-0.000 281 695 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 695(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
-0.000 281 695(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:

|-0.000 281 695| = 0.000 281 695


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: 0.000 281 695.

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) 0.000 281 695 × 2 = 0 + 0.000 563 39;
  • 2) 0.000 563 39 × 2 = 0 + 0.001 126 78;
  • 3) 0.001 126 78 × 2 = 0 + 0.002 253 56;
  • 4) 0.002 253 56 × 2 = 0 + 0.004 507 12;
  • 5) 0.004 507 12 × 2 = 0 + 0.009 014 24;
  • 6) 0.009 014 24 × 2 = 0 + 0.018 028 48;
  • 7) 0.018 028 48 × 2 = 0 + 0.036 056 96;
  • 8) 0.036 056 96 × 2 = 0 + 0.072 113 92;
  • 9) 0.072 113 92 × 2 = 0 + 0.144 227 84;
  • 10) 0.144 227 84 × 2 = 0 + 0.288 455 68;
  • 11) 0.288 455 68 × 2 = 0 + 0.576 911 36;
  • 12) 0.576 911 36 × 2 = 1 + 0.153 822 72;
  • 13) 0.153 822 72 × 2 = 0 + 0.307 645 44;
  • 14) 0.307 645 44 × 2 = 0 + 0.615 290 88;
  • 15) 0.615 290 88 × 2 = 1 + 0.230 581 76;
  • 16) 0.230 581 76 × 2 = 0 + 0.461 163 52;
  • 17) 0.461 163 52 × 2 = 0 + 0.922 327 04;
  • 18) 0.922 327 04 × 2 = 1 + 0.844 654 08;
  • 19) 0.844 654 08 × 2 = 1 + 0.689 308 16;
  • 20) 0.689 308 16 × 2 = 1 + 0.378 616 32;
  • 21) 0.378 616 32 × 2 = 0 + 0.757 232 64;
  • 22) 0.757 232 64 × 2 = 1 + 0.514 465 28;
  • 23) 0.514 465 28 × 2 = 1 + 0.028 930 56;
  • 24) 0.028 930 56 × 2 = 0 + 0.057 861 12;
  • 25) 0.057 861 12 × 2 = 0 + 0.115 722 24;
  • 26) 0.115 722 24 × 2 = 0 + 0.231 444 48;
  • 27) 0.231 444 48 × 2 = 0 + 0.462 888 96;
  • 28) 0.462 888 96 × 2 = 0 + 0.925 777 92;
  • 29) 0.925 777 92 × 2 = 1 + 0.851 555 84;
  • 30) 0.851 555 84 × 2 = 1 + 0.703 111 68;
  • 31) 0.703 111 68 × 2 = 1 + 0.406 223 36;
  • 32) 0.406 223 36 × 2 = 0 + 0.812 446 72;
  • 33) 0.812 446 72 × 2 = 1 + 0.624 893 44;
  • 34) 0.624 893 44 × 2 = 1 + 0.249 786 88;
  • 35) 0.249 786 88 × 2 = 0 + 0.499 573 76;
  • 36) 0.499 573 76 × 2 = 0 + 0.999 147 52;
  • 37) 0.999 147 52 × 2 = 1 + 0.998 295 04;
  • 38) 0.998 295 04 × 2 = 1 + 0.996 590 08;
  • 39) 0.996 590 08 × 2 = 1 + 0.993 180 16;
  • 40) 0.993 180 16 × 2 = 1 + 0.986 360 32;
  • 41) 0.986 360 32 × 2 = 1 + 0.972 720 64;
  • 42) 0.972 720 64 × 2 = 1 + 0.945 441 28;
  • 43) 0.945 441 28 × 2 = 1 + 0.890 882 56;
  • 44) 0.890 882 56 × 2 = 1 + 0.781 765 12;
  • 45) 0.781 765 12 × 2 = 1 + 0.563 530 24;
  • 46) 0.563 530 24 × 2 = 1 + 0.127 060 48;
  • 47) 0.127 060 48 × 2 = 0 + 0.254 120 96;
  • 48) 0.254 120 96 × 2 = 0 + 0.508 241 92;
  • 49) 0.508 241 92 × 2 = 1 + 0.016 483 84;
  • 50) 0.016 483 84 × 2 = 0 + 0.032 967 68;
  • 51) 0.032 967 68 × 2 = 0 + 0.065 935 36;
  • 52) 0.065 935 36 × 2 = 0 + 0.131 870 72;
  • 53) 0.131 870 72 × 2 = 0 + 0.263 741 44;
  • 54) 0.263 741 44 × 2 = 0 + 0.527 482 88;
  • 55) 0.527 482 88 × 2 = 1 + 0.054 965 76;
  • 56) 0.054 965 76 × 2 = 0 + 0.109 931 52;
  • 57) 0.109 931 52 × 2 = 0 + 0.219 863 04;
  • 58) 0.219 863 04 × 2 = 0 + 0.439 726 08;
  • 59) 0.439 726 08 × 2 = 0 + 0.879 452 16;
  • 60) 0.879 452 16 × 2 = 1 + 0.758 904 32;
  • 61) 0.758 904 32 × 2 = 1 + 0.517 808 64;
  • 62) 0.517 808 64 × 2 = 1 + 0.035 617 28;
  • 63) 0.035 617 28 × 2 = 0 + 0.071 234 56;
  • 64) 0.071 234 56 × 2 = 0 + 0.142 469 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:


0.000 281 695(10) =

0.0000 0000 0001 0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100(2)

6. Positive number before normalization:

0.000 281 695(10) =

0.0000 0000 0001 0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100(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:


0.000 281 695(10) =

0.0000 0000 0001 0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100(2) =

0.0000 0000 0001 0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100(2) × 20 =

1.0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100(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 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100


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 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100 =

0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100


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 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100


Decimal number -0.000 281 695 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0110 0000 1110 1100 1111 1111 1100 1000 0010 0001 1100

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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