-0.000 283 03 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 283 03(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
-0.000 283 03(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:

|-0.000 283 03| = 0.000 283 03


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: 0.000 283 03.

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) 0.000 283 03 × 2 = 0 + 0.000 566 06;
  • 2) 0.000 566 06 × 2 = 0 + 0.001 132 12;
  • 3) 0.001 132 12 × 2 = 0 + 0.002 264 24;
  • 4) 0.002 264 24 × 2 = 0 + 0.004 528 48;
  • 5) 0.004 528 48 × 2 = 0 + 0.009 056 96;
  • 6) 0.009 056 96 × 2 = 0 + 0.018 113 92;
  • 7) 0.018 113 92 × 2 = 0 + 0.036 227 84;
  • 8) 0.036 227 84 × 2 = 0 + 0.072 455 68;
  • 9) 0.072 455 68 × 2 = 0 + 0.144 911 36;
  • 10) 0.144 911 36 × 2 = 0 + 0.289 822 72;
  • 11) 0.289 822 72 × 2 = 0 + 0.579 645 44;
  • 12) 0.579 645 44 × 2 = 1 + 0.159 290 88;
  • 13) 0.159 290 88 × 2 = 0 + 0.318 581 76;
  • 14) 0.318 581 76 × 2 = 0 + 0.637 163 52;
  • 15) 0.637 163 52 × 2 = 1 + 0.274 327 04;
  • 16) 0.274 327 04 × 2 = 0 + 0.548 654 08;
  • 17) 0.548 654 08 × 2 = 1 + 0.097 308 16;
  • 18) 0.097 308 16 × 2 = 0 + 0.194 616 32;
  • 19) 0.194 616 32 × 2 = 0 + 0.389 232 64;
  • 20) 0.389 232 64 × 2 = 0 + 0.778 465 28;
  • 21) 0.778 465 28 × 2 = 1 + 0.556 930 56;
  • 22) 0.556 930 56 × 2 = 1 + 0.113 861 12;
  • 23) 0.113 861 12 × 2 = 0 + 0.227 722 24;
  • 24) 0.227 722 24 × 2 = 0 + 0.455 444 48;
  • 25) 0.455 444 48 × 2 = 0 + 0.910 888 96;
  • 26) 0.910 888 96 × 2 = 1 + 0.821 777 92;
  • 27) 0.821 777 92 × 2 = 1 + 0.643 555 84;
  • 28) 0.643 555 84 × 2 = 1 + 0.287 111 68;
  • 29) 0.287 111 68 × 2 = 0 + 0.574 223 36;
  • 30) 0.574 223 36 × 2 = 1 + 0.148 446 72;
  • 31) 0.148 446 72 × 2 = 0 + 0.296 893 44;
  • 32) 0.296 893 44 × 2 = 0 + 0.593 786 88;
  • 33) 0.593 786 88 × 2 = 1 + 0.187 573 76;
  • 34) 0.187 573 76 × 2 = 0 + 0.375 147 52;
  • 35) 0.375 147 52 × 2 = 0 + 0.750 295 04;
  • 36) 0.750 295 04 × 2 = 1 + 0.500 590 08;
  • 37) 0.500 590 08 × 2 = 1 + 0.001 180 16;
  • 38) 0.001 180 16 × 2 = 0 + 0.002 360 32;
  • 39) 0.002 360 32 × 2 = 0 + 0.004 720 64;
  • 40) 0.004 720 64 × 2 = 0 + 0.009 441 28;
  • 41) 0.009 441 28 × 2 = 0 + 0.018 882 56;
  • 42) 0.018 882 56 × 2 = 0 + 0.037 765 12;
  • 43) 0.037 765 12 × 2 = 0 + 0.075 530 24;
  • 44) 0.075 530 24 × 2 = 0 + 0.151 060 48;
  • 45) 0.151 060 48 × 2 = 0 + 0.302 120 96;
  • 46) 0.302 120 96 × 2 = 0 + 0.604 241 92;
  • 47) 0.604 241 92 × 2 = 1 + 0.208 483 84;
  • 48) 0.208 483 84 × 2 = 0 + 0.416 967 68;
  • 49) 0.416 967 68 × 2 = 0 + 0.833 935 36;
  • 50) 0.833 935 36 × 2 = 1 + 0.667 870 72;
  • 51) 0.667 870 72 × 2 = 1 + 0.335 741 44;
  • 52) 0.335 741 44 × 2 = 0 + 0.671 482 88;
  • 53) 0.671 482 88 × 2 = 1 + 0.342 965 76;
  • 54) 0.342 965 76 × 2 = 0 + 0.685 931 52;
  • 55) 0.685 931 52 × 2 = 1 + 0.371 863 04;
  • 56) 0.371 863 04 × 2 = 0 + 0.743 726 08;
  • 57) 0.743 726 08 × 2 = 1 + 0.487 452 16;
  • 58) 0.487 452 16 × 2 = 0 + 0.974 904 32;
  • 59) 0.974 904 32 × 2 = 1 + 0.949 808 64;
  • 60) 0.949 808 64 × 2 = 1 + 0.899 617 28;
  • 61) 0.899 617 28 × 2 = 1 + 0.799 234 56;
  • 62) 0.799 234 56 × 2 = 1 + 0.598 469 12;
  • 63) 0.598 469 12 × 2 = 1 + 0.196 938 24;
  • 64) 0.196 938 24 × 2 = 0 + 0.393 876 48;

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:


0.000 283 03(10) =

0.0000 0000 0001 0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110(2)

6. Positive number before normalization:

0.000 283 03(10) =

0.0000 0000 0001 0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110(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:


0.000 283 03(10) =

0.0000 0000 0001 0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110(2) =

0.0000 0000 0001 0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110(2) × 20 =

1.0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110(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 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110


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 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110 =

0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110


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 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110


Decimal number -0.000 283 03 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1100 0111 0100 1001 1000 0000 0010 0110 1010 1011 1110

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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