0.599 999 999 999 968 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.599 999 999 999 968 2(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
0.599 999 999 999 968 2(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: 0.599 999 999 999 968 2.

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) 0.599 999 999 999 968 2 × 2 = 1 + 0.199 999 999 999 936 4;
  • 2) 0.199 999 999 999 936 4 × 2 = 0 + 0.399 999 999 999 872 8;
  • 3) 0.399 999 999 999 872 8 × 2 = 0 + 0.799 999 999 999 745 6;
  • 4) 0.799 999 999 999 745 6 × 2 = 1 + 0.599 999 999 999 491 2;
  • 5) 0.599 999 999 999 491 2 × 2 = 1 + 0.199 999 999 998 982 4;
  • 6) 0.199 999 999 998 982 4 × 2 = 0 + 0.399 999 999 997 964 8;
  • 7) 0.399 999 999 997 964 8 × 2 = 0 + 0.799 999 999 995 929 6;
  • 8) 0.799 999 999 995 929 6 × 2 = 1 + 0.599 999 999 991 859 2;
  • 9) 0.599 999 999 991 859 2 × 2 = 1 + 0.199 999 999 983 718 4;
  • 10) 0.199 999 999 983 718 4 × 2 = 0 + 0.399 999 999 967 436 8;
  • 11) 0.399 999 999 967 436 8 × 2 = 0 + 0.799 999 999 934 873 6;
  • 12) 0.799 999 999 934 873 6 × 2 = 1 + 0.599 999 999 869 747 2;
  • 13) 0.599 999 999 869 747 2 × 2 = 1 + 0.199 999 999 739 494 4;
  • 14) 0.199 999 999 739 494 4 × 2 = 0 + 0.399 999 999 478 988 8;
  • 15) 0.399 999 999 478 988 8 × 2 = 0 + 0.799 999 998 957 977 6;
  • 16) 0.799 999 998 957 977 6 × 2 = 1 + 0.599 999 997 915 955 2;
  • 17) 0.599 999 997 915 955 2 × 2 = 1 + 0.199 999 995 831 910 4;
  • 18) 0.199 999 995 831 910 4 × 2 = 0 + 0.399 999 991 663 820 8;
  • 19) 0.399 999 991 663 820 8 × 2 = 0 + 0.799 999 983 327 641 6;
  • 20) 0.799 999 983 327 641 6 × 2 = 1 + 0.599 999 966 655 283 2;
  • 21) 0.599 999 966 655 283 2 × 2 = 1 + 0.199 999 933 310 566 4;
  • 22) 0.199 999 933 310 566 4 × 2 = 0 + 0.399 999 866 621 132 8;
  • 23) 0.399 999 866 621 132 8 × 2 = 0 + 0.799 999 733 242 265 6;
  • 24) 0.799 999 733 242 265 6 × 2 = 1 + 0.599 999 466 484 531 2;
  • 25) 0.599 999 466 484 531 2 × 2 = 1 + 0.199 998 932 969 062 4;
  • 26) 0.199 998 932 969 062 4 × 2 = 0 + 0.399 997 865 938 124 8;
  • 27) 0.399 997 865 938 124 8 × 2 = 0 + 0.799 995 731 876 249 6;
  • 28) 0.799 995 731 876 249 6 × 2 = 1 + 0.599 991 463 752 499 2;
  • 29) 0.599 991 463 752 499 2 × 2 = 1 + 0.199 982 927 504 998 4;
  • 30) 0.199 982 927 504 998 4 × 2 = 0 + 0.399 965 855 009 996 8;
  • 31) 0.399 965 855 009 996 8 × 2 = 0 + 0.799 931 710 019 993 6;
  • 32) 0.799 931 710 019 993 6 × 2 = 1 + 0.599 863 420 039 987 2;
  • 33) 0.599 863 420 039 987 2 × 2 = 1 + 0.199 726 840 079 974 4;
  • 34) 0.199 726 840 079 974 4 × 2 = 0 + 0.399 453 680 159 948 8;
  • 35) 0.399 453 680 159 948 8 × 2 = 0 + 0.798 907 360 319 897 6;
  • 36) 0.798 907 360 319 897 6 × 2 = 1 + 0.597 814 720 639 795 2;
  • 37) 0.597 814 720 639 795 2 × 2 = 1 + 0.195 629 441 279 590 4;
  • 38) 0.195 629 441 279 590 4 × 2 = 0 + 0.391 258 882 559 180 8;
  • 39) 0.391 258 882 559 180 8 × 2 = 0 + 0.782 517 765 118 361 6;
  • 40) 0.782 517 765 118 361 6 × 2 = 1 + 0.565 035 530 236 723 2;
  • 41) 0.565 035 530 236 723 2 × 2 = 1 + 0.130 071 060 473 446 4;
  • 42) 0.130 071 060 473 446 4 × 2 = 0 + 0.260 142 120 946 892 8;
  • 43) 0.260 142 120 946 892 8 × 2 = 0 + 0.520 284 241 893 785 6;
  • 44) 0.520 284 241 893 785 6 × 2 = 1 + 0.040 568 483 787 571 2;
  • 45) 0.040 568 483 787 571 2 × 2 = 0 + 0.081 136 967 575 142 4;
  • 46) 0.081 136 967 575 142 4 × 2 = 0 + 0.162 273 935 150 284 8;
  • 47) 0.162 273 935 150 284 8 × 2 = 0 + 0.324 547 870 300 569 6;
  • 48) 0.324 547 870 300 569 6 × 2 = 0 + 0.649 095 740 601 139 2;
  • 49) 0.649 095 740 601 139 2 × 2 = 1 + 0.298 191 481 202 278 4;
  • 50) 0.298 191 481 202 278 4 × 2 = 0 + 0.596 382 962 404 556 8;
  • 51) 0.596 382 962 404 556 8 × 2 = 1 + 0.192 765 924 809 113 6;
  • 52) 0.192 765 924 809 113 6 × 2 = 0 + 0.385 531 849 618 227 2;
  • 53) 0.385 531 849 618 227 2 × 2 = 0 + 0.771 063 699 236 454 4;

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:


0.599 999 999 999 968 2(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 1010 0(2)

5. Positive number before normalization:

0.599 999 999 999 968 2(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 1010 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:


0.599 999 999 999 968 2(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 1010 0(2) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 1010 0(2) × 20 =

1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100(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):
1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100


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. 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100 =

0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100


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) =
0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100


Decimal number 0.599 999 999 999 968 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 0001 0100

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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