20.333 333 333 333 339 58 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 339 58(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
20.333 333 333 333 339 58(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: 20.
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;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 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.

20(10) =

1 0100(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 339 58.

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) 0.333 333 333 333 339 58 × 2 = 0 + 0.666 666 666 666 679 16;
  • 2) 0.666 666 666 666 679 16 × 2 = 1 + 0.333 333 333 333 358 32;
  • 3) 0.333 333 333 333 358 32 × 2 = 0 + 0.666 666 666 666 716 64;
  • 4) 0.666 666 666 666 716 64 × 2 = 1 + 0.333 333 333 333 433 28;
  • 5) 0.333 333 333 333 433 28 × 2 = 0 + 0.666 666 666 666 866 56;
  • 6) 0.666 666 666 666 866 56 × 2 = 1 + 0.333 333 333 333 733 12;
  • 7) 0.333 333 333 333 733 12 × 2 = 0 + 0.666 666 666 667 466 24;
  • 8) 0.666 666 666 667 466 24 × 2 = 1 + 0.333 333 333 334 932 48;
  • 9) 0.333 333 333 334 932 48 × 2 = 0 + 0.666 666 666 669 864 96;
  • 10) 0.666 666 666 669 864 96 × 2 = 1 + 0.333 333 333 339 729 92;
  • 11) 0.333 333 333 339 729 92 × 2 = 0 + 0.666 666 666 679 459 84;
  • 12) 0.666 666 666 679 459 84 × 2 = 1 + 0.333 333 333 358 919 68;
  • 13) 0.333 333 333 358 919 68 × 2 = 0 + 0.666 666 666 717 839 36;
  • 14) 0.666 666 666 717 839 36 × 2 = 1 + 0.333 333 333 435 678 72;
  • 15) 0.333 333 333 435 678 72 × 2 = 0 + 0.666 666 666 871 357 44;
  • 16) 0.666 666 666 871 357 44 × 2 = 1 + 0.333 333 333 742 714 88;
  • 17) 0.333 333 333 742 714 88 × 2 = 0 + 0.666 666 667 485 429 76;
  • 18) 0.666 666 667 485 429 76 × 2 = 1 + 0.333 333 334 970 859 52;
  • 19) 0.333 333 334 970 859 52 × 2 = 0 + 0.666 666 669 941 719 04;
  • 20) 0.666 666 669 941 719 04 × 2 = 1 + 0.333 333 339 883 438 08;
  • 21) 0.333 333 339 883 438 08 × 2 = 0 + 0.666 666 679 766 876 16;
  • 22) 0.666 666 679 766 876 16 × 2 = 1 + 0.333 333 359 533 752 32;
  • 23) 0.333 333 359 533 752 32 × 2 = 0 + 0.666 666 719 067 504 64;
  • 24) 0.666 666 719 067 504 64 × 2 = 1 + 0.333 333 438 135 009 28;
  • 25) 0.333 333 438 135 009 28 × 2 = 0 + 0.666 666 876 270 018 56;
  • 26) 0.666 666 876 270 018 56 × 2 = 1 + 0.333 333 752 540 037 12;
  • 27) 0.333 333 752 540 037 12 × 2 = 0 + 0.666 667 505 080 074 24;
  • 28) 0.666 667 505 080 074 24 × 2 = 1 + 0.333 335 010 160 148 48;
  • 29) 0.333 335 010 160 148 48 × 2 = 0 + 0.666 670 020 320 296 96;
  • 30) 0.666 670 020 320 296 96 × 2 = 1 + 0.333 340 040 640 593 92;
  • 31) 0.333 340 040 640 593 92 × 2 = 0 + 0.666 680 081 281 187 84;
  • 32) 0.666 680 081 281 187 84 × 2 = 1 + 0.333 360 162 562 375 68;
  • 33) 0.333 360 162 562 375 68 × 2 = 0 + 0.666 720 325 124 751 36;
  • 34) 0.666 720 325 124 751 36 × 2 = 1 + 0.333 440 650 249 502 72;
  • 35) 0.333 440 650 249 502 72 × 2 = 0 + 0.666 881 300 499 005 44;
  • 36) 0.666 881 300 499 005 44 × 2 = 1 + 0.333 762 600 998 010 88;
  • 37) 0.333 762 600 998 010 88 × 2 = 0 + 0.667 525 201 996 021 76;
  • 38) 0.667 525 201 996 021 76 × 2 = 1 + 0.335 050 403 992 043 52;
  • 39) 0.335 050 403 992 043 52 × 2 = 0 + 0.670 100 807 984 087 04;
  • 40) 0.670 100 807 984 087 04 × 2 = 1 + 0.340 201 615 968 174 08;
  • 41) 0.340 201 615 968 174 08 × 2 = 0 + 0.680 403 231 936 348 16;
  • 42) 0.680 403 231 936 348 16 × 2 = 1 + 0.360 806 463 872 696 32;
  • 43) 0.360 806 463 872 696 32 × 2 = 0 + 0.721 612 927 745 392 64;
  • 44) 0.721 612 927 745 392 64 × 2 = 1 + 0.443 225 855 490 785 28;
  • 45) 0.443 225 855 490 785 28 × 2 = 0 + 0.886 451 710 981 570 56;
  • 46) 0.886 451 710 981 570 56 × 2 = 1 + 0.772 903 421 963 141 12;
  • 47) 0.772 903 421 963 141 12 × 2 = 1 + 0.545 806 843 926 282 24;
  • 48) 0.545 806 843 926 282 24 × 2 = 1 + 0.091 613 687 852 564 48;
  • 49) 0.091 613 687 852 564 48 × 2 = 0 + 0.183 227 375 705 128 96;
  • 50) 0.183 227 375 705 128 96 × 2 = 0 + 0.366 454 751 410 257 92;
  • 51) 0.366 454 751 410 257 92 × 2 = 0 + 0.732 909 502 820 515 84;
  • 52) 0.732 909 502 820 515 84 × 2 = 1 + 0.465 819 005 641 031 68;
  • 53) 0.465 819 005 641 031 68 × 2 = 0 + 0.931 638 011 282 063 36;

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:


0.333 333 333 333 339 58(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0(2)

5. Positive number before normalization:

20.333 333 333 333 339 58(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


20.333 333 333 333 339 58(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0(2) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0(2) × 20 =

1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0(2) × 24


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): 4

Mantissa (not normalized):
1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0001 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 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) =

1027(10) =

100 0000 0011(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. 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0 0010 =

0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111


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 0011

Mantissa (52 bits) =
0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111


Decimal number 20.333 333 333 333 339 58 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111

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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