0.107 044 474 88 × 2 = 0 + 0.214 088 949 76;
  • 51) 0.214 088 949 76 × 2 = 0 + 0.428 177 899 52;
  • 52) 0.428 177 899 52 × 2 = 0 + 0.856 355 799 04;
  • 53) 0.856 355 799 04 × 2 = 1 + 0.712 711 598 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).


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:


0.004 999 998 99(10) =

0.0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1(2)

5. Positive number before normalization:

1.004 999 998 99(10) =

1.0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.004 999 998 99(10) =

1.0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1(2) =

1.0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1(2) × 20


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): 0

Mantissa (not normalized):
1.0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000 1 =

0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000


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 1111

Mantissa (52 bits) =
0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000


Decimal number 1.004 999 998 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 0001 0100 0111 1010 1110 0001 0000 0010 0100 0101 1111 1000

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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