-0.000 280 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 280 57(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
-0.000 280 57(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:

|-0.000 280 57| = 0.000 280 57


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: 0.000 280 57.

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) 0.000 280 57 × 2 = 0 + 0.000 561 14;
  • 2) 0.000 561 14 × 2 = 0 + 0.001 122 28;
  • 3) 0.001 122 28 × 2 = 0 + 0.002 244 56;
  • 4) 0.002 244 56 × 2 = 0 + 0.004 489 12;
  • 5) 0.004 489 12 × 2 = 0 + 0.008 978 24;
  • 6) 0.008 978 24 × 2 = 0 + 0.017 956 48;
  • 7) 0.017 956 48 × 2 = 0 + 0.035 912 96;
  • 8) 0.035 912 96 × 2 = 0 + 0.071 825 92;
  • 9) 0.071 825 92 × 2 = 0 + 0.143 651 84;
  • 10) 0.143 651 84 × 2 = 0 + 0.287 303 68;
  • 11) 0.287 303 68 × 2 = 0 + 0.574 607 36;
  • 12) 0.574 607 36 × 2 = 1 + 0.149 214 72;
  • 13) 0.149 214 72 × 2 = 0 + 0.298 429 44;
  • 14) 0.298 429 44 × 2 = 0 + 0.596 858 88;
  • 15) 0.596 858 88 × 2 = 1 + 0.193 717 76;
  • 16) 0.193 717 76 × 2 = 0 + 0.387 435 52;
  • 17) 0.387 435 52 × 2 = 0 + 0.774 871 04;
  • 18) 0.774 871 04 × 2 = 1 + 0.549 742 08;
  • 19) 0.549 742 08 × 2 = 1 + 0.099 484 16;
  • 20) 0.099 484 16 × 2 = 0 + 0.198 968 32;
  • 21) 0.198 968 32 × 2 = 0 + 0.397 936 64;
  • 22) 0.397 936 64 × 2 = 0 + 0.795 873 28;
  • 23) 0.795 873 28 × 2 = 1 + 0.591 746 56;
  • 24) 0.591 746 56 × 2 = 1 + 0.183 493 12;
  • 25) 0.183 493 12 × 2 = 0 + 0.366 986 24;
  • 26) 0.366 986 24 × 2 = 0 + 0.733 972 48;
  • 27) 0.733 972 48 × 2 = 1 + 0.467 944 96;
  • 28) 0.467 944 96 × 2 = 0 + 0.935 889 92;
  • 29) 0.935 889 92 × 2 = 1 + 0.871 779 84;
  • 30) 0.871 779 84 × 2 = 1 + 0.743 559 68;
  • 31) 0.743 559 68 × 2 = 1 + 0.487 119 36;
  • 32) 0.487 119 36 × 2 = 0 + 0.974 238 72;
  • 33) 0.974 238 72 × 2 = 1 + 0.948 477 44;
  • 34) 0.948 477 44 × 2 = 1 + 0.896 954 88;
  • 35) 0.896 954 88 × 2 = 1 + 0.793 909 76;
  • 36) 0.793 909 76 × 2 = 1 + 0.587 819 52;
  • 37) 0.587 819 52 × 2 = 1 + 0.175 639 04;
  • 38) 0.175 639 04 × 2 = 0 + 0.351 278 08;
  • 39) 0.351 278 08 × 2 = 0 + 0.702 556 16;
  • 40) 0.702 556 16 × 2 = 1 + 0.405 112 32;
  • 41) 0.405 112 32 × 2 = 0 + 0.810 224 64;
  • 42) 0.810 224 64 × 2 = 1 + 0.620 449 28;
  • 43) 0.620 449 28 × 2 = 1 + 0.240 898 56;
  • 44) 0.240 898 56 × 2 = 0 + 0.481 797 12;
  • 45) 0.481 797 12 × 2 = 0 + 0.963 594 24;
  • 46) 0.963 594 24 × 2 = 1 + 0.927 188 48;
  • 47) 0.927 188 48 × 2 = 1 + 0.854 376 96;
  • 48) 0.854 376 96 × 2 = 1 + 0.708 753 92;
  • 49) 0.708 753 92 × 2 = 1 + 0.417 507 84;
  • 50) 0.417 507 84 × 2 = 0 + 0.835 015 68;
  • 51) 0.835 015 68 × 2 = 1 + 0.670 031 36;
  • 52) 0.670 031 36 × 2 = 1 + 0.340 062 72;
  • 53) 0.340 062 72 × 2 = 0 + 0.680 125 44;
  • 54) 0.680 125 44 × 2 = 1 + 0.360 250 88;
  • 55) 0.360 250 88 × 2 = 0 + 0.720 501 76;
  • 56) 0.720 501 76 × 2 = 1 + 0.441 003 52;
  • 57) 0.441 003 52 × 2 = 0 + 0.882 007 04;
  • 58) 0.882 007 04 × 2 = 1 + 0.764 014 08;
  • 59) 0.764 014 08 × 2 = 1 + 0.528 028 16;
  • 60) 0.528 028 16 × 2 = 1 + 0.056 056 32;
  • 61) 0.056 056 32 × 2 = 0 + 0.112 112 64;
  • 62) 0.112 112 64 × 2 = 0 + 0.224 225 28;
  • 63) 0.224 225 28 × 2 = 0 + 0.448 450 56;
  • 64) 0.448 450 56 × 2 = 0 + 0.896 901 12;

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:


0.000 280 57(10) =

0.0000 0000 0001 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000(2)

6. Positive number before normalization:

0.000 280 57(10) =

0.0000 0000 0001 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 280 57(10) =

0.0000 0000 0001 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000(2) =

0.0000 0000 0001 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000(2) × 20 =

1.0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000 =

0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 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 0011

Mantissa (52 bits) =
0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 0000


Decimal number -0.000 280 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0110 0011 0010 1110 1111 1001 0110 0111 1011 0101 0111 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