-0.000 282 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 26(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
-0.000 282 26(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:

|-0.000 282 26| = 0.000 282 26


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: 0.000 282 26.

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) 0.000 282 26 × 2 = 0 + 0.000 564 52;
  • 2) 0.000 564 52 × 2 = 0 + 0.001 129 04;
  • 3) 0.001 129 04 × 2 = 0 + 0.002 258 08;
  • 4) 0.002 258 08 × 2 = 0 + 0.004 516 16;
  • 5) 0.004 516 16 × 2 = 0 + 0.009 032 32;
  • 6) 0.009 032 32 × 2 = 0 + 0.018 064 64;
  • 7) 0.018 064 64 × 2 = 0 + 0.036 129 28;
  • 8) 0.036 129 28 × 2 = 0 + 0.072 258 56;
  • 9) 0.072 258 56 × 2 = 0 + 0.144 517 12;
  • 10) 0.144 517 12 × 2 = 0 + 0.289 034 24;
  • 11) 0.289 034 24 × 2 = 0 + 0.578 068 48;
  • 12) 0.578 068 48 × 2 = 1 + 0.156 136 96;
  • 13) 0.156 136 96 × 2 = 0 + 0.312 273 92;
  • 14) 0.312 273 92 × 2 = 0 + 0.624 547 84;
  • 15) 0.624 547 84 × 2 = 1 + 0.249 095 68;
  • 16) 0.249 095 68 × 2 = 0 + 0.498 191 36;
  • 17) 0.498 191 36 × 2 = 0 + 0.996 382 72;
  • 18) 0.996 382 72 × 2 = 1 + 0.992 765 44;
  • 19) 0.992 765 44 × 2 = 1 + 0.985 530 88;
  • 20) 0.985 530 88 × 2 = 1 + 0.971 061 76;
  • 21) 0.971 061 76 × 2 = 1 + 0.942 123 52;
  • 22) 0.942 123 52 × 2 = 1 + 0.884 247 04;
  • 23) 0.884 247 04 × 2 = 1 + 0.768 494 08;
  • 24) 0.768 494 08 × 2 = 1 + 0.536 988 16;
  • 25) 0.536 988 16 × 2 = 1 + 0.073 976 32;
  • 26) 0.073 976 32 × 2 = 0 + 0.147 952 64;
  • 27) 0.147 952 64 × 2 = 0 + 0.295 905 28;
  • 28) 0.295 905 28 × 2 = 0 + 0.591 810 56;
  • 29) 0.591 810 56 × 2 = 1 + 0.183 621 12;
  • 30) 0.183 621 12 × 2 = 0 + 0.367 242 24;
  • 31) 0.367 242 24 × 2 = 0 + 0.734 484 48;
  • 32) 0.734 484 48 × 2 = 1 + 0.468 968 96;
  • 33) 0.468 968 96 × 2 = 0 + 0.937 937 92;
  • 34) 0.937 937 92 × 2 = 1 + 0.875 875 84;
  • 35) 0.875 875 84 × 2 = 1 + 0.751 751 68;
  • 36) 0.751 751 68 × 2 = 1 + 0.503 503 36;
  • 37) 0.503 503 36 × 2 = 1 + 0.007 006 72;
  • 38) 0.007 006 72 × 2 = 0 + 0.014 013 44;
  • 39) 0.014 013 44 × 2 = 0 + 0.028 026 88;
  • 40) 0.028 026 88 × 2 = 0 + 0.056 053 76;
  • 41) 0.056 053 76 × 2 = 0 + 0.112 107 52;
  • 42) 0.112 107 52 × 2 = 0 + 0.224 215 04;
  • 43) 0.224 215 04 × 2 = 0 + 0.448 430 08;
  • 44) 0.448 430 08 × 2 = 0 + 0.896 860 16;
  • 45) 0.896 860 16 × 2 = 1 + 0.793 720 32;
  • 46) 0.793 720 32 × 2 = 1 + 0.587 440 64;
  • 47) 0.587 440 64 × 2 = 1 + 0.174 881 28;
  • 48) 0.174 881 28 × 2 = 0 + 0.349 762 56;
  • 49) 0.349 762 56 × 2 = 0 + 0.699 525 12;
  • 50) 0.699 525 12 × 2 = 1 + 0.399 050 24;
  • 51) 0.399 050 24 × 2 = 0 + 0.798 100 48;
  • 52) 0.798 100 48 × 2 = 1 + 0.596 200 96;
  • 53) 0.596 200 96 × 2 = 1 + 0.192 401 92;
  • 54) 0.192 401 92 × 2 = 0 + 0.384 803 84;
  • 55) 0.384 803 84 × 2 = 0 + 0.769 607 68;
  • 56) 0.769 607 68 × 2 = 1 + 0.539 215 36;
  • 57) 0.539 215 36 × 2 = 1 + 0.078 430 72;
  • 58) 0.078 430 72 × 2 = 0 + 0.156 861 44;
  • 59) 0.156 861 44 × 2 = 0 + 0.313 722 88;
  • 60) 0.313 722 88 × 2 = 0 + 0.627 445 76;
  • 61) 0.627 445 76 × 2 = 1 + 0.254 891 52;
  • 62) 0.254 891 52 × 2 = 0 + 0.509 783 04;
  • 63) 0.509 783 04 × 2 = 1 + 0.019 566 08;
  • 64) 0.019 566 08 × 2 = 0 + 0.039 132 16;

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:


0.000 282 26(10) =

0.0000 0000 0001 0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010(2)

6. Positive number before normalization:

0.000 282 26(10) =

0.0000 0000 0001 0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010(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:


0.000 282 26(10) =

0.0000 0000 0001 0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010(2) =

0.0000 0000 0001 0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010(2) × 20 =

1.0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010(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 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010


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 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010 =

0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010


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 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010


Decimal number -0.000 282 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1111 1000 1001 0111 1000 0000 1110 0101 1001 1000 1010

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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