-0.000 282 705 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 705(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
-0.000 282 705(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:

|-0.000 282 705| = 0.000 282 705


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: 0.000 282 705.

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) 0.000 282 705 × 2 = 0 + 0.000 565 41;
  • 2) 0.000 565 41 × 2 = 0 + 0.001 130 82;
  • 3) 0.001 130 82 × 2 = 0 + 0.002 261 64;
  • 4) 0.002 261 64 × 2 = 0 + 0.004 523 28;
  • 5) 0.004 523 28 × 2 = 0 + 0.009 046 56;
  • 6) 0.009 046 56 × 2 = 0 + 0.018 093 12;
  • 7) 0.018 093 12 × 2 = 0 + 0.036 186 24;
  • 8) 0.036 186 24 × 2 = 0 + 0.072 372 48;
  • 9) 0.072 372 48 × 2 = 0 + 0.144 744 96;
  • 10) 0.144 744 96 × 2 = 0 + 0.289 489 92;
  • 11) 0.289 489 92 × 2 = 0 + 0.578 979 84;
  • 12) 0.578 979 84 × 2 = 1 + 0.157 959 68;
  • 13) 0.157 959 68 × 2 = 0 + 0.315 919 36;
  • 14) 0.315 919 36 × 2 = 0 + 0.631 838 72;
  • 15) 0.631 838 72 × 2 = 1 + 0.263 677 44;
  • 16) 0.263 677 44 × 2 = 0 + 0.527 354 88;
  • 17) 0.527 354 88 × 2 = 1 + 0.054 709 76;
  • 18) 0.054 709 76 × 2 = 0 + 0.109 419 52;
  • 19) 0.109 419 52 × 2 = 0 + 0.218 839 04;
  • 20) 0.218 839 04 × 2 = 0 + 0.437 678 08;
  • 21) 0.437 678 08 × 2 = 0 + 0.875 356 16;
  • 22) 0.875 356 16 × 2 = 1 + 0.750 712 32;
  • 23) 0.750 712 32 × 2 = 1 + 0.501 424 64;
  • 24) 0.501 424 64 × 2 = 1 + 0.002 849 28;
  • 25) 0.002 849 28 × 2 = 0 + 0.005 698 56;
  • 26) 0.005 698 56 × 2 = 0 + 0.011 397 12;
  • 27) 0.011 397 12 × 2 = 0 + 0.022 794 24;
  • 28) 0.022 794 24 × 2 = 0 + 0.045 588 48;
  • 29) 0.045 588 48 × 2 = 0 + 0.091 176 96;
  • 30) 0.091 176 96 × 2 = 0 + 0.182 353 92;
  • 31) 0.182 353 92 × 2 = 0 + 0.364 707 84;
  • 32) 0.364 707 84 × 2 = 0 + 0.729 415 68;
  • 33) 0.729 415 68 × 2 = 1 + 0.458 831 36;
  • 34) 0.458 831 36 × 2 = 0 + 0.917 662 72;
  • 35) 0.917 662 72 × 2 = 1 + 0.835 325 44;
  • 36) 0.835 325 44 × 2 = 1 + 0.670 650 88;
  • 37) 0.670 650 88 × 2 = 1 + 0.341 301 76;
  • 38) 0.341 301 76 × 2 = 0 + 0.682 603 52;
  • 39) 0.682 603 52 × 2 = 1 + 0.365 207 04;
  • 40) 0.365 207 04 × 2 = 0 + 0.730 414 08;
  • 41) 0.730 414 08 × 2 = 1 + 0.460 828 16;
  • 42) 0.460 828 16 × 2 = 0 + 0.921 656 32;
  • 43) 0.921 656 32 × 2 = 1 + 0.843 312 64;
  • 44) 0.843 312 64 × 2 = 1 + 0.686 625 28;
  • 45) 0.686 625 28 × 2 = 1 + 0.373 250 56;
  • 46) 0.373 250 56 × 2 = 0 + 0.746 501 12;
  • 47) 0.746 501 12 × 2 = 1 + 0.493 002 24;
  • 48) 0.493 002 24 × 2 = 0 + 0.986 004 48;
  • 49) 0.986 004 48 × 2 = 1 + 0.972 008 96;
  • 50) 0.972 008 96 × 2 = 1 + 0.944 017 92;
  • 51) 0.944 017 92 × 2 = 1 + 0.888 035 84;
  • 52) 0.888 035 84 × 2 = 1 + 0.776 071 68;
  • 53) 0.776 071 68 × 2 = 1 + 0.552 143 36;
  • 54) 0.552 143 36 × 2 = 1 + 0.104 286 72;
  • 55) 0.104 286 72 × 2 = 0 + 0.208 573 44;
  • 56) 0.208 573 44 × 2 = 0 + 0.417 146 88;
  • 57) 0.417 146 88 × 2 = 0 + 0.834 293 76;
  • 58) 0.834 293 76 × 2 = 1 + 0.668 587 52;
  • 59) 0.668 587 52 × 2 = 1 + 0.337 175 04;
  • 60) 0.337 175 04 × 2 = 0 + 0.674 350 08;
  • 61) 0.674 350 08 × 2 = 1 + 0.348 700 16;
  • 62) 0.348 700 16 × 2 = 0 + 0.697 400 32;
  • 63) 0.697 400 32 × 2 = 1 + 0.394 800 64;
  • 64) 0.394 800 64 × 2 = 0 + 0.789 601 28;

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:


0.000 282 705(10) =

0.0000 0000 0001 0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 1010(2)

6. Positive number before normalization:

0.000 282 705(10) =

0.0000 0000 0001 0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 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:


0.000 282 705(10) =

0.0000 0000 0001 0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 1010(2) =

0.0000 0000 0001 0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 1010(2) × 20 =

1.0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 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 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 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 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 1010 =

0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 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 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 1010


Decimal number -0.000 282 705 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0111 0000 0000 1011 1010 1011 1010 1111 1100 0110 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