-0.000 282 055 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 055 6(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
-0.000 282 055 6(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:

|-0.000 282 055 6| = 0.000 282 055 6


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: 0.000 282 055 6.

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) 0.000 282 055 6 × 2 = 0 + 0.000 564 111 2;
  • 2) 0.000 564 111 2 × 2 = 0 + 0.001 128 222 4;
  • 3) 0.001 128 222 4 × 2 = 0 + 0.002 256 444 8;
  • 4) 0.002 256 444 8 × 2 = 0 + 0.004 512 889 6;
  • 5) 0.004 512 889 6 × 2 = 0 + 0.009 025 779 2;
  • 6) 0.009 025 779 2 × 2 = 0 + 0.018 051 558 4;
  • 7) 0.018 051 558 4 × 2 = 0 + 0.036 103 116 8;
  • 8) 0.036 103 116 8 × 2 = 0 + 0.072 206 233 6;
  • 9) 0.072 206 233 6 × 2 = 0 + 0.144 412 467 2;
  • 10) 0.144 412 467 2 × 2 = 0 + 0.288 824 934 4;
  • 11) 0.288 824 934 4 × 2 = 0 + 0.577 649 868 8;
  • 12) 0.577 649 868 8 × 2 = 1 + 0.155 299 737 6;
  • 13) 0.155 299 737 6 × 2 = 0 + 0.310 599 475 2;
  • 14) 0.310 599 475 2 × 2 = 0 + 0.621 198 950 4;
  • 15) 0.621 198 950 4 × 2 = 1 + 0.242 397 900 8;
  • 16) 0.242 397 900 8 × 2 = 0 + 0.484 795 801 6;
  • 17) 0.484 795 801 6 × 2 = 0 + 0.969 591 603 2;
  • 18) 0.969 591 603 2 × 2 = 1 + 0.939 183 206 4;
  • 19) 0.939 183 206 4 × 2 = 1 + 0.878 366 412 8;
  • 20) 0.878 366 412 8 × 2 = 1 + 0.756 732 825 6;
  • 21) 0.756 732 825 6 × 2 = 1 + 0.513 465 651 2;
  • 22) 0.513 465 651 2 × 2 = 1 + 0.026 931 302 4;
  • 23) 0.026 931 302 4 × 2 = 0 + 0.053 862 604 8;
  • 24) 0.053 862 604 8 × 2 = 0 + 0.107 725 209 6;
  • 25) 0.107 725 209 6 × 2 = 0 + 0.215 450 419 2;
  • 26) 0.215 450 419 2 × 2 = 0 + 0.430 900 838 4;
  • 27) 0.430 900 838 4 × 2 = 0 + 0.861 801 676 8;
  • 28) 0.861 801 676 8 × 2 = 1 + 0.723 603 353 6;
  • 29) 0.723 603 353 6 × 2 = 1 + 0.447 206 707 2;
  • 30) 0.447 206 707 2 × 2 = 0 + 0.894 413 414 4;
  • 31) 0.894 413 414 4 × 2 = 1 + 0.788 826 828 8;
  • 32) 0.788 826 828 8 × 2 = 1 + 0.577 653 657 6;
  • 33) 0.577 653 657 6 × 2 = 1 + 0.155 307 315 2;
  • 34) 0.155 307 315 2 × 2 = 0 + 0.310 614 630 4;
  • 35) 0.310 614 630 4 × 2 = 0 + 0.621 229 260 8;
  • 36) 0.621 229 260 8 × 2 = 1 + 0.242 458 521 6;
  • 37) 0.242 458 521 6 × 2 = 0 + 0.484 917 043 2;
  • 38) 0.484 917 043 2 × 2 = 0 + 0.969 834 086 4;
  • 39) 0.969 834 086 4 × 2 = 1 + 0.939 668 172 8;
  • 40) 0.939 668 172 8 × 2 = 1 + 0.879 336 345 6;
  • 41) 0.879 336 345 6 × 2 = 1 + 0.758 672 691 2;
  • 42) 0.758 672 691 2 × 2 = 1 + 0.517 345 382 4;
  • 43) 0.517 345 382 4 × 2 = 1 + 0.034 690 764 8;
  • 44) 0.034 690 764 8 × 2 = 0 + 0.069 381 529 6;
  • 45) 0.069 381 529 6 × 2 = 0 + 0.138 763 059 2;
  • 46) 0.138 763 059 2 × 2 = 0 + 0.277 526 118 4;
  • 47) 0.277 526 118 4 × 2 = 0 + 0.555 052 236 8;
  • 48) 0.555 052 236 8 × 2 = 1 + 0.110 104 473 6;
  • 49) 0.110 104 473 6 × 2 = 0 + 0.220 208 947 2;
  • 50) 0.220 208 947 2 × 2 = 0 + 0.440 417 894 4;
  • 51) 0.440 417 894 4 × 2 = 0 + 0.880 835 788 8;
  • 52) 0.880 835 788 8 × 2 = 1 + 0.761 671 577 6;
  • 53) 0.761 671 577 6 × 2 = 1 + 0.523 343 155 2;
  • 54) 0.523 343 155 2 × 2 = 1 + 0.046 686 310 4;
  • 55) 0.046 686 310 4 × 2 = 0 + 0.093 372 620 8;
  • 56) 0.093 372 620 8 × 2 = 0 + 0.186 745 241 6;
  • 57) 0.186 745 241 6 × 2 = 0 + 0.373 490 483 2;
  • 58) 0.373 490 483 2 × 2 = 0 + 0.746 980 966 4;
  • 59) 0.746 980 966 4 × 2 = 1 + 0.493 961 932 8;
  • 60) 0.493 961 932 8 × 2 = 0 + 0.987 923 865 6;
  • 61) 0.987 923 865 6 × 2 = 1 + 0.975 847 731 2;
  • 62) 0.975 847 731 2 × 2 = 1 + 0.951 695 462 4;
  • 63) 0.951 695 462 4 × 2 = 1 + 0.903 390 924 8;
  • 64) 0.903 390 924 8 × 2 = 1 + 0.806 781 849 6;

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:


0.000 282 055 6(10) =

0.0000 0000 0001 0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111(2)

6. Positive number before normalization:

0.000 282 055 6(10) =

0.0000 0000 0001 0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111(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:


0.000 282 055 6(10) =

0.0000 0000 0001 0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111(2) =

0.0000 0000 0001 0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111(2) × 20 =

1.0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111(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 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111


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 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111 =

0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111


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 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111


Decimal number -0.000 282 055 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 0001 1011 1001 0011 1110 0001 0001 1100 0010 1111

Share

» Convert decimal -0.000 282 057 7 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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