0.519 999 999 999 905 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.519 999 999 999 905(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
0.519 999 999 999 905(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: 0.519 999 999 999 905.

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) 0.519 999 999 999 905 × 2 = 1 + 0.039 999 999 999 81;
  • 2) 0.039 999 999 999 81 × 2 = 0 + 0.079 999 999 999 62;
  • 3) 0.079 999 999 999 62 × 2 = 0 + 0.159 999 999 999 24;
  • 4) 0.159 999 999 999 24 × 2 = 0 + 0.319 999 999 998 48;
  • 5) 0.319 999 999 998 48 × 2 = 0 + 0.639 999 999 996 96;
  • 6) 0.639 999 999 996 96 × 2 = 1 + 0.279 999 999 993 92;
  • 7) 0.279 999 999 993 92 × 2 = 0 + 0.559 999 999 987 84;
  • 8) 0.559 999 999 987 84 × 2 = 1 + 0.119 999 999 975 68;
  • 9) 0.119 999 999 975 68 × 2 = 0 + 0.239 999 999 951 36;
  • 10) 0.239 999 999 951 36 × 2 = 0 + 0.479 999 999 902 72;
  • 11) 0.479 999 999 902 72 × 2 = 0 + 0.959 999 999 805 44;
  • 12) 0.959 999 999 805 44 × 2 = 1 + 0.919 999 999 610 88;
  • 13) 0.919 999 999 610 88 × 2 = 1 + 0.839 999 999 221 76;
  • 14) 0.839 999 999 221 76 × 2 = 1 + 0.679 999 998 443 52;
  • 15) 0.679 999 998 443 52 × 2 = 1 + 0.359 999 996 887 04;
  • 16) 0.359 999 996 887 04 × 2 = 0 + 0.719 999 993 774 08;
  • 17) 0.719 999 993 774 08 × 2 = 1 + 0.439 999 987 548 16;
  • 18) 0.439 999 987 548 16 × 2 = 0 + 0.879 999 975 096 32;
  • 19) 0.879 999 975 096 32 × 2 = 1 + 0.759 999 950 192 64;
  • 20) 0.759 999 950 192 64 × 2 = 1 + 0.519 999 900 385 28;
  • 21) 0.519 999 900 385 28 × 2 = 1 + 0.039 999 800 770 56;
  • 22) 0.039 999 800 770 56 × 2 = 0 + 0.079 999 601 541 12;
  • 23) 0.079 999 601 541 12 × 2 = 0 + 0.159 999 203 082 24;
  • 24) 0.159 999 203 082 24 × 2 = 0 + 0.319 998 406 164 48;
  • 25) 0.319 998 406 164 48 × 2 = 0 + 0.639 996 812 328 96;
  • 26) 0.639 996 812 328 96 × 2 = 1 + 0.279 993 624 657 92;
  • 27) 0.279 993 624 657 92 × 2 = 0 + 0.559 987 249 315 84;
  • 28) 0.559 987 249 315 84 × 2 = 1 + 0.119 974 498 631 68;
  • 29) 0.119 974 498 631 68 × 2 = 0 + 0.239 948 997 263 36;
  • 30) 0.239 948 997 263 36 × 2 = 0 + 0.479 897 994 526 72;
  • 31) 0.479 897 994 526 72 × 2 = 0 + 0.959 795 989 053 44;
  • 32) 0.959 795 989 053 44 × 2 = 1 + 0.919 591 978 106 88;
  • 33) 0.919 591 978 106 88 × 2 = 1 + 0.839 183 956 213 76;
  • 34) 0.839 183 956 213 76 × 2 = 1 + 0.678 367 912 427 52;
  • 35) 0.678 367 912 427 52 × 2 = 1 + 0.356 735 824 855 04;
  • 36) 0.356 735 824 855 04 × 2 = 0 + 0.713 471 649 710 08;
  • 37) 0.713 471 649 710 08 × 2 = 1 + 0.426 943 299 420 16;
  • 38) 0.426 943 299 420 16 × 2 = 0 + 0.853 886 598 840 32;
  • 39) 0.853 886 598 840 32 × 2 = 1 + 0.707 773 197 680 64;
  • 40) 0.707 773 197 680 64 × 2 = 1 + 0.415 546 395 361 28;
  • 41) 0.415 546 395 361 28 × 2 = 0 + 0.831 092 790 722 56;
  • 42) 0.831 092 790 722 56 × 2 = 1 + 0.662 185 581 445 12;
  • 43) 0.662 185 581 445 12 × 2 = 1 + 0.324 371 162 890 24;
  • 44) 0.324 371 162 890 24 × 2 = 0 + 0.648 742 325 780 48;
  • 45) 0.648 742 325 780 48 × 2 = 1 + 0.297 484 651 560 96;
  • 46) 0.297 484 651 560 96 × 2 = 0 + 0.594 969 303 121 92;
  • 47) 0.594 969 303 121 92 × 2 = 1 + 0.189 938 606 243 84;
  • 48) 0.189 938 606 243 84 × 2 = 0 + 0.379 877 212 487 68;
  • 49) 0.379 877 212 487 68 × 2 = 0 + 0.759 754 424 975 36;
  • 50) 0.759 754 424 975 36 × 2 = 1 + 0.519 508 849 950 72;
  • 51) 0.519 508 849 950 72 × 2 = 1 + 0.039 017 699 901 44;
  • 52) 0.039 017 699 901 44 × 2 = 0 + 0.078 035 399 802 88;
  • 53) 0.078 035 399 802 88 × 2 = 0 + 0.156 070 799 605 76;

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:


0.519 999 999 999 905(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 1010 0110 0(2)

5. Positive number before normalization:

0.519 999 999 999 905(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 1010 0110 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:


0.519 999 999 999 905(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 1010 0110 0(2) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0110 1010 0110 0(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1101 0100 1100(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 0110 1101 0100 1100


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 0110 1101 0100 1100 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1101 0100 1100


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 0110 1101 0100 1100


Decimal number 0.519 999 999 999 905 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 0110 1101 0100 1100

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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