-0.080 000 003 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.080 000 003 8(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
-0.080 000 003 8(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:

|-0.080 000 003 8| = 0.080 000 003 8


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: 0.080 000 003 8.

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) 0.080 000 003 8 × 2 = 0 + 0.160 000 007 6;
  • 2) 0.160 000 007 6 × 2 = 0 + 0.320 000 015 2;
  • 3) 0.320 000 015 2 × 2 = 0 + 0.640 000 030 4;
  • 4) 0.640 000 030 4 × 2 = 1 + 0.280 000 060 8;
  • 5) 0.280 000 060 8 × 2 = 0 + 0.560 000 121 6;
  • 6) 0.560 000 121 6 × 2 = 1 + 0.120 000 243 2;
  • 7) 0.120 000 243 2 × 2 = 0 + 0.240 000 486 4;
  • 8) 0.240 000 486 4 × 2 = 0 + 0.480 000 972 8;
  • 9) 0.480 000 972 8 × 2 = 0 + 0.960 001 945 6;
  • 10) 0.960 001 945 6 × 2 = 1 + 0.920 003 891 2;
  • 11) 0.920 003 891 2 × 2 = 1 + 0.840 007 782 4;
  • 12) 0.840 007 782 4 × 2 = 1 + 0.680 015 564 8;
  • 13) 0.680 015 564 8 × 2 = 1 + 0.360 031 129 6;
  • 14) 0.360 031 129 6 × 2 = 0 + 0.720 062 259 2;
  • 15) 0.720 062 259 2 × 2 = 1 + 0.440 124 518 4;
  • 16) 0.440 124 518 4 × 2 = 0 + 0.880 249 036 8;
  • 17) 0.880 249 036 8 × 2 = 1 + 0.760 498 073 6;
  • 18) 0.760 498 073 6 × 2 = 1 + 0.520 996 147 2;
  • 19) 0.520 996 147 2 × 2 = 1 + 0.041 992 294 4;
  • 20) 0.041 992 294 4 × 2 = 0 + 0.083 984 588 8;
  • 21) 0.083 984 588 8 × 2 = 0 + 0.167 969 177 6;
  • 22) 0.167 969 177 6 × 2 = 0 + 0.335 938 355 2;
  • 23) 0.335 938 355 2 × 2 = 0 + 0.671 876 710 4;
  • 24) 0.671 876 710 4 × 2 = 1 + 0.343 753 420 8;
  • 25) 0.343 753 420 8 × 2 = 0 + 0.687 506 841 6;
  • 26) 0.687 506 841 6 × 2 = 1 + 0.375 013 683 2;
  • 27) 0.375 013 683 2 × 2 = 0 + 0.750 027 366 4;
  • 28) 0.750 027 366 4 × 2 = 1 + 0.500 054 732 8;
  • 29) 0.500 054 732 8 × 2 = 1 + 0.000 109 465 6;
  • 30) 0.000 109 465 6 × 2 = 0 + 0.000 218 931 2;
  • 31) 0.000 218 931 2 × 2 = 0 + 0.000 437 862 4;
  • 32) 0.000 437 862 4 × 2 = 0 + 0.000 875 724 8;
  • 33) 0.000 875 724 8 × 2 = 0 + 0.001 751 449 6;
  • 34) 0.001 751 449 6 × 2 = 0 + 0.003 502 899 2;
  • 35) 0.003 502 899 2 × 2 = 0 + 0.007 005 798 4;
  • 36) 0.007 005 798 4 × 2 = 0 + 0.014 011 596 8;
  • 37) 0.014 011 596 8 × 2 = 0 + 0.028 023 193 6;
  • 38) 0.028 023 193 6 × 2 = 0 + 0.056 046 387 2;
  • 39) 0.056 046 387 2 × 2 = 0 + 0.112 092 774 4;
  • 40) 0.112 092 774 4 × 2 = 0 + 0.224 185 548 8;
  • 41) 0.224 185 548 8 × 2 = 0 + 0.448 371 097 6;
  • 42) 0.448 371 097 6 × 2 = 0 + 0.896 742 195 2;
  • 43) 0.896 742 195 2 × 2 = 1 + 0.793 484 390 4;
  • 44) 0.793 484 390 4 × 2 = 1 + 0.586 968 780 8;
  • 45) 0.586 968 780 8 × 2 = 1 + 0.173 937 561 6;
  • 46) 0.173 937 561 6 × 2 = 0 + 0.347 875 123 2;
  • 47) 0.347 875 123 2 × 2 = 0 + 0.695 750 246 4;
  • 48) 0.695 750 246 4 × 2 = 1 + 0.391 500 492 8;
  • 49) 0.391 500 492 8 × 2 = 0 + 0.783 000 985 6;
  • 50) 0.783 000 985 6 × 2 = 1 + 0.566 001 971 2;
  • 51) 0.566 001 971 2 × 2 = 1 + 0.132 003 942 4;
  • 52) 0.132 003 942 4 × 2 = 0 + 0.264 007 884 8;
  • 53) 0.264 007 884 8 × 2 = 0 + 0.528 015 769 6;
  • 54) 0.528 015 769 6 × 2 = 1 + 0.056 031 539 2;
  • 55) 0.056 031 539 2 × 2 = 0 + 0.112 063 078 4;
  • 56) 0.112 063 078 4 × 2 = 0 + 0.224 126 156 8;

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:


0.080 000 003 8(10) =

0.0001 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100(2)

6. Positive number before normalization:

0.080 000 003 8(10) =

0.0001 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100(2)

7. Normalize the binary representation of the number.

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


0.080 000 003 8(10) =

0.0001 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100(2) =

0.0001 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100(2) × 20 =

1.0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100(2) × 2-4


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):
1.0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100


9. 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) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1019(10) =

011 1111 1011(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. 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100 =

0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100


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 1011

Mantissa (52 bits) =
0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100


Decimal number -0.080 000 003 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0100 0111 1010 1110 0001 0101 1000 0000 0000 0011 1001 0110 0100

» Convert decimal -0.080 000 007 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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