-0.000 286 21 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 286 21(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
-0.000 286 21(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:

|-0.000 286 21| = 0.000 286 21


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: 0.000 286 21.

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) 0.000 286 21 × 2 = 0 + 0.000 572 42;
  • 2) 0.000 572 42 × 2 = 0 + 0.001 144 84;
  • 3) 0.001 144 84 × 2 = 0 + 0.002 289 68;
  • 4) 0.002 289 68 × 2 = 0 + 0.004 579 36;
  • 5) 0.004 579 36 × 2 = 0 + 0.009 158 72;
  • 6) 0.009 158 72 × 2 = 0 + 0.018 317 44;
  • 7) 0.018 317 44 × 2 = 0 + 0.036 634 88;
  • 8) 0.036 634 88 × 2 = 0 + 0.073 269 76;
  • 9) 0.073 269 76 × 2 = 0 + 0.146 539 52;
  • 10) 0.146 539 52 × 2 = 0 + 0.293 079 04;
  • 11) 0.293 079 04 × 2 = 0 + 0.586 158 08;
  • 12) 0.586 158 08 × 2 = 1 + 0.172 316 16;
  • 13) 0.172 316 16 × 2 = 0 + 0.344 632 32;
  • 14) 0.344 632 32 × 2 = 0 + 0.689 264 64;
  • 15) 0.689 264 64 × 2 = 1 + 0.378 529 28;
  • 16) 0.378 529 28 × 2 = 0 + 0.757 058 56;
  • 17) 0.757 058 56 × 2 = 1 + 0.514 117 12;
  • 18) 0.514 117 12 × 2 = 1 + 0.028 234 24;
  • 19) 0.028 234 24 × 2 = 0 + 0.056 468 48;
  • 20) 0.056 468 48 × 2 = 0 + 0.112 936 96;
  • 21) 0.112 936 96 × 2 = 0 + 0.225 873 92;
  • 22) 0.225 873 92 × 2 = 0 + 0.451 747 84;
  • 23) 0.451 747 84 × 2 = 0 + 0.903 495 68;
  • 24) 0.903 495 68 × 2 = 1 + 0.806 991 36;
  • 25) 0.806 991 36 × 2 = 1 + 0.613 982 72;
  • 26) 0.613 982 72 × 2 = 1 + 0.227 965 44;
  • 27) 0.227 965 44 × 2 = 0 + 0.455 930 88;
  • 28) 0.455 930 88 × 2 = 0 + 0.911 861 76;
  • 29) 0.911 861 76 × 2 = 1 + 0.823 723 52;
  • 30) 0.823 723 52 × 2 = 1 + 0.647 447 04;
  • 31) 0.647 447 04 × 2 = 1 + 0.294 894 08;
  • 32) 0.294 894 08 × 2 = 0 + 0.589 788 16;
  • 33) 0.589 788 16 × 2 = 1 + 0.179 576 32;
  • 34) 0.179 576 32 × 2 = 0 + 0.359 152 64;
  • 35) 0.359 152 64 × 2 = 0 + 0.718 305 28;
  • 36) 0.718 305 28 × 2 = 1 + 0.436 610 56;
  • 37) 0.436 610 56 × 2 = 0 + 0.873 221 12;
  • 38) 0.873 221 12 × 2 = 1 + 0.746 442 24;
  • 39) 0.746 442 24 × 2 = 1 + 0.492 884 48;
  • 40) 0.492 884 48 × 2 = 0 + 0.985 768 96;
  • 41) 0.985 768 96 × 2 = 1 + 0.971 537 92;
  • 42) 0.971 537 92 × 2 = 1 + 0.943 075 84;
  • 43) 0.943 075 84 × 2 = 1 + 0.886 151 68;
  • 44) 0.886 151 68 × 2 = 1 + 0.772 303 36;
  • 45) 0.772 303 36 × 2 = 1 + 0.544 606 72;
  • 46) 0.544 606 72 × 2 = 1 + 0.089 213 44;
  • 47) 0.089 213 44 × 2 = 0 + 0.178 426 88;
  • 48) 0.178 426 88 × 2 = 0 + 0.356 853 76;
  • 49) 0.356 853 76 × 2 = 0 + 0.713 707 52;
  • 50) 0.713 707 52 × 2 = 1 + 0.427 415 04;
  • 51) 0.427 415 04 × 2 = 0 + 0.854 830 08;
  • 52) 0.854 830 08 × 2 = 1 + 0.709 660 16;
  • 53) 0.709 660 16 × 2 = 1 + 0.419 320 32;
  • 54) 0.419 320 32 × 2 = 0 + 0.838 640 64;
  • 55) 0.838 640 64 × 2 = 1 + 0.677 281 28;
  • 56) 0.677 281 28 × 2 = 1 + 0.354 562 56;
  • 57) 0.354 562 56 × 2 = 0 + 0.709 125 12;
  • 58) 0.709 125 12 × 2 = 1 + 0.418 250 24;
  • 59) 0.418 250 24 × 2 = 0 + 0.836 500 48;
  • 60) 0.836 500 48 × 2 = 1 + 0.673 000 96;
  • 61) 0.673 000 96 × 2 = 1 + 0.346 001 92;
  • 62) 0.346 001 92 × 2 = 0 + 0.692 003 84;
  • 63) 0.692 003 84 × 2 = 1 + 0.384 007 68;
  • 64) 0.384 007 68 × 2 = 0 + 0.768 015 36;

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:


0.000 286 21(10) =

0.0000 0000 0001 0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 1010(2)

6. Positive number before normalization:

0.000 286 21(10) =

0.0000 0000 0001 0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 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:


0.000 286 21(10) =

0.0000 0000 0001 0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 1010(2) =

0.0000 0000 0001 0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 1010(2) × 20 =

1.0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 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 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 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 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 1010 =

0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 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 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 1010


Decimal number -0.000 286 21 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1100 0001 1100 1110 1001 0110 1111 1100 0101 1011 0101 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