-0.000 280 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 280 69(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
-0.000 280 69(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:

|-0.000 280 69| = 0.000 280 69


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: 0.000 280 69.

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) 0.000 280 69 × 2 = 0 + 0.000 561 38;
  • 2) 0.000 561 38 × 2 = 0 + 0.001 122 76;
  • 3) 0.001 122 76 × 2 = 0 + 0.002 245 52;
  • 4) 0.002 245 52 × 2 = 0 + 0.004 491 04;
  • 5) 0.004 491 04 × 2 = 0 + 0.008 982 08;
  • 6) 0.008 982 08 × 2 = 0 + 0.017 964 16;
  • 7) 0.017 964 16 × 2 = 0 + 0.035 928 32;
  • 8) 0.035 928 32 × 2 = 0 + 0.071 856 64;
  • 9) 0.071 856 64 × 2 = 0 + 0.143 713 28;
  • 10) 0.143 713 28 × 2 = 0 + 0.287 426 56;
  • 11) 0.287 426 56 × 2 = 0 + 0.574 853 12;
  • 12) 0.574 853 12 × 2 = 1 + 0.149 706 24;
  • 13) 0.149 706 24 × 2 = 0 + 0.299 412 48;
  • 14) 0.299 412 48 × 2 = 0 + 0.598 824 96;
  • 15) 0.598 824 96 × 2 = 1 + 0.197 649 92;
  • 16) 0.197 649 92 × 2 = 0 + 0.395 299 84;
  • 17) 0.395 299 84 × 2 = 0 + 0.790 599 68;
  • 18) 0.790 599 68 × 2 = 1 + 0.581 199 36;
  • 19) 0.581 199 36 × 2 = 1 + 0.162 398 72;
  • 20) 0.162 398 72 × 2 = 0 + 0.324 797 44;
  • 21) 0.324 797 44 × 2 = 0 + 0.649 594 88;
  • 22) 0.649 594 88 × 2 = 1 + 0.299 189 76;
  • 23) 0.299 189 76 × 2 = 0 + 0.598 379 52;
  • 24) 0.598 379 52 × 2 = 1 + 0.196 759 04;
  • 25) 0.196 759 04 × 2 = 0 + 0.393 518 08;
  • 26) 0.393 518 08 × 2 = 0 + 0.787 036 16;
  • 27) 0.787 036 16 × 2 = 1 + 0.574 072 32;
  • 28) 0.574 072 32 × 2 = 1 + 0.148 144 64;
  • 29) 0.148 144 64 × 2 = 0 + 0.296 289 28;
  • 30) 0.296 289 28 × 2 = 0 + 0.592 578 56;
  • 31) 0.592 578 56 × 2 = 1 + 0.185 157 12;
  • 32) 0.185 157 12 × 2 = 0 + 0.370 314 24;
  • 33) 0.370 314 24 × 2 = 0 + 0.740 628 48;
  • 34) 0.740 628 48 × 2 = 1 + 0.481 256 96;
  • 35) 0.481 256 96 × 2 = 0 + 0.962 513 92;
  • 36) 0.962 513 92 × 2 = 1 + 0.925 027 84;
  • 37) 0.925 027 84 × 2 = 1 + 0.850 055 68;
  • 38) 0.850 055 68 × 2 = 1 + 0.700 111 36;
  • 39) 0.700 111 36 × 2 = 1 + 0.400 222 72;
  • 40) 0.400 222 72 × 2 = 0 + 0.800 445 44;
  • 41) 0.800 445 44 × 2 = 1 + 0.600 890 88;
  • 42) 0.600 890 88 × 2 = 1 + 0.201 781 76;
  • 43) 0.201 781 76 × 2 = 0 + 0.403 563 52;
  • 44) 0.403 563 52 × 2 = 0 + 0.807 127 04;
  • 45) 0.807 127 04 × 2 = 1 + 0.614 254 08;
  • 46) 0.614 254 08 × 2 = 1 + 0.228 508 16;
  • 47) 0.228 508 16 × 2 = 0 + 0.457 016 32;
  • 48) 0.457 016 32 × 2 = 0 + 0.914 032 64;
  • 49) 0.914 032 64 × 2 = 1 + 0.828 065 28;
  • 50) 0.828 065 28 × 2 = 1 + 0.656 130 56;
  • 51) 0.656 130 56 × 2 = 1 + 0.312 261 12;
  • 52) 0.312 261 12 × 2 = 0 + 0.624 522 24;
  • 53) 0.624 522 24 × 2 = 1 + 0.249 044 48;
  • 54) 0.249 044 48 × 2 = 0 + 0.498 088 96;
  • 55) 0.498 088 96 × 2 = 0 + 0.996 177 92;
  • 56) 0.996 177 92 × 2 = 1 + 0.992 355 84;
  • 57) 0.992 355 84 × 2 = 1 + 0.984 711 68;
  • 58) 0.984 711 68 × 2 = 1 + 0.969 423 36;
  • 59) 0.969 423 36 × 2 = 1 + 0.938 846 72;
  • 60) 0.938 846 72 × 2 = 1 + 0.877 693 44;
  • 61) 0.877 693 44 × 2 = 1 + 0.755 386 88;
  • 62) 0.755 386 88 × 2 = 1 + 0.510 773 76;
  • 63) 0.510 773 76 × 2 = 1 + 0.021 547 52;
  • 64) 0.021 547 52 × 2 = 0 + 0.043 095 04;

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:


0.000 280 69(10) =

0.0000 0000 0001 0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 1110(2)

6. Positive number before normalization:

0.000 280 69(10) =

0.0000 0000 0001 0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 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:


0.000 280 69(10) =

0.0000 0000 0001 0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 1110(2) =

0.0000 0000 0001 0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 1110(2) × 20 =

1.0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 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 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 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 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 1110 =

0010 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 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 0110 0101 0011 0010 0101 1110 1100 1100 1110 1001 1111 1110


Decimal number -0.000 280 69 converted to 64 bit double precision IEEE 754 binary floating point representation:

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