0.466 666 666 666 905 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.466 666 666 666 905(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
0.466 666 666 666 905(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: 0.466 666 666 666 905.

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) 0.466 666 666 666 905 × 2 = 0 + 0.933 333 333 333 81;
  • 2) 0.933 333 333 333 81 × 2 = 1 + 0.866 666 666 667 62;
  • 3) 0.866 666 666 667 62 × 2 = 1 + 0.733 333 333 335 24;
  • 4) 0.733 333 333 335 24 × 2 = 1 + 0.466 666 666 670 48;
  • 5) 0.466 666 666 670 48 × 2 = 0 + 0.933 333 333 340 96;
  • 6) 0.933 333 333 340 96 × 2 = 1 + 0.866 666 666 681 92;
  • 7) 0.866 666 666 681 92 × 2 = 1 + 0.733 333 333 363 84;
  • 8) 0.733 333 333 363 84 × 2 = 1 + 0.466 666 666 727 68;
  • 9) 0.466 666 666 727 68 × 2 = 0 + 0.933 333 333 455 36;
  • 10) 0.933 333 333 455 36 × 2 = 1 + 0.866 666 666 910 72;
  • 11) 0.866 666 666 910 72 × 2 = 1 + 0.733 333 333 821 44;
  • 12) 0.733 333 333 821 44 × 2 = 1 + 0.466 666 667 642 88;
  • 13) 0.466 666 667 642 88 × 2 = 0 + 0.933 333 335 285 76;
  • 14) 0.933 333 335 285 76 × 2 = 1 + 0.866 666 670 571 52;
  • 15) 0.866 666 670 571 52 × 2 = 1 + 0.733 333 341 143 04;
  • 16) 0.733 333 341 143 04 × 2 = 1 + 0.466 666 682 286 08;
  • 17) 0.466 666 682 286 08 × 2 = 0 + 0.933 333 364 572 16;
  • 18) 0.933 333 364 572 16 × 2 = 1 + 0.866 666 729 144 32;
  • 19) 0.866 666 729 144 32 × 2 = 1 + 0.733 333 458 288 64;
  • 20) 0.733 333 458 288 64 × 2 = 1 + 0.466 666 916 577 28;
  • 21) 0.466 666 916 577 28 × 2 = 0 + 0.933 333 833 154 56;
  • 22) 0.933 333 833 154 56 × 2 = 1 + 0.866 667 666 309 12;
  • 23) 0.866 667 666 309 12 × 2 = 1 + 0.733 335 332 618 24;
  • 24) 0.733 335 332 618 24 × 2 = 1 + 0.466 670 665 236 48;
  • 25) 0.466 670 665 236 48 × 2 = 0 + 0.933 341 330 472 96;
  • 26) 0.933 341 330 472 96 × 2 = 1 + 0.866 682 660 945 92;
  • 27) 0.866 682 660 945 92 × 2 = 1 + 0.733 365 321 891 84;
  • 28) 0.733 365 321 891 84 × 2 = 1 + 0.466 730 643 783 68;
  • 29) 0.466 730 643 783 68 × 2 = 0 + 0.933 461 287 567 36;
  • 30) 0.933 461 287 567 36 × 2 = 1 + 0.866 922 575 134 72;
  • 31) 0.866 922 575 134 72 × 2 = 1 + 0.733 845 150 269 44;
  • 32) 0.733 845 150 269 44 × 2 = 1 + 0.467 690 300 538 88;
  • 33) 0.467 690 300 538 88 × 2 = 0 + 0.935 380 601 077 76;
  • 34) 0.935 380 601 077 76 × 2 = 1 + 0.870 761 202 155 52;
  • 35) 0.870 761 202 155 52 × 2 = 1 + 0.741 522 404 311 04;
  • 36) 0.741 522 404 311 04 × 2 = 1 + 0.483 044 808 622 08;
  • 37) 0.483 044 808 622 08 × 2 = 0 + 0.966 089 617 244 16;
  • 38) 0.966 089 617 244 16 × 2 = 1 + 0.932 179 234 488 32;
  • 39) 0.932 179 234 488 32 × 2 = 1 + 0.864 358 468 976 64;
  • 40) 0.864 358 468 976 64 × 2 = 1 + 0.728 716 937 953 28;
  • 41) 0.728 716 937 953 28 × 2 = 1 + 0.457 433 875 906 56;
  • 42) 0.457 433 875 906 56 × 2 = 0 + 0.914 867 751 813 12;
  • 43) 0.914 867 751 813 12 × 2 = 1 + 0.829 735 503 626 24;
  • 44) 0.829 735 503 626 24 × 2 = 1 + 0.659 471 007 252 48;
  • 45) 0.659 471 007 252 48 × 2 = 1 + 0.318 942 014 504 96;
  • 46) 0.318 942 014 504 96 × 2 = 0 + 0.637 884 029 009 92;
  • 47) 0.637 884 029 009 92 × 2 = 1 + 0.275 768 058 019 84;
  • 48) 0.275 768 058 019 84 × 2 = 0 + 0.551 536 116 039 68;
  • 49) 0.551 536 116 039 68 × 2 = 1 + 0.103 072 232 079 36;
  • 50) 0.103 072 232 079 36 × 2 = 0 + 0.206 144 464 158 72;
  • 51) 0.206 144 464 158 72 × 2 = 0 + 0.412 288 928 317 44;
  • 52) 0.412 288 928 317 44 × 2 = 0 + 0.824 577 856 634 88;
  • 53) 0.824 577 856 634 88 × 2 = 1 + 0.649 155 713 269 76;
  • 54) 0.649 155 713 269 76 × 2 = 1 + 0.298 311 426 539 52;

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:


0.466 666 666 666 905(10) =

0.0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 1011 1010 1000 11(2)

5. Positive number before normalization:

0.466 666 666 666 905(10) =

0.0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 1011 1010 1000 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.466 666 666 666 905(10) =

0.0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 1011 1010 1000 11(2) =

0.0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 1011 1010 1000 11(2) × 20 =

1.1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011(2) × 2-2


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): -2

Mantissa (not normalized):
1.1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1021(10) =

011 1111 1101(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. 1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011 =

1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011


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 1101

Mantissa (52 bits) =
1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011


Decimal number 0.466 666 666 666 905 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 1101 1101 1101 1101 1101 1101 1101 1101 1101 1110 1110 1010 0011

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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