-0.000 282 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 92(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 92(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 92| = 0.000 282 92


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 92.

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 92 × 2 = 0 + 0.000 565 84;
  • 2) 0.000 565 84 × 2 = 0 + 0.001 131 68;
  • 3) 0.001 131 68 × 2 = 0 + 0.002 263 36;
  • 4) 0.002 263 36 × 2 = 0 + 0.004 526 72;
  • 5) 0.004 526 72 × 2 = 0 + 0.009 053 44;
  • 6) 0.009 053 44 × 2 = 0 + 0.018 106 88;
  • 7) 0.018 106 88 × 2 = 0 + 0.036 213 76;
  • 8) 0.036 213 76 × 2 = 0 + 0.072 427 52;
  • 9) 0.072 427 52 × 2 = 0 + 0.144 855 04;
  • 10) 0.144 855 04 × 2 = 0 + 0.289 710 08;
  • 11) 0.289 710 08 × 2 = 0 + 0.579 420 16;
  • 12) 0.579 420 16 × 2 = 1 + 0.158 840 32;
  • 13) 0.158 840 32 × 2 = 0 + 0.317 680 64;
  • 14) 0.317 680 64 × 2 = 0 + 0.635 361 28;
  • 15) 0.635 361 28 × 2 = 1 + 0.270 722 56;
  • 16) 0.270 722 56 × 2 = 0 + 0.541 445 12;
  • 17) 0.541 445 12 × 2 = 1 + 0.082 890 24;
  • 18) 0.082 890 24 × 2 = 0 + 0.165 780 48;
  • 19) 0.165 780 48 × 2 = 0 + 0.331 560 96;
  • 20) 0.331 560 96 × 2 = 0 + 0.663 121 92;
  • 21) 0.663 121 92 × 2 = 1 + 0.326 243 84;
  • 22) 0.326 243 84 × 2 = 0 + 0.652 487 68;
  • 23) 0.652 487 68 × 2 = 1 + 0.304 975 36;
  • 24) 0.304 975 36 × 2 = 0 + 0.609 950 72;
  • 25) 0.609 950 72 × 2 = 1 + 0.219 901 44;
  • 26) 0.219 901 44 × 2 = 0 + 0.439 802 88;
  • 27) 0.439 802 88 × 2 = 0 + 0.879 605 76;
  • 28) 0.879 605 76 × 2 = 1 + 0.759 211 52;
  • 29) 0.759 211 52 × 2 = 1 + 0.518 423 04;
  • 30) 0.518 423 04 × 2 = 1 + 0.036 846 08;
  • 31) 0.036 846 08 × 2 = 0 + 0.073 692 16;
  • 32) 0.073 692 16 × 2 = 0 + 0.147 384 32;
  • 33) 0.147 384 32 × 2 = 0 + 0.294 768 64;
  • 34) 0.294 768 64 × 2 = 0 + 0.589 537 28;
  • 35) 0.589 537 28 × 2 = 1 + 0.179 074 56;
  • 36) 0.179 074 56 × 2 = 0 + 0.358 149 12;
  • 37) 0.358 149 12 × 2 = 0 + 0.716 298 24;
  • 38) 0.716 298 24 × 2 = 1 + 0.432 596 48;
  • 39) 0.432 596 48 × 2 = 0 + 0.865 192 96;
  • 40) 0.865 192 96 × 2 = 1 + 0.730 385 92;
  • 41) 0.730 385 92 × 2 = 1 + 0.460 771 84;
  • 42) 0.460 771 84 × 2 = 0 + 0.921 543 68;
  • 43) 0.921 543 68 × 2 = 1 + 0.843 087 36;
  • 44) 0.843 087 36 × 2 = 1 + 0.686 174 72;
  • 45) 0.686 174 72 × 2 = 1 + 0.372 349 44;
  • 46) 0.372 349 44 × 2 = 0 + 0.744 698 88;
  • 47) 0.744 698 88 × 2 = 1 + 0.489 397 76;
  • 48) 0.489 397 76 × 2 = 0 + 0.978 795 52;
  • 49) 0.978 795 52 × 2 = 1 + 0.957 591 04;
  • 50) 0.957 591 04 × 2 = 1 + 0.915 182 08;
  • 51) 0.915 182 08 × 2 = 1 + 0.830 364 16;
  • 52) 0.830 364 16 × 2 = 1 + 0.660 728 32;
  • 53) 0.660 728 32 × 2 = 1 + 0.321 456 64;
  • 54) 0.321 456 64 × 2 = 0 + 0.642 913 28;
  • 55) 0.642 913 28 × 2 = 1 + 0.285 826 56;
  • 56) 0.285 826 56 × 2 = 0 + 0.571 653 12;
  • 57) 0.571 653 12 × 2 = 1 + 0.143 306 24;
  • 58) 0.143 306 24 × 2 = 0 + 0.286 612 48;
  • 59) 0.286 612 48 × 2 = 0 + 0.573 224 96;
  • 60) 0.573 224 96 × 2 = 1 + 0.146 449 92;
  • 61) 0.146 449 92 × 2 = 0 + 0.292 899 84;
  • 62) 0.292 899 84 × 2 = 0 + 0.585 799 68;
  • 63) 0.585 799 68 × 2 = 1 + 0.171 599 36;
  • 64) 0.171 599 36 × 2 = 0 + 0.343 198 72;

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 92(10) =

0.0000 0000 0001 0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010(2)

6. Positive number before normalization:

0.000 282 92(10) =

0.0000 0000 0001 0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010(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 92(10) =

0.0000 0000 0001 0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010(2) =

0.0000 0000 0001 0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010(2) × 20 =

1.0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010(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 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010


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 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010 =

0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010


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 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010


Decimal number -0.000 282 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1010 1001 1100 0010 0101 1011 1010 1111 1010 1001 0010

Share

» Convert decimal -0.000 282 95 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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