-0.000 278 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 278 32(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
-0.000 278 32(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:

|-0.000 278 32| = 0.000 278 32


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: 0.000 278 32.

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) 0.000 278 32 × 2 = 0 + 0.000 556 64;
  • 2) 0.000 556 64 × 2 = 0 + 0.001 113 28;
  • 3) 0.001 113 28 × 2 = 0 + 0.002 226 56;
  • 4) 0.002 226 56 × 2 = 0 + 0.004 453 12;
  • 5) 0.004 453 12 × 2 = 0 + 0.008 906 24;
  • 6) 0.008 906 24 × 2 = 0 + 0.017 812 48;
  • 7) 0.017 812 48 × 2 = 0 + 0.035 624 96;
  • 8) 0.035 624 96 × 2 = 0 + 0.071 249 92;
  • 9) 0.071 249 92 × 2 = 0 + 0.142 499 84;
  • 10) 0.142 499 84 × 2 = 0 + 0.284 999 68;
  • 11) 0.284 999 68 × 2 = 0 + 0.569 999 36;
  • 12) 0.569 999 36 × 2 = 1 + 0.139 998 72;
  • 13) 0.139 998 72 × 2 = 0 + 0.279 997 44;
  • 14) 0.279 997 44 × 2 = 0 + 0.559 994 88;
  • 15) 0.559 994 88 × 2 = 1 + 0.119 989 76;
  • 16) 0.119 989 76 × 2 = 0 + 0.239 979 52;
  • 17) 0.239 979 52 × 2 = 0 + 0.479 959 04;
  • 18) 0.479 959 04 × 2 = 0 + 0.959 918 08;
  • 19) 0.959 918 08 × 2 = 1 + 0.919 836 16;
  • 20) 0.919 836 16 × 2 = 1 + 0.839 672 32;
  • 21) 0.839 672 32 × 2 = 1 + 0.679 344 64;
  • 22) 0.679 344 64 × 2 = 1 + 0.358 689 28;
  • 23) 0.358 689 28 × 2 = 0 + 0.717 378 56;
  • 24) 0.717 378 56 × 2 = 1 + 0.434 757 12;
  • 25) 0.434 757 12 × 2 = 0 + 0.869 514 24;
  • 26) 0.869 514 24 × 2 = 1 + 0.739 028 48;
  • 27) 0.739 028 48 × 2 = 1 + 0.478 056 96;
  • 28) 0.478 056 96 × 2 = 0 + 0.956 113 92;
  • 29) 0.956 113 92 × 2 = 1 + 0.912 227 84;
  • 30) 0.912 227 84 × 2 = 1 + 0.824 455 68;
  • 31) 0.824 455 68 × 2 = 1 + 0.648 911 36;
  • 32) 0.648 911 36 × 2 = 1 + 0.297 822 72;
  • 33) 0.297 822 72 × 2 = 0 + 0.595 645 44;
  • 34) 0.595 645 44 × 2 = 1 + 0.191 290 88;
  • 35) 0.191 290 88 × 2 = 0 + 0.382 581 76;
  • 36) 0.382 581 76 × 2 = 0 + 0.765 163 52;
  • 37) 0.765 163 52 × 2 = 1 + 0.530 327 04;
  • 38) 0.530 327 04 × 2 = 1 + 0.060 654 08;
  • 39) 0.060 654 08 × 2 = 0 + 0.121 308 16;
  • 40) 0.121 308 16 × 2 = 0 + 0.242 616 32;
  • 41) 0.242 616 32 × 2 = 0 + 0.485 232 64;
  • 42) 0.485 232 64 × 2 = 0 + 0.970 465 28;
  • 43) 0.970 465 28 × 2 = 1 + 0.940 930 56;
  • 44) 0.940 930 56 × 2 = 1 + 0.881 861 12;
  • 45) 0.881 861 12 × 2 = 1 + 0.763 722 24;
  • 46) 0.763 722 24 × 2 = 1 + 0.527 444 48;
  • 47) 0.527 444 48 × 2 = 1 + 0.054 888 96;
  • 48) 0.054 888 96 × 2 = 0 + 0.109 777 92;
  • 49) 0.109 777 92 × 2 = 0 + 0.219 555 84;
  • 50) 0.219 555 84 × 2 = 0 + 0.439 111 68;
  • 51) 0.439 111 68 × 2 = 0 + 0.878 223 36;
  • 52) 0.878 223 36 × 2 = 1 + 0.756 446 72;
  • 53) 0.756 446 72 × 2 = 1 + 0.512 893 44;
  • 54) 0.512 893 44 × 2 = 1 + 0.025 786 88;
  • 55) 0.025 786 88 × 2 = 0 + 0.051 573 76;
  • 56) 0.051 573 76 × 2 = 0 + 0.103 147 52;
  • 57) 0.103 147 52 × 2 = 0 + 0.206 295 04;
  • 58) 0.206 295 04 × 2 = 0 + 0.412 590 08;
  • 59) 0.412 590 08 × 2 = 0 + 0.825 180 16;
  • 60) 0.825 180 16 × 2 = 1 + 0.650 360 32;
  • 61) 0.650 360 32 × 2 = 1 + 0.300 720 64;
  • 62) 0.300 720 64 × 2 = 0 + 0.601 441 28;
  • 63) 0.601 441 28 × 2 = 1 + 0.202 882 56;
  • 64) 0.202 882 56 × 2 = 0 + 0.405 765 12;

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:


0.000 278 32(10) =

0.0000 0000 0001 0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 1010(2)

6. Positive number before normalization:

0.000 278 32(10) =

0.0000 0000 0001 0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 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:


0.000 278 32(10) =

0.0000 0000 0001 0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 1010(2) =

0.0000 0000 0001 0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 1010(2) × 20 =

1.0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 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 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 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 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 1010 =

0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 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 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 1010


Decimal number -0.000 278 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0011 1101 0110 1111 0100 1100 0011 1110 0001 1100 0001 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