-0.129 999 999 999 967 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.129 999 999 999 967 5(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
-0.129 999 999 999 967 5(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. Start with the positive version of the number:

|-0.129 999 999 999 967 5| = 0.129 999 999 999 967 5


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.129 999 999 999 967 5.

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) 0.129 999 999 999 967 5 × 2 = 0 + 0.259 999 999 999 935;
  • 2) 0.259 999 999 999 935 × 2 = 0 + 0.519 999 999 999 87;
  • 3) 0.519 999 999 999 87 × 2 = 1 + 0.039 999 999 999 74;
  • 4) 0.039 999 999 999 74 × 2 = 0 + 0.079 999 999 999 48;
  • 5) 0.079 999 999 999 48 × 2 = 0 + 0.159 999 999 998 96;
  • 6) 0.159 999 999 998 96 × 2 = 0 + 0.319 999 999 997 92;
  • 7) 0.319 999 999 997 92 × 2 = 0 + 0.639 999 999 995 84;
  • 8) 0.639 999 999 995 84 × 2 = 1 + 0.279 999 999 991 68;
  • 9) 0.279 999 999 991 68 × 2 = 0 + 0.559 999 999 983 36;
  • 10) 0.559 999 999 983 36 × 2 = 1 + 0.119 999 999 966 72;
  • 11) 0.119 999 999 966 72 × 2 = 0 + 0.239 999 999 933 44;
  • 12) 0.239 999 999 933 44 × 2 = 0 + 0.479 999 999 866 88;
  • 13) 0.479 999 999 866 88 × 2 = 0 + 0.959 999 999 733 76;
  • 14) 0.959 999 999 733 76 × 2 = 1 + 0.919 999 999 467 52;
  • 15) 0.919 999 999 467 52 × 2 = 1 + 0.839 999 998 935 04;
  • 16) 0.839 999 998 935 04 × 2 = 1 + 0.679 999 997 870 08;
  • 17) 0.679 999 997 870 08 × 2 = 1 + 0.359 999 995 740 16;
  • 18) 0.359 999 995 740 16 × 2 = 0 + 0.719 999 991 480 32;
  • 19) 0.719 999 991 480 32 × 2 = 1 + 0.439 999 982 960 64;
  • 20) 0.439 999 982 960 64 × 2 = 0 + 0.879 999 965 921 28;
  • 21) 0.879 999 965 921 28 × 2 = 1 + 0.759 999 931 842 56;
  • 22) 0.759 999 931 842 56 × 2 = 1 + 0.519 999 863 685 12;
  • 23) 0.519 999 863 685 12 × 2 = 1 + 0.039 999 727 370 24;
  • 24) 0.039 999 727 370 24 × 2 = 0 + 0.079 999 454 740 48;
  • 25) 0.079 999 454 740 48 × 2 = 0 + 0.159 998 909 480 96;
  • 26) 0.159 998 909 480 96 × 2 = 0 + 0.319 997 818 961 92;
  • 27) 0.319 997 818 961 92 × 2 = 0 + 0.639 995 637 923 84;
  • 28) 0.639 995 637 923 84 × 2 = 1 + 0.279 991 275 847 68;
  • 29) 0.279 991 275 847 68 × 2 = 0 + 0.559 982 551 695 36;
  • 30) 0.559 982 551 695 36 × 2 = 1 + 0.119 965 103 390 72;
  • 31) 0.119 965 103 390 72 × 2 = 0 + 0.239 930 206 781 44;
  • 32) 0.239 930 206 781 44 × 2 = 0 + 0.479 860 413 562 88;
  • 33) 0.479 860 413 562 88 × 2 = 0 + 0.959 720 827 125 76;
  • 34) 0.959 720 827 125 76 × 2 = 1 + 0.919 441 654 251 52;
  • 35) 0.919 441 654 251 52 × 2 = 1 + 0.838 883 308 503 04;
  • 36) 0.838 883 308 503 04 × 2 = 1 + 0.677 766 617 006 08;
  • 37) 0.677 766 617 006 08 × 2 = 1 + 0.355 533 234 012 16;
  • 38) 0.355 533 234 012 16 × 2 = 0 + 0.711 066 468 024 32;
  • 39) 0.711 066 468 024 32 × 2 = 1 + 0.422 132 936 048 64;
  • 40) 0.422 132 936 048 64 × 2 = 0 + 0.844 265 872 097 28;
  • 41) 0.844 265 872 097 28 × 2 = 1 + 0.688 531 744 194 56;
  • 42) 0.688 531 744 194 56 × 2 = 1 + 0.377 063 488 389 12;
  • 43) 0.377 063 488 389 12 × 2 = 0 + 0.754 126 976 778 24;
  • 44) 0.754 126 976 778 24 × 2 = 1 + 0.508 253 953 556 48;
  • 45) 0.508 253 953 556 48 × 2 = 1 + 0.016 507 907 112 96;
  • 46) 0.016 507 907 112 96 × 2 = 0 + 0.033 015 814 225 92;
  • 47) 0.033 015 814 225 92 × 2 = 0 + 0.066 031 628 451 84;
  • 48) 0.066 031 628 451 84 × 2 = 0 + 0.132 063 256 903 68;
  • 49) 0.132 063 256 903 68 × 2 = 0 + 0.264 126 513 807 36;
  • 50) 0.264 126 513 807 36 × 2 = 0 + 0.528 253 027 614 72;
  • 51) 0.528 253 027 614 72 × 2 = 1 + 0.056 506 055 229 44;
  • 52) 0.056 506 055 229 44 × 2 = 0 + 0.113 012 110 458 88;
  • 53) 0.113 012 110 458 88 × 2 = 0 + 0.226 024 220 917 76;
  • 54) 0.226 024 220 917 76 × 2 = 0 + 0.452 048 441 835 52;
  • 55) 0.452 048 441 835 52 × 2 = 0 + 0.904 096 883 671 04;

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).


5. 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:


0.129 999 999 999 967 5(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1000 0010 000(2)

6. Positive number before normalization:

0.129 999 999 999 967 5(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1000 0010 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.129 999 999 999 967 5(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1000 0010 000(2) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1000 0010 000(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0001 0000(2) × 2-3


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): -3

Mantissa (not normalized):
1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


10. 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;
  • 1 020 ÷ 2 = 510 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


12. 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 1100 0001 0000 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0001 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0001 0000


Decimal number -0.129 999 999 999 967 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 0001 0000

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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