0.519 999 999 999 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.519 999 999 999 3(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
0.519 999 999 999 3(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: 0.519 999 999 999 3.

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) 0.519 999 999 999 3 × 2 = 1 + 0.039 999 999 998 6;
  • 2) 0.039 999 999 998 6 × 2 = 0 + 0.079 999 999 997 2;
  • 3) 0.079 999 999 997 2 × 2 = 0 + 0.159 999 999 994 4;
  • 4) 0.159 999 999 994 4 × 2 = 0 + 0.319 999 999 988 8;
  • 5) 0.319 999 999 988 8 × 2 = 0 + 0.639 999 999 977 6;
  • 6) 0.639 999 999 977 6 × 2 = 1 + 0.279 999 999 955 2;
  • 7) 0.279 999 999 955 2 × 2 = 0 + 0.559 999 999 910 4;
  • 8) 0.559 999 999 910 4 × 2 = 1 + 0.119 999 999 820 8;
  • 9) 0.119 999 999 820 8 × 2 = 0 + 0.239 999 999 641 6;
  • 10) 0.239 999 999 641 6 × 2 = 0 + 0.479 999 999 283 2;
  • 11) 0.479 999 999 283 2 × 2 = 0 + 0.959 999 998 566 4;
  • 12) 0.959 999 998 566 4 × 2 = 1 + 0.919 999 997 132 8;
  • 13) 0.919 999 997 132 8 × 2 = 1 + 0.839 999 994 265 6;
  • 14) 0.839 999 994 265 6 × 2 = 1 + 0.679 999 988 531 2;
  • 15) 0.679 999 988 531 2 × 2 = 1 + 0.359 999 977 062 4;
  • 16) 0.359 999 977 062 4 × 2 = 0 + 0.719 999 954 124 8;
  • 17) 0.719 999 954 124 8 × 2 = 1 + 0.439 999 908 249 6;
  • 18) 0.439 999 908 249 6 × 2 = 0 + 0.879 999 816 499 2;
  • 19) 0.879 999 816 499 2 × 2 = 1 + 0.759 999 632 998 4;
  • 20) 0.759 999 632 998 4 × 2 = 1 + 0.519 999 265 996 8;
  • 21) 0.519 999 265 996 8 × 2 = 1 + 0.039 998 531 993 6;
  • 22) 0.039 998 531 993 6 × 2 = 0 + 0.079 997 063 987 2;
  • 23) 0.079 997 063 987 2 × 2 = 0 + 0.159 994 127 974 4;
  • 24) 0.159 994 127 974 4 × 2 = 0 + 0.319 988 255 948 8;
  • 25) 0.319 988 255 948 8 × 2 = 0 + 0.639 976 511 897 6;
  • 26) 0.639 976 511 897 6 × 2 = 1 + 0.279 953 023 795 2;
  • 27) 0.279 953 023 795 2 × 2 = 0 + 0.559 906 047 590 4;
  • 28) 0.559 906 047 590 4 × 2 = 1 + 0.119 812 095 180 8;
  • 29) 0.119 812 095 180 8 × 2 = 0 + 0.239 624 190 361 6;
  • 30) 0.239 624 190 361 6 × 2 = 0 + 0.479 248 380 723 2;
  • 31) 0.479 248 380 723 2 × 2 = 0 + 0.958 496 761 446 4;
  • 32) 0.958 496 761 446 4 × 2 = 1 + 0.916 993 522 892 8;
  • 33) 0.916 993 522 892 8 × 2 = 1 + 0.833 987 045 785 6;
  • 34) 0.833 987 045 785 6 × 2 = 1 + 0.667 974 091 571 2;
  • 35) 0.667 974 091 571 2 × 2 = 1 + 0.335 948 183 142 4;
  • 36) 0.335 948 183 142 4 × 2 = 0 + 0.671 896 366 284 8;
  • 37) 0.671 896 366 284 8 × 2 = 1 + 0.343 792 732 569 6;
  • 38) 0.343 792 732 569 6 × 2 = 0 + 0.687 585 465 139 2;
  • 39) 0.687 585 465 139 2 × 2 = 1 + 0.375 170 930 278 4;
  • 40) 0.375 170 930 278 4 × 2 = 0 + 0.750 341 860 556 8;
  • 41) 0.750 341 860 556 8 × 2 = 1 + 0.500 683 721 113 6;
  • 42) 0.500 683 721 113 6 × 2 = 1 + 0.001 367 442 227 2;
  • 43) 0.001 367 442 227 2 × 2 = 0 + 0.002 734 884 454 4;
  • 44) 0.002 734 884 454 4 × 2 = 0 + 0.005 469 768 908 8;
  • 45) 0.005 469 768 908 8 × 2 = 0 + 0.010 939 537 817 6;
  • 46) 0.010 939 537 817 6 × 2 = 0 + 0.021 879 075 635 2;
  • 47) 0.021 879 075 635 2 × 2 = 0 + 0.043 758 151 270 4;
  • 48) 0.043 758 151 270 4 × 2 = 0 + 0.087 516 302 540 8;
  • 49) 0.087 516 302 540 8 × 2 = 0 + 0.175 032 605 081 6;
  • 50) 0.175 032 605 081 6 × 2 = 0 + 0.350 065 210 163 2;
  • 51) 0.350 065 210 163 2 × 2 = 0 + 0.700 130 420 326 4;
  • 52) 0.700 130 420 326 4 × 2 = 1 + 0.400 260 840 652 8;
  • 53) 0.400 260 840 652 8 × 2 = 0 + 0.800 521 681 305 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:


0.519 999 999 999 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1100 0000 0001 0(2)

5. Positive number before normalization:

0.519 999 999 999 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1100 0000 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:


0.519 999 999 999 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1100 0000 0001 0(2) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1010 1100 0000 0001 0(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 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):
1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 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. 0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 0010 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 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) =
0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 0010


Decimal number 0.519 999 999 999 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1010 0011 1101 0111 0000 1010 0011 1101 0101 1000 0000 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